Proof. The result automatically follows, using the well-known identities \[\Delta_g(f_1f_2) = f_1\Delta_g f_2 + f_2\Delta_g f_1 + 2 \langle \mathrm{grad}_{g}\,f_1, \mathrm{grad}_{g}\,f_2\rangle_g,\] and \[\mathrm{grad}(f_1f_2) = f_1\,\mathrm{grad}_{g}\,f_2 + f_2\, \mathrm{grad}_{g}\,f_1.\] ◻

Proof. Using 18 it holds that \[|\mathrm{grad}_{h}\,\tilde{F}(x)|_h \geq C_{\tilde{F}} d_h(x, \tilde{\mathcal{C}}).\] Thus, for every \(x \in B\) such that \(d_h(x, \tilde{\mathcal{C}}) \geq \frac{a^2}{4\beta}\) we obtain that \[\begin{align} \frac{\Delta_h \tilde{F}(x)}{2} - \frac{\beta}{4}|\mathrm{grad}_{h}\,\tilde{F}(x)|^2_h &\leq \frac{\Delta_h \tilde{F}(x)}{2} - \frac{\beta}{4}C_{\tilde{F}}^2 d_h(x, \tilde{\mathcal{C}})^2 \\&\leq \frac{\Delta_h \tilde{F}(x)}{2} - \frac{a^2}{16}C_{\tilde{F}}^2 \\&\leq \frac{A_2\dim(B)}{2} - \frac{1}{16}C_{\tilde{F}}^2 \frac{24A_2 \dim(B)}{C^2_{\tilde{F}}} \\&\leq -A_2 \dim(B), \end{align}\] where the second-last inequality follows by the assumption \(a^2 \geq \frac{24A_2 \dim(B)}{C^2_{\tilde{F}}}\). ◻

Proof. From the choice of \(\frac{a}{\sqrt{\beta}}\), we can decompose \[\overline{U}\Big(\frac{a}{2\sqrt{\beta}}, \tilde{\mathcal{S}}\Big)^c = \mathcal{B}\Big(\frac{a}{\sqrt{\beta}}, x^*\Big) \cup \overline{U}\Big(\frac{a}{2\sqrt{\beta}}, \tilde{\mathcal{C}}\Big)^c.\] Now, let us write \[\begin{align} \tilde{\operatorname{L}} W_1(x) &= \Big\langle -\mathrm{grad}_{h}\,\tilde{F}(x), \mathrm{grad}_{h}\,\exp\Big(\frac{\beta}{2}\tilde{F}(x)\Big)\Big\rangle_h + \frac{1}{\beta} \Delta_h \exp\left(\frac{\beta}{2}\tilde{F}(x)\right) \\&= -\frac{\beta}{2} W_1(x) |\mathrm{grad}_{h}\,\tilde{F}(x)|^2_h + \frac{1}{\beta} \left(\frac{\beta}{2}W_1(x) \Delta_h \tilde{F}(x) + \frac{\beta^2}{4} W_1(x) |\mathrm{grad}_{h}\,\tilde{F}(x)|^2_h\right) \\&= W_1(x) \Big(\frac{1}{2} \Delta_h \tilde{F}(x) - \frac{\beta}{4} |\mathrm{grad}_{h}\,\tilde{F}(x)|^2_h\Big). \end{align}\]

We can now apply 1 and the trivial bound involving the Lipschitz constant \(A_2\), \[\frac{1}{2} \Delta_h \tilde{F}(x) - \frac{\beta}{4} |\mathrm{grad}_{h}\,\tilde{F}(x)|^2_h \leq \frac{1}{2}\Delta_h \tilde{F}(x) \leq \frac{\dim(B) A_2}{2},\] which holds for every \(x \in B\), to conclude that \[\label{lyapunovW1} \frac{\tilde{\operatorname{L}} W_1(x)}{W_1(x)} \leq - A_2 \dim(B) + \frac{3}{2} A_2 \dim(B) \mathbb{1}_{\mathcal{B}\big(\frac{a}{\sqrt{\beta}}, x^*\big)}(x),\tag{5}\] for every \(x \in \overline{U}\Big(\frac{a}{2\sqrt{\beta}}, \tilde{\mathcal{S}}\Big)^c\). 5 proves that \(W_1\) is a Lyapunov function on \(\mathcal{B}\Big(\frac{a}{\sqrt{\beta}}, x^*\Big)\), as desired. ◻

Proof. Let \(f\) be the solution of the above Dirichlet problem. Note that \[(-\mathscr{L}f)(x) -\theta f(x) = 0,\] for every \(x \in V\), and \(Y_t\) solves the martingale problem associated with \(\mathscr{L}\), which implies that \[Z_t := e^{\theta t}f(Y_t)\] is a martingale [38]. Moreover, for every \(x \in V\) it holds that \[\mathbb{E}[|\exp(\theta \tau_{V^c})f(Y_{\tau_{V^c}})| \,|\, Y_0=x ] = \mathbb{E}[|\exp(\theta \tau_{V^c})|\,|\, Y_0=x] < \infty,\] where the equality follows from \(f(x)\) being 1 in \(V^c\), and finiteness follows from ?? . Therefore, it holds that \[\mathbb{E}[\exp(\theta \tau_{V^c}) | Y_0=x] = \mathbb{E}[Z_{\tau_{V^c}} | Y_0 = x]= Z_0 = f(x),\] where the second equality follows from [38] with \(T = \tau_{V^c}\) and \(S = 0\). ◻

Proof of 1. Let us first define the PDE \[\begin{cases} \tilde{\operatorname{L}}f = -\theta f, \quad &x \in U\Big(\frac{a}{\sqrt{\beta}}, \tilde{\mathcal{S}}\Big),\\ f = 1,\quad &x \in U\Big(\frac{a}{\sqrt{\beta}}, \tilde{\mathcal{S}}\Big)^c. \end{cases}\] Under the assumption that \(-\theta\) is not in the spectrum of \(\tilde{\operatorname{L}}\), the PDE has a unique solution. Indeed the boundary problem has a solution (cf. 12), which can be extended to \(U\Big(\frac{a}{\sqrt{\beta}}, \tilde{\mathcal{S}}\Big)^c\) by simply setting it to \(1\).

Thus, we can apply 13 to conclude that the solution \(f\) is of the form \[f(x) = \mathbb{E}[\exp(\theta \tau^*) | \tilde{X}_0 = x],\] which matches the definition of \(W_2\). This implies that \(W_2 \geq 1\), and, using the assumption on the regularity of the boundary of \(U\Big(\frac{a}{\sqrt{\beta}}, \tilde{\mathcal{S}}\Big)\), we can conclude that \[W_2 \in C^{\infty}\Big(\overline{U}\Big(\frac{a}{\sqrt{\beta}}, \tilde{\mathcal{S}}\Big)\Big) \cap C^\infty\Big(U\Big(\frac{a}{\sqrt{\beta}}, \tilde{\mathcal{S}}\Big)^c\Big)\] (cf. 12.3). ◻

Proof. Let \(a > 0\) and \(T > 0\). Then \[\mathbb{E}[e^{a|X|}\mathbb{1}_{\{e^{a|X|} \leq e^{aT}\}}] \leq \int_0^{e^{aT}} \mathbb{P}[e^{a|X|} \geq t] dt \leq 1 + \int_1^{e^{aT}} \mathbb{P}\left[|X| \geq \frac{\log t}{a}\right]dt.\] Since we are assuming that ?? holds, we can conclude that \[\mathbb{E}[e^{a|X|}\mathbb{1}_{\{e^{a|X|} \leq e^{aT}\}}] \leq 1 + c_1 \int_1^{e^{aT}} e^{-\frac{c_2 \log t}{a}} dt = 1 + c_1 \int_1^{e^{aT}} t^{-c_2/a} dt.\] Thus, for any \(a \in [0, \frac{c_2}{2}]\), it holds that \[\mathbb{E}[e^{a|X|}\mathbb{1}_{\{e^{a|X|} \leq e^{aT}\}}] \leq 1 + c_1(1 - e^{-aT}) \leq 1 + c_1.\] Taking the limit as \(T \rightarrow \infty\), we conclude that \(\mathbb{E}[e^{a|X|}]\) is finite for all \(a \in [0, \frac{c_2}{2}]\). Since both \(e^{aX}\) and \(e^{-aX}\) are upper bounded by \(e^{|a||X|}\), it follows that \(\mathbb{E}[e^{aX}]\) is finite for all \(|a|\leq \frac{c_2}{2}\). ◻

Proof. The proof follows automatically by applying 3, using the fact that \(\mathrm{grad}_{h}\,\tilde{F}(y) = 0\) and that \(\tilde{F}\) has a \(K_3\)-Lipschitz Hessian. ◻

Proof of 15. First of all, note that we are assuming that \(\frac{a}{\sqrt{\beta}} \leq \frac{D}{3}\). Thus, as the critical points of \(\tilde{F}\) are isolated by 15 we can write \(U\Big(\frac{a}{\sqrt{\beta}}, \tilde{\mathcal{S}}\Big)\) as a disjoint union \[U\Big(\frac{a}{\sqrt{\beta}}, \tilde{\mathcal{S}}\Big) = \bigsqcup_{y_i \in \tilde{\mathcal{S}}} B\big(y_i, \frac{a}{\sqrt{\beta}}\big),\] where each ball \(B\big(y_i, \frac{a}{\sqrt{\beta}}\big)\) is a connected component of \(U\Big(\frac{a}{\sqrt{\beta}}, \tilde{\mathcal{S}}\Big)\). Now, let \(x \in U\Big(\frac{a}{\sqrt{\beta}}, \tilde{\mathcal{S}}\Big)\). This decomposition implies that there exists one single \(y_i \in \tilde{\mathcal{S}}\) such that \(x \in \mathcal{B}\Big(\frac{a}{\sqrt{\beta}}, y_i\Big)\). Let us denote \(y := y_i\) for simplicity.

Moreover, as we are also assuming that \(\frac{a}{\sqrt{\beta}} \leq i(B)\), it holds that \[\tau^* = \tau_{\mathcal{B}(\frac{a}{\sqrt{\beta}}, y)^c} \leq \tau_{\mathcal{B}(i(B), y)^c}\] and so we can always study \(\tilde{X}_t\) in normal coordinates centered at \(y\).

Let \(v \in T_y B\) be the unitary eigenvector of \(\nabla^2 \tilde{F}(y)\) that corresponds to the minimum eigenvalue of \(\nabla^2 \tilde{F}(y)\). By 16 we know that \(\nabla^2 \tilde{F}(y)[v, v] \leq -\lambda_*\).

As mentioned earlier, we will study the process \(\tilde{r}_{y, v}(\tilde{X}_t)\) instead of \(d_h(y, \tilde{X}_t)\). We wish to show that with high probability, \(\tilde{r}_{y, v}(\tilde{X}_t)\) is larger than \(\frac{a}{\sqrt{\beta}}\). By 8 , escape of \(\tilde{X}_t\) from \(B\big(y, \frac{a}{\sqrt{\beta}}\big)\) thus occurs with high probability.

For the sake of clarity, we will use \(\tilde{r}\) as a shorthand for \(\tilde{r}_{y, v}\). Using Itô’s formula (see 98) and the standard chain rule, we know that before \(\tilde{X}_t\) leaves \(\mathcal{B}(i(B), y)\), \(\frac{1}{2}\tilde{r}(\tilde{X}_t)^2\) solves the following SDE; \[\begin{align} \label{originalSDE} \begin{aligned} d\left[\frac{1}{2}\tilde{r}(\tilde{X}_t)^2\right] &= \tilde{\operatorname{L}}\left[\frac{1}{2}\tilde{r}(\tilde{X}_t)^2\right]dt + \sqrt{\frac{2}{\beta}} \Big|\mathrm{grad}_{h}\,\frac{1}{2}\tilde{r}(\tilde{X}_t)^2\Big|_hdB_t \\&= \left(\langle -\mathrm{grad}_{h}\,\tilde{F}(\tilde{X}_t), \frac{1}{2} \mathrm{grad}_{h}\,\tilde{r}(\tilde{X}_t)^2\rangle_h + \frac{1}{2\beta} \Delta_h \tilde{r}(\tilde{X}_t)^2\right)dt \\&\quad + \sqrt{\frac{2}{\beta}} \Big|\mathrm{grad}_{h}\,\frac{1}{2}\tilde{r}(\tilde{X}_t)^2\Big|_hdB_t \\&= \left(\langle -\mathrm{grad}_{h}\,\tilde{F}(\tilde{X}_t), \tilde{r}(\tilde{X}_t)\mathrm{grad}_{h}\,\tilde{r}(\tilde{X}_t) \rangle_h \right. \\&\quad\quad \left. + \frac{1}{\beta}\left(\tilde{r}(\tilde{X}_t)\Delta_h \tilde{r}(\tilde{X}_t) + |\mathrm{grad}_{h}\,\tilde{r}(\tilde{X}_t)|^2_h\right)\right)dt \\&\quad + \sqrt{\frac{2}{\beta}} |\tilde{r}(\tilde{X}_t)||\mathrm{grad}_{h}\,\tilde{r}(\tilde{X}_t)|_h dB_t, \end{aligned} \end{align}\tag{9}\] where \(B_t\) is a one-dimensional Brownian motion in \(\mathbb{R}\).

With the aim of controlling the first escaping time of \(\tilde{X}_t\) from \(U\Big(\frac{a}{\sqrt{\beta}}, \tilde{\mathcal{S}}\Big)\), it would be useful to obtain an upper bound on \[\label{eq:probtilderpeq} \mathbb{P}\Big[\frac{1}{2}\tilde{r}(\tilde{X}_t)^2 < \frac{a^2}{2\beta}\Big],\tag{10}\] since whenever \(\frac{1}{2}\tilde{r}(\tilde{X}_t)^2 \geq \frac{a^2}{2\beta}\), it holds that \(\frac{1}{2}d_h(\tilde{X}_t, \tilde{\mathcal{S}}) \geq \frac{a^2}{2\beta}\), and so it must be the case that \(\tau^* \leq t\). Therefore, having an upper bound for the probability shown in 10 which decays exponentially with respect to \(t\) would allow us to finish the proof, as we will see later.

Nevertheless, to the best of our knowledge, with the SDE shown in 9 it is far from obvious to obtain upper bounds for 10 . To circumvent this problem, we will obtain a stochastic process which lower bounds \(\frac{1}{2}\tilde{r}(\tilde{X}_t)^2\). To find such a process, we will use standard comparison results for SDEs (cf. 11.3). Indeed, if we are able to lower bound the drift term of 9 , we will be able to define a new SDE for a process which lower bounds \(\frac{1}{2}\tilde{r}(\tilde{X}_t)^2\).

The drift term of the SDE shown in 9 is given by \[\label{eq:originaldrift} \mathcal{D}(\tilde{X}_t) := \langle -\mathrm{grad}_{h}\,\tilde{F}(\tilde{X}_t), \tilde{r}(\tilde{X}_t)\mathrm{grad}_{h}\,\tilde{r}(\tilde{X}_t) \rangle_h + \frac{1}{\beta}\left(\tilde{r}(\tilde{X}_t)\Delta_h \tilde{r}(\tilde{X}_t) + |\mathrm{grad}_{h}\,\tilde{r}(\tilde{X}_t)|^2_h\right).\tag{11}\] Let us first lower bound the first summand of this expression. Because parallel transport is an isometry, we can write \[\begin{align} \label{eq:auxbound1} \begin{aligned} \langle -\mathrm{grad}_{h}\,\tilde{F}(x), \mathrm{grad}_{h}\,\tilde{r}(x)\rangle_h &= \langle -\mathrm{P}_{\log_y x}\,H(x), \mathrm{grad}_{h}\,\tilde{r}(x)\rangle_h \\&\quad -\langle \mathrm{grad}_{h}\,\tilde{F}(x) - \mathrm{P}_{\log_y x}\, H(x), \mathrm{grad}_{h}\,\tilde{r}(x)\rangle_h \\&\geq \langle -H(x), \mathrm{P}^{-1}_{\log_y x}\, \mathrm{grad}_{h}\,\tilde{r} (x)\rangle_h \\&\quad - |\mathrm{P}^{-1}_{\log_y x}\,\mathrm{grad}_{h}\,\tilde{F}(x) - H(x)|_h |\mathrm{grad}_{h}\,\tilde{r}(x)|_h \\&\geq \langle -H(x), \mathrm{P}^{-1}_{\log_y x}\,\mathrm{grad}_{h}\,\tilde{r}(x)\rangle_h - \frac{A_3 a^2}{2\beta} |\mathrm{grad}_{h}\,\tilde{r}(x)|_h, \end{aligned} \end{align}\tag{12}\] for every \(x \in \mathcal{B}(\frac{a}{\sqrt{\beta}}, y)\), where \(\mathrm{P}_{\log_y x}\) denotes the parallel transport from \(y\) to \(x\) along the unique geodesic joining the two points. The last inequality follows from 4.

Moreover, since \(\frac{a}{\sqrt{\beta}} \leq \min\{i(M), i(B), \frac{1}{72 \dim(M)^{5/2} \mathbf{K}}\}\) by assumption, we can apply 32 to conclude that⁸ \[\begin{align} \label{eq:auxbound2} \begin{aligned} |\mathrm{grad}_{h}\,\tilde{r}(x)|^2_h &\leq 2,\\ |\mathrm{grad}_{h}\,\tilde{r}(x)|^2_h + \tilde{r}(x)\Delta_h\tilde{r}(x) &\geq \frac{1}{2}, \end{aligned} \end{align}\tag{13}\] for every \(x \in \mathcal{B}(\frac{a}{\sqrt{\beta}}, y)\), and so, inserting 12 13 into 11 shows that \[\mathcal{D}(x) \geq \tilde{r}(x) \left(\langle -H(x), \mathrm{P}^{-1}_{\log_y x}\,\mathrm{grad}_{h}\,\tilde{r}(x)\rangle_h - \frac{A_3 a^2}{\beta}\right) + \frac{1}{2\beta},\] for every \(x \in \mathcal{B}(\frac{a}{\sqrt{\beta}}, y)\).

Lastly, as \(\frac{a}{\sqrt{\beta}} \leq \frac{1}{\sqrt{2}\dim(M)^{3/2} \mathbf{K}}\), using 33 we can further bound \[\langle -H(x), \mathrm{P}^{-1}_{\log_y x}\, \mathrm{grad}_{h}\,\tilde{r}(x)\rangle_h \geq \lambda_* \tilde{r}(x) - 4\dim(M)^5 A_2 \mathbf{K} \frac{a^3}{\beta\sqrt{\beta}},\] for every \(x \in \mathcal{B}(\frac{a}{\sqrt{\beta}}, y)\), and conclude that \[\begin{align} \mathcal{D}(x) &\geq \tilde{r}(x) \left(\lambda_* \tilde{r}(x) - 4\dim(M)^5 A_2 \mathbf{K} \frac{a^3}{\beta\sqrt{\beta}}- \frac{A_3 a^2}{\beta} \right) + \frac{1}{2\beta} \\&= \lambda_* \tilde{r}(x)^2 - \tilde{r}(x)\left(4\dim(M)^5 A_2 \mathbf{K} \frac{a^3}{\beta\sqrt{\beta}}+ \frac{A_3 a^2}{\beta}\right) + \frac{1}{2\beta}, \end{align}\] for every \(x \in \mathcal{B}(\frac{a}{\sqrt{\beta}},y)\).

At this point, we can use the fact that \(\tilde{r}(x) < \frac{a}{\sqrt{\beta}}\) for every \(x \in \mathcal{B}(\frac{a}{\sqrt{\beta}},y)\) to reach \[\frac{1}{2\beta}- \tilde{r}(x)\left(4\dim(M)^5 A_2 \mathbf{K} \frac{a^3}{\beta\sqrt{\beta}}+ \frac{A_3 a^2 }{\beta}\right) > \frac{1}{\beta} \left(\frac{1}{2}- \left(4\dim(M)^5 A_2 \mathbf{K} \frac{a^4}{\beta}+ \frac{A_3 a^3 }{\sqrt{\beta}}\right)\right).\]

As we are also choosing \[\beta \geq 72^2 \dim(M)^5 A_2 A^2_3 \mathbf{K}^2 a^6 \geq \max\{32 \dim(M)^5 A_2 \mathbf{K}a^4, 64 A_3^2a^6\},\] we find that \[\left(\frac{1}{2}- \left(4\dim(M)^5 A_2 \mathbf{K} \frac{a^4}{\beta}+ \frac{A_3 a^3 }{\sqrt{\beta}}\right)\right) \geq \frac{1}{2} - \Big(\frac{1}{8} + \frac{1}{8}\Big) = \frac{1}{4}.\] This bound implies that, before \(\tilde{X}_t\) leaves \(U\Big(\frac{a}{\sqrt{\beta}}, \tilde{\mathcal{S}}\Big)\), it holds that \[\label{ineqdttermsSDE} \mathcal{D}(\tilde{X}_t) \geq \lambda_* \tilde{r}(\tilde{X}_t)^2 + \frac{1}{4\beta}.\tag{14}\]

Using the comparison result for SDEs (see 106), we can obtain a process \(z_t\) such that its square lower bounds \(\tilde{r}(\tilde{X}_t)^2\) before \(X_t\) leaves \(\mathcal{B}(\frac{a}{\sqrt{\beta}}, y)\), i.e. \[\label{eq:comparisonprocesses} \mathbb{P}\Big[\tilde{r}^2(\tilde{X}_t) \geq z_t^2 \, |\, t < \tau_{\mathcal{B}(\frac{a}{\sqrt{\beta}},y)^c}\Big] = 1,\quad \forall t \geq 0.\tag{15}\] The process \(z_t\) is such that \(\frac{1}{2}(z_t)^2\) verifies the SDE \[d\left[\frac{1}{2}(z_t)^2\right] = \left[2 \lambda_* \frac{1}{2} (z_t)^2 + \frac{1}{4\beta} \right]dt + \sqrt{\frac{2}{\beta}} |z_t| \theta_t\, dB_t,\] where \(\theta_t = |\mathrm{grad}_{h}\,\tilde{r}(\tilde{X}_t)|_h\). Choosing \(Y_t = \frac{1}{2}(z_t)^2\), we can rewrite this SDE as \[dY_t = \Big(2\lambda_* Y_t + \frac{1}{4\beta}\Big)dt + 2\sqrt{\frac{Y_t}{\beta}} \theta_t\, dB_t.\]

Recall that in 13 we showed that \(\theta^2_t \leq 2\). Thus, by 102 we know that for every \(A > 0\) and every \(0 < \theta \leq \frac{\lambda_* \beta}{2}\), it holds that \[P[Y_t \leq A] \leq e^{\theta (A - Y_0)} e^{-\frac{\theta}{4\beta}t},\quad \forall t \geq 0.\] Choosing \(\theta = \frac{\lambda_* \beta}{2}\) and \(A = \frac{a^2}{2\beta}\) we can conclude that \[P\Big[Y_t \leq \frac{a^2}{2\beta}\Big] \leq C e^{-\frac{\lambda_*}{8}t},\] with \(C := \exp\Big(\frac{\lambda_* \beta}{2} (\frac{a^2}{2\beta} - Y_0)\Big)\). This bound allows us to conclude the proof: since \[\tau^* \geq t \iff \sup_{s \in [0, t]} \frac{1}{2}d_h(\tilde{X}_s, \tilde{\mathcal{S}})^2 < \frac{a^2}{2\beta},\] we know that \[\mathbb{P}[\tau^* \geq t | \tilde{X}_0 = x] = \mathbb{P}\left[\sup_{s \in [0, t]} \frac{1}{2}d_h(\tilde{X}_s, \tilde{\mathcal{S}})^2 < \frac{a^2}{2\beta} \bigg| \tilde{X}_0 = x\right].\] Now, \[\mathbb{P}\left[\sup_{s \in [0, t]} \frac{1}{2}d_h(\tilde{X}_s, \tilde{\mathcal{S}})^2 < \frac{a^2}{2\beta} \bigg| \tilde{X}_0 = x\right] \leq \mathbb{P}\left[\frac{1}{2}d_h(\tilde{X}_t, \tilde{\mathcal{S}})^2 < \frac{a^2}{2\beta} \bigg| \tilde{X}_0 = x\right].\] Therefore, since with probability one \[Y_t \leq \frac{1}{2}\tilde{r}(\tilde{X}_t)^2 \leq \frac{1}{2}d_h(\tilde{X}_t, S)^2,\] we can conclude that \[\begin{align} \mathbb{P}\left[ \frac{1}{2}d_h(\tilde{X}_t, \tilde{\mathcal{S}})^2 < \frac{a^2}{2\beta} \bigg| \tilde{X}_0 = x\right] &\leq \mathbb{P}\left[\frac{1}{2}\tilde{r}_{y}(\tilde{X}_t)^2 < \frac{a^2}{2\beta} \bigg| \tilde{X}_0 = x \right] \\&\leq \mathbb{P}\left[\left. Y_t < \frac{a^2}{2 \beta} \right| Y_0 = \frac{1}{2}\tilde{r}^2(x) \right] \\&\leq e^{-\frac{\lambda_*}{8} t}C, \end{align}\] where the second inequality follows from the bound of 15 , and \(\lambda_*\) is the bound on the eigenvalues of \(\nabla^2 F\) from 14. ◻

Proof. In 15 we proved that for every \(x \in U\Big(\frac{a}{\sqrt{\beta}}, \tilde{\mathcal{S}}\Big)\), there exists some \(C >0\) such that \[\mathbb{P}[\tau^* \geq t | \tilde{X}_0 = x] \leq C e^{-\lambda_* t / 8},\quad \forall\, t \geq 0\] and so by 2 we know that \[\mathbb{E}\Big[\exp\Big(\frac{\lambda_*}{16} \tau^*\Big) \Big| \tilde{X}_0 = x\Big] < \infty,\] for every \(x \in U\Big(\frac{a}{\sqrt{\beta}}, \tilde{\mathcal{S}}\Big)\).

Furthermore, by the choice of \(\beta\), it holds that \(\frac{a}{\sqrt{\beta}} \leq i(B)\) and so, as we saw in 14 it holds without loss of generality that \(\partial U\Big(\frac{a}{\sqrt{\beta}}, \tilde{\mathcal{S}}\Big)\) is smooth.

Finally, it also holds without loss of generality that \(-\lambda_*/16 \notin \sigma(\tilde{\operatorname{L}})\), and so all of the conditions of 1 are satisfied, which allows us to conclude the proof. ◻

Proof. Let \(x^*\) be the unique minimum of \(\tilde{F}\). From the definition of \(\tilde{F}\) having an \(A_3\)-Lipschitz Hessian (cf. 43) we know that for every \(x \in G_3\) \[\max_{\substack{s \in T_{x^*}B\\|s|_h=1}} |\mathrm{P}^{-1}_{\log_{x^*}x} \circ \nabla^2 \tilde{F}(x) \circ \mathrm{P}_{\log_{x^*}x}[s] - \nabla^2 \tilde{F}(x^*)[s]|_h \leq A_3 d_h(x,x^*),\] where \(\mathrm{P}_{\log_{x^*}x}\) denotes the parallel transport from \(x^*\) to \(x\) along the unique geodesic joining the two points. Since the parallel transport is an isometry, we can use Weyl’s inequality to conclude that for every \(x \in G_3\), \[\begin{align} |\lambda_1(\nabla^2 \tilde{F}(x)) - \lambda_1(\nabla^2 \tilde{F}(x^*))| &\leq \max_{\substack{s \in T_{x^*}M\\|s|_h=1}} |\mathrm{P}^{-1}_{\log_{x^*}x} \circ \nabla^2 \tilde{F}(x) \circ \mathrm{P}_{\log_{x^*}x}[s] - \nabla^2 \tilde{F}(x^*)[s]|_h \\&\leq A_3 d_h(x,x^*), \end{align}\] where \(\lambda_1(\nabla^2\tilde{F}(x))\) denotes the first eigenvalue of the Hessian of \(\tilde{F}\) at \(x\). This way, we can apply 17 to conclude that \[\lambda_1(\nabla^2 \tilde{F}(x)) \geq \lambda_1(\nabla^2 \tilde{F}(x^*)) - A_3 d_h(x, x^*) \geq \frac{\lambda_*}{2},\quad \mathrm{whenever } d_h(x, x^*) \leq \frac{\lambda_*}{2A_3}.\] The above inequality is fulfilled on \(G_3 = \{x \in B : d_h(x, x^*)^2 < \frac{a^2}{\beta}\}\), as by assumption \(\frac{a^2}{\beta} \leq \frac{\lambda^2_*}{4 A^2_3}\).

For this reason, and the fact that we assumed that \(\mathrm{Ric}_h \geq -R_B\) where \(R_B \geq 0\), then, given our choice of \(\beta\), \[\nabla^2 \tilde{F}(x) + \frac{1}{\beta}\mathrm{Ric}_h \geq \left(\frac{\lambda_*}{2} - \frac{1}{\beta}R_B\right)h \geq \frac{\lambda_*}{4}h.\]

Since we are ensuring that \(\frac{a}{\sqrt{\beta}} \leq \mathit{conv}(B)\), we know that \((G_3, \left.\tilde{\nu}\right|_{G_3}, \tilde{\Gamma}|_{G_3})\) satisfies a \(\mathrm{PI}(\lambda_*/8)\). See 97 for more details. ◻

Proof. First, we can define a smooth bump function \(\psi: \mathbb{R} \to [0, 1]\) such that \[\psi(x) = \begin{cases} 0, \quad & x \leq r/2,\\ \mathit{increasing}, \quad &x\in (r/2, r),\\ 1,\quad & x\geq r, \end{cases}\] with \(\norm{\psi'}_{\infty} \leq \frac{2}{r} + \varepsilon\) following standard methods (cf. [18]). Now, we define \(\chi: B \to [0, 1]\) as \[\chi(x) := \psi(d_h(x,\tilde{\mathcal{S}})).\] Note that \(\psi(x)\) is constant in the interior of \(G_1\) and in the complement of \(G_2\). Furthermore, since we are assuming that \(r \leq i(B)\), it holds by 6 that the distance function is smooth in \(G_2 \setminus \tilde{\mathcal{S}}\), and so \(\chi\) is smooth in \(B\) and \[\mathrm{grad}_{h}\,\chi = \mathrm{grad}_{h}\,\psi(d_h(x, \tilde{\mathcal{S}})) = \psi'(d_h(x, S)) \mathrm{grad}_{h}\,d_h(x, \tilde{\mathcal{S}}),\] for every \(x \in B\) such that \(r/2 < d_h(x, \tilde{\mathcal{S}}) < r\) and \(\mathrm{grad}_{h}\,\chi \equiv 0\) elsewhere. Again, by 6, the distance function has a gradient with constant norm one, and so it holds that \[|\mathrm{grad}_{h}\,\chi(x)|_h = |\psi'(d_h(x, \tilde{\mathcal{S}}))|_h\] on \(B\).

For this reason, for every \(x \in B\) such that \(r/2 < d_h(x, \tilde{\mathcal{S}}) < r\), \[\lVert\tilde{\Gamma}(\chi(x))\rVert_{\infty} = \frac{1}{\beta} \lVert\mathrm{grad}_{h}\,\chi(x)\rVert^2_\infty = \frac{1}{\beta}\lVert\psi'(d_h(x, \tilde{\mathcal{S}}))\rVert^2_\infty \leq \frac{1}{\beta}\Big(\frac{2}{r} + \varepsilon\Big)^2.\] ◻

Proof. By the definition of \(\tilde{\nu}\) and \(\tilde{\operatorname{L}}\), \[\begin{align} \int_N \phi \tilde{\operatorname{L}}W \,d\tilde{\nu} &= \int_N \phi\langle-\mathrm{grad}_{h}\,\tilde{F}, \mathrm{grad}_{h}\,W\rangle_h \,d\tilde{\nu} + \frac{1}{\beta} \int_N \phi \Delta_h W \,d\tilde{\nu} \\&= \int_N \phi\langle-\mathrm{grad}_{h}\,\tilde{F}, \mathrm{grad}_{h}\,W\rangle_h \,d\tilde{\nu} + \frac{1}{\beta} \int_N \phi e^{-\beta \tilde{F}}\Delta_h W\; \mathrm{dVol}_h. \end{align}\] Using Green’s identity, since we are assuming that \(\phi\) vanishes at \(\partial N\), it holds that \[\begin{align} \int_N \phi \tilde{\operatorname{L}}W \,d\tilde{\nu} &= \int_N \phi\langle-\mathrm{grad}_{h}\,\tilde{F}, \mathrm{grad}_{h}\,W\rangle_h \,d\tilde{\nu} - \frac{1}{\beta} \int_N \langle \mathrm{grad}_{h}\,(\phi e^{-\beta \tilde{F}}),\mathrm{grad}_{h}\,W\rangle_h\; \mathrm{dVol}_h \\&= \int_N \phi\langle-\mathrm{grad}_{h}\,\tilde{F}, \mathrm{grad}_{h}\,W\rangle_h \,d\tilde{\nu} - \frac{1}{\beta} \int_N \langle \mathrm{grad}_{h}\,\phi,\mathrm{grad}_{h}\,W\rangle_h \,d\tilde{\nu} \\&\quad + \int_N \phi \langle \mathrm{grad}_{h}\,\tilde{F},\mathrm{grad}_{h}\,W\rangle_h \,d\tilde{\nu} \\&= - \frac{1}{\beta} \int_N \langle \mathrm{grad}_{h}\,\phi,\mathrm{grad}_{h}\,W\rangle_h \,d\tilde{\nu}. \end{align}\] ◻

Proof of 10. Let \(f \in C^2(B)\). Without loss of generality, we can assume that \(\int_{G_3} f \, d\tilde{\nu} = 0\), as \[\int_B f^2 \, d\tilde{\nu} - \Big(\int_B f \,d\tilde{\nu} \Big)^2 \leq \int_B (f + c)^2 \, d\tilde{\nu},\] for every \(c \in \mathbb{R}\). Our goal will be to upper bound the right-hand side of this inequality by some multiple of \(\int_B |\mathrm{grad}_{h}\,f|^2_h \,d\tilde{\nu}\) in order to obtain a Poincaré inequality.

Let \(\chi\) be the bump function from 7. Since \(\chi \equiv 0\) in \(G_1\) and \(\chi \equiv 1\) in \(G_2^c\), we know that \[\begin{align} \label{firstinequality} \begin{aligned} \int_B f^2 \,d\tilde{\nu} &= \int_B (f(1 - \chi) + f \chi)^2 \,d\tilde{\nu} \\&\leq 2\int_B f^2(1-\chi)^2 \, d\tilde{\nu} + 2\int_B f^2\chi^2 \, d\tilde{\nu} \\&= 2\int_{G_2} f^2(1-\chi)^2 \, d\tilde{\nu} + 2\int_{G_1^c} f^2\chi^2 \, d\tilde{\nu}. \end{aligned} \end{align}\tag{17}\]

In 2 we proved that \(W_2 \in C^\infty(\overline{G_2}) \cap C^\infty(G_2^c)\), \(W_2 \geq 1\) and \[\label{eq:lyapunovW2} \frac{\tilde{\operatorname{L}}W_2(x)}{W_2(x)} = -\frac{\lambda_*}{16},\quad \forall x \in G_2.\tag{18}\] Using 18 and the Neumann boundary condition from 8 we can bound the first summand of 17 as follows: \[\begin{align} \int_{G_2} f^2(1-\chi)^2 \,d\tilde{\nu} &= -\frac{16}{\lambda_*}\int_{G_2} \frac{\tilde{\operatorname{L}}W_2}{W_2}f^2(1-\chi)^2 \,d\tilde{\nu} \\&= \frac{16}{\lambda_* \beta} \int_{G_2} \Big\langle \mathrm{grad}_{h}\,\frac{f^2(1-\chi)^2}{W_2}, \mathrm{grad}_{h}\,W_2\Big\rangle_h \,d\tilde{\nu}, \end{align}\] since \((1-\chi)\) vanishes at \(\partial G_2\). Using the chain rule we can further bound this expression, \[\begin{align} \label{eq:ineqchainrule} \begin{aligned} \frac{16}{\lambda_* \beta} \int_{G_2} \Big\langle \mathrm{grad}_{h}\,\frac{f^2(1-\chi)^2}{W_2}, \mathrm{grad}_{h}\,&W_2\Big\rangle_h \,d\tilde{\nu} \\&= \frac{32}{\lambda_*\beta} \int_{G_2} \frac{f(1-\chi)}{W_2}\langle \mathrm{grad}_{h}\,f(1-\chi), \mathrm{grad}_{h}\,W_2\rangle_h \,d\tilde{\nu} \\&\quad - \frac{16}{\lambda_*\beta}\int_{G_2} \frac{f^2(1-\chi)^2}{W_2^2} |\mathrm{grad}_{h}\,W_2|^2_h \,d\tilde{\nu} \\&= \frac{16}{\lambda_* \beta} \int_{G_2} |\mathrm{grad}_{h}\,f(1-\chi)|^2_h \,d\tilde{\nu} \\&\quad- \frac{16}{\lambda_* \beta} \int_{G_2} \Big|\mathrm{grad}_{h}\,f(1-\chi)- \frac{f(1-\chi)}{W_2}\mathrm{grad}_{h}\,W_2\Big|^2_h \,d\tilde{\nu} \\&\leq \frac{16}{\lambda_* \beta} \int_{G_2} |\mathrm{grad}_{h}\,f(1-\chi)|^2_h \,d\tilde{\nu} \\&= \frac{16}{\lambda_*}\int_{G_2} \tilde{\Gamma}(f(1-\chi)) \,d\tilde{\nu}. \end{aligned} \end{align}\tag{19}\] Lastly, using the fact that \[\label{productGamma} \tilde{\Gamma}(fg) \leq 2(f^2 \tilde{\Gamma}(g) + g^2 \tilde{\Gamma}(f)),\quad \forall f,g \in C^2(B),\tag{20}\] it holds that \[\begin{align} \frac{16}{\lambda_*}\int_{G_2} \tilde{\Gamma}(f(1-\chi)) \,d\tilde{\nu} &\leq \frac{32}{\lambda_*} \left(\int_{G_2} f^2 \tilde{\Gamma}(\chi) \,d\tilde{\nu} + \int_{G_2} (1-\chi)^2 \tilde{\Gamma}(f) \,d\tilde{\nu}\right) \\&\leq \frac{32\lVert\tilde{\Gamma}(\chi)\rVert_\infty}{\lambda_*} \int_{G_2} f^2 \,d\tilde{\nu} + \frac{32}{\lambda_*}\int_{G_2} \tilde{\Gamma}(f) \,d\tilde{\nu} \\&\leq \frac{32\lVert\tilde{\Gamma}(\chi)\rVert_\infty}{\lambda_*} \int_B f^2 \,d\tilde{\nu} + \frac{32}{\lambda_*}\int_B \tilde{\Gamma}(f) \,d\tilde{\nu}. \end{align}\]

Thus, we have concluded that \[\label{conclusion1} 2\int_{G_2} f^2(1-\chi)^2 d\tilde{\nu} \leq \frac{64\lVert\tilde{\Gamma}(\chi)\rVert_\infty}{\lambda_*} \int_B f^2 d\tilde{\nu} + \frac{64}{\lambda_*}\int_B \tilde{\Gamma}(f)d\tilde{\nu}.\tag{21}\]

Moreover, in 12 we proved that \(W_1\) is smooth, \(W_1 \geq 1\) and \[\label{eq:lyapunovW1} \frac{\tilde{\operatorname{L}}W_1(x)}{W_1(x)} \leq -A_2 \dim(B) + \frac{3}{2}A_2 \dim(B) \mathbb{1}_{G_3}(x), \quad \forall x \in G_1^c \cup G_3.\tag{22}\] Thus, the second summand of 17 can be bounded similarly, using 22 and 8, since \(\chi\) vanishes on \(\partial G_1\). Recall that \(G_1^c = G_3 \cup G_4\). Thus, \[\begin{align} \int_{G_1^c} f^2 \chi^2 \,d\tilde{\nu} &\leq \frac{1}{A_2 \dim(B)} \int_{G_1^c} \frac{-\tilde{\operatorname{L}}W_1}{W_1} f^2 \chi^2 \,d\tilde{\nu} + \frac{3}{2} \int_{G_3} f^2 \,d\tilde{\nu} \\&= \frac{1}{A_2 \beta \dim(B)} \int_{G_1^c} \Big\langle \mathrm{grad}_{h}\,\frac{f^2 \chi^2}{W_1}, \mathrm{grad}_{h}\,W_1\Big\rangle_h \,d\tilde{\nu} + \frac{3}{2} \int_{G_3} f^2 \,d\tilde{\nu} \\&\leq \frac{1}{A_2 \dim(B)} \int_{G_1^c} \tilde{\Gamma}(f\chi) \,d\tilde{\nu} + \frac{3}{2}\int_{G_3} f^2 \,d\tilde{\nu}, \end{align}\] where the last inequality is obtained following an analogous procedure as in 19 . Now, since we already proved that \((G_3, \left.\tilde{\nu}\right|_{G_3}, \tilde{\Gamma}|_{G_3})\) satisfies a \(\mathrm{PI}(\frac{\lambda_*}{8})\), it holds that \[\int_{G_3} f^2 \,d\tilde{\nu} = \frac{Z_{G_3}}{Z_B} \int_{G_3} f^2 \,\left.d\tilde{\nu}\right|_{G_3} \leq \frac{Z_{G_3}}{Z_B} \frac{8}{\lambda_*}\int_{G_3} \tilde{\Gamma}(f) \, \left.d\tilde{\nu}\right|_{G_3} = \frac{8}{\lambda_*}\int_{G_3} \tilde{\Gamma}(f) \,d\tilde{\nu}.\] Thus, \[\begin{align} \int_{G_1^c} f^2 \chi^2 \,d\tilde{\nu} &= \frac{1}{A_2 \dim(B)} \int_{G_1^c} \tilde{\Gamma}(f\chi) \,d\tilde{\nu} + \frac{3}{2}\int_{G_3} f^2 \,d\tilde{\nu} \\&\leq \frac{1}{A_2 \dim(B)} \int_{G_1^c} \tilde{\Gamma}(f\chi) \,d\tilde{\nu} + \frac{12}{\lambda_*}\int_{G_3} \tilde{\Gamma}(f) \,d\tilde{\nu} \\&\leq \frac{1}{A_2 \dim(B)} \int_{B} \tilde{\Gamma}(f\chi) \,d\tilde{\nu} + \frac{12}{\lambda_*}\int_B \tilde{\Gamma}(f) \,d\tilde{\nu}. \end{align}\] Using 20 we can upper bound this expression, \[\begin{align} \frac{1}{A_2 \dim(B)} \int_{B} \tilde{\Gamma}(f\chi) \,d\tilde{\nu} + \frac{12}{\lambda_*}\int_B \tilde{\Gamma}(f) \,d\tilde{\nu} &\leq \frac{2\lVert\tilde{\Gamma}(\chi)\rVert_\infty}{A_2 \dim(B)} \int_{B} f^2 \,d\tilde{\nu} \\&\quad + \left(\frac{2}{A_2 \dim(B)}+ \frac{12}{\lambda_*}\right) \int_{B} \tilde{\Gamma}(f) \,d\tilde{\nu}, \end{align}\] which ultimately leads to \[\label{conclusion2} 2\int_{G_1^c} f^2 \chi^2 \,d\tilde{\nu} \leq \frac{4\lVert\tilde{\Gamma}(\chi)\rVert_\infty}{A_2 \dim(B)} \int_{B} f^2 \,d\tilde{\nu} + \left(\frac{4}{A_2 \dim(B)} + \frac{24}{\lambda_*}\right) \int_{B} \tilde{\Gamma}(f) \,d\tilde{\nu}.\tag{23}\]

Inserting 21 23 into 17 , we can conclude that \[\begin{align} \int_B f^2 \,d\tilde{\nu} \leq 4\lVert\tilde{\Gamma}(\chi)\rVert_\infty\left(\frac{16}{\lambda_*} + \frac{1}{A_2 \dim(B)} \right)\int_{B} f^2 \,d\tilde{\nu} + \left(\frac{4}{A_2 \dim(B)} + \frac{88}{\lambda_*}\right) \int_{B} \tilde{\Gamma}(f) \,d\tilde{\nu}. \end{align}\]

Recall that \(\chi\) is such that \[\lVert\tilde{\Gamma}(\chi)\rVert_{\infty} \leq \frac{1}{\beta}\Big(\frac{2}{r} + \varepsilon\Big)^2,\] for every \(\varepsilon > 0\), where \(r = \frac{a}{\sqrt{\beta}}\). Taking \(\varepsilon \to 0\), we get \[\lVert\Gamma(\chi)\rVert_{\infty} \leq \frac{4}{a^2}.\] Therefore, since \(A_2 \dim(B) \geq 1 \geq \lambda_*\), it holds that \[\begin{align} 1 - 4\norm{\Gamma(\chi)}_{\infty} \left(\frac{16}{\lambda_*} + \frac{1}{A_2 \dim(B)} \right) &\geq 1 - 4\norm{\Gamma(\chi)}_{\infty} \frac{17}{\lambda_*} \\&\geq 1 - \frac{272}{a^2 \lambda_*} \\&\geq \frac{1}{2}, \end{align}\] where the last inequality follows from the assumption that \(a^2 \geq \frac{544}{\lambda_*}\). Therefore, we know that \[\int_B f^2 \,d\tilde{\nu} \leq \Big(\frac{8}{A_2 \dim(B)} + \frac{176}{\lambda_*}\Big) \int_{B} \tilde{\Gamma}(f) \,d\tilde{\nu}.\] Lastly, using again that \(A_2 \dim(B) \geq 1 \geq \lambda_*\) we conclude that \[\int_B f^2 \,d\tilde{\nu} \leq \frac{184}{\lambda_*} \int_{B} \tilde{\Gamma}(f) \,d\tilde{\nu},\] and so the claim follows. ◻

Proof. Let \(\eta = \mathrm{dVol}_{h(\pi(x))}\), and \(\omega = \mathrm{dVol}_{\hat{g}(x)}\). Since \(\pi\) is a Riemannian submersion, it has no critical points and \[\mathrm{dVol}_{\hat{g}(x)} \wedge \pi^* \mathrm{dVol}_{h(\pi(x))} = \mathrm{dVol}_{g(x)},\] which follows from the decomposition of \(g\) into its horizontal and vertical components (cf. 8.4). ◻

Proof. It follows from 16 that \(\hat{F}\) is well-defined and integrable, so it only remains to prove that \(\hat{F} \in C^k(M/G)\). For every \(x_0 \in M/G\) we can consider an open neighborhood \(U\) around \(x_0\) and a local trivialization, so that in local coordinates \[\hat{F}(x) = \int_{\pi^{-1}(x)} f(y) \,\mathrm{dVol}_{\hat{g}}(y) = \int_{G} f(x, y)\, \mathrm{dVol}_{\hat{g}}(y),\] for every \(x \in U\). Now, since \(G\) is compact, if \(f\) is continuous on \(M\), we can apply the dominated convergence theorem to conclude that \(\hat{F}\) is continuous at \(x_0\). Thus, \(\hat{F}\) is continuous in \(M/G\). Furthermore, if \(f\) is assumed to be in \(C^k(M)\), then, using the compactness of \(G\) and the regularity of \(f\), we can differentiate under the integral sign to conclude that \(\hat{F}\) is \(C^k(M/G)\). ◻

Proof. The result simply follows by rewriting \[\begin{align} \int_M f(x) d\nu(x) &= \frac{1}{Z}\int_M f(x)e^{-\beta F(x)}\; \mathrm{dVol}_{g}(x) \\&= \frac{1}{Z}\int_{M/G} \left(\int_{\pi^{-1}(x)} f(y)e^{-\beta F(y)}\; \mathrm{dVol}_{\hat{g}}(y) \right)\; \mathrm{dVol}_{h}(x) \\&= \frac{1}{Z}\int_{M/G} \left(\int_{\pi^{-1}(x)} f(y)\; \mathrm{dVol}_{\hat{g}}(y) \right)\; e^{-\beta \tilde{F}(x)}\; \mathrm{dVol}_{h}(x) \\&= \frac{\tilde{Z}}{Z}\int_{M/G} \left(\int_{\pi^{-1}(x)} f(y)\; \mathrm{dVol}_{\hat{g}}(y) \right)\; d\tilde{\nu}(x) \\&= \frac{\tilde{Z}}{Z}\int_{M/G} \tilde{f}(x)\, d\tilde{\nu}(x). \end{align}\] ◻

Proof. First, recall that as \((G, \hat{g})\) has finite volume and \(f \in C^2(M)\), it holds that \(|f|\) and \(|(\mathrm{grad}_g\, f)^{\mathcal{H}}|^2_g\) are both integrable in \(\pi^{-1}(x)\) for every \(x \in M/G\), so the above expression is well-defined.

Let us now fix \(x \in M/G\). We choose coordinates \((z^1, \dotsc, z^k)\) in \(M/G\) centered at \(x = (0, \dotsc, 0)\) and such that \(\{\frac{\partial}{\partial z^1}, \dotsc, \frac{\partial}{\partial z^k}\}\) is an orthonormal basis for \(T_x M/G\), i.e. \(\left.h_{ij}\right|_x = \delta_{ij}\).

Furthermore, we can extend these coordinates to \(\pi^{-1}(x) \setminus \mathcal{X} (\simeq G \setminus \mathcal{X})\), where \(\mathcal{X}\) is a measure-zero set. By a slight abuse of notation, we denote such coordinates as \[(z^1, \dotsc, z^k, y^1, \dotsc, y^{n-k}).\] This way, every point \(y \in \pi^{-1}(x) \setminus \mathcal{X}\) can be written as \((0, \dotsc, 0, y^1, \dotsc, y^{n-k})\). Furthermore, in these coordinates, the local frame \[\left\{\frac{\partial}{\partial z^1}, \dotsc, \frac{\partial}{\partial z^k}, \frac{\partial}{\partial y^1}, \dotsc, \frac{\partial}{\partial y^{n-k}}\right\}\] is such that \(\left\{\frac{\partial}{\partial z^1}, \dotsc, \frac{\partial}{\partial z^k}\right\}\) are orthonormal with respect to \(g\) and horizontal with respect to \(\pi\) (cf. 26), \(\left\{\frac{\partial}{\partial y^1}, \dotsc, \frac{\partial}{\partial y^{n-k}}\right\}\) are vertical and \(g(\frac{\partial}{\partial z^i}, \frac{\partial}{\partial y^j}) = 0\) for every \(i \in \{1, \dotsc, k\}\), \(j \in \{1, \dotsc, n-k\}\).

From our choice of coordinates, we know that on \(\pi^{-1}(x) \setminus \mathcal{X}\), \(g\) is of the form \[\label{matrixmetric} \left(\begin{array}{ c | c } \mathbb{1} & 0 \\ \hline 0 & A \end{array}\right),\tag{25}\] where \(\mathbb{1}\) is the identity matrix of size \(k \times k\) and \(A\) is an \((n-k) \times (n-k)\) matrix.

In such coordinates, since the metric \(h\) at \(x\) is simply the Euclidean metric, \[\left.\mathrm{grad}_{h}( \cdot)\right|_x = h^{ij}(x)\frac{\partial }{\partial z^i}(\cdot)\frac{\partial }{\partial z^j} = \delta^{ij}\frac{\partial }{\partial z^i}(\cdot)\frac{\partial }{\partial z^j},\] and so \[\begin{gather} \left|\frac{1}{\mathrm{Vol}_{\hat{g}}(G)} \mathrm{grad}_{h} \int_{\pi^{-1}(x)} f(y)\; \mathrm{dVol}_{\hat{g}}(y)\right|_{h}^2 = \frac{1}{\mathrm{Vol}_{\hat{g}}(G)^2} \sum_{i = 1}^k \left(\frac{\partial}{\partial z^i} \int_{\pi^{-1}(x)} f(y)\; \mathrm{dVol}_{\hat{g}}(y)\right)^2 \\= \frac{1}{\mathrm{Vol}_{\hat{g}}(G)^2} \sum_{i = 1}^k \left(\frac{\partial}{\partial z^i} \int_{G \setminus \mathcal{X}} f(0, \dotsc, 0, y)\; \mathrm{dVol}_{\hat{g}}(y)\right)^2 \\= \frac{1}{\mathrm{Vol}_{\hat{g}}(G)^2} \sum_{i = 1}^k \left(\int_{G \setminus \mathcal{X}} \frac{\partial}{\partial z^i} f(0, \dotsc, 0, y)\; \mathrm{dVol}_{\hat{g}}(y)\right)^2 \\\leq \frac{1}{\mathrm{Vol}_{\hat{g}}(G)} \sum_{i = 1}^k \int_{G \setminus \mathcal{X}} \left(\frac{\partial}{\partial z^i} f(0, \dotsc, 0, y)\right)^2 \mathrm{dVol}_{\hat{g}}(y), \end{gather}\] where the last inequality is Cauchy-Schwarz’s inequality: \[\left(\int_G g(z)\; \mathrm{dVol}_{\hat{g}}(z)\right)^2 \leq \int_G g(z)^2\; \mathrm{dVol}_{\hat{g}}(z) \int_{G} \mathrm{dVol}_{\hat{g}}(z).\]

On the other hand, let us rewrite \[\label{intofgrad} \frac{1}{\mathrm{Vol}_{\hat{g}}(G)} \int_{\pi^{-1}(x)} |(\mathrm{grad}_g\, f(y))^{\mathcal{H}}|^2_g\; \mathrm{dVol}_{\hat{g}}(y)\tag{26}\] using the coordinates from the beginning of this proof, for some point in \(\pi^{-1}(x) \setminus \mathcal{X}\), \[(\mathrm{grad}_g\, f)^{\mathcal{H}} = g^{ij}\frac{\partial f}{\partial x^i}\frac{\partial }{\partial z^j},\] where \(x^i = z^i\) for \(i \in \{1, \dotsc, k\}\) and \(x^i = y^{i-k}\) for \(i \in \{k+1, \dotsc, n\}\). Moreover, since \(g\) in \(\pi^{-1}(x) \setminus \mathcal{X}\) is of the form shown in 25 we can conclude that \[(\mathrm{grad}_g\, f)^{\mathcal{H}} = \delta^{ij}\frac{\partial f}{\partial z^i}\frac{\partial }{\partial z^j}.\] Therefore, its norm with respect to \(g\) is simply \[|(\mathrm{grad}_g\, f(y))^{\mathcal{H}}|^2_g = g\left(\delta^{ij}\frac{\partial f}{\partial z^i}\frac{\partial }{\partial z^j}, \delta^{ij}\frac{\partial f}{\partial z^i}\frac{\partial }{\partial z^j}\right) = \delta_{ij}\frac{\partial f}{\partial z^i}\frac{\partial f}{\partial z^j}.\] This allows us to write 26 as \[\frac{1}{\mathrm{Vol}_{\hat{g}}(G)} \int_{G \setminus \mathcal{X}} \sum_{i = 1}^k \left(\frac{\partial}{\partial z^i}f(0,\dotsc,0, y)\right)^2 \mathrm{dVol}_{\hat{g}}(y),\] finishing the proof. ◻

Proof. Fix \(x \in M/G\). Working in the same coordinates as in the proof of 9 we know that on \(\pi^{-1}(x) \setminus \mathcal{X}\), \(g\) is of the form \[\left(\begin{array}{ c | c } \mathbb{1} & 0 \\ \hline 0 & A \end{array}\right).\] Using the same argument as before, we obtain \[\left(\mathrm{grad}_{g}\, f\right)^{\mathcal{V}} = g^{ij} \frac{\partial f}{\partial y^i}\frac{\partial }{\partial y^j},\] finishing the proof. ◻

Proof. The fact that \(\hat{F}\) belongs to \(C^2(M/G)\) follows from the regularity of \(f\) by 17. Now, using Cauchy-Schwarz’s inequality \[\hat{F}(x)^2 = \left(\int_{\pi^{-1}(x)} f(y)\, \mathrm{dVol}_{\hat{g}}(y)\right)^2 \leq \mathrm{Vol}_{\hat{g}}(G)\int_{\pi^{-1}(x)} f^2(y)\, \mathrm{dVol}_{\hat{g}}(y),\] and so by 18 \[\begin{align} \int_{M/G} \hat{F}(x)^2\, d\tilde{\nu}(x) &\leq \mathrm{Vol}_{\hat{g}}(G) \int_{M/G} \left(\int_{\pi^{-1}(x)} f^2(y)\, \mathrm{dVol}_{\hat{g}}(y)\right) d\tilde{\nu} (x) \\&= \frac{Z}{\tilde{Z}}\mathrm{Vol}_{\hat{g}}(G) \int_M f^2(x) \,d\nu(x) < \infty, \end{align}\] as \(f \in H^1(M, \nu)\), which implies that \(\hat{F} \in L^2(M/G, \tilde{\nu})\). Moreover, from 9, we know that for every \(x \in M/G\) \[|\mathrm{grad}_{h}\, \hat{F}(x)|^2_{h} \leq \mathrm{Vol}_{\hat{g}}(G) \int_{\pi^{-1}(x)} \left|\left(\mathrm{grad}_{g}\, f(y)\right)^{\mathcal{H}}\right|_{g}^2 \,\mathrm{dVol}_{\hat{g}}(y).\] Using again 18 we can conclude that \[\begin{align} \int_{M/G} |\mathrm{grad}_{h}\, \hat{F}(x)|^2_{h}\, d\tilde{\nu}(x) &\leq \mathrm{Vol}_{\hat{g}}(G) \int_{M/G} \left( \int_{\pi^{-1}(x)} \left|\left(\mathrm{grad}_{g} \,f(y)\right)^{\mathcal{H}}\right|_{g}^2 \,\mathrm{dVol}_{\hat{g}}(y)\right) d\tilde{\nu}(x) \\&= \mathrm{Vol}_{\hat{g}}(G) \frac{Z}{\tilde{Z}} \int_M \left|\left(\mathrm{grad}_{g} \,f(x)\right)^{\mathcal{H}}\right|_{g}^2\, d\nu(x) \\&\leq \mathrm{Vol}_{\hat{g}}(G) \frac{Z}{\tilde{Z}} \int_M \left|\mathrm{grad}_{g} \,f(x)\right|_{g}^2 \,d\nu(x) < \infty, \end{align}\] as \(f \in H^1(M, \nu)\), which allows us to conclude that \(\hat{F} \in H^1(M/G, \tilde{\nu})\). ◻

Proof of 20. First, recall that \((M/G, \tilde{\nu}, \tilde{\Gamma})\) satisfies a \(\mathrm{PI}(\tilde{\kappa})\) iff for every \(f \in C^2(M/G) \cap H^1(M/G, \tilde{\nu})\) \[\int_{M/G} f^2\; d\tilde{\nu} - \left(\int_{M/G} f \;d\tilde{\nu} \right)^2 \leq \frac{1}{\tilde{\kappa}\beta} \int_{M/G} |\mathrm{grad}_{h}\, f|_{h}^2\, d\tilde{\nu}.\] Let \(f \in C^2(M) \cap H^1(M, \nu)\). Without loss of generality, we can assume that \(\int_M f d\nu = 0\). Using 18 we can write \[\label{eq:initialeqPoincare} \int_M f^2\; d\nu = \frac{\tilde{Z}}{Z} \int_{M/G} \left(\int_{\pi^{-1}(x)} f^2(y) \,\mathrm{dVol}_{\hat{g}}(y)\right) d\tilde{\nu}(x).\tag{27}\] As \(f \in C^2(\pi^{-1}(x))\) for every \(x \in M/G\), we can now use the fact that the fibers satisfy a Poincaré inequality; we apply 19 to 27 and obtain \[\begin{align} \label{eq:firstauxineqpoincare} \begin{aligned} \int_M f^2\; d\nu &\leq \frac{\tilde{Z}}{Z}\left\{\frac{1}{\mathrm{Vol}_{\hat{g}}(G)}\int_{M/G} \left(\int_{\pi^{-1}(x)} f(y)\, \mathrm{dVol}_{\hat{g}}(y) \right)^2 d\tilde{\nu}(x)\right. \\&\quad\left. + C \int_{M/G} \left(\int_{\pi^{-1}(x)} \left|\mathrm{grad}_{\hat{g}} \,f(y)\right|_{\hat{g}}^2 \,\mathrm{dVol}_{\hat{g}}(y)\right) d\tilde{\nu}(x) \right\}, \end{aligned} \end{align}\tag{28}\] where \(C := \frac{\mathrm{diam}(G)^2}{\pi^2}\).

Using 10, we know that \[\left|\mathrm{grad}_{\hat{g}}\, f\right|_{\hat{g}}^2 = \left|\left(\mathrm{grad}_{g} \,f\right)^{\mathcal{V}}\right|_{g}^2,\] and so, using the fact that \(f \in H^1(M, \nu)\), we can apply 18 to the second term on the right-hand side of 28 to conclude that \[\int_M f^2\; d\nu = \frac{\tilde{Z}}{Z\mathrm{Vol}_{\hat{g}}(G)}\int_{M/G} \left(\int_{\pi^{-1}(x)} f(y)\, \mathrm{dVol}_{\hat{g}}(y) \right)^2 d\tilde{\nu}(x) + C \int_{M} \left|\left(\mathrm{grad}_{g}\, f\right)^{\mathcal{V}}\right|_{g}^2 d\nu.\]

Now, it follows from 11 that \[x \mapsto \int_{\pi^{-1}(x)} f(y) \,\mathrm{dVol}_{\hat{g}}(y)\] is a function in \(C^2(M/G) \cap H^1(M/G, \tilde{\nu})\). Since we are assuming that \((M/G, \tilde{\nu}, \tilde{\Gamma})\) satisfies a Poincaré inequality and that \(\int_M f d\nu = 0\) we can further bound \[\begin{align} \int_M f^2\; d\nu &\leq \frac{\tilde{Z}}{Z\mathrm{Vol}_{\hat{g}}(G)}\Biggl\{\left( \int_{M/G} \left(\int_{\pi^{-1}(x)} f(y)\, \mathrm{dVol}_{\hat{g}}(y)\right)\; d\tilde{\nu}(x)\right)^2 \\&\quad + \frac{1}{\tilde{\kappa}\beta} \int_{M/G} \left|\mathrm{grad}_{h} \int_{\pi^{-1}(x)} f(y) \,\mathrm{dVol}_{\hat{g}}(y) \right|^2_{h} d\tilde{\nu}(x)\Biggr\} \\&\quad + C \int_{M} \left|\left(\mathrm{grad}_{g} \,f\right)^{\mathcal{V}}\right|_{g}^2 d\nu \\&= \frac{\tilde{Z}}{Z\mathrm{Vol}_{\hat{g}}(G)}\Biggl\{\frac{Z^2}{\tilde{Z}^2}\left(\int_M f\, d\nu\right)^2 + \frac{1}{\tilde{\kappa}\beta}\int_{M/G} \left|\mathrm{grad}_{h} \int_{\pi^{-1}(x)} f(y) \,\mathrm{dVol}_{\hat{g}}(y) \right|^2_{h} d\tilde{\nu}(x)\Biggr\} \\&\quad + C \int_{M} \left|\left(\mathrm{grad}_{g} \,f\right)^{\mathcal{V}}\right|_{g}^2 d\nu \\&= \frac{\tilde{Z}}{Z}\frac{1}{\mathrm{Vol}_{\hat{g}}(G)\tilde{\kappa}\beta}\int_{M/G} \left|\mathrm{grad}_{h} \int_{\pi^{-1}(x)} f(y) \, \mathrm{dVol}_{\hat{g}}(y) \right|^2_{h} d\tilde{\nu}(x) \\&\quad + C \int_{M} \left|\left(\mathrm{grad}_{g} \, f\right)^{\mathcal{V}}\right|_{g}^2 d\nu \\&= \frac{1}{\mathrm{Vol}_{\hat{g}}(G)^2 \tilde{\kappa}\beta}\int_{M/G} \left|\mathrm{grad}_{h} \int_{\pi^{-1}(x)} f(y)\, \mathrm{dVol}_{\hat{g}}(y) \right|^2_{h} d\tilde{\nu}(x) \\&\quad + C \int_{M} \left|\left(\mathrm{grad}_{g} \, f\right)^{\mathcal{V}}\right|_{g}^2 d\nu, \end{align}\] where the last equality follows from 24 . Lastly, we apply 9 and 18 to conclude that \[\begin{align} \int_M f^2\; d\nu &\leq \frac{1}{\mathrm{Vol}_{\hat{g}}(G)\tilde{\kappa}\beta }\int_{M/G} \left(\int_{\pi^{-1}(x)} |\left(\mathrm{grad}_{g} \,f(y)\right)^{\mathcal{H}}|_{g}^2 \,\mathrm{dVol}_{\hat{g}}(y)\right) d\tilde{\nu}(x) \\&\quad + C \int_{M} \left|\left(\mathrm{grad}_{g}\, f\right)^{\mathcal{V}}\right|_{g}^2 d\nu \\&= \frac{\tilde{Z}}{Z}\frac{1}{\tilde{\kappa}\beta}\int_{M/G} \left(\int_{\pi^{-1}(x)} |\left(\mathrm{grad}_{g}\, f(y)\right)^{\mathcal{H}}|_{g}^2 \,\mathrm{dVol}_{\hat{g}}(y)\right) d\tilde{\nu}(x) \\&\quad+ C \int_{M} \left|\left(\mathrm{grad}_{g} \,f\right)^{\mathcal{V}}\right|_{g}^2 d\nu \\&= \frac{1}{\tilde{\kappa}\beta}\int_{M} \left|(\mathrm{grad}_{g}\, f)^{\mathcal{H}} \right|_{g}^2 d\nu + C \int_{M} \left|\left(\mathrm{grad}_{g}\, f\right)^{\mathcal{V}}\right|_{g}^2 d\nu. \end{align}\]

Therefore, if \((M/G, \tilde{\nu}, \tilde{\Gamma})\) satisfies a \(\mathrm{PI}(\tilde{\kappa})\), \((M, \nu, \Gamma)\) satisfies a \(\mathrm{PI}(\kappa)\) verifying \[\frac{1}{\kappa} = \max \Big\{\frac{1}{\tilde{\kappa}}, \beta C\Big\}= \max\Big\{\frac{1}{\tilde{\kappa}}, \frac{\mathrm{diam}(G)^2 \beta }{\pi^2}\Big\},\] where \(\mathrm{diam}(G)\) is the diameter of the fibers with the restricted metric \((G, \hat{g})\). ◻

Proof of 21. Let \(\tilde{f} \in C^2(M/G)\cap H^1(M/G, \tilde{\nu})\), and let \(f: M \to \mathbb{R}\) be the unique function such that \(\tilde{f} \circ \pi = f\). In particular, \(f\) is constant in the fibers of \(\pi\) and belongs to \(C^2(M)\). Using 3 and the fact that \(\tilde{f} \in H^1(M/G, \tilde{\nu})\) we know that \[x \mapsto \int_{\pi^{-1}(x)} f^2(y) e^{-\beta F(y)}\,\mathrm{dVol}_{\hat{g}}(y) = \tilde{f}^2(x) e^{-\beta \tilde{F}(x)}\, \mathrm{Vol(G)}\] is integrable on \(M/G\). Thus \(f^2 e^{-\beta F}\) is integrable on \(M\)—i.e. \(f \in L^2(M, \nu)\). Substituting \(f^2\) by \(|\mathrm{grad}_g\, f|^2_g\) in the above expression we can similarly conclude that \(|\mathrm{grad}_g\, f|^2_g\, e^{-\beta F}\) is integrable on \(M\), which implies that \(f \in H^1(M, \nu)\).

Without loss of generality, assume that \(\int_{M/G} \tilde{f}\; d\tilde{\nu} = 0\). Thus, by 18 we know that \[\begin{align} \int_M f(x)\, d\nu(x) &= \frac{\tilde{Z}}{Z} \int_{M/G} \left(\int_{\pi^{-1}(x)} f(y)\, \mathrm{dVol}_{\hat{g}}(y)\right) d\tilde{\nu}(x) \\&= \frac{\tilde{Z}}{Z} \int_{M/G} \left(\tilde{f}(x) \int_{\pi^{-1}(x)} \mathrm{dVol}_{\hat{g}}(y)\right) d\tilde{\nu}(x) \\&= \frac{\tilde{Z}}{Z} \mathrm{Vol}(G) \int_{M/G} \tilde{f}(x)\, d\tilde{\nu}(x) = 0. \end{align}\]

Furthermore, using again 18, we know that \[\begin{align} \int_M f^2(x) \; d\nu(x) = \frac{\tilde{Z}}{Z} \mathrm{Vol}(G) \int_{M/G} \tilde{f}^2(x) \;d\tilde{\nu}(x). \end{align}\]

Thus, since \((M, \nu, \Gamma)\) satisfies a \(\mathrm{PI}(\kappa)\) and \(f \in C^2(M) \cap H^1(M, \nu)\), we know that \[\begin{align} \int_{M/G} \tilde{f}^2(x)\; d\tilde{\nu}(x) &= \frac{Z}{\tilde{Z} \mathrm{Vol}(G)} \int_M f^2(x)\; d\nu(x) \\&\leq \frac{Z}{\tilde{Z} \mathrm{Vol}(G)} \frac{1}{\kappa \beta} \int_M |\mathrm{grad}_g \,f(x)|^2_g\; d\nu(x) \\&= \frac{1}{\mathrm{Vol}(G)}\frac{1}{\kappa \beta} \int_{M/G} \left(\int_{\pi^{-1}(x)} |\mathrm{grad}_g\, f(y)|^2_g \,\mathrm{dVol}_{\hat{g}}(y)\right)d\tilde{\nu}(x). \end{align}\]

Moreover, by 74 we can conclude that for every \(x \in M\), it holds that \[|\mathrm{grad}_g\, f(x)|^2_g = |\mathrm{grad}_{h}\, \tilde{f}(\pi(x))|^2_{h},\] and so \[\begin{align} \int_{M/G} \tilde{f}^2(x)\; d\tilde{\nu}(x) &\leq \frac{1}{\mathrm{Vol}(G)}\frac{1}{\kappa \beta} \int_{M/G} \left(\int_{\pi^{-1}(x)} |\mathrm{grad}_{h}\, \tilde{f}(x)|^2_{h}\, \mathrm{dVol}_{\hat{g}}(y)\right)d\tilde{\nu}(x) \\&= \frac{1}{\kappa \beta} \int_{M/G} |\mathrm{grad}_{h}\, \tilde{f}(x)|^2_{h}\; d\tilde{\nu}(x), \end{align}\] finishing the proof. ◻

Proof. Indeed, by [12] we know that \[\label{eq:boundKL} H(\rho_t | \nu) \leq e^{-2\alpha t} H(\rho_0 | \nu), \quad \forall t \geq 0.\tag{29}\] Using the Pinsker-Csizsár-Kullback inequality, we get \[\norm{\rho_t - \nu}^2_{\mathrm{TV}} \leq \frac{1}{2} H(\rho_t | \nu) \leq \frac{1}{2} e^{-2\alpha t} H(\rho_0 | \nu).\] ◻

Proof. The fact that \(\mu \ll \nu\) follows immediately. Now, since \(d\mu = \frac{1}{\mathrm{Vol}(M)} \mathrm{dVol}_g(x)\) and \(d\nu = \frac{1}{Z}e^{-\beta F(x)} \mathrm{dVol}_g(x)\), \[\frac{d\mu}{d\nu} = \frac{Z}{\mathrm{Vol}(M)} e^{\beta F(x)},\] which implies that \[\log \frac{d\mu}{d\nu} = \log (\frac{Z}{\mathrm{Vol}(M)}) + \beta F(x).\] This expression allows us to write the relative entropy of \(\mu\) with respect to \(\nu\) as \[H(\mu | \nu) = \frac{1}{\mathrm{Vol}(M)} \int_M \log \frac{d\mu}{d\nu} \mathrm{dVol}_g(x) = \log (\frac{Z}{\mathrm{Vol}(M)}) + \frac{\beta}{\mathrm{Vol}(M)} \int_M F(x) \mathrm{dVol}_g(x).\] Now, note that since \(F \geq 0\), \[Z = \int_M e^{-\beta F(x)} \mathrm{dVol}_g(x) \leq \int_M \mathrm{dVol}_g(x) = \mathrm{Vol}_g(M),\] and so we can upper bound the relative entropy as \[H(\mu | \nu) \leq \log(1) + \frac{\beta}{\mathrm{Vol}(M)} \int_M \max_{x \in M} F(x) \mathrm{dVol}_g(x) \leq \beta \max_{x \in M} F(x).\] ◻

Proof. Setting \(p = 2\) in 12 we obtain \[W_2(\mu, \eta)^2 \leq 2\, \mathrm{diam}(M)^2 \int_{M} d|\mu - \eta|(x) = 2\,\mathrm{diam}(M)^2 \norm{\mu - \eta}_{\text{TV}}.\] Now \[\norm{\mu - \eta}_{\text{TV}} = \sup\{|\mu(A) - \eta(A)| : A \in \mathcal{F}\} \leq 2,\] yielding \[W_2(\mu, \eta)^2 \leq 4\, \mathrm{diam}(M)^2.\] ◻

Proof of 24. Let \(\nu\) be the Gibbs distribution, and let \(\mu\) be any measure on \(M\) satisfying \(\mu \ll \nu\). Consider the relative entropy between \(\mu\) and \(\nu\), \(H(\mu|\nu)\). In order to apply 27, we have to rewrite \(\nu\) as \(\nu = e^{-V}\) for a suitable \(C^2\) function \(V\). It suffices to consider \(V = \beta F + \log(Z)\). For this choice of \(V\), the curvature-dimension condition of 27 is fulfilled. Indeed, \[\begin{align} \nabla^2 V + \mathrm{Ric}_g = \nabla^2(\beta F + \log(Z)) + \mathrm{Ric}_g = \beta \nabla^2 F + \mathrm{Ric}_g \geq -\beta \kappa_1 g, \end{align}\] where the inequality follows by multiplying the curvature-dimension condition written in ?? by \(\beta\). Therefore, applying 27 we conclude that \[\label{ineqrelativeentropy} H(\mu | \nu) \leq W_2(\mu, \nu) \sqrt{\mathbf{I}(\mu | \nu)} + \frac{\beta \kappa_1}{2} W_2(\mu, \nu)^2,\tag{30}\] where \[\mathbf{I}(\mu | \nu) = \beta I(\mu | \nu),\] and \(I(\mu | \nu)\) is the Fisher information. Furthermore, we can apply Young’s inequality \[ab \leq \frac{\varepsilon}{2}a^2 + \frac{1}{2\varepsilon}b^2, \quad \forall \varepsilon > 0,\;\forall a, b \geq 0.\] to 30 with \(a = \sqrt{\mathbf{I}(\mu | \nu)}\) and \(b = W_2(\mu, \nu)\) to obtain \[\begin{align} H(\mu | \nu) &\leq \frac{\varepsilon}{2}\mathbf{I}(\mu | \nu) + \left(\frac{1}{2\varepsilon} + \frac{\beta \kappa_1}{2}\right)W_2(\mu, \nu)^2\\ &\leq \frac{\varepsilon}{2}\mathbf{I}(\mu | \nu) + \left(\frac{1}{2\varepsilon}+ \frac{\beta \kappa_1}{2}\right)4\,\mathrm{diam}(M)^2\\ &= \frac{\varepsilon\beta}{2}I(\mu | \nu) + \left(\frac{1}{2\varepsilon}+ \frac{\beta \kappa_1}{2}\right)4\, \mathrm{diam}(M)^2, \end{align}\] where the second inequality follows from 28.

The above inequality is a defective logarithmic Sobolev inequality (cf. ?? ) with constants \(\alpha = \frac{1}{\varepsilon \beta}\) and \(A = \left(\frac{1}{2\varepsilon}+ \frac{\beta \kappa_1}{2}\right)4\, \mathrm{diam}(M)^2\). Thus, we can apply the tightening result from 29 to conclude that \((M, \nu, \Gamma)\) satisfies a tight logarithmic Sobolev inequality with constant \[\frac{1}{\varepsilon \beta + \frac{1}{\kappa_2}\left(\left(\frac{1}{2\varepsilon}+ \frac{\beta \kappa_1}{2}\right)2\,\mathrm{diam}(M)^2 + 1\right)},\] i.e. \[H(\mu | \nu) \leq \frac{1}{2}\left(\varepsilon\beta + \frac{1}{\kappa_2}\left\{\left(\frac{1}{2\varepsilon}+ \frac{\beta \kappa_1}{2}\right)2\,\mathrm{diam}(M)^2 + 1\right\}\right)I(\mu | \nu).\] Since this inequality works for all \(\varepsilon > 0\) we can choose \(\varepsilon = \frac{\mathrm{diam}(M)}{\sqrt{\beta \kappa_2}}\) to obtain the least upper bound, i.e. \[\frac{1}{\tilde{\alpha}} := 2\,\mathrm{diam}(M)\sqrt{\frac{\beta}{\kappa_2}} + \frac{\beta \kappa_1 \mathrm{diam}(M)^2 + 1}{\kappa_2}.\]

Let us conclude by obtaining a more readable log-Sobolev inequality constant. Indeed as we are assuming without loss of generality that \(\beta \geq 1\), \(0 < \kappa_2 \leq 1\), \(\mathrm{diam}(M) \geq 1\) and \(\kappa_1 > 1\) we can bound \[\begin{align} 2\,\mathrm{diam}(M)\sqrt{\frac{\beta}{\kappa_2}} + \frac{\beta \kappa_1 \mathrm{diam}(M)^2 + 1}{\kappa} &\leq \frac{2\beta \,\mathrm{diam}(M) + \beta \kappa_1 \mathrm{diam}(M)^2 + 1}{\kappa_2}\\ &\leq \frac{4\beta \kappa_1 \mathrm{diam}(M)^2}{\kappa_2}, \end{align}\] finishing the proof. ◻

Proof of 30. First, recall that under the assumptions of the theorem, we can apply 10 and conclude that the Markov triple \((M/G, \tilde{\nu}, \tilde{\Gamma})\) satisfies a \(\mathrm{PI}(\kappa_{M/G})\) with \[\kappa_{M/G} = \frac{\lambda_*}{184}.\] Now, we can lift this Poincaré inequality from \(M/G\) to \(M\) by simply applying 20, and claim that \((M, \nu, \Gamma)\) satisfies a \(\mathrm{PI}(\kappa_M)\) with \[\begin{align} \frac{1}{\kappa_{M}} &= \max \Big\{\frac{1}{\kappa_{M/G}}, \frac{\mathrm{diam}(G)^2 \beta}{\pi^2}\Big\}\\ &= \max \Big\{\frac{184}{\lambda_*}, \frac{\mathrm{diam}(G)^2 \beta}{\pi^2}\Big\}. \end{align}\]

After obtaining a Poincaré inequality for \((M, \nu, \Gamma)\) and \((M/G, \tilde{\nu}, \tilde{\Gamma})\), it only remains to tighten them to logarithmic Sobolev inequalities. This can be done using 24. Indeed, note that by 4 it holds that \[\begin{align} \nabla^2_g F &\geq -A_2 g,\\ \nabla^2_h \tilde{F} &\geq -A_2 h, \end{align}\] and so, as we are assuming that \(\mathrm{Ric}_g \geq -R_M\), \(\mathrm{Ric}_h \geq -R_{M/G}\) for some constants \(R_M, R_{M/G} \geq 0\), we can conclude that \[\begin{align} \nabla_g^2 F + \frac{1}{\beta}\mathrm{Ric}_g &\geq -(A_2 + R_M),\\ \nabla_h^2 \tilde{F} + \frac{1}{\beta}\mathrm{Ric}_h &\geq -(A_2 + R_{M/G}). \end{align}\] Thus, by 24 we can conclude that \((M, \nu, \Gamma)\) satisfies an \(\mathrm{LSI}(\alpha_M)\) with \[\begin{align} \frac{1}{\alpha_M} &= \frac{4\beta (A_2 + R_M) \mathrm{diam}(M)^2}{\kappa_M}\\ &= 4\beta (A_2 + R_M) \mathrm{diam}(M)^2\max \Big\{\frac{184}{\lambda_*}, \frac{\mathrm{diam}(G)^2 \beta}{\pi^2}\Big\}. \end{align}\] and \((M/G, \tilde{\nu}, \tilde{\Gamma})\) satisfies an \(\mathrm{LSI}(\alpha_{M/G})\) with \[\begin{align} \frac{1}{\alpha_{M/G}} = \frac{4\beta (A_2 + R_{M/G}) \mathrm{diam}(M/G)^2}{\kappa_{M/G}} = \frac{736\beta (A_2 + R_{M/G}) \mathrm{diam}(M/G)^2}{\lambda_*}. \end{align}\] ◻

Proof. In 30 we proved that \((M, \nu, \Gamma)\) satisfies an \(\mathrm{LSI}(\alpha_M)\) (see ?? ) and the Markov triple \((M/G, \tilde{\nu}, \tilde{\Gamma})\) satisfies an \(\mathrm{LSI}(\alpha_{M/G})\) (see ?? ). Using 22 we can obtain an upper bound on the total variation distance between the densities of \(X_t\) and \(\tilde{X}_t\) and the Gibbs measures \(\nu\) and \(\tilde{\nu}\), \[||\rho_t - \nu||_{\mathrm{TV}}^2 \leq \frac{1}{2} e^{-2\alpha_M t} H(\rho_0|\nu),\quad \mathrm{and}\quad ||\tilde{\rho}_t - \tilde{\nu}||_{\mathrm{TV}}^2 \leq \frac{1}{2} e^{-2\alpha_{M/G}\, t} H(\tilde{\rho}_0|\tilde{\nu}),\] for every \(t \geq 0\). Moreover, by 23, we can bound the relative entropy between the uniform distribution and the Gibbs distribution on \(M\) and \(M/G\) as \[H(\rho_0 | \nu) \leq \beta \max_{y \in M} F(y),\quad \mathrm{and}\quad H(\tilde{\rho}_0 | \tilde{\nu}) \leq \beta \max_{y \in M/G} \tilde{F}(y) = \beta \max_{y \in M} F(y),\] concluding the proof. ◻

Proof. By the Gauss lemma, since \(R \leq i(M)/2\), \[\label{eq:rewrittenintegral} \int_{\mathcal{B}(R, y)} e^{-\beta \frac{A_2}{2} d_g(y, x)^2} \; \mathrm{dVol}_g(x) = \int_0^R e^{-\beta\frac{A_2}{2}\rho^2} \left(\int_{S_{y}(\rho)} d\sigma \right)\; d\rho,\tag{31}\] where \[S_{y}(\rho) := \{\text{exp}_{y}(v)\, : \, |v|_g = \rho\}\] is the geodesic sphere of radius \(\rho\) centered at \(y\), \(\exp_{y}\) denotes the exponential map centered at \(y\), and \(d\sigma\) is the induced volume form on \(S_{y}(\rho)\). Now, since \(M\) is compact, by [50], it holds that \[\mathrm{Vol}(S_{y}(\rho)) \geq 2^{d-1} \frac{\mathrm{Vol}_{g_{\mathit{round}}}(\mathbb{S}^{d-1})^d}{\mathrm{Vol}_{g_{\mathit{round}}}(\mathbb{S}^{d})^{d-1} d^{d-1}} \rho^{d-1},\] for all \(\rho \leq i(M)/2\), where \(\mathrm{Vol}_{g_{\mathit{round}}}(\mathbb{S}^{d})\) denotes the volume of the \(d\)-dimensional round sphere of radius \(1\), which is given by (cf. [51]) \[\mathrm{Vol}_{g_{\mathit{round}}}(\mathbb{S}^{d}) = \frac{2 \pi^{(d+1)/2}}{\Gamma(\frac{d+1}{2})}.\] Thus, \[\begin{align} \mathrm{Vol}(S_{y}(\rho)) &\geq 2^{d-1} \frac{\mathrm{Vol}_{g_{\mathit{round}}}(\mathbb{S}^{d-1})^d}{\mathrm{Vol}_{g_{\mathit{round}}}(\mathbb{S}^{d})^{d-1} d^{d-1}} \rho^{d-1} \\&= 2^{d-1} \frac{\left(\frac{2 \pi^{d/2}}{\Gamma(\frac{d}{2})}\right)^d}{\left(\frac{2 \pi^{(d+1)/2}}{\Gamma(\frac{d+1}{2})}\right)^{d-1} d^{d-1}} \rho^{d-1}\\ &=2^{d} \pi^{1/2} \frac{\Gamma(\frac{d+1}{2})^{d-1}}{\Gamma(\frac{d}{2})^d d^{d-1}} \rho^{d-1} \\&= \mathcal{A}_d\, \rho^{d-1}, \end{align}\] where \[\mathcal{A}_d := 2^{d} \pi^{1/2} \frac{\Gamma(\frac{d+1}{2})^{d-1}}{\Gamma(\frac{d}{2})^d d^{d-1}}.\]

This way, we can lower-bound the right-hand side of 31 as \[\int_0^R e^{-\beta\frac{A_2}{2}\rho^2} \left(\int_{S_{y}(\rho)}\; d\sigma \right)\; d\rho \geq \mathcal{A}_d \int_0^R e^{-\beta\frac{A_2}{2}\rho^2} \rho^{d-1}\; d\rho.\] applying the change of variable \(r = \sqrt{\beta A_2}\rho\) we get \[\begin{align} \label{eq1} \nonumber \int_0^R e^{-\beta\frac{A_2}{2}\rho^2} \rho^{d-1}\; d\rho &= \int_0^{\sqrt{\beta A_2}R} e^{-\frac{r^2}{2}} \left(\frac{1}{\beta A_2}\right)^{d/2} r^{d-1}\; dr \\&= \left(\frac{1}{\beta A_2}\right)^{d/2} \int_0^{\sqrt{\beta A_2}R} e^{-\frac{r^2}{2}} r^{d-1}\; dr. \end{align}\tag{32}\] Now, since \(d \geq 2\) and \(R = \sqrt{\frac{2\varepsilon}{A_2}}\), if we choose \(\beta\) such that \[\beta \geq \frac{1}{A_2 R^2} = \frac{1}{2\varepsilon},\] it holds that \[\sqrt{\beta A_2}R \geq 1,\] and so we can lower bound 32 as follows: \[\begin{align} \left(\frac{1}{\beta A_2}\right)^{d/2} \int_0^{\sqrt{\beta A_2}R} e^{-\frac{r^2}{2}} r^{d-1}\; dr &\geq \left(\frac{1}{\beta A_2}\right)^{d/2} \int_1^{\sqrt{\beta A_2}R} e^{-\frac{r^2}{2}} r^{d-1}\; dr\\ &\geq \left(\frac{1}{\beta A_2}\right)^{d/2} \int_1^{\sqrt{\beta A_2}R} e^{-\frac{r^2}{2}} r \; dr \\ &= \left(\frac{1}{\beta A_2}\right)^{d/2} \left.\left(-e^{-\frac{r^2}{2}}\right)\right|_{1}^{\sqrt{\beta A_2} R}\\ &= \left(\frac{1}{\beta A_2}\right)^{d/2} \left(e^{-1/2} - e^{-\frac{\beta A_2 R^2}{2}}\right)\\ &= \left(\frac{1}{\beta A_2}\right)^{d/2} \left(e^{-1/2} - e^{-\beta \varepsilon}\right). \end{align}\]

Furthermore, \[e^{-1/2} - e^{-\beta \varepsilon} \geq \frac{1}{2}e^{-1/2} \iff \log\frac{1}{2} - \frac{1}{2} \geq -\beta \varepsilon \iff \beta \geq \frac{1}{2\varepsilon} + \frac{\log 2}{\varepsilon} = \frac{1+2\log 2}{2\varepsilon}.\] Therefore, for such values of \(\beta\) \[\left(\frac{1}{\beta A_2}\right)^{d/2} \int_0^{\sqrt{\beta A_2}R} e^{-\frac{r^2}{2}} r^{d-1}\; dr \geq \frac{1}{2} \left(\frac{1}{\beta A_2}\right)^{d/2}e^{-1/2},\] which allows us to conclude that \[\int_{\mathcal{B}(R, y)} e^{-\beta \frac{A_2}{2} d_g(y, x)^2} \; \mathrm{dVol}_g(x) \geq 2^{d-1} \pi^{1/2} \frac{\Gamma(\frac{d+1}{2})^{d-1}}{\Gamma(\frac{d}{2})^d d^{d-1}} \left(\frac{1}{\beta A_2}\right)^{d/2}e^{-1/2} .\] ◻

Proof of 31. From the previous lemmas we obtained the following bound: \[\nu(F \geq \varepsilon) \leq \frac{e^{-\beta \varepsilon} \mathrm{Vol}(M)}{2^{d-1} \pi^{1/2} \frac{\Gamma(\frac{d+1}{2})^{d-1}}{\Gamma(\frac{d}{2})^d d^{d-1}} \left(\frac{1}{\beta A_2}\right)^{d/2}e^{-1/2}},\] for \(\beta \geq \frac{1+2\log 2}{2\varepsilon}\). If we now define \[\mathscr{B}_d := \frac{e^{1/2}\Gamma(\frac{d}{2})^d d^{d-1}A_2^{d/2} \mathrm{Vol}(M)}{2^{d-1} \pi^{1/2} \Gamma(\frac{d+1}{2})^{d-1} },\] it only remains to find a value of \(\beta\) such that \[\label{goalineq} \beta^{d/2} e^{-\beta \varepsilon} \mathscr{B}_d \leq \delta.\tag{33}\] Taking logarithms in this expression we obtain \[\frac{d}{2}\log \beta - \beta\varepsilon + \log \mathscr{B}_d \leq \log \delta,\] which can be rewritten as \[\label{eq:inequalitybeta} \beta\varepsilon - \frac{d}{2} \log \beta \geq \log \frac{1}{\delta} + \log \mathscr{B}_d.\tag{34}\] Now, note that \[\frac{\beta\varepsilon}{2} - \frac{d}{2} \log \beta\] is an increasing function of \(\beta\) for \(\beta \geq \frac{d}{\varepsilon}\). Therefore, for such values \[\frac{\beta\varepsilon}{2} - \frac{d}{2}\log \beta \geq \frac{d}{2}\Big(1 - \log \frac{d}{\varepsilon}\Big).\] Inserting this inequality into 34 , it is easy to see that we only need to find the values of \(\beta\) for which \[\frac{\beta\varepsilon}{2} + \frac{d}{2}\Big(1 - \log \frac{d}{\varepsilon}\Big) \geq \log \frac{1}{\delta} + \log \mathscr{B}_d,\] that is, \[\beta \geq \frac{2}{\varepsilon}\left(\log \frac{1}{\delta} + \log \mathscr{B}_d + \frac{d}{2}\Big(\log \frac{d}{\varepsilon} - 1\Big)\right).\]

Let us express the logarithm of \(\mathscr{B}_d\) explicitly: \[\begin{align} \log \mathscr{B}_d &= \log \left(\frac{e^{1/2}\Gamma(\frac{d}{2})^d d^{d-1}A_2^{d/2} \mathrm{Vol}(M)}{2^{d-1} \pi^{1/2} \Gamma(\frac{d+1}{2})^{d-1} }\right) \\&= \frac{1}{2} + d \log\Gamma\left(\frac{d}{2}\right) + (d-1)\log d + \frac{d}{2}\log A_2 + \log \mathrm{Vol}(M) \\&\quad - (d-1)\log 2 -\frac{1}{2}\log\pi - (d-1)\log\Gamma\left(\frac{d+1}{2}\right) \\&\leq \frac{1}{2} + d \log\Gamma\left(\frac{d}{2}\right) + (d-1)\log d + \frac{d}{2}\log A_2 + \log \mathrm{Vol}(M) \\&\leq \frac{1}{2} + d \log\Gamma\left(\frac{d}{2}\right) + d\log (dA_2 \mathrm{Vol}(M)). \end{align}\] Therefore, we can choose \[\begin{align} \beta &\geq \frac{2}{\varepsilon}\left(\frac{1}{2} + d \log\Gamma\left(\frac{d}{2}\right) + d\log\frac{d^2A_2 \mathrm{Vol}(M)}{\delta\varepsilon}\right)\\ &\geq \frac{2}{\varepsilon}\left(\log \frac{1}{\delta} + \log \mathscr{B}_d + \frac{d}{2}(\log \frac{d}{\varepsilon} - 1)\right). \end{align}\]

Now, for \(d > 2\) we can use the following upper bound on the Gamma function (see [52]) \[\Gamma\left(\frac{d}{2}\right) < \sqrt{2\pi} \left(\frac{d - 1}{2e}\right)^{(d-1)/2},\] and so \[\log \Gamma\left(\frac{d}{2}\right) < \log \sqrt{2\pi} + \frac{d-1}{2}\log \frac{d-1}{2e} < d\log(d\sqrt{2\pi}).\] Thus, a suitable lower bound for \(\beta\) is \[\beta \geq \frac{2}{\varepsilon}\left( \frac{1}{2} + d^2 \log(d\sqrt{2\pi}) +d\log\frac{d^2A_2 \mathrm{Vol}(M)}{\delta\varepsilon} \right).\] Lastly, to simplify this, we note that whenever \[\beta \geq \frac{2}{\varepsilon}\left( \frac{1}{2} +d^2\log\frac{d^3A_2 \mathrm{Vol}(M)\sqrt{2\pi}}{\delta\varepsilon} \right),\] 33 holds, finishing the proof.

In particular, if \(d = 2\), since \(\Gamma(1) = 1\), we can bound \[\log \mathscr{B}_d \leq \frac{1}{2} + \log 2 + \log A_2 + \log \mathrm{Vol}(M) = \frac{1}{2} + \log 2 + \log(A_2 \mathrm{Vol}(M)),\] and so it suffices to choose \[\beta \geq \frac{2}{\varepsilon}\left(2\log \frac{2A_2\mathrm{Vol}(M)}{\varepsilon\delta} + \frac{1}{2} + \log 2\right).\] In particular, the general bound found for \(d > 2\) is also sufficient. ◻

Proof. As we saw earlier, the critical points of \(\tilde{f}\) correspond to the set of rank-one projectors \(\ketbra{\varphi}\) that satisfy \[[\ketbra{\varphi}, \Phi(\ketbra{\varphi})] = 0.\] This condition can be rewritten as \[\Phi(\ketbra{\varphi}) \ketbra{\varphi} = \ketbra{\varphi} \Phi(\ketbra{\varphi}).\] Now, since \(\braket{\varphi} = 1\), it holds that \[\begin{align} \Phi(\ketbra{\varphi}) \ket{\varphi} &=\ketbra{\varphi} \Phi(\ketbra{\varphi}) \ket{\varphi}\\ &= \bra{\varphi}(A - \tilde{f}(\ketbra{\varphi})B)\ket{\varphi}\ket{\varphi} \\&= (\bra{\varphi}A\ket{\varphi} - \tilde{f}(\ketbra{\varphi})\bra{\varphi}B\ket{\varphi})\ket{\varphi} \\&= 0, \end{align}\] and so \[\label{eq:psieig} A\ket{\varphi} = \tilde{f}(\ketbra{\varphi}) B \ket{\varphi}.\tag{40}\]

Using 35 we know that there exists some non-singular matrix \(S\) such that \(A = S\Lambda S^\dagger\) and \(B = SS^\dagger\), where \(\Lambda\) is a diagonal matrix with diagonal entries equal to the eigenvalues of \(A^{-1}B\). Therefore, 40 can be rewritten as \[S\Lambda S^\dagger \ket{\varphi} = \tilde{f}(\ketbra{\varphi}) S S^\dagger \ket{\varphi},\] and so \[\Lambda S^\dagger \ket{\varphi} = \tilde{f}(\ketbra{\varphi}) S^\dagger \ket{\varphi},\] which implies that \(S^\dagger \ket{\varphi}\) is an eigenvector of \(\Lambda\). Since we are assuming that \(\Lambda\) has \(n\) distinct eigenvalues, then, up to a complex phase, there are only \(n\) possible choices for \(S^\dagger \ket{\varphi}\), which in turn implies that \(\tilde{f}\) has \(n\) critical points. ◻

Proof. When \(B = \mathbb{1}\), any critical point \(P\) of \(\tilde{f}\) satisfies \[[P, A - \tilde{f}(P)] = 0,\] which is equivalent to the condition that \[[P, A] = 0.\] This way, for every critical point \(P\) of \(\tilde{f}\), we know that \(P\) and \(A\) can be diagonalized in the same basis. Assuming again that \(A\) has \(n\) distinct eigenvalues with associated eigenvectors \(\{\ket{\varphi_i}\}_{i = 1}^n\), it must be the case that we can write \(P\) as \[P = \sum_{j = i_1}^{i_k} \ketbra{\varphi_{j}},\] where \(\{i_1, \dotsc, i_k\} \subset \{1, \dotsc, n\}\). Thus, there are \(\binom{n}{k}\) choices of \(P\).

Using the above description of the critical points, we know that every two critical points will differ in at least one orthogonal direction. Thus, by 42 , any two critical points \(P\) and \(Q\) will lay in each other’s cut locus. Therefore, their distance is lower bounded by the injectivity radius of \(\mathrm{Gr}_k(\mathbb{C}^n)\) which in turn is greater than \(\frac{\sqrt{2}\pi}{2}\) (cf. 7). ◻

Proof. In order to prove this result, it suffices to bound the injectivity radius and the volume of \(\mathrm{U}(n)\), \(\mathrm{V}_k(\mathbb{C}^n)\) and \(\mathrm{Gr}_k(\mathbb{C}^n)\), and apply 31. The bounds obtained in 7 for the injectivity radii of these manifolds are \[i(\mathrm{U}(n)) \geq \pi, \quad i(\mathrm{V}_k(\mathbb{C}^n)) \geq \frac{\sqrt{2}\pi}{2}, \quad \mathrm{and}\quad i(\mathrm{Gr}_k(\mathbb{C}^n)) \geq \frac{\sqrt{2}\pi}{2},\] and the bounds on their volume can be deduced from the expressions shown in 8.6.3, i.e. \[\mathrm{Vol}(\mathrm{U}(n)) \leq (2\pi)^{n(n+1)/2}, \quad \mathrm{Vol}(\mathrm{V}_k(\mathbb{C}^n)) \leq (2\pi)^{nk - \frac{k(k-1)}{2}},\quad \mathrm{Vol}(\mathrm{Gr}_k(\mathbb{C}^n)) \leq (2\pi)^{nk -k^2}.\] Lastly, the dimensions of \(\mathrm{U}(n)\), \(\mathrm{V}_k(\mathbb{C}^n)\) and \(\mathrm{Gr}_k(\mathbb{C}^n)\) can be substituted by the explicit expressions found in 8.6.3, i.e. \[\dim(\mathrm{U}(n)) = n^2,\quad \dim(\mathrm{V}_k(\mathbb{C}^n)) =2nk-k^2, \quad \dim(\mathrm{Gr}_k(\mathbb{C}^n)) = 2k(n-k).\] ◻

Proof. We begin by applying 30 to the Markov triples \((\mathrm{U}(n), \nu_F, \Gamma_g)\) and
\((\mathrm{Gr}_k(\mathbb{C}^n), \nu_{\tilde{f}}, \Gamma_{h_2})\). We showed in 57 that the coefficients of the Riemann curvature tensor of \(\mathrm{U}(n)\) in normal coordinates when endowed with the bi-invariant metric are upper bounded by \(\frac{1}{2}\)—and so we can take \(\mathbf{K} = 1\) in 30. Furthermore, we also showed in 55 73 that the Ricci curvatures of the unitary group and the Grassmann manifold are non-negative, and so the constants \(R_{\mathrm{U}(n)}\) and \(R_{\mathrm{Gr}_k(\mathbb{C}^n)}\) from 30 can be taken to be \(0\).

Also, the Riemannian submersion \(\pi_1 : (\mathrm{U}(n), g) \to (\mathrm{Gr}_k(\mathbb{C}^n), h_2)\) has totally geodesic fibers, which implies by 21 22 that the sectional curvature of its fibers coincides with that of the total space. Since \((\mathrm{U}(n), g)\) has non-negative sectional curvature (cf. 55), we conclude that the sectional curvature of the fibers is also non-negative and so their Ricci curvature is non-negative as well.

The result now follows by substituting the manifold-dependent constants in the statement of 30 with the known bounds derived in 8. We use the lower bounds for the injectivity radii of \((\mathrm{U}(n), g)\) and \((\mathrm{Gr}_k(\mathbb{C}^n), h_2)\), along with the lower bound on the convexity radius of \((\mathrm{Gr}_k(\mathbb{C}^n), h_2)\) obtained in 7, i.e. \[\mathit{conv}(\mathrm{Gr}_k(\mathbb{C}^n)) \geq \frac{\sqrt{2}\pi}{4}.\] We also use the upper bound on the diameter of \((\mathrm{U}(n), g)\), and \((\mathrm{Gr}_k(\mathbb{C}^n), h_2)\) found in 9 56 , i.e. \[\mathrm{diam}(\mathrm{Gr}_k(\mathbb{C}^n)) \leq \mathrm{diam}(\mathrm{U}(n)) \leq \pi \sqrt{n}(1 + \sqrt{2}),\] and substitute the dimensions of \(\mathrm{U}(n)\) and \(\mathrm{Gr}_k(\mathbb{C}^n)\).

Lastly, we obtain a bound on the diameter of the fibers \(\mathrm{U}(k) \times \mathrm{U}(n-k)\) with the induced metric, which can be derived via 86 and 9, i.e. \[\mathrm{diam}(\mathrm{U}(k) \times \mathrm{U}(n-k)) \leq \sqrt{\pi^2 k (1+ \sqrt{2})^2 + \pi^2 (n-k) (1+\sqrt{2})^2} = \pi (1 + \sqrt{2}) \sqrt{n}.\]

All of the above bounds allow us to obtain the log-Sobolev inequality constants for the Markov triples \((\mathrm{U}(n), \nu_F, \Gamma_g)\) and \((\mathrm{Gr}_k(\mathbb{C}^n), \nu_{\tilde{f}}, \Gamma_{h_2})\).

To obtain the log-Sobolev inequality of the triple \((\mathrm{V}_k(\mathbb{C}^n), \nu_f, \Gamma_{h_1})\), we proceed as in the proof of 30: we first apply 10 to \((\mathrm{Gr}_k(\mathbb{C}^n), \nu_{\tilde{f}}, \Gamma_{h_2})\) to find that it satisfies a \(\mathrm{PI}(\kappa)\) with \[\kappa = \frac{\lambda_*}{184}.\] Then using the lifting theorem for Poincaré inequalities shown in 20, we lift the Poincaré inequality via \(\pi_2\) to ensure that \((\mathrm{V}_k(\mathbb{C}^n), \nu_f, \Gamma_{h_1})\) satisfies a \(\mathrm{PI}(\tilde{\kappa})\) where \[\frac{1}{\tilde{\kappa}} = \max\Big\{\frac{184}{\lambda_*}, (1+\sqrt{2})^2 k \beta\Big\},\] which can be then tightened using 24 to obtain an \(\mathrm{LSI}(\hat{\alpha})\), where \[\frac{1}{\hat{\alpha}} = \frac{4\beta A_2 \mathrm{diam}(\mathrm{V}_k(\mathbb{C}^n))^2}{\tilde{\kappa}}.\] Lastly, using the bound on the diameter of the Stiefel manifold shown in 56 we see that \[\mathrm{diam}(\mathrm{V}_k(\mathbb{C}^n)) \leq \pi \sqrt{n}(1 + \sqrt{2}),\] and so the result follows. ◻

Proof. First, note that \[\mathrm{SU}(2) = \left\{\begin{pmatrix} a & -\overline{b} \\ b & \overline{a} \end{pmatrix}\;:\;a, b \in \mathbb{C},\; |a|^2+|b|^2=1\right\} \simeq \mathbb{S}^3.\] Recall from 48 that its lie algebra \(\mathfrak{su}(2)\) consists of \(2 \times 2\) anti-Hermitian matrices with zero trace, \[\mathfrak{su}(2) = T_{\mathbb{1}} \mathrm{SU}(2) = \left\{ \begin{pmatrix} ix & z\\ -\overline{z} & -ix \end{pmatrix}\;:\;x \in \mathbb{R},\;z \in \mathbb{C}\right\}.\]

Moreover, the bi-invariant of \(\mathrm{SU}(2)\) at the identity is given by \[\require{physics} \Tr \Bigg(\begin{pmatrix} -ix_2 & -z_2\\ \overline{z_2} & ix_2 \end{pmatrix} \begin{pmatrix} ix_1 & z_1\\ -\overline{z_1} & -ix_1 \end{pmatrix}\Bigg) = x_1x_2 + z_2 \overline{z_1} + z_1 \overline{z_2} + x_1 x_2\] for any two tangent vectors \(\begin{pmatrix} ix_2 & z_2\\ -\overline{z_2} & -ix_2 \end{pmatrix}\), \(\begin{pmatrix} ix_1 & z_1\\ -\overline{z_1} & -ix_1 \end{pmatrix} \in T_{\mathbb{1}}\mathrm{SU}(2)\), which corresponds to \(2g_{\mathit{Eucl}}\) in \(\mathbb{R}^3\).

Now, the metric \(g_{\mathit{bi}}\) is extended from \(T_\mathbb{1} \mathrm{SU}(2)\) to \(T\mathrm{SU}(2)\) as \[\require{physics} g_{\mathit{bi}}(AG, BG) := \Tr(G^\dagger B^\dagger A G) = \Tr (B^\dagger A).\]

Note that the map \[\begin{align} \mathbb{R}^4 \rightarrow \mathbb{R}^4,\quad \begin{pmatrix} x & y\\ -\overline{y} & \overline{x} \end{pmatrix} \mapsto \begin{pmatrix} x & y\\ -\overline{y} & \overline{x} \end{pmatrix}G \end{align}\] where \(x, y \in \mathbb{C}\), is an isometry in \(\mathbb{R}^4\), and so the extension of the of the scalar product coincides with twice the usual scalar product in the round sphere \(\mathbb{S}^3\), i.e. \(2g_{\mathit{round}}\). ◻

Proof. Let us give an expression for the differential of \(f\) at any point \(U\) in the direction of a tangent vector \(iXU\), where \(X\) is some Hermitian matrix. To do so, we parametrize \(f\) along a geodesic starting at \(U\) in the direction of \(iXU\), \[\gamma(t) = e^{itX} U,\] yielding \[f(t) := f(\gamma(t)) = \bra{0} U^\dagger e^{-itX} \hat{H} e^{itX} U \ket{0}.\] Computing its derivative, we obtain \[\begin{align} f'(t) &= -i \bra{0} U^\dagger e^{-itX}X\hat{H}e^{itX}U\ket{0} + i\bra{0} U^\dagger e^{-itX}\hat{H}Xe^{itX}U\ket{0} \\&= i\bra{0} U^\dagger e^{-itX}[\hat{H}, X]e^{itX}U\ket{0}. \end{align}\] Using the notation \(\ket{U} := U\ket{0}\) we can rewrite this expression as \[f'(t) = i\bra{U} e^{-itX}[\hat{H}, X]e^{itX}\ket{U},\] which allows us to conclude that \[df|_U(iXU) = f'(0) = i\bra{U}[\hat{H}, X]\ket{U}.\]

Assume now that \(U\ket{0}\) is an eigenvector of \(\hat{H}\), i.e. \(\hat{H}\ket{U} = E \ket{U}\) for some \(E \in \mathbb{R}\). Thus, for every \(iX \in \mathfrak{su}(n)\) \[df|_U (iXU) = \bra{U}[\hat{H}, X]\ket{U} = \bra{U}\hat{H}X - X\hat{H}\ket{U} = E \bra{U}X - X\ket{U} = 0.\] To prove the converse, let \(U \in \mathrm{SU}(n)\) and suppose that \[\label{eq:eq1} \bra{U}[\hat{H}, X]\ket{U} = 0,\quad \forall iX \in \mathfrak{su}(n).\tag{46}\] We can write \(\ket{U}\) in the (orthonormal) eigenbasis of \(\hat{H}\); \(\ket{U} = \sum_{i = 1}^n \alpha_i \ket{\varphi_i}\), where \(\hat{H} \ket{\varphi_i} = E_i \ket{\varphi_i}\). With this expression, 46 becomes \[\left(\sum_{i = 1}^n \overline{\alpha_i} \bra{\varphi_i}\right)\hat{H}X\left(\sum_{i = 1}^n \alpha_i \ket{\varphi_i}\right) = \left(\sum_{i = 1}^n \overline{\alpha_i} \bra{\varphi_i}\right)X\hat{H}\left(\sum_{i = 1}^n \alpha_i \ket{\varphi_i}\right),\] which implies that \[\left(\sum_{i = 1}^n \overline{\alpha_i} E_i \bra{\varphi_i}\right)X\left(\sum_{i = 1}^n \alpha_i \ket{\varphi_i}\right) = \left(\sum_{i = 1}^n \overline{\alpha_i} \bra{\varphi_i}\right)X\left(\sum_{i = 1}^n \alpha_i E_i\ket{\varphi_i}\right),\] and so \[\sum_{i, j = 1}^n \overline{\alpha_i} \alpha_j E_i \bra{\varphi_i}X\ket{\varphi_j} = \sum_{i, j = 1}^n \overline{\alpha_i} \alpha_j E_j \bra{\varphi_i}X\ket{\varphi_j}.\]

Grouping all the terms on the left-hand side we obtain \[\sum_{i, j= 1}^n \overline{\alpha_i} \alpha_j (E_i-E_j) \bra{\varphi_i}X\ket{\varphi_j} = 0.\] If we define the matrix \(A = \left(\overline{\alpha_j}\alpha_i (E_j - E_i)\right)_{ij}\) we can rewrite the expression as \[\sum_{i, j = 1}^n \overline{\alpha_i} \alpha_j (E_i-E_j) \bra{\varphi_i}X\ket{\varphi_j} = \sum_{i, j = 1}^n A_{ji} X_{ij} = \mathrm{Tr}(AX) = 0\] for every \(iX \in \mathfrak{su}(n)\).

In order to apply 15, take \(Y\) to be any Hermitian matrix. Therefore, we can write \(Y = \Lambda + X\) where \(\Lambda\) is diagonal and \(X\) is such that \(iX \in \mathfrak{su}(n)\), thus \[\mathrm{Tr}(AY) = \mathrm{Tr}(A\Lambda) + \mathrm{Tr}(AX) = \mathrm{Tr}(A\Lambda) = 0,\] where the last equality follows from the fact that \(\Lambda\) is diagonal. Indeed \[\mathrm{Tr}(A\Lambda) = \sum_{i = 1}^n A_{ii}\Lambda_{ii} = \sum_{i = 1}^n |\alpha_i|^2 (E_i-E_i)\Lambda_{ii} = 0.\] Using 15 we can conclude that \[\overline{\alpha_j}\alpha_i (E_j - E_i) = 0,\quad \forall i, j \in \{1, \dotsc, n\}.\] This implies that, if all of the eigenvalues are different, at most one \(\alpha_i\) is non-zero, implying that \(\ket{U}\) is an eigenvector of \(\hat{H}\). In the case where some eigenvalues are equal, we can conclude that \(\ket{U}\) has to be a linear combination of eigenvectors associated with the same eigenvalue, implying that \(\ket{U}\) is too an eigenvector of \(\hat{H}\). ◻

Proof. Let \((x, y) \in M_1 \times M_2\), let \(W \in T_{(x,y)} (M_1 \times M_2)\) and let \(\gamma(t)\) be the geodesic such that \(\gamma(0) = x\) and \(\gamma'(0) = W\). Writing both \(W\) and \(\gamma(t)\) in coordinates, we know that \(W = (U, V)\) with \(U \in T_x M_1\) and \(V \in T_y M_2\), and \(\gamma(t) = (\gamma_1(t), \gamma_2(t))\) with \(\gamma_1(0) = x\), \(\gamma_1'(0) = U\), \(\gamma_2(0) = y\), and \(\gamma_2'(0) = V\).

Using the definition of the differential of \(\hat{f}\) and the product rule in \(\mathbb{R}\) we conclude that \[\begin{align} d\hat{f}|_{(x,y)}(U, V) = \left.\frac{d}{dt}\right|_{t = 0} \hat{f}(\gamma(t)) &= \left.\frac{d}{dt}\right|_{t = 0} \hat{f}(\gamma_1(t), \gamma_2(t)) \\&= \left.\frac{d}{dt}\right|_{t = 0} f_1(\gamma_1(t))f_2(\gamma_2(t))\\ &= f_2(y)\left.\frac{d}{dt}\right|_{t = 0} f_1(\gamma_1(t)) + f_1(y)\left.\frac{d}{dt}\right|_{t = 0} f_2(\gamma_2(t)) \\&=f_2(y)df_1|_{x}(U) + f_1(x)df_2|_{y}(V). \end{align}\] Therefore, \[\begin{align} g(\mathrm{grad}_g\, \hat{f}, W) = d\hat{f}(W) &= f_2(y)df_1(U) + f_1(x)df_2(V) \\&= f_2(y)g_1(\mathrm{grad}_{g_1}\,f_1, U) + f_1(x)g_{2}(\mathrm{grad}_{g_2}\,f_2, V) \\&= g\left((f_2(y)\mathrm{grad}_{g_1}\,f_1, f_1(x)\mathrm{grad}_{g_2}\,f_2), (U, V)\right), \end{align}\] and the claim follows. ◻

Proof of 43. Assume that we are working on an \(n \times n\) lattice, with \(n \geq 2\). We will prove that every critical point corresponds to an eigenvector of \(H\). To do so, we will consider a critical point and assume that at least one of the \(\nabla f_{(i, j)}\) is non-zero. This will be done in three steps:

1. We will first assume that \(\nabla f_{(1,1)}\) is non-zero, and reach a contradiction. Therefore, by the symmetry of the problem we can conclude that every critical point must have \[\nabla f_{(1, 1)} = \nabla f_{(1, n)} = \nabla f_{(n, 1)} = \nabla f_{(n, n)} = 0.\]

2. For the next step, assuming that \(\nabla f_{(1,1)} = \nabla f_{(1, n)} = 0\) we will prove that if \(\nabla f_{(1, k)} \neq 0\) then we cannot be at a critical point.

   Again, by the symmetry of the problem, for the third step we may assume that \(\nabla f_{(1,i)} = \nabla f_{(i, 1)} = 0\) for every \(i = 1, \dotsc, n\).

3. For the last step of the proof, we will assume that \(\nabla f_{(k, l)} \neq 0\) but \(\nabla f_{(k-1, l)} = \nabla f_{(k, l-1)} = 0\) for some \((k, l)\) with \(k \in \{2, \dots, n-1\}, l \in \{2, n-1\}\) and prove that we cannot be at a critical point.

7.2.1.1 First step: corners.

Assume that \(\nabla f_{(1,1)} \neq 0\). Therefore, by 47 \[f_{(1,2)} + f_{(2,1)} + \varepsilon_{(1,1)} + 2 = 0,\] which cannot happen, as \(-1 \leq f_{(i, j)} \leq 1\) for every \((i, j) \in V\). Therefore, we may now assume that the gradients \(\nabla f_{(i, j)}\) of the corners vanish, and continue with the second step of our proof.

7.2.1.2 Second step: first row.

Assume that, for some \(1 < k < n\), \(\nabla f_{(1,k)} \neq 0\), and \(\nabla f_{(1, k-1)} = 0\). This implies that \[f_{(1,k-1)} + f_{(1,k+1)} + f_{(2,k)} + 3 + \varepsilon_{(1, k)} = 0.\] Since \(\nabla f_{(1,k-1)} = 0\) by assumption, it has to be the case that \(f_{(1, k-1)} = \pm1\), and so \[f_{(1,k+1)} + f_{(2,k)} + 2 + \varepsilon_{(1,k)} = 0,\quad \mathrm{or}\quad f_{(1,k+1)} + f_{(2,k)} + 4 + \varepsilon_{(1,k)} = 0,\] which in either case is a contradiction.

Again, by the symmetry of the problem, we can assume that all of the \(f_{(i,j)}\) laying on the outer layer of our lattice have vanishing gradients.

7.2.1.3 Third step: inner spins.

It only remains to prove the case when \(\nabla f_{(i, j)} \neq 0\) for some \(1 < i, j < n\), and \[\nabla f_{(i-1, j)} = \nabla f_{(i, j-1)} = 0.\] In this case \[f_{(i-1, j)} + f_{(i, j-1)} + f_{(i+1, j)} + f_{(i, j+1)} + 4 + \varepsilon_{(i, j)} = 0,\] which cannot happen, as \[f_{(i-1, j)} + f_{(i, j-1)} \in \{-2, 0, 2\}.\] ◻

Proof. For any two eigenvectors \(\ket{x} = \ket{x_1, \dotsc, x_{n^2}}\), \(\ket{x'} = \ket{x'_1, \dotsc, x'_{n^2}}\) of \(H\) there exists some \(j \in \{1, \dotsc, n^2\}\) such that \(x_j = 1\) and \(x'_j = 0\). Since any point in \(\mathbb{S}^2\) can be written as \[\cos\frac{\theta}{2} \ket{0} + \sin \frac{\theta}{2} e^{i \varphi} \ket{1},\] it is clear that the points in the Bloch sphere \(\mathbb{S}^2\) corresponding to \(x_j\) and \(x'_j\) are antipodal. Therefore, the distance between any two distinct critical points, \(\ket{x}\) and \(\ket{x'}\), when considering the round metric in \(\mathbb{S}^2\) scaled by \(\frac{1}{2}\), is lower bounded by \(\frac{\sqrt{2}}{2}\pi\). Indeed, the distance between two antipodal points in \((\mathbb{S}^2,g_{\mathit{round}})\) is \(\pi\), and when scaling the metric by \(\frac{1}{2}\), the distances get scaled by a factor of \(\frac{1}{\sqrt{2}}\). ◻

Proof. Let us parametrize \(F\) along a geodesic \[\gamma(t) = (e^{itX_1}U_1, \dotsc, e^{itX_{n^2}}U_{n^2}).\] Taking derivatives, we may conclude that \[\begin{align} \nabla^2 F|_{\overrightarrow{U}}(i\overrightarrow{XU},i\overrightarrow{XU}) &= - \sum_{i, j = 1}^{n^2} \bra{U} (X_i \otimes X_j) H \ket{U} - \sum_{i, j = 1}^{n^2} \bra{U} H (X_i \otimes X_j) \ket{U} \\&\quad + 2\sum_{i, j = 1}^{n^2} \bra{U} X_i H X_j \ket{U}, \end{align}\] where we denote \[\overrightarrow{U} := (U_1, \dotsc, U_{n^2}),\;\ket{U} := U_1 \otimes \dotsm \otimes U_{n^2}\ket{0}^{\otimes n^2}\;\mathrm{and}\;\overrightarrow{XU} := (X_1U_1, X_2U_2, \dotsc, X_{n^2}U_{n^2}),\] for simplicity. Furthermore, if \(\ket{U}\) is an eigenvector of \(H\) with eigenvalue \(E\), then \[\label{eqHessianMultiGate} \left.\nabla^2 F\right|_{\overrightarrow{U}}(i\overrightarrow{XU},i\overrightarrow{XU}) = 2\left(\sum_{i, j = 1}^d \bra{U} X_i H X_j \ket{U} - E\sum_{i, j = 1}^d \bra{U} (X_i \otimes X_j) \ket{U}\right).\tag{48}\]

Writing each \(X_j\) in the eigenbasis \(\{\ket{0}, \ket{1}\}\), we obtain \[X_j = \mathbb{1} \otimes \dotsm \otimes \mathbb{1} \otimes \left(\sum_{k,l = 0}^1 \alpha_{k, l}^{(j)} \ket{k}\bra{l}\right)\otimes \mathbb{1} \otimes \dotsm \otimes \mathbb{1}.\] Moreover, writing \(\ket{U}\) as \[\ket{U} = \ket{l_1} \otimes \ket{l_2} \otimes \dotsm \otimes \ket{l_{n^2}},\] where \(l_i \in \{0, 1\}\) for every \(i \in \{1, \dotsc, n^2\}\), we can conclude that \[\begin{align} X_j \ket{U} &= \ket{l_1} \otimes \overset{j)}{\dotsm} \otimes \left(\sum_{k = 0}^1 \alpha_{k, l_j}^{(j)} \ket{k}\right)\otimes \dotsm \otimes \ket{l_{n^2}} \\&= \sum_{k = 0}^1 \alpha_{k, l_j}^{(j)} \ket{l_1} \otimes \overset{j)}{\dotsm} \otimes \ket{k}\otimes \dotsm \otimes \ket{l_{n^2}}. \end{align}\] If \(i \neq j\) we can write the second term on the right-hand side of 48 as \[\begin{align} \bra{U}\left(X_i \otimes X_j\right) \ket{U}&= \bra{U}X_i X_j \ket{U} \\&= \left(\sum_{k = 0}^1 \overline{\alpha}_{k, l_i}^{(i)} \bra{l_1} \otimes \overset{i)}{\dotsm} \otimes \bra{k}\otimes \dotsm \otimes \bra{l_{n^2}}\right) \\&\quad \times \left(\sum_{k = 0}^1 \alpha_{k, l_j}^{(j)} \ket{l_1} \otimes \overset{j)}{\dotsm} \otimes \ket{k}\otimes \dotsm \otimes \ket{l_{n^2}}\right) \\&= \left(\sum_{k = 0}^1 \overline{\alpha}_{k, l_i}^{(i)}\bra{k}\right)\ket{l_i}\bra{l_j} \left(\sum_{k = 0}^1 \alpha_{k, l_j}^{(j)} \ket{k}\right) \\&= \overline{\alpha}^{(i)}_{l_i, l_i}\alpha^{(j)}_{l_j, l_j}. \end{align}\] If \(i = j\) we can write \[\bra{U} X_i^2 \ket{U} = \sum_{k = 0}^1 |\alpha^{(i)}_{k, l_i}|^2.\]

On the other hand, if \(i \neq j\) we can write the first term on the right-hand side of 48 as \[\begin{align} \bra{U}X_i H X_j\ket{U}&= \bra{U} X_i H \left(\sum_{k = 0}^1 \alpha_{k, l_j}^{(j)} \ket{l_1} \otimes \overset{j)}{\dotsm} \otimes \ket{k}\otimes \dotsm \otimes \ket{l_d}\right) \\&= \bra{U} X_i \left(\sum_{k = 0}^1 \alpha_{k, l_j}^{(j)}E_{l_1, \overset{j)}{\dotsc}, k, \dotsc, l_{n^2}} \ket{l_1} \otimes \overset{j)}{\dotsm} \otimes \ket{k}\otimes \dotsm \otimes \ket{l_d}\right) \\&= \overline{\alpha}^{(i)}_{l_i, l_i}\alpha_{l_j, l_j}^{(j)}E_{l_1, \dotsc, l_j, \dotsc, l_{n^2}}, \end{align}\] where \(E_{l_1, \dotsc, l_{n^2}}\) denotes the eigenvalue of \(H\) associated with the eigenstate \(\ket{l_1, \dotsc, l_{n^2}}\).

If \(i = j\) then \[\begin{align} \bra{U}X_i H X_i\ket{U}&= \bra{U} X_i H \left(\sum_{k = 0}^1 \alpha_{k, l_i}^{(i)} \ket{l_1} \otimes \overset{i)}{\dotsm} \otimes \ket{k}\otimes \dotsm \otimes \ket{l_{n^2}}\right) \\&= \bra{U} X_i \left(\sum_{k = 0}^1 \alpha_{k, l_i}^{(i)}E_{l_1, \overset{i)}{\dotsc}, k, \dotsc, l_{n^2}} \ket{l_1} \otimes \overset{i)}{\dotsm} \otimes \ket{k}\otimes \dotsm \otimes \ket{l_{n^2}}\right) \\&= \sum_{k = 1}^n |\alpha^{(i)}_{k, l_i}|^2 E_{l_1, \overset{i)}{\dotsc}, k, \dotsc, l_{n^2}}. \end{align}\]

Substituting these expressions into 48 we obtain \[\begin{align} \label{expressionhessiano} \begin{aligned} \frac{1}{2}\left.\nabla^2 F\right|_{\overrightarrow{U}}(i\overrightarrow{XU},i\overrightarrow{XU}) &= \sum_{i, j = 1}^{n^2} \bra{U} X_i H X_j \ket{U} - E\sum_{i, j = 1}^{n^2} \bra{U} (X_i \otimes X_j) \ket{U} \\&= \sum_{\substack{i, j = 1\\i \neq j}}^{n^2} \left(\overline{\alpha}^{(i)}_{l_i, l_i}\alpha_{l_j, l_j}^{(j)}E_{l_1, \dotsc, l_j, \dotsc, l_d} - E \overline{\alpha}^{(i)}_{l_i, l_i}\alpha^{(j)}_{l_j, l_j}\right) \\&\quad + \sum_{i = 1}^{n^2} \left(\sum_{k = 1}^n |\alpha^{(i)}_{k, l_i}|^2 E_{l_1, \dotsc, k, \dotsc, l_d} - E \sum_{k = 1}^n |\alpha^{(i)}_{k, l_i}|^2\right) \\&= \sum_{i = 1}^{n^2} \sum_{k = 0}^1 \left(E_{l_1, \overset{i)}{\dotsc}, k, \dotsc, l_d} - E\right) |\alpha^{(i)}_{k, l_i}|^2 \\&= \sum_{i = 1}^{n^2} \left(E_{l_1, \overset{i)}{\dotsc}, l_i + 1 \mathrm{ mod } 2, \dotsc, l_d} - E\right) |\alpha^{(i)}_{l_i + 1 \mathrm{ mod } 2, l_i}|^2. \end{aligned} \end{align}\tag{49}\] This way, if \(E\) corresponds to the lowest (resp. highest) eigenvalue of \(H\), the Hessian of \(F\) is positive-semidefinite (resp. negative-semidefinite).

Observe that in 49 \(E_{l_1, \overset{i)}{\dotsc}, (l_i + 1 \mod 2), \dotsc, l_d}\) corresponds to the eigenvalue obtained when flipping a single spin—in the \(i\)-th position. Thus, if the eigenvector is not that with minimal eigenvalue of \(H\) we can flip at least one spin from \(\ket{1}\) to \(\ket{0}\), in which case the associated eigenvalue decreases at least by \(\min_{I \in V}\varepsilon_I\). ◻

Proof. We want to conclude that the kernel of \(\nabla^2 F\) at its global minimum is included in the vertical tangent space of \(\mathrm{SU}(2)^{\times n^2}\) with respect to the submersion \(\pi: \mathrm{SU}(2)^{\times n^2} \to (\mathbb{S}^2)^{\times n^2}\). This way, using 78, we will conclude that \(\tilde{F}\) has a positive-definite Hessian at the global minimum.

Let us use the expression for the Hessian shown in 49 and let us denote by \(E\) the smallest (non-degenerate) eigenvalue of \(H\), which corresponds to the eigenvector \(\ket{0}^{\otimes n^2}\). Then at any \(U\) for which \(U\ket{0}^{\otimes n^2} = e^{i\theta} \ket{0}^{\otimes n^2}\), it holds that \[\frac{1}{2}\left.\nabla^2 F\right|_{\overrightarrow{U}}(i\overrightarrow{XU},i\overrightarrow{XU}) = \sum_{i = 1}^{n^2} \left(E_{0, \overset{i)}{\dotsc}, 1, \dotsc, 0} - E\right) |\alpha^{(i)}_{1, 0}|^2,\] where \(E_{0, \overset{i)}{\dotsc}, 1, \dotsc, 0}\) denotes the eigenvalue associated with \(|0, \overset{i)}{\dotsc}, 1, \dotsc, 0\rangle\) and the terms \(\alpha^{(i)}_{j,k}\) are the coordinates of the \(i\)-th component of the vector \(\overrightarrow{X}\).

Since \(E\) is the unique lowest eigenvalue of \(H\), the only directions \(i\overrightarrow{XU}\) that lie in the kernel of \(\left.\nabla^2 F\right|_{\overrightarrow{U}}\) correspond to those such that \(\alpha^{(i)}_{k, li} = 0\) for every \(i\) and every \(k \neq l_i\). Otherwise, we would have a non-zero term in the sum: \(E_{l_1, \overset{i)}{\dotsc}, k, \dotsc, l_n} - E > 0\).

For this reason, every \(X_j\) is of the form \[X_j = \mathbb{1} \otimes \dotsm \otimes \mathbb{1} \otimes \left(\alpha_{0, 0}^{(j)} \ketbra{0} - \alpha_{1, 1}^{(j)} \ketbra{1}\right)\otimes \mathbb{1} \otimes \dotsm \otimes \mathbb{1},\] and so it can be written as \[X_j = \mathbb{1} \otimes \dotsm \otimes \mathbb{1} \otimes \begin{pmatrix} ia & 0\\ 0 & -ia \end{pmatrix}\otimes \mathbb{1} \otimes \dotsm \otimes \mathbb{1},\] for some \(a\). These directions are vertical with respect to the action shown in 45 , and so the Hessian of \(\tilde{F}\) at its global minimum is positive definite. ◻

Proof of 46. To prove the suboptimality result for the Gibbs measures \(\nu_F\) and \(\nu_{\tilde{F}}\), we need to bound the injectivity radii and the volume of \((\mathrm{SU}(2)^{\times n^2}, g)\) and \(((\mathbb{S}^{2})^{\times n^2}, h)\). First, as we saw in 41 that \[(\mathrm{SU}(2), g_{\mathit{bi}}) \simeq (\mathbb{S}^3, 2g_{\mathit{round}}),\] where \(g_{\mathit{bi}}\) denotes the bi-invariant metric of \(\mathrm{SU}(2)\), we know that \[\mathrm{Vol}(\mathrm{SU}(2)) = 2^{3/2} \mathrm{Vol}(\mathbb{S}^3) = 2^{3/2} \frac{2\pi^2}{\Gamma(2)} = 2^{5/2} \pi^2,\] where we used the expression for the volume of the sphere endowed with the round metric seen in 8.6.3, and so \[\mathrm{Vol}(\mathrm{SU}(2)^{\times n^2}) = (2^{5/2} \pi^2)^{n^2}.\] Also \[\mathrm{Vol}\Big(\Big(\mathbb{S}^2, \frac{1}{2}g_{\mathit{round}}\Big)\Big) = \frac{1}{2}\mathrm{Vol}(\mathbb{S}^2) = \frac{\pi^{3/2}}{\Gamma(3/2)},\] and so \[\mathrm{Vol}(((\mathbb{S}^2)^{\times n^2}, h))) = \Big(\frac{\pi^{3/2}}{\Gamma(3/2)}\Big)^{n^2}.\] For the injectivity radii, we simply use the scaling property of the distance, along with the bounds obtained in 7 to conclude that \[i(\mathrm{SU}(2)^{\times n^2}, g) = i(\mathrm{SU}(2), g_{\mathit{bi}}) = i(\mathbb{S}^3, 2g_{\mathit{round}}) = \sqrt{2}i(\mathbb{S}^3) \geq \sqrt{2}\pi.\] \[i(((\mathbb{S}^2)^{\times n^2}, h)) = i(\mathbb{S}^2, \frac{1}{2}g_{\mathit{round}}) = \frac{\sqrt{2}}{2}i(\mathbb{S}^2) \geq \frac{\sqrt{2}\pi}{2}.\] Substituting these bounds into 31, the suboptimality for the Gibbs measures follows. ◻

Proof. In order to obtain the log-Sobolev inequalities, recall from 43 and 17 that the critical points of \(\tilde{F}\) are isolated and separated by a distance of at least \(\frac{\sqrt{2}\pi}{2}\). Moreover, we proved in 18 45 that 8 9 hold with \(\lambda_*\) given by \[\min_{I \in V} \varepsilon_I.\]

The Ricci curvatures of both \((\mathrm{SU}(2)^{\times n^2}, g)\) and \(((\mathbb{S}^2)^{\times n^2}, h)\) are non-negative (cf. 58 56 82). Furthermore, it follows by 57 58 that the coefficients of the Riemann curvature tensor of \((\mathrm{SU}(2)^{\times n^2}, g)\) in normal coordinates when endowed with the product of bi-invariant metrics are bounded by \(\frac{1}{2}\).

Also, it holds that the Riemannian submersion \(\pi_1 : (\mathrm{SU}(2)^{\times n^2}, g) \to ((\mathbb{S}^2)^{\times n^2}, h)\) has totally geodesic fibers, which means by 21 22 that the sectional curvature of the fibers coincides with that of the total space. Since \((\mathrm{SU}(2)^{\times n^2}, g)\) has non-negative sectional curvature by 55 [curvatureproduct], we conclude that the sectional curvature of the fibers is also non-negative and so their Ricci curvature is non-negative as well.

Now, the result follows by substituting the manifold-dependent constants in the statement of 30 with the known bounds found in 8. In particular, the dimensions of \(\mathrm{SU}(2)^{\times n^2}\) and \((\mathbb{S}^2)^{\times n^2}\) are known to be \(3n^2\) and \(2n^2\), respectively. Also, using 86, 8 and the scaling property of the distances, we know that \[\begin{align} \mathrm{diam}(((\mathbb{S}^2)^{\times n^2}, h)) &= n\, \mathrm{diam}(\mathbb{S}^2, \frac{1}{2}g_{\mathit{round}}) = n\frac{\sqrt{2}\pi}{2},\\ \mathrm{diam}(\mathrm{SU}(2)^{\times n^2}, g) &= n\, \mathrm{diam}(\mathrm{SU}(2), g_{\mathit{bi}}) \leq 2\pi n. \end{align}\] Lastly, the diameter of the fibers with the induced metric can be bounded using 86 and 9 as \[\mathrm{diam}(\mathrm{U}(1)^{\times n^2}) = n\, \mathrm{diam}(\mathrm{U}(1)) \leq n\pi(1 + \sqrt{2}),\] and by 7 and the scaling of the metric, the convexity radius of \(((\mathbb{S}^2)^{\times n^2}, g)\) is lower bounded by \[\mathit{conv}(((\mathbb{S}^2)^{\times n^2}, g)) = \mathit{conv}((\mathbb{S}^2, \frac{1}{2}g_{\mathit{round}})) = \frac{\sqrt{2}}{2}\mathit{conv}((\mathbb{S}^2, g_{\mathit{round}})) \geq \frac{\pi\sqrt{2}}{4},\] so the result follows. ◻

Proof. By 53 54 it holds that \[\require{physics} \mathrm{Ric}_{g}(X, X) = -\frac{1}{2}\left(n\Tr(X^2) - \Tr(X)^2\right) \geq 0,\] for any \(p \in \mathrm{U}(n)\) and any \(X \in T_p \mathrm{U}(n)\).

Now, given two orthonormal vectors \(X,Y \in T_p \mathrm{U}(n)\), using 53 we obtain \[\begin{align} 0\leq K_{g}(X, Y) &= \frac{1}{4} \left|[X, Y]\right|_g^2. \end{align}\] Furthermore, it is known [61] that \[\label{ineqliebracket} \left|[X, Y]\right|_g \leq 2 |X|_g^2 |Y|_g^2,\tag{50}\] which allows us to conclude that \[0\leq K_{g}(X, Y) \leq \frac{2}{4} \left|X\right|_g^2 \left|Y\right|_g^2 \leq \frac{1}{2}.\] ◻

Proof. The bound of the sectional curvature can be proven following the same steps as in 55.

In order to obtain the equation for the Ricci curvature, note that by 53 54 \[\require{physics} \mathrm{Ric}_{g}(X, Y) = -\frac{2}{4}n\Tr(XY) = \frac{n}{2}g(X, Y),\] for every \(p \in \mathrm{SU}(n)\) and every pair \(X, Y \in T_p \mathrm{SU}(n)\). ◻

Proof. Let \(x \in G\). Taking normal coordinates centered at \(x\) (cf. 60), we know that \[|R_{ijk}^l(x)| = |\delta^{la}R_{ijka}(x)|,\] for every \(i, j, k, l \in \{1, \dotsc, d\}\). Now, using 53 we know that \[R(u, v)w = \frac{1}{4}[[u, v],w],\] for every \(u, v, w \in \mathfrak{g}\). Let \(\{e_n\}_{n = 1}^d\) be an orthonormal basis of \(T_x G\), and consider four vectors \(e_i, e_j, e_k, e_l \in \{e_n\}_{n = 1}^d\). Then \[R_{ijkl}(x) = \frac{1}{4}g([[e_i, e_j],e_k], e_l) \leq \frac{1}{4} |[[e_i, e_j],e_k]|_g |e_l|_g.\] Now, we can apply 50 along with the fact that the vectors have norm one, to conclude that \[|R_{ijk}^l(x)| = |\delta^{la}R_{ijka}(x)| \leq \frac{\sqrt{2}}{4} |[e_i, e_j]|_g |e_k|_g \leq \frac{1}{2}.\] ◻

Proof. From the previous Lemma, we know that for any point \((x, y) \in M \times N\) and any four tangent vectors \(X, Y, Z, W \in T_{(x, y)} M \times N\), it holds that \[R_{g \oplus h}(X, Y, Z, W) = R_g(X_1, Y_1, Z_1, W_1) + R_h(X_2, Y_2, Z_2, W_2),\] where we are using the same notation as in 19 for the decomposition of the tangent vectors. This identity allows us to conclude that, in normal coordinates centered at any point \((x, y) \in M \times N\), \[|(R_{g \oplus h})_{ijkl}(x,y)| \leq \max\{\Lambda_1, \Lambda_2\},\] for every \(i, j, k, l \in \{1, \dotsc, d_M + d_N\}\), by simply considering an orthonormal basis of \(T_{(x, y)} M \times N\). Thus, \[|(R_{g \oplus h})_{ijkl}(x,y)| = |R_{g \oplus h}(e_i, e_j, e_k, e_l)(x,y)| \neq 0\] if and only if \(e_i, e_k, e_k\) and \(e_l\) all belong to either \(T_x M\) or \(T_y N\).

Now, let \(\Pi\) be a plane in \(T_{(x, y)} M \times N\). If \(\Pi \subset T_x M\), then \(\Pi = \mathit{span}\{X, Y\}\), for some orthonormal vectors \(X, Y \in T_x M\). Thus \[K_{g \oplus h}(\Pi) = K_{g \oplus h}(X, Y) = R_{g \oplus h}(X, Y, Y, X) = R_g(X, Y, Y, X) = K_g(X, Y).\] If \(\Pi \subset T_y N\) it follows that \(K_{g \oplus h}(\Pi) = K_h(\tilde{X}, \tilde{Y})\) for some orthonormal vectors \(\tilde{X}, \tilde{Y} \in T_y N\). If \(\Pi \not\subset T_x M\), \(\Pi \not\subset T_y N\) then we can take \(X \in T_x M\), \(Y \in T_y N\) such that \(\Pi = \mathit{span}\{X, Y\}\). In this case, by 20 \[K_{g \oplus h}(X, Y) = R_{g \oplus h}(X, Y, Y, X) = R_g(X, 0, 0, X) + R_h(0, Y, Y, 0) = 0.\]

For the last inequality, let \(X \in T_{(x, y)} M \times N\). Then \[\begin{align} \mathrm{Ric}_{g \oplus h}(X, X) &= \sum_{i = 1}^{d_M+d_N} R_{g \oplus h}(e_i, X, X, e_i) \\&= \sum_{i = 1}^{d_M} R_g(e_i, X_1, X_1, e_i) + \sum_{i = d_M+1}^{d_M+d_N} R_h(e_i, X_2, X_2, e_i) \\&= \mathrm{Ric}_g(X_1, X_1) + \mathrm{Ric}_h(X_2, X_2), \end{align}\] where \(\{e_i\}_{i = 1}^{d_M+d_N}\) is an orthonormal basis for \(T_{(x, y)} M \times N\), which can be divided into an orthonormal basis for \(T_x M\), and an orthonormal basis for \(T_y N\), namely \(\{e_i\}_{i = 1}^{d_M}\) and \(\{e_i\}_{i = d_M+1}^{d_M+d_N}\), respectively. ◻

Proof. First, let \(y \in \mathcal{B}(r,x)\). Let \(\gamma\) be the unique—as \(d_g(x, y) < i(M)\)—geodesic in \(\mathcal{B}(r, x)\) such that \(\gamma(0) = x\) and \(\gamma(1) = y\). Then \[r > d_g(x, y) = \int_0^1 |\gamma'(t)|_g \,dt \geq \int_0^1 |\gamma'(t)^\mathcal{H}|_g \,dt = \int_{0}^1 |(\pi \circ \gamma)'(t)|_h\, dt \geq d_h(\pi(x), \pi(y)),\] and so \(\pi(y) \in \mathcal{B}(r, \pi(x))\).

Now, let \(y \in \mathcal{B}(r, \pi(x))\), and let \(\tilde{\gamma}\) be the unique geodesic in \(\mathcal{B}(r, \pi(x))\) such that \(\tilde{\gamma}(0) = y\) and \(\tilde{\gamma}(1) = \pi(x)\). Since \(M\) is complete, by 64 we know that there exists a horizontal geodesic \(\gamma\) starting at \(x\) and such that \(\tilde{\gamma} = \pi \circ \gamma\), \(l(\tilde{\gamma}) = l(\gamma) < r\). Therefore, \(\gamma(1) \in \mathcal{B}(r, x)\) and so \(y = \tilde{\gamma}(1) = \pi(\gamma(1))\), where \(\gamma(1) \in \mathcal{B}(r, x)\). ◻

Proof. Taking coordinates on a neighborhood of \(p\), we can write \[X = X^i(x) \left.\frac{\partial}{\partial h^i}\right|_x,\quad Y = Y^i(x) \left.\frac{\partial}{\partial h^i}\right|_x,\] where \(\{\left.\frac{\partial}{\partial h^i}\right|_x\}\) form an orthonormal basis of \(H_x M\), for every \(x\) in a neighborhood of \(p\). In these coordinates, \[\left.[X, Y]\right|_p = \sum_{i, j} X^i(p) Y^j(p)\Big[\frac{\partial}{\partial h^i}\bigg|_p,\, \frac{\partial}{\partial h^j}\bigg|_p\Big] + \sum_{j} X(Y^j(p))\frac{\partial}{\partial h^j}\bigg|_p - \sum_{i} Y(X^i(p)) \frac{\partial}{\partial h^i}\bigg|_p.\] Note that the two last summands in this expression are horizontal, thus \[\left(\left.[X, Y]\right|_p\right)^\mathcal{V} = \left(\sum_{i, j} X^i(p) Y^j(p)\Big[\frac{\partial}{\partial h^i}\bigg|_p, \frac{\partial}{\partial h^j}\bigg|_p\Big]\right)^\mathcal{V},\] which only depends on the point \(p\).

Working in the same coordinates, and using the definition of \(T\) \[(T_U V)|_p = \left.(\nabla_{U} V)^\mathcal{H}\right|_p = (U(V^k(p)) + U^i(p) V^j(p) \Gamma^k_{ij}(p)) \left.\frac{\partial}{\partial h^k}\right|_p.\] Note that, since \(V\) is vertical, \(V^k(x) \equiv 0\) for every \(k\) (we are only summing over the horizontal components). Thus \[(T_U V)|_p = (U^i(p) V^j(p) \Gamma^k_{ij}(p)) \left.\frac{\partial}{\partial h^k}\right|_p,\] which only depends on \(p\). ◻

Proof. We want to show that the O’Neill tensor \(T\) associated with \(\pi_2\) vanishes identically. To do so, we will show that for any two vertical vector fields with respect to \(\pi_2\), \(U, V \in \mathfrak{X}(M_2)\), it holds that \[\label{eq:vanishingverticalcon} (\nabla^{M_2}_U V)^{\mathcal{H}} = 0.\tag{51}\] Should this hold, the result would follow: using the compatibility of the Levi-Civita connection \(\nabla^{M_2}\), we know that for every two vertical vector fields with respect to \(\pi_2\), \(U, V \in \mathfrak{X}(M_2)\) and any horizontal vector field with respect to \(\pi_2\), \(X \in \mathfrak{X}(M_2)\), \[0 = U g_2(V, X) = g_2(\nabla^{M_2}_U V, X) + g_2(V, \nabla^{M_2}_U X) = g_2(V, \nabla^{M_2}_U X),\] where the first equality follows from \(V\) begin vertical and \(X\) horizontal, and the last equality follows from the assumption that 51 holds. This allows us to conclude that \(\nabla^{M_2}_U X\) is always horizontal, and so \(T\) vanishes (cf. 36).

Let us then prove that 51 holds. Given \(U, V \in \mathfrak{X}(M_2)\)—which are vertical with respect to \(\pi_2\) by assumption—we can lift them to \(M_1\), and obtain \(\tilde{U}, \tilde{V} \in \mathfrak{X}(M_1)\) which are such that \(d\pi_1(\tilde{U}) = U\) and \(d\pi_1(\tilde{V}) = V\). Note that \(\tilde{U}\) and \(\tilde{V}\) are vertical with respect to \(\pi_3\), \[\begin{align} d\pi_3(\tilde{U}) = d\pi_2(d\pi_1(\tilde{U})) = d\pi_2(U) = 0,\\ d\pi_3(\tilde{V}) = d\pi_2(d\pi_1(\tilde{V})) = d\pi_2(V) = 0, \end{align}\] where the last equality follows from the assumption that \(U\) and \(V\) are vertical with respect to \(\pi_2\). Now, from \(\pi_3\) having totally geodesic fibers, we know that \(\nabla^{M_1}_{\tilde{U}} \tilde{V}\) is vertical with respect to \(\pi_3\). Moreover, it is known [66] that \(\nabla^{M_1}_{\tilde{U}} \tilde{V}\) is such that \(\pi_1(\nabla^{M_1}_{\tilde{U}} \tilde{V}) = \nabla^{M_2}_U V\), and so \[0 = d\pi_3(\nabla^{M_1}_{\tilde{U}} \tilde{V}) = d\pi_2(d\pi_1(\nabla^{M_1}_{\tilde{U}} \tilde{V})) = d\pi_2(\nabla^{M_2}_U V),\] finishing the proof. ◻

Proof. Note that \[\begin{align} L_g: G \to G\\ x \mapsto gx \end{align}\] is an isometry for every \(g \in G\). Furthermore, \(L_g\) maps fibers to fibers: \(L_g(xH) = (gx)H\). Therefore, if \(X\) is horizontal at \(e\)—i.e. perpendicular to the fiber \(H\)—it follows that \((L_g)_* X\) is perpendicular to \((dL_g)_* H\), where \((L_g)_* X = gX\) and \((dL_g)_* H\) is the tangent space of the fiber \(gH\). ◻

Proof. Recall that \(\mathrm{V}_k(\mathbb{C}^{n})\) is the quotient space associated with the action shown in 52 . Now, let \(p \in \mathrm{U}(n)\). We can use 21 71 to conclude that, given two horizontal orthonormal vectors \(X, Y \in T_p \mathrm{U}(n)\), we can extend them as left-invariant vector fields \(\tilde{X}, \tilde{Y}\) which remain horizontal. Thus, \[\begin{align} 0 \leq K_{h}(\pi_* \tilde{X}(p), \pi_* \tilde{Y}(p)) &= K_{g}(\tilde{X}(p), \tilde{Y}(p)) + \frac{3}{4}|[\tilde{X}, \tilde{Y}]^\mathcal{V}|^2_g \\&\leq K_{g}(X, Y) + \frac{3}{4}|[\tilde{X}, \tilde{Y}]|^2_g \\&= |[X, Y]|^2_g \\&\leq 2. \end{align}\] where the last equality follows from 49 50 53, and the last inequality follows from 50 .

Furthermore, since the sectional curvature is non-negative, it holds that its Ricci curvature is non-negative as well. Indeed, let \(V \in T_p \mathrm{V}_k(\mathbb{C}^n)\) be a unitary vector, we can extend \(V\) to a orthonormal basis of \(T_p \mathrm{V}_k(\mathbb{C}^n)\), \(\{V, e_2, \dotsc, e_n\}\). This way, writing \(\mathrm{Ric}_{h}(V, V)\) as \[\mathrm{Ric}_{h}(V, V) = \sum_{i = 2}^n K_{h}(V, e_k) \geq 0,\] the result follows. ◻

Proof. Assume that \(x \in M\) is some critical point of \(f\), and let \(v \in T_x M\). Using 77 and the linearity of \(\nabla\) we conclude that \[\nabla_v \, \mathrm{grad}_g\, f|_x = \nabla_{v^{\mathcal{V}}} \,\mathrm{grad}_g\, f|_x + \nabla_{v^{\mathcal{H}}} \,\mathrm{grad}_g\, f|_x = \nabla_{v^{\mathcal{H}}} \,\mathrm{grad}_g\, f|_x.\]

Therefore, \[\begin{align} \nabla^2_g\, f(x)[v, v] = \langle \nabla_v \,\mathrm{grad}_g\, f, v\rangle_g &= \langle \nabla_{v^{\mathcal{H}}} \,\mathrm{grad}_g\, f, v\rangle_g \\&= \langle \nabla_{v^{\mathcal{H}}} \,\mathrm{grad}_g\, f, v^{\mathcal{H}}\rangle_g + \langle \nabla_{v^{\mathcal{H}}} \,\mathrm{grad}_g\, f, v^{\mathcal{V}}\rangle_g \\&= \langle \nabla_{v^{\mathcal{H}}} \,\mathrm{grad}_g\, f, v^{\mathcal{H}}\rangle_g + \langle \nabla_{v^{\mathcal{V}}} \,\mathrm{grad}_g\, f, v^{\mathcal{H}}\rangle_g \\&= \langle \nabla_{v^{\mathcal{H}}} \,\mathrm{grad}_g\, f, v^{\mathcal{H}}\rangle_g \\&= \nabla^2_g\, f(x)[v^{\mathcal{H}}, v^{\mathcal{H}}], \end{align}\] where the third-last equality follows from the symmetry of the Hessian. Thus, \[\begin{align} \nabla^2_g\, f(x)[v, v] = \nabla^2_g \, f(x)[v^{\mathcal{H}}, v^{\mathcal{H}}] = \langle \nabla^2_g\, f(x)[v^H], v^{\mathcal{H}}\rangle_g &= \langle (\nabla^2_g\, f(x)[v^H])^{\mathcal{H}}, v^{\mathcal{H}}\rangle_g \\ &= \langle \nabla^2_h\, \tilde{f}(\pi(x))[\pi_* v^H], \pi_* v^{\mathcal{H}}\rangle_h \\&= \nabla^2_h\, \tilde{f}(\pi(x))[\pi_* v, \pi_* v], \end{align}\] where the second-last equality follows from 76 and from \(\pi\) being a Riemannian submersion. This expression also allows us to conclude that, for every critical point \(p \in B\) of \(\tilde{f}\), every \(x \in \pi^{-1}(p)\), and every \(v \in T_p B\), \[\nabla^2_h\, \tilde{f}(p)[v, v] = \nabla^2_g\, f(x)[\mathrm{lift}_x\, v, \mathrm{lift}_x\, v].\] ◻

Proof. For every \(x \in M\), we pick an orthonormal basis of \(T_x M = H_x M \oplus V_xM\), given by \(\{H_i\} \cup \{V_i\}\). Then \[\begin{align} \Delta_g\, f(x) = \sum_{V_i \in V_x M} \langle \nabla_{V_i}\, \mathrm{grad}_g\, f, V_i \rangle_g + \sum_{H_i \in H_x M} \langle \nabla_{H_i}\, \mathrm{grad}_g\, f, H_i \rangle_g. \end{align}\] Note that the gradient of \(f\) is horizontal (cf. 74) and so for every \(V_i \in V_xM\) it holds that \[\langle \nabla_{V_i}\, \mathrm{grad}_g\, f, V_i \rangle_g = \langle (\nabla_{V_i}\, \mathrm{grad}_g\, f)^{\mathcal{V}}, V_i \rangle_g = \langle T_{V_i} \,\mathrm{grad}_g\, f, V_i\rangle_g = 0,\] since \(T \equiv 0\) by 22. Therefore, \[\begin{align} \Delta_g\, f(x) = \sum_{H_i \in H_x M} \langle \nabla_{H_i}\, \mathrm{grad}_g \,f, H_i \rangle_g &= \sum_{H_i \in H_x M} \langle \nabla^2_g\, f(x)[H_i], H_i \rangle_g \\&= \sum_{H_i \in H_x M} \langle (\nabla^2_g\, f(x)[H_i])^{\mathcal{H}}, H_i \rangle_g \\&= \sum_{H_i \in H_x M} \langle \mathrm{lift}_x \nabla^2_h\, \tilde{f}(\pi(x))[\pi_* H_i], H_i \rangle_g \\&= \sum_{H_i \in H_x M} \langle \nabla^2_h\, \tilde{f}(\pi(x))[\pi_* H_i], \pi_* H_i \rangle_h \\&= \Delta_h\, \tilde{f}(\pi(x)), \end{align}\] where the third-last equality follows from 76 and the second-last equality follows from \(\pi\) being a Riemannian submersion, which in turn implies that \(\{\pi_* H_i\}_{H_i \in H_x M}\) form an orthonormal basis of \(T_{\pi(x)} B\). ◻

Proof. Let \(p, q \in B\), and let \(x \in \pi^{-1}(p)\), \(y \in \pi^{-1}(q)\) be such that \(d_h(p, q) = d_g(x, y)\). Then \[|\tilde{f}(p) - \tilde{f}(q)| = |f(x) - f(y)| \leq A_1d_g(x, y) = A_1 d_h(p, q).\]

Now, recall that the gradient of \(f\) is \(A_2\)-Lipschitz if \[|\mathrm{P}^{-1}_{v}\,\mathrm{grad}_g\, f(\exp_x(v)) - \mathrm{grad}_g\, f(x)|_g \leq A_2 |v|_g,\] for every \(x \in M\) and every \(v \in T_xM\), where \(\mathrm{P}^{-1}_{v}\) is the inverse of the parallel transport in \(M\) from \(x\) to \(\exp_x(v)\) along \(\gamma(t) = \exp^M_x(t v)\).

We will denote by \(\exp^M\) and \(\exp^B\) the exponential maps in \(M\) and \(B\), respectively.

Let \(p \in B\), \(x \in \pi^{-1}(p)\), \(w \in T_p B\), and let us denote by \(\tilde{\mathrm{P}}^{-1}_{w}\) the inverse of the parallel transport in \(B\) from \(p\) to \(\exp_p^B(v)\) along \(\tilde{\gamma}(t) = \exp^B_p(t w)\). Then relating the gradients of \(f\) and \(\tilde{f}\) via 74 and 64 we can write \[\begin{align} |\tilde{\mathrm{P}}^{-1}_{w}\,\mathrm{grad}_h\, \tilde{f}(\exp^B_p(w)) - \mathrm{grad}_h\, &\tilde{f}(p)|_h \\&= |\tilde{\mathrm{P}}^{-1}_{w}\, \pi_*(\mathrm{grad}_g\, f(\exp^M_x(\mathrm{lift}_x w))) - \pi_*(\mathrm{grad}_g\, f(x))|_h \\&= |\pi_*\, \mathrm{P}^{-1}_{\mathrm{lift}_x w}\, \mathrm{grad}_g\, f(\exp_x^M(\mathrm{lift}_x w)) - \pi_*(\mathrm{grad}_g\, f(x))|_h, \end{align}\] where the second equality follows from 23. Furthermore, as \(\pi\) is a Riemannian submersion, it holds that \[\begin{align} |\pi_*\, \mathrm{P}^{-1}_{\mathrm{lift}_x w}\, \mathrm{grad}_g\, f(\exp_x^M(\mathrm{lift}_x w)) - &\pi_*(\mathrm{grad}_g\, f(x))|_h \\&=|(\mathrm{P}^{-1}_{\mathrm{lift}_x w}\, \mathrm{grad}_g\, f(\exp_x^M(\mathrm{lift}_x w)))^{\mathcal{H}} - (\mathrm{grad}_g\, f(x))^{\mathcal{H}}|_g \\&\leq |\mathrm{P}^{-1}_{\mathrm{lift}_x w}\, \mathrm{grad}_g\, f(\exp_x^M(\mathrm{lift}_x w)) - \mathrm{grad}_g\, f(x)|_g \\&\leq A_2 |\mathrm{lift}_x w|_g \\&= A_2 |w|_h. \end{align}\]

Lastly, the Hessian of \(f\) is \(A_3\) Lipschitz if \[\label{eq:HessianLipschitzcond} \max_{\substack{s \in T_xM\\|s|_g=1}} |\mathrm{P}^{-1}_{v} \circ \nabla^2_g\, f(\exp^M_x(v)) \circ \mathrm{P}_{v}[s] - \nabla^2_g\, f(x)[s]|_g \leq A_3 |v|_g,\tag{53}\] for every \(x \in M\) and every \(v \in T_xM\).

Let \(p \in B\), \(x\in \pi^{-1}(p)\), and \(v \in H_x M\) be some horizontal tangent vector at \(x\). We can lower bound 53 by \[\begin{align} A_3 |\pi_* v|_h = A_3 |v|_g &\geq \max_{\substack{s \in T_xM\\|s|_g=1}} |\mathrm{P}^{-1}_{v} \circ \nabla^2_g\, f(\exp^M_x(v)) \circ \mathrm{P}_{v}[s] - \nabla^2_g\, f(x)[s]|_g \\&\geq \max_{\substack{s \in T_xM\\|s|_g=1}} |(\mathrm{P}^{-1}_{v} \circ \nabla^2_g\, f(\exp^M_x(v)) \circ \mathrm{P}_{v}[s])^{\mathcal{H}} - (\nabla^2_g\, f(x)[s])^{\mathcal{H}}|_g \\&= \max_{\substack{s \in T_xM\\|s|_g=1}} |\pi_*(\mathrm{P}^{-1}_{v} \circ \nabla^2_g\, f(\exp^M_x(v)) \circ \mathrm{P}_{v}[s]) - \pi_* (\nabla^2_g\, f(x)[s])|_h, \end{align}\] where the last equality follows from the fact that \(\pi\) is a Riemannian submersion. We can use 76 and 23 to conclude that \[\begin{align} \max_{\substack{s \in T_xM\\|s|_g=1}} &|\pi_*(\mathrm{P}^{-1}_{v} \circ \nabla^2_g\, f(\exp^M_x(v)) \circ \mathrm{P}_{v}[s]) - \pi_* (\nabla^2_g f(x)[s])|_h \\&= \max_{\substack{s \in T_xM\\|s|_g=1}} |\tilde{\mathrm{P}}^{-1}_{\pi_* v} \circ \pi_* \circ \nabla^2_g\, f(\exp^M_x(v)) \circ \mathrm{P}_{v}[s] - \nabla^2_h\, \tilde{f}(\pi(x))[\pi_* s])|_h \\&= \max_{\substack{s \in T_xM\\|s|_g=1}} |\tilde{\mathrm{P}}^{-1}_{\pi_* v} \circ \nabla^2_h\, \tilde{f}(\exp^B_p(\pi_* v))[\pi_* \mathrm{P}_{v}(s)] - \nabla^2_h\, \tilde{f}(\pi(x))[\pi_* s])|_h \\&= \max_{\substack{s \in T_xM\\|s|_g=1}} |\tilde{\mathrm{P}}^{-1}_{\pi_* v} \circ \nabla^2_h\, \tilde{f}(\exp^B_p(\pi_* v)) \circ \tilde{\mathrm{P}}_{\pi_* v}[\pi_* s] - \nabla^2_h \tilde{f}(\pi(x))[\pi_* s])|_h. \end{align}\]

Putting everything together, we obtain that \[\max_{\substack{s \in T_xM\\|s|_g=1}} |\tilde{\mathrm{P}}^{-1}_{\pi_* v} \circ \nabla^2_h\, \tilde{f}(\exp^B_p(\pi_* v)) \circ \tilde{\mathrm{P}}_{\pi_* v}[\pi_* s] - \nabla^2_h\, \tilde{f}(p)[\pi_* s])|_h \leq A_3 |\pi_* v|_h,\] for every \(p \in B\), \(x \in \pi^{-1}(p)\), and \(v \in H_x M\). Since there is a one-to-one correspondence between horizontal vectors in \(T_xM\) and tangent vectors in \(T_p B\), the result follows. ◻

Proof. Let us fix \(X \in \mathfrak{u}(n)\) with norm one, i.e. \(\require{physics} -\Tr(X^2) = 1\), and a geodesic starting in \(U \in \mathrm{U}(n)\) in the direction of \(X\), i.e. \(\gamma(t) = e^{tX}U\). We want to find \[\min\{t > 0 : e^{tX}U = U\} = \min\{t > 0 : e^{tX} = \mathbb{1}\}.\]

Since \(X \in \mathfrak{u}(n)\) we know that \(X^\dagger = -X\), therefore all its eigenvalues are imaginary. Furthermore, \(X\) is normal (i.e. \(X^\dagger X = X X^\dagger\)). Therefore, without loss of generality, we can consider \(X\) to be in its diagonal form \[\Sigma = \begin{bmatrix} i\lambda_1 & & & \\ & i\lambda_2 & & \\ & & \ddots & \\ & & & i\lambda_n \end{bmatrix},\] where \(\lambda_j \in \mathbb{R}\) for every \(j \in \{1, \dotsc, n\}\). Otherwise, there would exist a unitary matrix \(V\) such that \(X = V \Sigma V^\dagger\) and \(e^{tX} = V e^{t\Sigma} V^\dagger\), which yields \[e^{tX} = \mathbb{1} \iff V e^{t\Sigma} V^\dagger = \mathbb{1} \iff e^{t\Sigma} = \mathbb{1}.\] Now, \[e^{tX} = \begin{bmatrix} e^{it\lambda_1} & & \\ & e^{it\lambda_2} & & \\ & & \ddots &\\ & & & e^{it\lambda_n} \end{bmatrix},\] where all the off-diagonal elements are zero. In order for \(e^{tX}\) to be the identity it has to hold that \[e^{it\lambda_j} = 1,\quad t > 0, \text{ for each } j \in \{1, \dotsc, n\},\] which implies that \[t\lambda_j = 2\pi k_j\quad \text{for some } k_j \in \mathbb{Z},\quad \text{for every }j \in \{1,\dotsc,n\}.\] Since for every eigenvalue \(i\lambda_j\) it holds that \(|\lambda_j|^2 \leq 1\), we can conclude that \(t\) has to be greater than \(2\pi\) and so \(l(\mathrm{U}(n)) \geq 2\pi\). ◻

Proof. Let us fix some \(P \in \mathrm{Gr}_k(\mathbb{C}^n)\), and some unitary tangent vector \(X \in T_P \mathrm{Gr}_k(\mathbb{C}^n)\). It is known (cf. [56]) that \(X\) corresponds to a Hermitian matrix such that \(X = PX + XP\) and that the geodesic starting at \(P\) in the direction \(X\) is given by \[\gamma_{P, X}(t) := e^{t[X, P]}P e^{-t[X, P]}.\]

Without loss of generality, we may assume that \(P = \begin{pmatrix} \mathbb{1}_k & 0 \\ 0 & 0_{n-k} \end{pmatrix}\), and so the condition \(X = XP + PX\) implies that \(X\) is of the form \[X = \begin{pmatrix} 0_k & A\\ A^\dagger & 0_{n-k} \end{pmatrix},\] where \(A\) is an \(k \times (n-k)\)-complex matrix.

Standard calculations allow us to conclude that \[\begin{align} [X, P]^{2k} &= \begin{pmatrix} (-1)^k (AA^\dagger)^k & 0\\ 0 & (-1)^k (A^\dagger A)^k \end{pmatrix},\\ [X, P]^{2k+1} &= \begin{pmatrix} 0 & (-1)^{k+1}(A A^\dagger)^k A\\ (-1)^{k}(A^\dagger A)^k A^\dagger & 0 \end{pmatrix}, \end{align}\] for every \(k \in \mathbb{N}\cup\{0\}\). Thus, \[\begin{align} e^{t[X, P]} &= \begin{pmatrix} \cos(t\sqrt{AA^\dagger}) & -\frac{\sin(t\sqrt{AA^\dagger})}{\sqrt{AA^\dagger}} A\\ -\frac{\sin(t\sqrt{A^\dagger A})}{\sqrt{A^\dagger A}} A^\dagger & \cos(t\sqrt{A^\dagger A}) \end{pmatrix}. \end{align}\]

This way, in order for \(P = e^{t[X, P]} P e^{-t[X, P]}\) to hold, it must be the case that \[\label{eq:identitycosines} \cos^2(t\sqrt{AA^\dagger}) = \mathbb{1}_k.\tag{54}\]

Let us assume without loss of generality that \(A A^\dagger\) is diagonal with eigenvalues \(\{\lambda_i\}_{i = 1}^k\). If 54 holds, every \(\lambda_i\) must be non-negative and \[\label{eq:conditionclosedgeodGr} t\sqrt{\lambda_i} = \pi k,\quad k \in \mathbb{Z},\tag{55}\] for every \(i \in \{1, \dotsc, k\}\).

Finally, note that, as \(X\) has norm one, \[\require{physics} 1 = \Tr(X^\dagger X) = \Tr(AA^\dagger) + \Tr(A^\dagger A) = 2\sum_{i = 1}^k \lambda_i,\] and so, as the eigenvalues are assumed to be non-negative, it holds that \(\lambda_i \leq \frac{1}{2}\) for every \(i \in \{1, \dotsc, k\}\), and \(\sqrt{\lambda_i} \leq \frac{\sqrt{2}}{2}\). Thus, in order for 55 to hold, it must be the case that \[t \geq \frac{2\pi}{\sqrt{2}} = \sqrt{2}\pi,\] finishing the proof. ◻

Proof. To obtain the bounds, we can use 61 62, along with the bounds on the sectional curvature and the length of their shortest non-trivial periodic geodesic shown in 1 2.

For the unitary group endowed with the bi-invariant metric, we obtained that \[K \leq \frac{1}{2},\quad \text{and}\quad l(\mathrm{U}(n)) \geq 2\pi.\] Thus \[i(\mathrm{U}(n)) \geq \min\{\sqrt{2}\pi, \pi\} = \pi,\quad \mathrm{and}\quad \mathit{conv}(\mathrm{U}(n)) \geq \frac{\pi}{2}.\]

For the sphere endowed with the round metric, we showed that \[K \leq 1,\quad \mathrm{and}\quad l(\mathbb{S}^n) \leq 2\pi,\] so \[i(\mathbb{S}^n) \geq \min\{\pi, \pi\} = \pi,\quad \mathrm{and}\quad \mathit{conv}(\mathbb{S}^n) \geq \frac{\pi}{2}.\]

For the Stiefel manifold \(\mathrm{V}_k(\mathbb{C}^n)\) endowed with the metric \(h\) making \(\pi: (\mathrm{U}(n), g) \to (\mathrm{V}_k(\mathbb{C}^n), h)\) a Riemannian submersion, where \(g\) is the bi-invariant metric of \(\mathrm{U}(n)\), we proved that \[K \leq 2,\quad \mathrm{and}\quad l(\mathrm{V}_k(\mathbb{C}^n)) \geq 2\sqrt{2}\pi,\] which implies that \[i(\mathrm{V}_k(\mathbb{C}^n)) \geq \min\{\frac{\sqrt{2}\pi}{2}, \sqrt{2}\pi\} = \frac{\sqrt{2}\pi}{2},\quad \mathrm{and}\quad \mathit{conv}(\mathrm{V}_k(\mathbb{C}^n)) \geq \frac{\sqrt{2}\pi}{4}.\]

Lastly, for the Grassmann manifold also endowed with its induced metric, we showed that \[K \leq 2,\quad \mathrm{and}\quad l(\mathrm{Gr}_k(\mathbb{C}^n)) \geq \sqrt{2}\pi,\] yielding \[i(\mathrm{Gr}_k(\mathbb{C}^n)) \geq \min\{\frac{\sqrt{2}\pi}{2}, \frac{\sqrt{2}\pi}{2}\} = \frac{\sqrt{2}\pi}{2},\quad \mathrm{and}\quad \mathit{conv}(\mathrm{Gr}_k(\mathbb{C}^n)) \geq \frac{\sqrt{2}\pi}{4}.\] ◻

Proof. Since \(M\) is compact, the distance between \(x\) and \(y\) is attained by a geodesic. We denote such a geodesic by \(\gamma\), and assume that \(\gamma(0) = x\), \(\gamma(1) = y\). Then \[\begin{align} d_g(x, y) = \int_0^1 |\gamma'(s)|_g\, ds \geq \int_0^1 |(\gamma')(s)^\mathcal{H}|_g\, ds = \int_0^1 |\pi_* (\gamma')(s)^\mathcal{H}|_h\, ds \geq d_h(\pi(x), \pi(y)), \end{align}\] concluding the proof. ◻

Proof. First, using 82 we know that the round sphere \(\mathbb{S}^n\) is an Einstein manifold with constant \(n-1\), and so it satisfies the Bonnet-Myers theorem with constant \(c = 1\).

For the special unitary group, recall from 56 that \(\mathrm{SU}(n)\) is an Einstein manifold with constant \(\frac{n}{2}\). Moreover, \(\mathrm{SU}(n)\) is connected, compact and has dimension \(n^2-1\). Therefore \(\mathrm{Ric}(v,v) = ((n^2-1) -1)c\) with \[c = \frac{n}{2(n^2 -2)}.\] This way, \(R = \sqrt{\frac{2(n^2-2)}{n}}\) and so we can conclude that \[\begin{align} \mathrm{diam}(\mathrm{SU}(n)) &\leq \pi \sqrt{\frac{2(n^2-2)}{n}} \leq \pi \sqrt{2n}. \end{align}\] ◻

Proof. Let \(A, B \in \mathrm{U}(n)\). We can rewrite \(A\) and \(B\) as \(A = n_A h_A\), \(B = n_B h_B\), where \(n_A, n_B \in \mathrm{U}(1)\) and \(h_A, h_B \in \mathrm{SU}(n)\). Using this decomposition we can write \[d_{\mathrm{U}(n)}(A, B) = d_{\mathrm{U}(n)}(n_A h_A, n_B h_B).\] Moreover, since the metric of \(\mathrm{U}(n)\) is bi-invariant, it follows that \[d_{\mathrm{U}(n)}(n_A h_A, n_B h_B) = d_{\mathrm{U}(n)}(h_A, \overline{n}_A n_B h_B).\] Now, applying the triangular inequality \[d_{\mathrm{U}(n)}(h_A, \overline{n}_A n_B h_B) \leq d_{\mathrm{U}(n)}(h_A, h_B) + d_{\mathrm{U}(n)}(h_B, \overline{n}_A n_B h_B).\]

Using the fact that \(d_{\mathrm{U}(n)}(h_A, h_B) \leq d_{\mathrm{SU}(n)}(h_A, h_B)\) and the bi-invariance of the metric again, \[\begin{align} d_{\mathrm{U}(n)}(h_A, h_B) + d_{\mathrm{U}(n)}(h_B, \overline{n}_A n_B h_B) &\leq d_{\mathrm{SU}(n)}(h_A, h_B) + d_{\mathrm{U}(n)} (h_B, \overline{n}_A n_B h_B) \\&\leq \mathrm{diam}(\mathrm{SU}(n)) + d_{\mathrm{U}(n)}(\mathbb{1}, \overline{n}_A n_B). \end{align}\]

Finally, since \(n_A, n_B \in \mathrm{U}(1)\) and \(d_{\mathrm{U}(n)}(\mathbb{1}, \overline{n}_A n_B) = \sqrt{n}d_{\mathrm{U}(1)}(1, \overline{n}_A n_B)\), \[\begin{align} \mathrm{diam}(\mathrm{SU}(n)) + d_{\mathrm{U}(n)}(\mathbb{1}, \overline{n}_A n_B) &\leq \mathrm{diam}(\mathrm{SU}(n)) + d_{\mathrm{U}(1)}(1, \overline{n}_A n_B) \\&\leq \mathrm{diam}(\mathrm{SU}(n)) + \pi\sqrt{n}, \end{align}\] where the last inequality follows from \(\mathrm{U}(1) \simeq \mathbb{S}^1\) having a diameter upper bounded by \(\pi\). ◻

Proof. Let us only prove \(1\), as it implies the rest of the equalities. Recall that \(d\mu(x) = D(x) \mathrm{dVol}_g(x)\), where \(D(x)\) is assumed to be smooth and positive. Thus, since \(M\) is compact, there exist two constants \(k_1\) and \(k_2\) such that \[0 < k_1 \leq D(x) \leq k_2 < \infty,\] and so \[k_1 \int_U |f|^2 \mathrm{dVol}_g \leq \int_U |f|^2 d\mu|_U \leq k_2\int_U |f|^2 \mathrm{dVol}_g,\] which finishes the proof. ◻

Proof. Assume first that ?? holds. Using the explicit expression for the carré du champ operator \(\Gamma\) associated with \(\operatorname{L}\), which is given by \[\Gamma(f, h) = \frac{1}{\beta} \langle \mathrm{grad}_{g}\,f, \mathrm{grad}_{g}\,h\rangle_g,\] and using 25 we know that \[\begin{align} \Gamma_2(f, f) &= \frac{1}{\beta^2}|\nabla^2 f|^2_g + \frac{1}{\beta^2}\mathrm{Ric}_g(\mathrm{grad}_{g}\,f,\mathrm{grad}_{g}\,f) + \frac{1}{\beta}\nabla^2 F(\mathrm{grad}_{g}\,f, \mathrm{grad}_{g}\,f) \\&\geq \frac{1}{\beta^2}\mathrm{Ric}_g(\mathrm{grad}_{g}\,f,\mathrm{grad}_{g}\,f) + \frac{1}{\beta}\nabla^2 F(\mathrm{grad}_{g}\,f, \mathrm{grad}_{g}\,f) \\&\geq \frac{\kappa}{\beta}\langle\mathrm{grad}_{g}\,f, \mathrm{grad}_{g}\,f \rangle_g \\&= \kappa \Gamma(f, f). \end{align}\]

To prove the converse result, let \(p \in M\) and \(X \in T_pM\). Let \(f\) be a function such that \(\mathrm{grad}_{g}\,f(p) = X\) and \(\nabla^2 f(p) = 0\). Then \[\begin{align} \Gamma_2(f, f) &= \frac{1}{\beta^2}|\nabla^2 f|^2_g + \frac{1}{\beta^2} \mathrm{Ric}_g(\mathrm{grad}_{g}\,f, \mathrm{grad}_{g}\,f) + \frac{1}{\beta} \nabla^2F(\mathrm{grad}_{g}\,f, \mathrm{grad}_{g}\,f) \\&= \frac{1}{\beta^2} \mathrm{Ric}_g(X, X) + \frac{1}{\beta} \nabla^2F(X, X), \end{align}\] and \[\Gamma(f, f) = \frac{1}{\beta} |\mathrm{grad}_{g}\,f|^2_g = \frac{1}{\beta} |X|^2_g.\] Lastly, since \[\Gamma_2(f, f) \geq \kappa \Gamma(f, f),\] holds by assumption, we obtain that \[\frac{1}{\beta^2} \mathrm{Ric}_g(X, X) + \frac{1}{\beta} \nabla^2F(X, X) \geq \frac{\kappa}{\beta} |X|^2_g,\] finishing the proof. ◻

Proof. Let \(f \in C^2(U) \cap H^1(U,\mu|_U)\) and consider \[\tilde{f} := \frac{f - \left.\mu\right|_U(f)}{\left.\mu\right|_U((f - \left.\mu\right|_U(f))^2)^{\frac{1}{2}}}.\] Since \(f \in L^2(U, \mu|_U)\) and \(\mu|_U\) is a probability measure on \(U\), 24 ensures that \(f \in L^1(U, \mu|_U)\), and so \(\tilde{f}\) is well-defined and belongs to \(C^2(U) \cap H^1(U, \mu|_U)\).

It is clear that \(\left.\mu\right|_U(\tilde{f}) = 0\) and \(\left.\mu\right|_U(\tilde{f}^2) = 1\). Thus by ?? it holds that \[\lambda_{1, \hat{\mathcal{L}}}(U) \leq \left.\mu\right|_U(|\mathrm{grad}_{g}\,\tilde{f}|_g^2) = \frac{\left.\mu\right|_U(|\mathrm{grad}_{g}\,f|_g^2)}{\left.\mu\right|_U((f - \left.\mu\right|_U(f))^2)},\] i.e. \[\label{eq:preliminaryineqpoincare} \left.\mu\right|_U((f - \left.\mu\right|_U(f))^2) \leq \frac{1}{\lambda_{1, \hat{\mathcal{L}}}(U)}\left.\mu\right|_U(|\mathrm{grad}_{g}\,f|_g^2).\tag{59}\]

Moreover, since \(\mu|_U\) is a probability measure on \(U\), it holds that \[\left.\mu\right|_U((f - \left.\mu\right|_U(f))^2) = \left.\mu\right|_U(f^2) - 2\left.\mu\right|_U(f)^2 + \left.\mu\right|_U(f)^2 = \left.\mu\right|_U(f^2) - \left.\mu\right|_U(f)^2,\] and so we can rewrite 59 as \[\left.\mu\right|_U(f^2) - \left.\mu\right|_U(f)^2 \leq \frac{1}{\lambda_{1, \hat{\mathcal{L}}}(U)}\left.\mu\right|_U(|\mathrm{grad}_{g}\,f|_g^2)\] for every \(f \in C^2(U) \cap H^1(U, \mu|_U)\). ◻

Proof. First, let \(V\) be defined as \[\label{eq:defofV} V(x) := -\beta F(x) - \log(Z),\tag{60}\] where \(Z = \int_U e^{-\beta F(x)} dx\), and define \[\mathrm{\boldsymbol{L}} := \beta \langle - \mathrm{grad}_{g}\,F, \cdot\rangle_g + \Delta_g.\]

Let us obtain a lower bound on the first Neumann eigenvalue \(\lambda_{1, \mathrm{\boldsymbol{L}}}(U)\). In order to apply 96, let us begin by rewriting the curvature-dimension condition.

The curvature-dimension condition from ?? can be rewritten as \[\label{eq:compare1} \beta \nabla^2 F +\mathrm{Ric}_g \geq \beta \kappa g.\tag{61}\]

If we now rewrite the left-hand side of ?? for \(V(x)\) as defined in 60 , using the fact that \[\nabla^2 F(X, Y) = \langle \nabla_X \mathrm{grad}_{g}\,F, Y\rangle_g,\] we arrive at \[\require{upgreek} \label{eq:compare2} \mathrm{Ric}_g(X,X) +\beta \nabla^2 F(X,X) \geq -\upkappa |X|^2_g.\tag{62}\]

Comparing 61 62 , we conclude that 62 holds in our case for \(\require{upgreek} -\upkappa := \beta\kappa\). Thus, we can apply 96 to find that \[\lambda_{1, \mathrm{\boldsymbol{L}}}(U) \geq \frac{8}{D^2} + \frac{1}{2}\beta \kappa \geq \frac{\beta \kappa}{2},\] where \(D\) is the diameter of \(U\). And so, using 95, since \(\nu(x)|_U := \mathbb{1}_U\frac{1}{Z} e^{-\beta F}\) is a probability measure on \(U\), it holds that \[\int_U f^2\, d\nu|_U - \Big(\int_U f \, d\nu|_U \Big)^2 \leq \frac{2}{\kappa \beta} \int_U |\mathrm{grad}_{g}\,f|_g^2 \, d\nu|_U,\] for every \(f \in C^2(U) \cap H^1(U, \nu|_U)\). ◻

Proof. First, it is known (cf. [36]) that the square of any continuous real semi-martingale \(Y_t\) satisfies the SDE given by \[Y^2_t = Y_0^2 + 2\int_0^t Y_s dY_s + \langle Y, Y \rangle_t.\] Substituting \(f(X_t)\) in this equation we obtain that \[\label{eq:squareIto} f^2(X_t) = f^2(X_0) + 2\int_0^t f(X_s)df(X_s) + \langle f(X), f(X)\rangle_t.\tag{64}\]

Now, the quadratic variation of \(f(X_t)\) can be rewritten in terms of the quadratic variation of \(M^f\). Indeed, since \(f(X_t)\) satisfies the SDE from 63 , we can use 99 to conclude that \[\label{eq:quadraticvariation} \langle f(X), f(X)\rangle_t = \langle M^f, M^f\rangle_t,\tag{65}\] and so, substituting 65 and the expression for \(df(X_t)\) shown in 63 into 64 , we conclude that \[\label{eq:firstidentityfsquared} f^2(X_t) = f^2(X_0) + 2\int_0^t f(X_s)dM^f_s + 2\int_0^t f(X_s) \operatorname{L}f(X_s) ds + \langle M^f, M^f\rangle_t.\tag{66}\]

On the other hand, we can apply 63 to \(f^2\) to conclude that there exists a local martingale \(M^{f^2}_t\) such that \[\label{eq:secondidentityfsquared} f^2(X_t) = f^2(X_0) + M^{f^2}_t + \int_0^t \operatorname{L}f^2(X_s) ds.\tag{67}\]

Since the decomposition of a semimartingale into a local martingale and a process with finite variation is unique, we can equate the bounded variation parts in 66 67 to conclude that \[2\int_0^t f(X_s) \operatorname{L}f(X_s) ds + \langle M^f, M^f\rangle_t = \int_0^t \operatorname{L}f^2(X_s) ds,\] which allows us to rewrite \[\begin{align} \langle M^f, M^f\rangle_t = \int_0^t \operatorname{L}f^2(X_s) ds - 2 f(X_s) \operatorname{L}f(X_s) ds &= 2\int_0^t \Gamma(f(X_s)) ds \\&= \frac{2}{\beta} \int_0^t |\mathrm{grad}_{g}\,f(X_s)|^2_g\; ds, \end{align}\] finishing the proof. ◻

Proof of 98. Recall that we know that for any \(f \in C^\infty(M)\), there exists a local martingale satisfying \[f(X_t) = f(X_0) + M^f_t + \int_0^t \operatorname{L}f(X_s) ds,\] where by 100, the quadratic variation of \(M^f\) can be written as \[\langle M^f, M^f\rangle_t = \frac{2}{\beta}\int_0^t |\mathrm{grad}_{g}\,f(X_s)|_g^2 ds.\] Since \(M\) is compact, it is clear that \[d\langle M^f, M^f\rangle_t = \frac{2}{\beta} |\mathrm{grad}_{g}\,f(X_t)|_g^2 dt\] is equivalent to the Lebesgue measure \(dt\). Thus, we apply 101 to conclude that \[dM_t = \sqrt{\frac{2}{\beta}} |\mathrm{grad}_{g}\,f(X_t)|_gdB_s,\] finishing the proof. ◻

Proof. As the SDE is defined in \(\mathbb{R}\), it suffices to ensure that the diffusion coefficient \[c \sigma_t \sqrt{|Y_t|}\] is Hölder and the drift coefficient \[aY_t + b\] is Lipschitz. To this end, we define \[\begin{align} F(x) &:= ax + b,\\ G(t, \omega, x) &:= c \sigma_t(\omega) \sqrt{x}. \end{align}\] Then for any \(T > 0\), \(t \in [0, T]\), \(\omega \in \Omega\) and \(x, y \geq 0\), it holds that \[\begin{align} |F(x) - F(y)| &\leq a|x-y|,\\ |G(t, \omega, x) - G(t, \omega, y)| &\leq c |\sigma_t(\omega)| |\sqrt{x} - \sqrt{y}| \leq c \sqrt{\sigma} |x - y|^\frac{1}{2}, \end{align}\] as \(x \mapsto \sqrt{x}\) is Hölder \(1/2\). Thus, we can apply [78] to ensure that the SDE has a unique strong solution.

Furthermore, note that whenever \(\xi \geq 0\), \(Y_t\) will be non-negative. Indeed, by continuity of the process \(Y_t\), before becoming negative it has to have value \(0\). In this case, whenever \(Y_s(\omega) = 0\) for some \(s > 0\) and \(\omega \in \Omega\), ?? implies that the dynamics of \(Y_t(\omega)\) at time \(s\) are deterministic and given by \(dY_t = b\,dt\). Now, since \(b > 0\) this forces \(Y_t\) to grow, bouncing it back to the positive values. ◻

Proof of 102. First of all, recall from 103 that the unique solution of 68 is such that \(Y_t \geq 0\) for every \(t \geq 0\) whenever \(Y_0 \geq 0\). Now, let \(0 < \theta \leq \frac{2a}{c^2 \sigma}\) be fixed, and let us define \[M_t := f(t, Y_t) = \exp(\theta b t - \theta Y_t),\] where \(f(t, x) := \exp(\theta b t - \theta x)\). Thus, we can apply Itô’s formula from 104, using the identities \[\begin{align} \frac{\partial f}{\partial t}(t, Y_t) &= \theta b M_t,\\ \frac{\partial f}{\partial x}(t, Y_t) &= -\theta M_t,\\ \frac{\partial^2 f}{\partial x^2}(t, Y_t) &= \theta^2 M_t, \end{align}\] to conclude that \(M_t\) satisfies the SDE \[\begin{align} dM_t &= M_t\Big( \big(\theta b -\theta (a Y_t + b) + \frac{1}{2} c^2 \sigma_t^2 Y_t \theta^2\big)dt - c\sqrt{Y_t} \sigma_t dB_t\Big)\\ &= M_t\Big( \big(-\theta a + \frac{1}{2} c^2 \sigma_t^2 \theta^2\big)Y_t dt - c\sqrt{Y_t} \sigma_t dB_t\Big). \end{align}\]

Note that, as \(Y_t \geq 0\) and \(\sigma_t^2 \leq \sigma\) for every \(t \geq 0\), it holds that \[\label{eq:negativedrift} \big(-\theta a + \frac{1}{2} c^2 \sigma_t^2 \theta^2\big)Y_t \leq \big(-\theta a + \frac{1}{2} c^2 \sigma \theta^2\big)Y_t \leq 0,\quad \forall t\geq 0,\tag{69}\] since by definition, \(\theta\) is such that \[0< \theta \leq \frac{2a}{c^2 \sigma}.\]

69 implies that \(M_t\) is a supermartingale: \[\begin{align} \mathbb{E}[M_t] &= \mathbb{E}[M_0] + \mathbb{E}\Big[\int_0^t M_s \big(-\theta a + \frac{1}{2} c^2 \sigma_s^2 \theta^2\big)Y_s ds\Big] - \mathbb{E}\Big[\int_0^t M_s c\sqrt{Y_s} \sigma_s dB_s\Big]\\ &= \mathbb{E}[M_0] + \mathbb{E}\Big[\int_0^t M_s \big(-\theta a + \frac{1}{2} c^2 \sigma_s^2 \theta^2\big)Y_s ds\Big]\\ &\leq \mathbb{E}[M_0], \end{align}\] where the second identity follows from the fact that \[\int_0^t M_s c\sqrt{Y_s} \sigma_s dB_s\] is a martingale by [80]—and so its expected value vanishes—and the inequality follows from 69 and the fact that \(M_t\) is non-negative by definition. Therefore, for every \(t \geq 0\), \[\mathbb{E}[M_t | Y_0 = \xi] \leq M_0 = e^{-\theta \xi},\] which can be rewritten as \[\mathbb{E}[\exp(\theta b t - \theta Y_t) | Y_0 = \xi] = e^{\theta b t} \mathbb{E}[\exp(- \theta Y_t) | Y_0 = \xi] \leq e^{-\theta \xi},\] and so \[\mathbb{E}[\exp(- \theta Y_t) | Y_0 = \xi] \leq e^{-\theta \xi}e^{-\theta b t},\quad \forall t\geq 0.\]

Lastly, we can use Markov’s inequality from 105 to conclude that for every \(t \geq 0\), \[\mathbb{P}[Y_t \leq A | Y_0 = \xi] = \mathbb{P}[e^{-\theta Y_t} \geq e^{-\theta A} | Y_0 = \xi ] \leq e^{\theta A} \mathbb{E}[e^{-\theta Y_t} | Y_0 = \xi] \leq e^{\theta(A - \xi)} e^{-\theta b t},\] finishing the proof. ◻

Proof. In order to see this, let us first define two auxiliary functions, \[\begin{align} \Phi(t, x) &:= \langle -\mathrm{grad}_{h}\,\tilde{F}(x), \tilde{r}(x)\mathrm{grad}_{h}\,\tilde{r}(x) \rangle_h + \frac{1}{\beta}\left(\tilde{r}(x)\Delta_h\tilde{r}(x) + |\mathrm{grad}_{h}\,\tilde{r}(x)|_h^2\right),\\ \tilde{\Phi}(t, z) &:= 2 \lambda_* \frac{1}{2} z^2 + \frac{1}{4\beta}, \end{align}\] where \(x \in B\) and \(z \in \mathbb{R}\). In 14 we showed that \[\label{eq:ineqPhis} \Phi(t, \tilde{X}_t) \geq \tilde{\Phi}(t, \tilde{r}(\tilde{X}_t)),\tag{72}\] for every \(t \geq 0\) before \(\tilde{X}_t\) escapes from \(\mathcal{B}(i(B), y)\).

Let us now define the following auxiliary processes; \[\begin{align} U_t &:= \frac{1}{2}\left(\tilde{r}(\tilde{X}_t)^2 - (z_t)^2\right),\\ a_t &:= \begin{cases} \frac{1}{U_t}(\tilde{\Phi}(t, \tilde{r}(\tilde{X}_t)) - \tilde{\Phi}(t, z_t)), \quad &\text{if } U_t \neq 0,\\ 0,\quad &\text{if } U_t = 0, \end{cases}\\ b_t &:= \Phi(t, \tilde{X}_t) - \tilde{\Phi}(t, \tilde{r}(\tilde{X}_t)),\\ c_t &:= \begin{cases} \frac{1}{U_t} \sqrt{\frac{2}{\beta}}(|\tilde{r}(\tilde{X}_t)| - |z_t|)|\mathrm{grad}_{h}\,\tilde{r}(\tilde{X}_t)|_h, \quad &\text{if } U_t \neq 0,\\ 0,\quad &\text{if } U_t = 0. \end{cases} \end{align}\] These processes satisfy \[\begin{align} b_t + a_t U_t &= \Phi(t, \tilde{X}_t) - \tilde{\Phi}(t, z_t),\\ U_t c_t &= \sqrt{\frac{2}{\beta}}(|\tilde{r}(\tilde{X}_t)| - |z_t|)|\mathrm{grad}_{h}\,\tilde{r}(\tilde{X}_t)|_h. \end{align}\] Then, subtracting 71 from 70 we obtain that \(U_t\) satisfies \[U_t = U_0 + \int_0^t (a_s U_s + b_s)ds + \int_0^t U_s c_s dB_s.\]

The process \(U_t\) is continuous—as \(\tilde{r}(\tilde{X}_t)\) and \(z_t\) are continuous. Furthermore, \(U_t\) is non-negative. This follows from the same reasoning used to prove that the solution of the modified CIR process from 68 was non-negative. Indeed, \(b_t \geq 0\) by 72 and so whenever \(U_t\) is zero, its associated SDE at that time is of the form \[dU_t = b_t dt,\] which does not allow for \(U_t\) to attain negative values. ◻

Sketch of the proof. The proof relies on 88 and 89 to define a bilinear form \(\mathfrak{B}_\mu\) associated with the operator \(\mathfrak{L} + \mu\). The operator \(\mathfrak{B}_\mu\) is defined on \(H^1_0(U)\) and is proven to be bounded and coercive. Then using 110 the result follows. ◻

Sketch of the proof. The proof relies on 112. A bounded and linear operator on \(L^2(U)\) is defined, which is proven to be compact by the Rellich-Kondrachov theorem (107). Finally, the Fredholm alternative (108) is applied to the operator, and its spectrum is studied using 109. ◻

Sketch of the proof. The proof relies on 112 113, and uses the same linear, bounded and compact operator from the proof of 113. To conclude, its eigenvalues are studied using 109. ◻

Proof of 111. Note that the operator considered in the statement of the theorem is not elliptic but rather \(-\operatorname{L}\) is. Therefore, instead of considering the problem from 73 , we can consider \(\tilde{f} = f - 1\) and the associated PDE \[\begin{cases} -\operatorname{L}\tilde{f} = -\theta \tilde{f} - \theta&\quad x \in U,\\ \tilde{f} = 0&\quad x \in \partial U. \end{cases}\]

Now, we know that this problem has a solution for almost every \(\theta \in \mathbb{R}\)—except for at most a countable subset of \(\mathbb{R}\)—by 114. Thus, the problem from the statement has a unique solution for almost every \(\theta \in \mathbb{R}\). ◻

Proof. Without loss of generality, we can assume that \(p \in \mathbb{S}^2\) has coordinates \((0, 0, 1)\), and therefore \(v = (v_1, v_2, 0)\) with \(v_1^2 + v_2^2 = 1\). In these coordinates, \[\log_p(x) = (\theta \cos\varphi, \theta\sin\varphi, 0)\] where \(\theta \in (0, \pi)\) and \(\varphi \in [0, 2\pi)\). Therefore, \[\tilde{r}_{p, v}(x) = \theta(v_1\cos\varphi + v_2 \sin\varphi).\]

It is a standard fact that the round metric of \(\mathbb{S}^2\) in normal coordinates is of the form \[g_{\mathit{round}}(\theta, \varphi) = d\theta^2 + \sin^2 \theta d\varphi^2,\] which implies that \[\Gamma^\theta_{\varphi \varphi} = - \sin\theta\cos\theta,\quad \Gamma^\varphi_{\varphi \theta} = \Gamma^\varphi_{\theta\varphi} = \cot\theta,\] while the rest of the Christoffel symbols vanish identically.

Since the Laplacian can be written in local coordinates as \[\label{eq:Laplacianincoords} \Delta_g = g^{ij}(x) \frac{\partial^2}{\partial x^i \partial x^j} - g^{kl}(x) \Gamma^n_{kl}(x) \frac{\partial}{\partial x^n},\tag{75}\] we can use the expressions for the Christoffel symbols, the metric and its inverse to conclude that \[\begin{align} \Delta_{g_\mathit{round}} &= g^{\theta \theta}_{\mathit{round}}(x) \frac{\partial^2}{\partial \theta^2} + g^{\varphi \varphi}_{\mathit{round}}(x) \frac{\partial^2}{\partial \varphi^2} - g^{\varphi\varphi}_{\mathit{round}}(x) \Gamma^\theta_{\varphi\varphi}(x) \frac{\partial}{\partial \theta} \\&= \frac{\partial^2}{\partial \theta^2} + \frac{1}{\sin^2 \theta} \frac{\partial^2}{\partial \varphi^2} + \cot \theta \frac{\partial}{\partial \theta}, \end{align}\] and so \[\begin{align} \Delta_{g_\mathit{round}} \tilde{r}_{p, v}(x) &= \Big(\frac{\partial^2}{\partial \theta^2} + \frac{1}{\sin^2 \theta} \frac{\partial^2}{\partial \varphi^2} + \cot \theta \frac{\partial}{\partial \theta}\Big)\theta(v_1\cos\varphi + v_2 \sin\varphi) \\&= -\frac{\theta}{\sin^2 \theta}(v_1\cos\varphi+v_2\sin\varphi) + \cot\theta (v_1\cos\varphi + v_2 \sin\varphi) \\&= (v_1\cos\varphi + v_2 \sin\varphi)\Big(\cot\theta -\frac{\theta}{\sin^2 \theta}\Big) . \end{align}\] Observing this identity we can conclude that, even if the Laplacian of \(\tilde{r}_{p, v}\) tends to zero as \(\theta\) becomes small, this does not imply that \(\Delta_{g_\mathit{round}}\, \tilde{r}_{p, v}(x) \geq 0\) near \(p\), as the expression on the right-hand side has opposite signs for \(\varphi = 0\) and \(\varphi = \pi\) when \(\theta > 0\). ◻

Proof. Using the expressions for \(\tilde{r}_{p, v}\) and the round metric in normal coordinates obtained in the previous remark, along with the expression of \(\mathrm{grad}_{g_{\mathit{round}}}\, \tilde{r}_{p, v}\) in local coordinates, namely \[\mathrm{grad}_{g_{\mathit{round}}}\, \tilde{r}_{p, v}(x) = g_{\mathit{round}}^{ij}(x) \frac{\partial \tilde{r}_{p, v}}{\partial x^i} \partial_j,\] we conclude that \[\mathrm{grad}_{g_{\mathit{round}}}\, \tilde{r}_{p, v}(x) = (v_1 \cos \varphi + v_2 \sin \varphi) \partial_{\theta} + \frac{1}{\sin^2\theta} \theta (v_2 \cos\varphi - v_1 \sin \varphi)\partial_{\varphi}.\] Therefore \[|\mathrm{grad}_{g_{\mathit{round}}}\, \tilde{r}_{p, v}(x)|^2_{g_{\mathit{round}}} = (v_1 \cos \varphi + v_2 \sin \varphi)^2 + \frac{1}{\sin^2\theta} \theta^2 (v_2 \cos\varphi - v_1 \sin \varphi)^2.\] Although it is easy to see that the norm of the gradient tends to \(1\) as \(\theta\) approaches zero, it is clear that the expression is not constant in \(\mathcal{B}(R, p)\) for any value of \(R\). ◻