Proof. Suppose \(h\in\mathcal{P}\) and consider its Ricci flow \(h(\ell)\) with \(h(0) = h\) and \(\ell\in (L_-,L_+)\), where \(L_-<0\) and \(L_+>0\). Assume for a contradiction that \(h(\ell_0) \notin\mathcal{P}\) for some \(\ell_0\in (0,L_+)\). Because \(\mathcal{P}\) is open, there is minimum such time; without loss of generality we assume \(\ell_0\) denotes this time. It follows that \(h(\ell_0)\in \partial\mathcal{P}.\)

Let \(g:=h(\ell_0)\) and let \(g(\ell)\) denote its Ricci flow. Then by uniqueness of the Ricci flow, \(g(\ell) = h(\ell+\ell_0)\) for all \(\ell\in(L_--\ell_0,L_+-\ell_0)\). By assumption, as \(g(0)= h(\ell_0)\in \partial \mathcal{P}\), there is some \(\varepsilon>0\) such that \(g(\ell)\notin\mathcal{P}\) for all \(\ell\in (-\varepsilon,0)\). Then, for any \(\ell\in(-\varepsilon,0)\) with \(|\ell|<\ell_0\), we have \(0< \ell+\ell_0 < \ell_0\) and \(h(\ell + \ell_0) \notin\mathcal{P}\). This implies that the Ricci flow of \(h\) leaves \(\mathcal{P}\) at a time earlier than \(\ell_0\), contradicting the choice of \(\ell_0\). ◻

Proof. As mentioned above, the metric \(g_{t,1}\) on \(\mathbb{S}^{4n+3}\) has \(\sec >0\) for \(t < 4/3\). It follows from the Gray-O’Neill formulas [15], [16] applied to the Riemannian submersion \((\mathbb{S}^{4n+3},g_{t,1}) \to(\mathbb{C}P^{2n+1},g^{\mathbb{C}P}_{t,1})\) that \(g^{\mathbb{C}P}_{t,1}\) has \(\sec >0\) when \(t < 4/3\).

Conversely, we know from [10] that there are \(2\)-planes whose sectional curvature for the metric \(g^{\mathbb{C}P}_{t,1}\) equals \(4-3t\). In the notation of [10], these \(2\)-planes are spanned by the orthonormal vectors \(Y_\alpha\) and \(Y_{\alpha i}\), where \(1\leq \alpha\leq n\) and \(2\leq i \leq 3\), both of which lie in the horizontal space \(H\) of the decomposition from 2 . Clearly, it follows that \(g^{\mathbb{C}P}_{t,1}\) does not have \(\sec>0\) for \(t\geq \frac{4}{3}\). ◻

Proof of Theorem 4. Following Proposition 3, we consider an arbitrary metric \(g_{t,s}\) in \(\partial\mathcal{P}\). From 7 , we obtain that \(g_{t,s}\in \partial \mathcal{P}\) if and only if \(g_{t,s}=sg_{\frac{4}{3}, 1}\) for some \(s>0\). In fact, for our purposes we may assume that \(s=1\) because of the following reasons. First, the condition \(\sec>0\) is invariant under scalings. Second, it is a general fact that the Ricci flow of a scaled metric \(sg\) is \(sg\left(\frac{\ell}{s}\right)\), where \(g(\ell)\) is the Ricci flow of \(g\), see [17]. Altogether we conclude that a metric \(sg\in\partial\mathcal{P}\) with \(s>0\) satisfies the hypothesis in Proposition 3 if and only if \(g\) does.

Thus we consider \(g_{\frac{4}{3},1}\in\partial\mathcal{P}\) and its Ricci flow \(g(\ell)\), i.e. \(g(0)=g_{\frac{4}{3},1}\). We shall show that \[\label{eq:positive1} \frac{d}{d\ell}\Big\vert_{\ell=0} F(g(\ell))>0.\tag{8}\] Since \(F(g(0))=0\), the positivity of the derivative implies that there is some small \(\varepsilon>0\) such that \(F(g(\ell))<0\) for all \(\ell\in (-\varepsilon,0)\). Consequently, \(F(g(\ell))\notin\mathcal{P}\) for all \(\ell\in (-\varepsilon,0)\). Proposition 3 will then complete the proof.

It remains to show that 8 holds. To compute the derivative we use the chain rule: \[\frac{d}{d\ell}\Big\vert_{\ell=0} F(g(\ell)) =\nabla F(g(0)) g'(0),\] where we interpret \(\nabla F(g(0))\) as a \(1\times 2\) vector and \(g'(0)\) as a \(2\times 1\) vector. Using the Ricci flow equation given in 3 we find that \(g(\ell)\) is the family \(g_{t(\ell),s(\ell)}\) determined by \((t',s')=(-2r_t t,-2r_s s)\). For \(\ell=0\) we have \((t,s)=(\frac{4}{3},1)\), so \(g'(0)=(-\frac{8}{3}r_t,-2r_s)^T\), where \(T\) denotes the transpose vector.

The gradient of \(F\) is \(\nabla F=\left( -\frac{1}{s}, \frac{t}{s^2} \right)\), so \(\nabla F(g(0))=\left( -1, \frac{4}{3} \right)\). Thus \[\frac{d}{d\ell}\Big\vert_{\ell=0} F(g(\ell))=\left( -1, \frac{4}{3} \right)\left(-\frac{8}{3}r_t,-2r_s\right)^T=\frac{8}{3}(r_t-r_s).\] For the case of a sphere, the Ricci eigenvalues of \(g(0)=g_{\frac{4}{3},1}\) reduce to: \[r_t=\frac{3}{4}(d_V-1)+ \frac{4}{3} d_H,\qquad r_s=d_H -1 + \frac{1}{3} d_V.\] Then \[\frac{d}{d\ell}\Big\vert_{\ell=0} F(g(\ell))=\frac{8}{3}(r_t-r_s)=\frac{2}{9} ( 5 d_V + 4 d_H + 3)>0.\] For complex projective spaces, the Ricci eigenvalues of \(g(0)=g_{\frac{4}{3},1}\) reduce to \[r_t = \frac{16}{3}n + 3,\qquad r_s=4n+\frac{8}{3}.\] Then \[\frac{d}{d\ell}\Big\vert_{\ell=0} F(g(\ell)) = \frac{8}{3}(r_t-r_s) = \frac{8}{9}\left(4n+1\right) > 0.\] ◻

Strategy of the proof for Theorem 7. We will prove Theorem 7 by appealing to Proposition 3. First we need to analyze the boundary \(\partial \mathcal{P}\). As in the proof of Theorem 4, note that preservation of \(\sec>0\) under the Ricci flow is not affected by scalings of the initial metric, so it is enough to study the behavior of metrics of the form \(g_{x,y,z,1}\). Note that a metric \(g_{x,y,z,1}\) lies in \(\partial\mathcal{P}\) if and only if the nine functions \(H_i, V_i, P_i\) for \(i=1,2,3\) are non-negative and at least one of them is zero.

Our main approach to showing that the hypotheses of Proposition 3 are satisfied is similar in spirit to the proof of Theorem 4, although it is inevitably more involved since in this case the set \(\mathcal{P}\) is determined by the positivity of nine functions instead of just one. Let us be more precise. As a first step we need to consider each metric of the form \(g_{x,y,z,1}\in\partial\mathcal{P}\). Then at least one of the nine functions \(H_i,V_i,P_i\) vanishes at \((x,y,z)\). Let \(g(\ell)\) be the Ricci flow of \(g_{x,y,z,1}\). It will then be sufficient to show that, among the vanishing functions, we can find one, say \(H_1\), which is smooth at \((x,y,z)\) and for which the corresponding extended function \(\tilde{H}_1(g(\ell))\) has positive derivative at \(\ell=0\). Because \(\tilde{H}_1(g(0))=0\), the positivity of the derivative implies that there is some small \(\varepsilon>0\) such that \(\tilde{H}_1(g(\ell))<0\) for all \(\ell\in (-\varepsilon,0)\). Hence, \(\tilde{H}_1(g(\ell))\notin\mathcal{P}\) for all \(\ell\in (-\varepsilon,0)\). Proposition 3 completes the proof.

Let us remark that, while the functions \(H_i,V_i\) are smooth on the whole \(\mathbb{R}_+^3\), it will turn out, however, that there are some metrics on the boundary \(\partial\mathcal{P}\) at which some \(P_i\) is non-differentiable. Fortunately, for these metrics it turns out that either \(H_i\) or \(V_i\) vanishes as well and the derivative of the corresponding extended function has the right sign, which is enough for our purposes.

In principle, one would have to study nine cases, in each of which one assumes that one of the nine functions vanishes while the others are non-negative. However, the functions \(H_i\), \(V_i\), and \(P_i\) are symmetric in the three variables. This implies that, for instance, if a metric \(g\in\partial\mathcal{P}\) satisfies \(H_2=0\) or \(H_3=0\), then there exists a metric \(\overline{g}\in\partial\mathcal{P}\) with \(H_1=0\) having the same Ricci flow behavior as \(g\) with respect to the hypotheses of Proposition 3; and analogous properties hold for \(V_i\) and \(P_i\). Therefore, it suffices to consider only three cases, corresponding to the vanishing of \(H_1\), \(V_1\), and \(P_1\). For convenience, we define the sets \[\begin{align} \mathcal{B}_H &:=\{g_{x,y,z,1}\in\partial \mathcal{P}\; :\; H_1(x,y,z)=0\}, \\ \mathcal{B}_V &:=\{g_{x,y,z,1}\in\partial \mathcal{P}\; :\; V_1(x,y,z)=0\}, \\ \mathcal{B}_P &:=\{g_{x,y,z,1}\in\partial \mathcal{P}\; :\; P_1(x,y,z)=0\}. \end{align}\] Note that \(\mathcal{B}_H\), \(\mathcal{B}_V\) and \(\mathcal{B}_P\) are not disjoint. By the discussion above it follows that Theorem 7 will be proven if we show that for each metric \(g_{x,y,z,1}\in\mathcal{B}_H\cup \mathcal{B}_V\cup\mathcal{B}_P\) it holds that the derivative of the extended function \(\tilde{H}_1(g(\ell))\), \(\tilde{V}_1(g(\ell))\) or \(\tilde{P}_1(g(\ell))\) is positive for at least one of the functions \(H_1\), \(V_1\) or \(P_1\) which vanishes at \((x,y,z)\). This plan will be carried out in Propositions 10, 12 and 14, where we will consider, respectively, metrics in \(\mathcal{B}_H\), \(\mathcal{B}_V\) and \(\mathcal{B}_P\setminus (\mathcal{B}_H \cup \mathcal{B}_V)\). ◻

Proof. We first observe that since \(y,z\in \left(0,\frac{4}{3}\right]\), it easily follows that \(a = y+z \in \left(0,\frac{8}{3}\right]\). For each value of \(a \in \left(0,\frac{8}{3}\right]\), we will determine the set of \(b\)-values which are obtainable with this choice of \(a\).

Assume initially that \(a\in \left(0,\frac{4}{3}\right]\). We observe that for any \(y<a\), there is a \(z\in \left(0,\frac{4}{3}\right]\) for which \(y+z = a\), but that this statement is false for any \(y\geq a\). Thus, for fixed \(a\in \left(0,\frac{4}{3}\right]\), we conclude that \(y\in (0,a)\).

Since \(b= yz\) and \(z= a-y\), we find \(b = y(a-y)\) with \(y\in (0,a)\). For fixed \(a\), the formula \(b= y(a-y)\) defines a parabola. The value of \(b\) at \(y=0\) and \(y=a\) is \(0\), while the value at the vertex \(\frac{a}{2}\in (0,a)\) is \(\frac{a^2}{4} > 0\). It follows that, as \(y\) varies over \((0,a)\), the value of \(b\) varies in \(\left(0,\frac{a^2}{4}\right]\).

We next assume that \(a\in \left(\frac{4}{3},\frac{8}{3}\right]\). Here, for a given \(y\in \left(0,\frac{4}{3}\right]\), there is a \(z\in \left(0,\frac{4}{3}\right]\) with \(y+z = a\) if and only if \(y\in \left[a-\frac{4}{3}, \frac{4}{3}\right]\). Then we have \(b = y(a-y)\) with \(y\in\left[a-\frac{4}{3},\frac{4}{3}\right]\). The value of \(b\) at both endpoints \(y=a-\frac{4}{3}\) and \(y = \frac{4}{3}\) is \(\frac{4}{3}\left(a-\frac{4}{3}\right)\), and the value at the vertex \(\frac{a}{2}\) is \(\frac{a^2}{4}\). Since the coefficient of \(y^2\) is negative, we deduce that \(\frac{4}{3} \left(a-\frac{4}{3}\right) \leq \frac{a^2}{4}\) for \(a\in\left(\frac{4}{3},\frac{8}{3}\right]\). It follows that, for \(a\in\left(\frac{4}{3},\frac{8}{3}\right]\), the region \(T\) ranges between the graphs of \(b = \frac{4}{3}\left(a-\frac{4}{3}\right)\) and \(b = \frac{a^2}{4}\). ◻

Proof of Proposition 11 assuming Lemma 2. Take an arbitrary metric \(g_{\frac{4}{3},y,z,1}\) with \((y,z)\in\left(0,\frac{4}{3}\right]^2\), and let \(g(\ell)\) denote its Ricci flow, i.e. \(g(0)=g_{\frac{4}{3},y,z,1}\). To compute the derivative we use the chain rule: \[\frac{d}{d\ell}\Big\vert_{\ell=0} \tilde{H}_1(g(\ell)) =\nabla \tilde{H}_1(g(0)) g'(0),\] where, analogously to the proof of Theorem 4, we interpret \(\nabla\tilde{H}_1(g(0))\) as a \(1\times 4\) vector and \(g'(0)\) as a \(4\times 1\) vector. From the Ricci flow equation 6 we know that \(g(\ell)\) is the family \(g_{x(\ell),y(\ell),z(\ell),s(\ell)}\) determined by \[(x',y',z',s')=(-2r_x x,-2r_y y,-2r_z z, -2r_s s).\] For \(\ell=0\) we have \((x,y,z,s)=(\frac{4}{3},y,z,1)\), so \(g'(0)=(-\frac{8}{3}r_x,-2r_y y,-2r_z z,-2r_s)^T\).

The gradient of \(\tilde{H}_1\) is \(\nabla\tilde{H}_1 = \left( -\frac{3}{s}, 0, 0, \frac{3x}{s^2} \right)\), which at \(g_{\frac{4}{3},y,z,1}\) equals \(\nabla\tilde{H}_1(g(0)) = \left( -3, 0, 0, 4 \right)\). Then: \[\frac{d}{d\ell}\Big\vert_{\ell=0} \tilde{H}_1(g(\ell))=\left( -3, 0, 0, 4 \right) \left( -\frac{8}{3}r_x, -2r_y y, -2r_z z, -2r_s \right)^T=8(r_x-r_s).\] In order to analyze the sign of this expression we use the variables \((a,b)\in T\) from 9 instead of \((y,z)\in \left(0,\frac{4}{3}\right]^2\). The Ricci eigenvalues \(r_x,r_s\) are given in Lemma 1. At the metric \(g_{\frac{4}{3},y,z,1}\), they equal \[r_x = 3+\frac{16n}{3}+3\frac{\frac{16}{9}-a^2 + 2b}{2b},\qquad r_s = 4(n+2) -\frac{8}{3}-2a.\] After substituting, we get \[\frac{d}{d\ell}\Big\vert_{\ell=0} \tilde{H}_1(g(\ell))=8(r_x-r_s)=\frac{4}{b}\left( \frac{16}{3} - 3a^2 + \left(\frac{4}{3} + \frac{8}{3}n\right)b+4ab\right).\]

Now, by Lemma 2, we know that \(F^H(a,b)>0\) for all \((a,b)\in T\). Multiplying by \(\frac{4}{b} > 0\), we get the following chain of inequalities on \(T\) \[\begin{align} 0& < \frac{4}{b}F^H(a,b)\\ &= \frac{4}{b}\left(\frac{16}{3}-3a^2 + 4b+4ab\right)\\ &= \frac{4}{b}\left( \frac{16}{3} - 3a^2 + \left(\frac{4}{3}+ \frac{8}{3}(1)\right)b + 4ab\right)\\ &\leq\frac{4}{b}\left( \frac{16}{3} - 3a^2 + \left(\frac{4}{3} + \frac{8}{3}n\right)b+4ab\right)\\ &= \frac{d}{d\ell}\Big\vert_{\ell=0}\tilde{H}_1(g(\ell)). \end{align}\] It follows that \(\frac{d}{d\ell}\Big\vert_{\ell=0}\tilde{H}_1(g(\ell)) > 0\) on \(T\). ◻

Proof of Lemma 2. We will show that the graph determined by the equation \(F^H = 0\) does not intersect the region \(T\). As \(T\) is obviously connected and \(F^H\) is continuous on \(\mathbb{R}^2\), it will follow that the sign of \(F^H\) is constant on \(T\). Taking the sample point \(\left(1,\frac{1}{8}\right)\in T\) we see that \(F^H\left(1,\frac{1}{8}\right) = \frac{10}{3} > 0\), so this will complete the proof that \(F_H > 0\) on \(T\).

Hence let us show that \(F^H = 0\) does not intersect \(T\). The equation \(F^H=0\) is linear in \(b\). Solving for \(b\), we find \[b = \frac{9a^2-16}{12(a+1)}.\]

We first observe that for \(a \leq \frac{4}{3}\) this formula gives \(b\leq 0\), so \((a,b)\notin T\). Thus, any point in the intersection of \(T\) and the graph of \(F^H=0\) must have \(a \in \left(\frac{4}{3},\frac{8}{3}\right]\). For these values of \(a\), the corresponding \(b\)-values satisfy \(\frac{4}{3}\left(a-\frac{4}{3}\right)\leq b\leq \frac{a^2}{4}\). Hence, to see that the graph of \(F_H=0\) does not intersect the region \(T\), it is sufficient to show that \(\frac{9a^2-16}{12(a+1)}< \frac{4}{3}\left(a - \frac{4}{3}\right)\) for all \(a\in\left(\frac{4}{3}, \frac{8}{3}\right]\). This holds if and only if \(9a^2 - 16 < 16a^2-\frac{16}{3}a -\frac{64}{3}\), which holds if and only if \(0 < 7a^2 -\frac{16}{3}a -\frac{16}{3} = \frac{1}{3}(7a+4)(3a-4)\). For \(a > \frac{4}{3}\) both factors \((7a+4)\) and \((3a-4)\) are positive, which implies the desired inequality. ◻

Proof. Note that \(x_-(a,b)\) defines a continuous function on the region \(b\leq\frac{a^2}{3}\), which contains the region \(T\). We will show that the graph of \(x_-(a,b) = 0\) only intersects \(T\) along the boundary and that \(x_- < 0\) at one point in the interior of \(T\). This implies that \(x_-\leq 0\) on \(T\). That is, \(x_-\notin \left(0,\frac{4}{3}\right]\), hence Remark 8 implies that \(g_{x_-,y,z,1}\notin \mathcal{B}_V\).

As a sample point in \(T\) we take \((a,b) = \left(1,\frac{5}{27}\right)\), where \(\sqrt{a^2-3b}=\frac{2}{3}\). We compute \(x_-\left(1,\frac{5}{27}\right)=-\frac{1}{9}<0\).

Now we study the intersection of the set \(x_-(a,b)=0\) and \(T\). Note that the equation \(x_- = 0\) is equivalent to \(a = 2\sqrt{a^2 - 3b}\), i.e., they have precisely the same solution set. Because \(a>0\) on \(T\), it is equivalent to \(a^2 = 4(a^2 - 3b)\). This simplifies to \(b= \frac{a^2}{4}\), which is a boundary component of \(T\). This completes the proof. ◻

Proof of Proposition 13. Take an arbitrary metric \(g_{x_+,y,z,1}\) with \((y,z)\in\left(0,\frac{4}{3}\right]^2\), and let \(g(\ell)\) denote its Ricci flow, i.e. \(g(0)=g_{x_+,y,z,1}\). As in the proof of Proposition 11, to compute the derivative of \(\tilde{V}_1(g(\ell))\) at \(\ell=0\) we use the chain rule: \[\label{EQ:chain95for95V1} \frac{d}{d\ell}\Big\vert_{\ell=0} \tilde{V}_1(g(\ell)) =\nabla \tilde{V}_1(g(0)) g'(0).\tag{12}\] It follows from the Ricci flow equation 6 that \(g (\ell)\) is the family \(g_{x(\ell),y(\ell),z(\ell),s(\ell)}\) determined by \[(x',y',z',s')=(-2r_x x,-2r_y y,-2r_z z, -2r_s s).\] At \(\ell=0\) we have \((x,y,z,s)=(x_+,y,z,1)\), so \(g'(0)=(-2r_x x_+,-2r_y y,-2r_z z,-2r_s)^T\).

Now we compute the gradient of \(\tilde{V}_1\). Observe that \(\tilde{V}_1=\frac{V_1}{s}\). Then the gradient is \[\nabla\tilde{V}_1=\left( \frac{1}{s}\frac{\partial V_1}{\partial x} , \frac{1}{s}\frac{\partial V_1}{\partial y} , \frac{1}{s}\frac{\partial V_1}{\partial z} , \frac{1}{s^2}\left(s\frac{\partial V_1}{\partial s}-V_1 \right) \right)\] Note that \(\frac{\partial V_1}{\partial s}=0\) because \(V_1\) does not depend on \(s\). Also, at \(g(0)=g_{x_+,y,z,1}\) we have \(s=1\) and \(V_1=0\). Thus, the gradient reduces to \[\begin{align} \nabla\tilde{V}_1(g(0))&=\left( \frac{\partial V_1}{\partial x} , \frac{\partial V_1}{\partial y} , \frac{\partial V_1}{\partial z} , 0 \right)\Big\vert_{g(0)} \nonumber\\ & =\left( \frac{(-6x+2y+2z)x-xV_1}{x^2}, \frac{2(y+x-z)}{x} , \frac{2(z+x-y)}{x},0 \right)\label{EQ:gradient95V} \\ & = \frac{2}{x}\left( -3x+y+z, y+x-z , z+x-y,0 \right).\nonumber \end{align}\tag{13}\] Inserting the corresponding values in 12 we find that: \[\label{eq:deriv95V1} \frac{d}{d\ell}\Big\vert_{\ell=0} \tilde{V}_1(g(\ell))= -\frac{4}{x}\Bigl( (-3x+y+z)r_x x + (y+x-z)r_y y + (z+x-y)r_z z \Bigr).\tag{14}\] Observe that in the equations 13 and 14 we have dropped the subindex \(+\) from \(x_+\) in order to simplify the notation.

Now we make the following changes in 14 . First, we substitute \(r_x,r_y,r_z\) by their values from 5 at the metric \(g_{x,y,z,1}\). This gives an expression in the variables \(x,y,z\) which is symmetric in \(y,z\). Second, we make the change of variables from \((y,z)\in \left(0,\frac{4}{3}\right]^2\) to \((a,b)\in T\), thus obtaining an expression in the variables \(x,a,b\). Third and last, we substitute \(x\) by \(x_+=\frac{a + 2\sqrt{a^2 -3b}}{3}\), which gives an expression in \(a,b\). These tedious computations give the following identity: \[\frac{d}{d\ell}\Big\vert_{\ell=0} \tilde{V}_1(g(\ell)) = -\frac{4}{x_+}\left( \frac{F_n^V(a,b)}{9b\left(a + 2\sqrt{a^2-3b}\right)}\right).\] Since \(-\frac{4}{x_+} < 0\) and both \(a\) and \(b\) are positive, it now easily follows that \(\frac{d}{d\ell}\Big\vert_{\ell=0} \tilde{V}_1(g(\ell)) > 0\) if and only if \(F_n^V(a,b) <0\) for all \((a,b)\in T\). The latter holds by Lemma 4. ◻

Proof. Assume for a contradiction that \(b-a+x = 0\). Substituting this into the equation \(P_1 = 0\), we find \(\sqrt{H_1 V_1} + b = 0\). This is a contradiction as the left hand side is positive, being a sum of a non-negative term and a positive term. ◻

Proof. We argue by contradiction, so suppose there exists such a metric. By Lemma 7, the parameter \(x\) must satisfy \(Bx+C=0\). We distinguish two cases: \(B=0\) and \(B\neq 0\). If \(B=0\), then \(C=0\). Substituting \(b=a-1\) (which corresponds to \(A=0\)) into \(C=0\) we find that \(a=2\), and consequently \(b=1\). Using 15 we find that \(x\geq\frac{4}{3}\) (or equivalently \(H_1\leq 0\)), which yields a contradiction because any metric in \(\mathcal{B}_P\setminus(\mathcal{B}_H\cup\mathcal{B}_V)\) must satisfy \(H_1 > 0\).

If \(B\neq 0\), then \(x=-\frac{C}{B}\). Substituting \(b=a-1\) into this expression we obtain \(x=1\). However, for \(x=1\) we find that \(b-a+x=0\), which is a contradiction with the assumption \(b-a+x>0\). ◻

Proof. We will show that the absolute minimum of \(\Delta\) on the closure of \(T\) is \(0\). This absolute minimum must either occur on the boundary of \(T\) or it must be a critical point. We begin by analyzing the behavior of \(\Delta\) on the boundary of \(T\).

Along the boundary curve \(b=\frac{a^2}{4}\), we find \(\Delta\left(a,\frac{a^2}{4}\right) = \frac{1}{256} a^2 (a-8)^2(a-2)^4\), which is obviously non-negative. Along the boundary curve \(b=0\), we find \(\Delta(a,0) = a^2(-4 + 3a)^2\) which is also obviously non-negative. Along the boundary curve \(b = \frac{4}{3}\left(a-\frac{4}{3}\right)\), we find \(\Delta\left(a, \frac{4}{3}\left(a-\frac{4}{3}\right)\right) =\frac{\left(441 a^{2}-2040 a+2368\right) \left(-4+3 a\right)^{2}}{6561}\). Since the factor \((441 a^{2}-2040 a+2368)\) has no real roots, we conclude that \(\Delta \geq 0\) on this boundary curve as well.

Having shown that \(\Delta \geq 0\) on the boundary of \(T\), we now turn our attention towards interior critical points. We will show that \((a,b)=\left( \frac{17}{6} - \frac{\sqrt{465}}{18}, \frac{23}{3}-\frac{\sqrt{465}}{3}\right)\) is the only possible critical point in the interior of \(T\). Believing this momentarily, evaluating \(\Delta\) at this point we get \[\frac{17825\sqrt{465}}{162} - \frac{128125}{54} = \frac{17825 \sqrt{\frac{236391585}{508369}}}{162} - \frac{128125}{54} > \frac{17825 \sqrt{\frac{236390625}{508369}}}{162} - \frac{128125}{54} = 0,\] which implies \(\Delta \geq 0\) on \(T\), as claimed.

Now let us analyze the interior critical points of \(\Delta\), i.e. the common zeros \((a_0,b_0)\) of the partial derivatives \(\Delta_a\) and \(\Delta_b\). Recall from Section 2 that both of the resultants \(\operatorname{res}_a(\Delta_a,\Delta_b)\in\mathbb{Z}[b]\) and \(\operatorname{res}_b(\Delta_a,\Delta_b)\in\mathbb{Z}[a]\) must vanish at \((a_0,b_0)\). Computing \(\Delta_a\) and \(\Delta_b\) and the corresponding resultants one finds that: \[\operatorname{res}_b:=\operatorname{res}_b(\Delta_a,\Delta_b) = -2304(a-2)^4(3a-4)(27a^2-153a+178)\] and \[\operatorname{res}_a:=\operatorname{res}_a(\Delta_a,\Delta_b) = 62208b(b-1)^4(9b^2 - 138b + 64).\] The roots of \(\operatorname{res}_b\) are \(2,\frac{4}{3}, \frac{17}{6} - \frac{\sqrt{465}}{18}\), and \(\frac{17}{6} + \frac{\sqrt{465}}{18}\), but the last root is larger than \(\frac{16}{6} = \frac{8}{3}\), so lies outside of \(T\). Since \(b > 0\) on the interior of \(T\), the relevant roots of \(\operatorname{res}_a\) are \(1, \frac{23}{3} - \frac{\sqrt{465}}{3},\) and \(\frac{23}{3} + \frac{\sqrt{465}}{3}\), where, similarly, the last root is larger than \(\frac{16}{9}\), so lies out of \(T\). In summary, any interior critical point \((a_0,b_0)\) must have \[\label{EQ:possibilities95for95a95b} a_0\in\left\{2,\tfrac{4}{3}, \tfrac{17}{6} - \tfrac{\sqrt{465}}{18}\right\}\qquad\text{and}\qquad b_0\in \left\{1, \tfrac{23}{3} - \tfrac{\sqrt{465}}{3}\right\}.\tag{17}\] Assume initially that \(a_0=2\). Then \(\Delta_a(2,b) = -2(b-1)(3b^2-34b+32)\), whose roots are \(b\in\{1, \frac{17}{3} \pm \frac{\sqrt{193}}{3}\}\). By comparing them to the possibilities in 17 , we find \(b_0=1\). But \((a_0,b_0)=(2,1)\) is not in the interior of \(T\), so \(a_0=2\) yields no interior critical points.

Next, assume that \(a_0= \frac{4}{3}\). We have \(\Delta_b(\frac{4}{3},b) = \frac{2}{3} b (6b^2 -63b + 19)\), with roots \(b\in \{0, \frac{21}{4} \pm \frac{\sqrt{3513}}{12}\}\). None of them match with the possibilities in 17 , so \(a_0=\frac{4}{3}\) also yields no interior critical points. It follows that any interior critical point \((a_0,b_0)\) must have \(a_0 = \frac{17}{6} - \frac{\sqrt{465}}{18}\).

Now, we rule out the possibility that \(b_0=1\), and thus conclude that the only possible interior critical point is \((a_0,b_0) = \left(\frac{17}{6} - \frac{\sqrt{465}}{18}, \frac{23}{3}-\frac{\sqrt{465}}{3}\right).\) We have \(\Delta_a(a,1) = 2(a-2)(18a^2-27a-17)\), with roots \(a\in\{2, \frac{3}{4} \pm \frac{\sqrt{217}}{12}\}\). Since neither of these is equal to \(\frac{17}{6} - \frac{\sqrt{465}}{18}\), we deduce that \(b_0\neq 1\). ◻

Proof. Suppose there exists such a metric. By Lemma 7, the value of \(x\) must be \(x_1\) or \(x_2\). We will see that both cases yield a contradiction. Recall throughout the proof that both \(x_1\) and \(x_2\) are continuous functions on \(T\cap\{A\neq 0\}\), see Lemma 9. Note that \(T\cap\{A\neq 0\}\) has three connected components, namely \[U_1:=T\cap\{A> 0\},\qquad U_2:=T\cap\{A< 0 \text{ and } a < 2\}, \qquad U_3:=T\cap\{A<0 \text{ and } a > 2\},\] see Figures 2 and [fig:TA0zoom]. When dealing with \(x_1\) and \(x_2\), we will analyze their behavior in each \(U_i\) separately.

In order to rule out \(x_1\), we will show that, under the given assumptions, \(x_1\geq\frac{4}{3}\) on \(U_1\) and \(x_1\leq 0\) on both \(U_2\) and \(U_3\). Note that \(x_1\geq\frac{4}{3}\) is equivalent to \(H_1\leq 0\), which is a contradiction because any metric in \(\mathcal{B}_P\setminus(\mathcal{B}_H \cup \mathcal{B}_V)\) must have \(H_1>0\). Moreover, \(x_1\leq 0\) means that \(g_{x_1,y,z,1}\) does not even represent a metric, so is obviously a contradiction.

We begin by showing that \(x_1\leq 0\) on \(U_2\) and \(U_3\). Because \(A<0\) on \(U_2\) and \(U_3\), we only need to prove that \(-B+\sqrt{\Delta}\geq 0\). Since \(C=a^2-4b\geq 0\) on \(T\), we have \[\sqrt{\Delta}=\sqrt{B^2-4AC}\geq \vert B\vert\geq B\] on both \(U_2\) and \(U_3\). This implies that \(-B+\sqrt{\Delta}\geq 0\), as desired.

Next, we show that \(x_1 \geq \frac{4}{3}\) on \(U_1\). Consider the equation \(x_1=\frac{4}{3}\). Since \(A\neq 0\), this is equivalent to \(-3B+3\sqrt{\Delta}=8A\). Following the approach in Remark 15, we arrive at the equation \(9B^2-36AC=64A^2+9B^2+48AB\). Dividing by \(A\neq 0\) we get \(64A+48B+36C=0\), which can be seen to be equivalent to \(\left(3 a-2 b-4\right)^{2}=0\). Thus, \(x_1=\frac{4}{3}\) can occur only along the line \(3 a-2 b-4=0\). This line does not intersect \(U_2\) or \(U_3\) and it divides \(U_1\) into two connected components (see Figure 3, where the dotted line depicts \(-2b+3a-4=0\)). Altogether we conclude that, on each subregion of \(U_1\), the value of \(x_1\) is either larger than \(\frac{4}{3}\) or smaller than \(\frac{4}{3}\). We take a sample point in each subregion: \[\left(\frac{3}{2}, \frac{1}{3}\right)\in U_1\cap \left\{{ b>\frac{3}{2}}a-2\right\},\qquad \left(2, \frac{32}{35}\right)\in U_1\cap \left\{{ b<\frac{3}{2}}a-2\right\}.\] We compute that \(x_1\left(\frac{3}{2},\frac{1}{3}\right) = \frac{3}{2} > \frac{4}{3}\) and \(x_1\left(2,\frac{32}{35}\right) = \frac{10}{7} > \frac{4}{3}\), so \(x_1\geq\frac{4}{3}\) on \(U_1\), as claimed.

Now we focus on \(x_2\). We will show that \(x_2\) satisfies \(2b+3x_2-3a \leq 0\) on \(T\cap\{A\neq 0\}\), so Lemma 7 implies that \(x_2\) does not yield a metric in \(\mathcal{B}_P\setminus(\mathcal{B}_H\cup\mathcal{B}_V)\).

To prove that \(2b+3x_2-3a{<}0\) on \(T\cap\{ A\neq 0\}\), we will see that the graph of \(2b+3x_2-3a=0\) does not intersect \(T\cap\{ A\neq 0\}\). Assuming this, at the sample points \[\left(\frac{3}{2}, \frac{1}{3}\right)\in U_1, \qquad \left(\frac{2}{3}, \frac{1}{19}\right)\in U_2,\qquad \left(\frac{43}{18}, \frac{17}{12}\right)\in U_3,\] we compute that \(2b + 3x_2 - 3a\) is given by respectively \(-\frac{1}{6}, -\frac{35}{209}, -\frac{13}{12},\) so all three are negative. Hence \(2b+3x_2-3a<0\) on \(T\cap\{A\neq 0\}\).

Consider the equation \(2b+3x_2-3a=0\). Substitute \(x_2=\frac{-B-\sqrt{\Delta}}{2A}\) and multiply both sides by \(\frac{2A}{3}\). This gives \(\frac{4}{3}bA-B-\sqrt{\Delta}-2A = 0\), which can be rearranged as \(\left(\frac{4}{3}b-2\right)A-B= \sqrt{\Delta}\). Following the approach in Remark 15, we square both sides and work out the expression, and we arrive at the equation \[0 = A b \left(-2 b+3 a-4\right) \left(-b-3+2 a\right).\] Since \(Ab\neq 0\) on \(T\cap\{A\neq 0\}\), it follows that \(2b+3x_2-3a\) can only vanish if \(-2 b+3 a-4=0\) or \(-b-3+2 a=0\). Let us show that neither of these possibilities occur.

Suppose \(-2 b+3 a-4=0\). As mentioned in the case of \(x_1\) and illustrated in Figure 3, this line does not intersect \(U_2\) and \(U_3\) and divides \(U_1\) into two connected components. Moreover, the intersection of \(-2 b+3 a-4=0\) and \(U_1\) occurs only for \(a\in \left(\frac{4}{3},2\right]\). Substituting \(b=\frac{3}{2}a-2\), we get \(A=-\frac{3}{2}(a-2)\), \(B=-\frac{1}{4}(a-2)(3a-20)\), \(C=(a-2)(a-4)\), and \(\Delta=\frac{1}{16}(a-2)^2(3a-4)^2\). Since \(a\in \left(\frac{4}{3},2\right]\) along the intersection, we conclude that \(\sqrt{\Delta} = - \frac{1}{4}(a-2)(3a-4)\), and hence \(x_2=2-\frac{a}{2}\). Consequently, \(2b+3x_2-3a=-\frac{3}{2}a+2\) equals \(-b\), and is therefore negative on all of \(T\).

Suppose \(-b-3+2 a=0\). This line also does not intersect \(U_2\) and \(U_3\) and divides \(U_1\) into two connected regions, see Figure [fig:feasible295A95pos]. This intersection of \(-b-3+2 a=0\) and \(U_1\) occurs for \(a\in\left( \frac{11}{6} ,2 \right)\). Substituting \(b=-3+2a\), we obtain \(A=-3(a-2)\), \(B=-(a-2)(a-9)\), \(C=(a-2)(a-6)\), and \(\Delta=(a-2)^2(a-3)^2\). For \(a\in (0,2)\) we find that \(x_2=1\). Hence \(2b+3x_2-3a=a-3\), which is negative for \(a\in\left( \frac{11}{6} ,2 \right)\). This finishes the proof. ◻

Proof. Suppose for a contradiction that there exists such a metric. By Lemma 11, the parameter \(x\) must satisfy \(Ex+C=0\). We will show that any solution to \(Ex+C=0\) satisfies \(-4b+3a-3x<0\), hence giving the desired contradiction.

As a first step we study the set \(T\cap\{D=0\}\). Note that \(D=0\) is equivalent to \(b = \frac{1}{2}(a-1)\). Since \(b>0\) in \(T\), the points in the intersection must have \(a > 1\). Moreover, the line \(b = \frac{1}{2}(a-1)\) intersects the line \(b =\frac{4}{3}\left(a-\frac{4}{3}\right)\) (which is a bounding curve of \(T\)) when \(a = \frac{23}{15}\). Thus, we find \(a \in (1,\frac{23}{15}]\).

Now consider the equation \(Ex+C=0\). We observe that in this case \(E\) cannot vanish; indeed, if we substitute \(b = \frac{1}{2}(a-1)\) in \(E\) we obtain \(E=-\frac{1}{2}(2a^2-5a+5)\), which is a polynomial with no real roots. It follows that \(x=-\frac{C}{E}\). Substituting \(b = \frac{1}{2}(a-1)\) into \(C\) gives \(C=a^2-2a+2\), hence \[x = \frac{2(a^2-2a+2)}{2a^2-5a+5}.\]

Finally, substituting \(b = \frac{1}{2}(a-1)\) and \(x = \frac{2(a^2-2a+2)}{2a^2-5a+5}\) in \(-4b+3a-3x\) gives: \[-4b+3a-3x=-\frac{(1-a)(a-2)(2a-1)}{2a^2-5a+5}.\] The denominator equals \(-2E\) which, as we have observed above, is never zero. Moreover, \(2a^2-5a+5\) is positive for all \(a\in\mathbb{R}\). As for the numerator, it is straightforward to check that it is positive for all \(a\in \left(1,\frac{23}{15}\right]\). The minus sign in front of the fraction implies that \(-4b+3a-3x<0\), as claimed at the beginning of the proof. ◻

Proof. It is straight forward to verify that \[\Gamma = \sqrt{ \frac{1}{16}a^2 (a^3-6a^2+9a-8)} = \frac{1}{4}| a (a^3-6a^2+9a-8)|.\] The maximum value of the cubic factor on \([0,\frac{8}{3}]\) is \(-4\), occurring when \(a=1\). This follows since its derivative is \(3(a-1)(a-3)\). Thus, as \(a\in(0,\frac{8}{3}]\), it follows that \(|a (a^3-6a^2 +9a-8| = - a(a^3-6a^2+9a-8)\neq 0\), completing the proof. ◻

Proof. In order to prove that \(\Gamma\geq 0\) on \(T\), we will show that the zero set of \(\Gamma\) intersects the closure of \(T\) only in isolated points. Since \(\Gamma\) is a continuous function of \((a,b)\), it will then follow that on \(T\) either \(\Gamma \geq 0\) or \(\Gamma \leq 0\). Since \(\Gamma\left(1,\frac{1}{8}\right) = \frac{385}{256} > 0\), we conclude that \(\Gamma \geq 0\) on \(T\).

We begin by showing that the zero set of \(\Gamma\) intersects the boundary of \(T\) only at the points \((0,0)\) and \(\left(\frac{4}{3},0\right)\). First consider the boundary curve \(b = \frac{a^2}{4}\) with \(a\in \left[0,\frac{8}{3}\right]\). When \(a=0\) we get the obvious solution \((0,0)\). For \(a\in(0,\frac{8}{3}]\), Lemma 14 implies that there are no further points in the zero set. Second, along the boundary curve \(b=0\) with \(a\in \left(0,\frac{4}{3}\right]\) the equation \(\Gamma=0\) simplifies to \(a^2(3a-4)^2 = 0\), which yields the point \(\left(\frac{4}{3},0\right)\). Finally, along the curve \(b = \frac{4}{3}\left(a-\frac{4}{3}\right)\) the equation \(\Gamma = 0\) is equivalent to \[(3a-4)^2 ( 3249a^2 - 20760a + 43072)=0.\] The quadratic factor has a discriminant of \(-128786112 < 0\), so it has no real zeros. Since the boundary curve \(b = \frac{4}{3}\left(a - \frac{4}{3}\right)\) has domain \(a\in \left(\frac{4}{3},\frac{8}{3}\right]\), we conclude that there are no additional intersections in this case.

Now we analyze the behavior of \(\Gamma\) in small neighborhoods of the points \((0,0)\) and \(\left(\frac{4}{3},0\right)\). Consider initially the point \((0,0)\). We claim that the zero set of \(\Gamma\) is disjoint from \(T\) in a deleted neighborhood of \((0,0)\). Since the partial derivative \(\Gamma_b(0,0)\neq 0\), by the implicit function theorem, there is a neighborhood of \(a =0\in \mathbb{R}\) for which the set \(\Gamma = 0\) equals the graph of some function \(b = g(a)\). We claim that \(g(a) > \frac{a^2}{4}\) for all \(a\) in a possibly smaller deleted neighborhood of \(a=0\), so that \((a,g(a))\notin T\). To verify this, we compute that the first derivative of \(g\) at \(a=0\) is equal to \(-\frac{\Gamma_a}{\Gamma_b}=0\), and the second derivative is \(\frac{2}{3}\). For the curve \(b = \frac{a^2}{4}\), the first derivative at \(a=0\) is \(0\) and the second derivative is \(\frac{1}{2}\). Since \(\frac{2}{3} > \frac{1}{2}\), we conclude that \(g(a) > \frac{a^2}{4}\) for small \(a>0\). It follows that in a deleted neighborhood of \((0,0)\), the zero set of \(\Gamma\) is disjoint from \(T\).

We next claim that \(\left(\frac{4}{3},0\right)\) is an isolated point of the zero set of \(\Gamma\). We observe that \(\left(\frac{4}{3},0\right)\) is a critical point of \(\Gamma\) in the sense that both partial derivatives, \(\Gamma_a\) and \(\Gamma_b\), vanish at \(\left(\frac{4}{3},0\right)\). In addition, the Hessian \(\left(\begin{smallmatrix}\Gamma_{aa} & \Gamma_{ba}\\\Gamma_{ab}&\Gamma_{bb} \end{smallmatrix}\right)\) of \(\Gamma\) at \(\left(\frac{4}{3},0\right)\) is \(\left(\begin{smallmatrix} 32 & -52 \\ -52 & \frac{278}{3}\end{smallmatrix}\right)\), which is positive definite. It follows that \(\left(\frac{4}{3},0\right)\) is a strict local minimum of \(\Gamma\), so that \(\left(\frac{4}{3},0\right)\) is an isolated point of the zero set.

To conclude the proof, we will show that the zero set of \(\Gamma\) does not intersect the interior of \(T\). Our first step is to show that the intersection would necessarily be a smooth \(1\)-manifold of the interior of \(T\). From the implicit function theorem, it is sufficient to verify that at least one of the partial derivatives of \(\Gamma\) is non-zero at every point in the interior of \(T\).

Towards that end, we first observe that \(C\geq 0\) for all \((a,b)\in T\), and moreover \(C>0\) if \(b\neq \frac{a^2}{4}\). Thus, in order to have \(E^2 - 4DC = 0\), it is necessary that \(D > 0\), or equivalently \(b < \frac{a-1}{2}\). We note that \((a,b)\in T\) and \(D > 0\) implies \(a\in \left(1,\frac{23}{15}\right]\) and \(b\in \left(0,\frac{4}{15}\right]\).

Now, in order to find common zeros of \(\Gamma_a\) and \(\Gamma_b\), we compute the resultant \(\operatorname{res}_b(\Gamma_a,\Gamma_b)\), see Section 2 for details: \[\operatorname{res}_b(\Gamma_a,\Gamma_b) = -1179648 (3a - 4)(a + 1)(216a^5 - 1116a^4 + 2240a^3 - 2220a^2 + 933a - 58).\] Applying Sturm’s Theorem (see also Remark 2) to the last factor, we see that it has no zeros with \(a\in \left(1,\frac{23}{15}\right]\). The factor \((a+1)\) obviously has no zeros with \(a\in \left(1,\frac{23}{15}\right]\). Thus, we conclude that any common zero \((a_0,b_0)\) of \(\Gamma_a\) and \(\Gamma_b\) must have \(a_0 = \frac{4}{3}\). Substituting \(a = \frac{4}{3}\) into the equation \(\Gamma = 0\), we find \(\frac{1}{3} b^2 ( 48b^2 -264b + 139) = 0\), so \(b_0\) must be either equal to \(0\) or a root of the quadratic factor. The case \(b_0=0\) corresponds to a boundary point of \(T\), while the roots of the quadratic factor are \(\frac{11}{4} \pm \frac{\sqrt{42}}{3}\), both of which are larger than \(\frac{4}{15}\).

It follows that the intersection of the zero set of \(\Gamma\) and the interior of \(T\) is a relatively closed embedded submanifold. Since we have already shown that the intersection of the zero set of \(\Gamma\) and the boundary consists of isolated points, it follows that, in fact, the intersection of the zero set of \(\Gamma\) and the interior of \(T\) can contain no components homeomorphic to an interval. In other words, the intersection must be, up to diffeomorphism, a disjoint union of a finite collection of circles. We conclude the proof by showing there are no such circles.

To see this, assume for a contradiction that there is some circle. Then there are at least two points (the top most and the bottom most) where the tangent lines are horizontal, so that \(\Gamma_a = 0\). To find common zeros \((a_0,b_0)\) of \(\Gamma\) and \(\Gamma_a\) we compute the resultant: \[\operatorname{res}_b(\Gamma,\Gamma_a) = 1179648 a(72a^5 - 588a^4 + 1772a^3 - 2940a ^2 + 2631a - 1096)(a + 1)^2(3a - 4)^2.\] Since \(a_0\in \left(1,\frac{23}{15}\right]\), we conclude that either \(a_0 = \frac{4}{3}\) or \(a_0\) is a root of \((72a^5 - 588a^4 + 1772a^3 - 2940a^2 + 2631a - 1096)\). In the first case, as argued above, we find \(b_0 = 0\), obtaining a boundary point instead of an interior point. In the second case, Sturm’s Theorem shows it has no zeros in \(\left(1,\frac{23}{15}\right]\), completing the proof. ◻

Proof. Suppose there exists such a metric. Recall from Lemma 14 that \(x_3\) defines a continuous function on \(T\cap \{D\neq 0\}\). Note that \(T\cap\{D\neq 0\}\) has two connected components, namely \(T\cap\{D> 0\}\) and \(T\cap\{D< 0\}\), see Figure 4. We will denote \[W_1:=T\cap\{D> 0\},\qquad W_2:=T\cap\{D< 0\}.\] We will show that, under the given assumptions, \(x_3>\frac{4}{3}\) on \(W_1\) and \(x_3\leq 0\) on \(W_2\). This concludes the proof, since any metric \(g_{x_3,y,z,1}\in\mathcal{B}_P\setminus(\mathcal{B}_H \cup \mathcal{B}_V)\) must have \(x_3\in (0,\frac{4}{3})\).

We first show that \(x_3\leq 0\) on \(W_2\). This is equivalent to \(-E+\sqrt{\Gamma}\geq 0\) because \(D<0\) on \(W_2\). Since \(C=a^2-4b\geq 0\) on \(T\), we have \[\sqrt{\Gamma}=\sqrt{E^2-4DC}\geq \vert E\vert\geq E\] on \(W_2\). Thus we get that \(-E+\sqrt{\Gamma}\geq 0\), as desired.

Now we show that \(x_3>\frac{4}{3}\) on \(W_1\). A simple calculation shows that \(x_3 = \frac{4}{3}\) if and only if \(4D(-4+3a-4b)^2 = 0\). As \(D\neq 0\), we find that \(x_3 = \frac{4}{3}\) if and only if \(b = -1 + \frac{3}{4}a\). We claim that the graph of \(b = -1 + \frac{3}{4} a\) does not intersect the region \(T\), and in particular it does not intersect \(W_1\). To prove this, consider a point \((a,b)\) with \(b = -1 + \frac{3}{4} a\). If \(a\leq \frac{4}{3}\), we find \(b\leq 0\), so \((a,b)\notin T\). If otherwise \(a>\frac{4}{3}\), then we observe that \(b<\frac{4}{3}a-\frac{16}{9}\), so \((a,b)\notin T\) either.

Since \(x_3\) is continuous on \(W_1\), it follows that either \(x_3<\frac{4}{3}\) or \(x_3>\frac{4}{3}\) for all \((a,b)\in W_1\). Taking the sample point \((a,b) = (\frac{7}{5}, \frac{1}{10})\in W_1\), we see that \(x_3\left(\frac{7}{5}, \frac{1}{10}\right) = 2> \frac{4}{3}.\) Consequently \(x_3 > \frac{4}{3}\) on \(W_1\), as claimed. ◻

Proof of Proposition 16. Let \(g_{x,y,z,1}\) be a metric satisfying the hypotheses in the statement. Since \(H_2\) and \(H_3\) are non-negative, we know from Remark 8 that \((y,z)\in(0,\frac{4}{3}]^2\) and hence \((a,b)\in T\). Since \(P_1=0\), Lemmas 12 and 15 imply that \(D\neq 0\) and \(x=x_4\). Recall from the proof of Lemma 15 and Figure 4 that \(T\cap \{D\neq 0\}\) consists of two connected components \(W_1\) and \(W_2\), with \(W_2\) containing the graph of \(b= \frac{a^2}{4}\). We shall show that \(-4b+3a-3x_4<0\) on \(W_1\), which is a contradiction to Lemma 11.

Indeed, recall from Lemma 14 that \(x_4(a,b)\) defines a continuous function on \(T\cap \{D\neq 0\}\). Consider the equation \(-4b+3a-3x_4 = 0\). Substituting \(x_4\) by its value and solving it for \(b\), we find \(b = 0\), \(b=\frac{3}{4}a-1\) and \(b =a-\frac{3}{4}\) are the only solutions. The graph of \(b=0\) is not part of \(T\), while \(b = \frac{3}{4}a-1\) intersects the closure of \(T\) only at \((\frac{4}{3},0)\notin T\). The line \(b=a-\frac{3}{4}\) intersects the interior of \(T\). However, it does not intersect \(W_1\), thus the sign of \(-4b+3a-3x_4\) is constant on \(W_1\). Using the sample point \((a,b)=\left(\frac{7}{5},\frac{1}{10}\right)\in W_1\) we find that \(-4b+3a-3x_4 = - \frac{1}{10} <0\) on \(W_1\).

It follows that \((a,b)\) must lie in \(W_2\). Note that the line \(b=a-\frac{3}{4}\) divides \(W_2\) in two components and that one of the components is precisely the subset \(S\), see Figure 5. Thus, the sign of \(-4b+3a-3x_4\) on \(S\) is constant and it is also constant on \(W_2\setminus S\). We take the sample points \((a,b)=\left(\frac{4}{5}, \frac{1}{7}\right)\in S\) and \((a,b)=\left(\frac{4}{3},\frac{1}{3}\right)\in W_2\setminus S\). For \(\left(\frac{4}{5}, \frac{1}{7}\right)\) we find \(-4b+3a-3x_4 = \frac{4}{35}>0\). Similarly, for \(\left(\frac{4}{3}, \frac{1}{3}\right)\) we find \(-4b+3a-3x_4 = -\frac{1}{3} < 0\). Thus, \(-4b+3a-3x_4 \leq 0\) on \(W_2\setminus S\). Altogether, we have shown that \(-4b+3a-3x_4 > 0\) implies that \((a,b)\in S\).

Now we prove the converse. Let \(g_{x,y,z,1}\) be a metric with \((a,b)\in S\) and \(x=x_4\). Since \(S\subset T\), we know that \(H_2,H_3\geq 0\). Also, from the previous paragraph it is clear that \(-4b+3a-3x_4{\color{teal}>}{\color{gray}\geq} 0\) on \(S\). It remains to prove that \(b-a+x_4<0\), \(H_1(x_4,a,b)> 0\), and \(V_1(x_4,a,b)> 0\) on \(S\).

To show that \(b-a+x_4<0\), note that \(b-a+x_4 = 0\) if and only if \(b=0\), \(b= \frac{3}{2}a-1\), or \(b=a-1\). None of these lines intersect \(S\), so the sign of \(b-a+x_4\) is constant on \(S\). At \((a,b) = \left(\frac{4}{5}, \frac{1}{7}\right)\), we find \(b-a+x_4 = - \frac{3}{35} < 0\), so \(b-a+x_4<0\) on all of \(S\).

We similarly find that \(H_1(x_4,a,b) = 4-3x_4 = 0\) if and only if \(b = \frac{3}{4}a-1\), whose graph does not intersect \(S\). At \(\left(\frac{4}{5}, \frac{1}{7}\right)\), we find \(4-3x_4 = \frac{16}{7} > 0\), so we conclude that \(H_1(x_4,a,b)>0\) for all \((a,b)\in S\).

Finally, following the procedure of Remark 15, we find that \(V_1(x_4,a,b) = 0\) implies \[b (-4b-3+4a) (a-2b-1)(a^2-4b) = 0.\] The first three factors yield curves which do not intersect \(S\). Finally, we argue \(V_1(x_4,a,b)\neq 0\) when \(b = \frac{a^2}{4}\). Substituting \(b=\frac{a^2}{4}\) in \(D\) and \(E\) and using the value of \(\Gamma\) given in Lemma 13, we find that \(x_4= \frac{-E-\sqrt{\Gamma}}{2D}= -\frac{a(a^3 - 6a^2 + 9a - 8)}{6a^2 - 12a + 12}\), from which it easily follows that \(V_1 = \frac{a^2 (a-1)^2}{2(a^2-2a+2)}\). Since the numerator is positive on \((0,1)\) and the denominator is always positive, it follows that \(V_1 > 0\) along the boundary curve \(b = \frac{a^2}{4}\). It now follows that \(V_1 > 0\) on all of \(S\). ◻

Proof. Let \(g\) be an arbitrary metric in \(\hat{S}\) and let \(g(\ell)\) be its Ricci flow, i.e. \(g(0)=g\). As we have done in the proofs of Propositions 11 and 13, to compute the derivative of \(\tilde{P}_1(g(\ell))\) at \(\ell=0\) we use the chain rule: \[\label{EQ:chain95for95P1} \frac{d}{d\ell}\Big\vert_{\ell=0} \tilde{P}_1(g(\ell)) =\nabla \tilde{P}_1(g(0)) g'(0).\tag{23}\] From 6 we know that the Ricci flow \(g(\ell)\) is the family \(g_{x(\ell),y(\ell),z(\ell),s(\ell)}\) determined by \((x',y',z',s')=(-2r_x x,-2r_y y,-2r_z z, -2r_s s)\). At \(\ell=0\), we have \((x,y,z,s)=(x_4,y,z,1)\), so \(g'(0)=(-2r_x x_4,-2r_y y,-2r_z z,-2r_s)^T\).

The gradient \(\nabla\tilde{P}_1=\left(\frac{\partial\tilde{P}_1}{\partial x} , \frac{\partial\tilde{P}_1}{\partial y} , \frac{\partial\tilde{P}_1}{\partial z} , \frac{\partial\tilde{P}_1}{\partial s} \right)\) can be computed using the expression in 22 . When evaluating it at a metric of the form \(g_{x,y,z,1}\) we find that: \[\begin{align} \frac{\partial\tilde{P}_1}{\partial x} &= {( H_1 V_1)^{-1/2}} \left(\frac{-6 x^{2}-2 \left(y-z\right)^{2}+9 x^{2} \left(x-\frac{y}{3}-\frac{z}{3}\right)}{ x^{2}}\right)+3, \\ \frac{\partial\tilde{P}_1}{\partial y} &= 4( H_1 V_1)^{-1/2} \left(\frac{ \left(1-\frac{3 x}{4}\right) \left(y+x-z\right)}{x }\right)+4 z-3, \\ \frac{\partial\tilde{P}_1}{\partial z} &= 4( H_1 V_1)^{-1/2} \left(\frac{ \left(1-\frac{3 x}{4}\right) \left(z+x-y\right)}{x }\right)+4 y-3, \\ \frac{\partial\tilde{P}_1}{\partial s} &= 6( H_1 V_1)^{-1/2}\left(\frac{ \left(-\frac{3 x}{2}+1\right) \left(x^{2}-\frac{2 \left(y+z\right) x}{3}-\frac{\left(y-z\right)^{2}}{3}\right)}{ x}\right)-8 y z+3 y+3 z-3 x. \end{align}\]

Now, we compute \(\nabla\tilde{P}_1(g(0)) (-2r_x x,-2r_y y,-2r_z z,-2r_s)^T\) and substitute \(r_x,r_y,r_z,r_s\) by their values from 5 at the metric \(g_{x,y,z,1}\). This gives an expression in the variables \(x,y,z\) which is symmetric in \(y,z\).

We make the change of variables from \((y,z)\) to \((a,b)\) given in 9 and obtain the following expression in the variables \(x,a,b\): \[\begin{align} \frac{d}{d\ell}\Big\vert_{\ell=0} \tilde{P}_1(g(\ell))=\frac{1}{x^2 b\sqrt{H_1V_1}} D_n(x, a,b), \end{align}\] where \[D_n(x, a, b)=Fn+G,\] with \[\begin{align} F:=& 16x b \big(\left(-3 a+6 b+3\right) x^{3}+\left(-3 a^{2}+6 a b+6 a-12 b-3\right) x^{2}\\ &+\left(6 a^{3}-12 a^{2} b+8 a \,b^{2}-7 a^{2}+3 a b-4 b^{2}+2 a+6 b\right) x-2 a^{3}+a^{2}+8 a b-4b\big), \end{align}\] and \[\begin{align} G:=& (-48ab + 144b^2 - 24b + 24)x^4\\ &+ (128b^3 - 48a^2 + 440ab - 752b^2 + 16a - 144b)x^3\\ &+ (48a^3b - 144a^2b^2 + 128ab^3 - 360a^2 b + 656ab^2 - 512b^3 + 208ab + 224b^2 - 64b)x^2\\ &+ (48a^4 - 56a^3 b - 16a^3 - 160a^2 b + 224ab^2 + 64ab - 128b^2)x\\ &- 24a^4 + 128a^2 b - 128b^2. \end{align}\] Note that, in obtaining theses expressions, we used the fact that \(\sqrt{H_1V_1}=-4b+3a-3x\).

Recall that our goal is to show that \(\frac{d}{d\ell}\mid_{\ell=0} \tilde{P}_1(g(\ell)) > 0\) for any \(g(0)=g\in\hat{S}\). This obviously holds if and only if \(D_n(x, a, b)>0\) for any \((a,b)\in S\) and for \(x=x_4(a,b)\). Therefore, it suffices to prove that \(F(x_4,a,b)n + G(x_4,a,b) > 0\) for all \((a,b)\in S\). This will follow from the following two facts:

• \(F(x_4,a,b)>0\) for all \((a,b)\in S\), which is the content of Proposition 19,

• \(-\frac{1}{2} F(x_4,a,b) + G(x_4,a,b) \geq 0\) for all \((a,b)\in S\), which is done in Proposition 20.

These two results imply that \(G\geq \frac{1}{2} F > 0\), so clearly \(Fn+G > 0\) for all \(n\geq 1\). ◻

Proof. We will show that the zero set of \(F(x_4,a,b)\) does not intersect the region \(S\), so its sign is constant on \(S\). We will also show that \(F(x_4,a,b)\) is positive on the boundary curve \(a^2-4b=0\) of \(S\). It will follow that \(F(x_4,a,b)\) is positive on \(S\).

Let us consider the zero set of \(F(x_4,a,b)\). In order to simplify the computations we consider instead the zero set of \(\frac{F(x_4,a,b)}{16x_4b}\), which is the same since both \(x_4 > 0\) and \(b>0\) on \(S\). To do so, we substitute \(x=x_4\) in the expression of \(F(x,a,b)\) and follow the strategy of Remark 15 for \(\frac{F(x_4,a,b)}{16x_4b}=0\). After rearranging the expression, we find that the zero set of \(\frac{F(x_4,a,b)}{16x_4 b}\) is a subset of the zero set of the following equation: \[108 b^{2} \left(a+1\right) \left(-b-1+a\right) \left(a-2 b-1\right)^{2}\left(a^{2}-4 b\right) \left(4 a^{2}+8 a b-16 b^{2}-4 a-16 b-1\right) =0.\] For all the factors except the last two, it is clear that their zero set does not intersect \(S\). Thus, we need only analyze the last two factors.

We begin with the zero set of the factor \((a^2-4b)\). We will show that, in fact, if \(a^2 - 4b = 0\) and \((a,b)\in S\), then \(F(x_4,a,b)>0\). As we have computed in the proof of Proposition 16, along the boundary curve \(b=\frac{a^2}{4}\) we have \(x_4 = -\frac{a(a^3-6a^2+9a-8)}{6(a^2-2a+2)}\).

Substituting this into \(F(x_4,a,b)\) and then substituting \(b = \frac{a^2}{4}\) into the resulting expression, we obtain \[\label{eqn:boundarypos} \frac{a^{6} \left(a^{3}-6 a^{2}+9 a-8\right)^{2} \left(a+1\right) \left(a^{3}-3 a^{2}+3 a+1\right) \left(a-2\right)^{2}}{216 \left(a^{2}-2 a+2\right)^{3}}.\tag{24}\] The factor \((a^{2}-2 a+2)\) in the denominator has negative discriminant, so it is positive for all \(a\). Therefore, the sign of 24 is determined by the sign of the numerator. Since \(a\in (0,1)\) on \(S\), the factors \(a^6\), \((a+1)\) and \((a-2)^2\) are positive. Moreover, from the proof of Lemma 13, we know that \((a^3-6a^2 + 9a-8)< 0\) on \((0,1)\), so \((a^3-6a^2 + 9a-8)^2 > 0\). Finally, the factor \((a^3 -3a^2 + 3a + 1)\) has the value \(1> 0\) at \(a=0\) and derivative \(3(a-1)^2 \geq 0\), so it is also positive on \((0,1)\). Thus, 24 is positive on \(S\), as claimed.

To complete the proof that \(F>0\) on \(S\), it remains to investigate the last factor \((4a^2 + 8ab - 16b^2 -4a-16b-1)\). We shall show that its zero set does not intersect \(S\). Note that the graph of \(4 a^{2}+8 a b-16 b^{2}-4 a-16 b-1=0\) is a hyperbola. In particular, no component of its graph is contained in any compact set. It follows that if the graph intersects \(S\), it must intersect the boundary. We now show that, in fact, it does not intersect the boundary. Observe that \(\partial S\) consists of three arcs with equations \(b=\frac{a^2}{4}\) with \(a\in [0,1]\), \(b=0\) with \(a\in \left[0,\frac{3}{4}\right]\), and \(b=a-\frac{3}{4}\) with \(a\in\left[\frac{3}{4},1\right]\).

First, substituting \(b= \frac{a^2}{4}\) into \(4 a^{2}+8 a b-16 b^{2}-4 a-16 b-1 = 0\), we obtain \(-(a+1)(a^3-3a^2+3a+1)=0\). This has no solutions on \([0,1]\), so there the graph does not intersect the boundary of \(S\) along this component.

Second, substituting \(b= 0\) into \(4 a^{2}+8 a b-16 b^{2}-4 a-16 b-1 = 0\), we find \(4a^2 - 4a - 1 = 0\), which has no solutions on \(\left[0,\frac{3}{4}\right]\).

Finally, substituting \(b = a-\frac{3}{4}\) into \(4 a^{2}+8 a b-16 b^{2}-4 a-16 b-1 = 0\), we obtain \(-2(a+1)(2a-1)= 0\). This has no solutions in \(\left[\frac{3}{4},1\right]\). ◻

Proof of Proposition 20 assuming Proposition 21. The zero set of \(-\frac{1}{2} F+ G\) is a subset of the zero set of \(H\), so \(-\frac{1}{2}F + G\) has a constant sign on each component of \(S\setminus Z_H\). We substitute each of the two points of Proposition 21 into \(-\frac{1}{2} F(x_4,a,b) + G(x_4,a,b)\), and we obtain the positive values \(\frac{9294208}{6591796875}\) and \(\frac{87381}{250000}\), respectively. By continuity, it follows that \(-\frac{1}{2}F + G\) is positive on \(S\setminus Z_H\), and hence non-negative on \(S\). ◻

Proof. The singular points are exactly the common solutions to \(j = 0\), \(j_a = 0\) and \(j_b = 0\). Since the lowest degree term of \(j\) has total degree \(2\), it follows that \((0,0)\) is a singular point.

We now show that there are no other singular points in the appropriate region. To do so, let \((a_0,b_0)\in \left[0\vphantom{\frac{1}{4}},1\right]\times \left[0,\frac{1}{4}\right]\) be a common zero of \(j\) and \(j_a\). From Section 2 we know that the resultant \(\operatorname{res}_a(j,j_a)\in\mathbb{Z} [b]\) must vanish at \(b_0\). Applying Sturm’s Theorem 1 (see also Remark 2) to \(\operatorname{res}_a(j,j_a)\) on the slightly larger interval \(\left(-\frac{1}{100}, \frac{1}{4}\right]\), we find that there is exactly one value of \(b\) in this interval for which \(\operatorname{res}_a(j,j_a)\) vanishes. Since we know that \(b=0\) is such a zero (with \(a=0\)), it follows that \(b_0\) must be equal to \(0\). By continuity, it follows that there is a number \(\delta_b \in \left(0,\frac{1}{100}\right)\) with the property that, on the interval \(\left[-\delta_b, \frac{1}{4} + \delta_b\right]\), the only common zero \((a_0,b_0)\) of \(j\) and \(j_a\) has \(b_0=0\), so it is of the form \((a_0,0)\).

Next, suppose that \((a_0,0)\) with \(a_0\in [0,1]\) satisfies \(j_b=0\). Applying Sturm’s Theorem to \(j_b(a,0)\), which is a single-variable polynomial in \(a\), shows that it has exactly one real root on the larger interval \(\left(-\frac{1}{100},1\right]\). Since \(a=0\) is one such zero, it must be the only one, so \(a_0=0\). Again, by continuity, there is a number \(\delta_a\in \left(0,\frac{1}{100}\right)\) with the property that on the interval \([-\delta_a, 1 + \delta_a]\) the only zero \((a_0,0)\) of \(j_b\) is \((0,0)\). In particular, the only common zero \((a_0,0)\) to \(j\), \(j_a\) and \(j_b\) is \((0,0)\).

Taking \(\delta = \min\{\delta_a,\delta_b\}\) completes the proof. ◻

Proof. We will show that there exists a neighborhood \(U\) of \((0,0)\) such that \(j\) has no zeros with both coordinates non-negative, other than \((0,0)\). In other words, the only point of \(Z_j\cap\mathbb{R}^2_{\geq 0}\cap U\) is \((0,0)\), where \(\mathbb{R}^2_{\geq 0}\) consists of the points \((a,b)\) with \(a,b\geq 0\). This implies the claim in the statement because of the obvious inclusion \(\overline{S}\subset\mathbb{R}^2_{\geq 0}\).

In this proof we use the following notation. Given non-negative integers \(\alpha,\beta\), let \(c_{\alpha,\beta}\) denote the coefficient of \(a^\alpha b^\beta\) in the polynomial \(j\). Let \(j_2=\sum_{\alpha + \beta=2} c_{\alpha,\beta} a^\alpha b^\beta\) be the degree \(2\) part of \(j\). Further, let \(M= \sum_{\alpha+\beta>2} |c_{\alpha,\beta}|\).

The behavior of \(j\) near \((0,0)\) is roughly controlled by its lowest degree term \(j_2\), which equals \[j_2(a,b)=-256 a^{2}-768 a b-576 b^{2}=-64(2a+3b)^2.\] This polynomial vanishes along the graph of \(2a+3b=0\), which only intersects \(\overline{S}\) at \((0,0)\). Moreover, we have that \[\label{EQ:bound95on95j2} |j_2(a,b)|>64(2a+2b)^2=256(a+b)^2>256(a^2+b^2)\quad\text{for all }(a,b)\in\mathbb{R}_{\geq 0}^2.\tag{25}\] Now observe that \(|j|=|-j_2 -j +j_2| \geq |j_2| - |j-j_2|\) by the triangle inequality, so we need to estimate \(|j-j_2|\). We first claim that \(|j(a,b)-j_2(a,b)| \leq M\sqrt{a^2+b^2}^3\) for all \((a,b)\) with \(\sqrt{a^2 +b^2}\leq 1\). Indeed, we have: \[\begin{align} |j(a,b)-j_2(a,b)| &\leq \sum_{\alpha+\beta > 2} \left|c_{\alpha+\beta} a^\alpha b^\beta\right|\\ &\leq \sum_{\alpha+\beta>2} \left|c_{\alpha,\beta}\right| \sqrt{a^2+b^2}^\alpha \sqrt{a^2+b^2}^\beta\\ &= \sum_{\alpha+\beta>2} |c_{\alpha,\beta}| \sqrt{a^2+b^2}^{\alpha+\beta}\\ &\leq \sum_{\alpha+\beta>2} |c_{\alpha,\beta}| \sqrt{a^2+b^2}^3\\ &= M \sqrt{a^2+b^2}^3. \end{align}\] Here, the first line follows from the triangle inequality, the second follows because both \(a,b\leq \sqrt{a^2+b^2}\), and the fourth follows from the fact that \(\alpha+\beta \geq 3\) and \(\sqrt{a^2+b^2}\leq 1\).

Now, take any positive number \(r\) with \(r < \frac{256}{M}\) and let \(U\) be the open ball around \((0,0)\) with radius \(r\). Note that \(M>256\), so \(r<1\) and thus \(U\) is contained in the unit disk \(\sqrt{a^2 +b^2}\leq 1\). In particular \(|j-j_2|\leq M\sqrt{a^2+b^2}^3\) on \(U\). Using that \(r < \frac{256}{M}\), we conclude that \[\label{EQ:bound95on95j95minus95j2} j(a,b)-j_2(a,b)< 256(a^2+b^2 ),\quad\text{for all }(a,b)\in U\setminus \{(0,0)\}.\tag{26}\] Now we use 25 and 26 in the following chain of inequalities on \(\mathbb{R}_{\geq 0}^2\cap (U\setminus\{(0,0)\})\): \[\begin{align} |j(a,b)| &\geq |j_2(a,b)| - |j(a,b)-j_2(a,b)|\\ &> 256(a^2+b^2) - 256(a^2+b^2)\\ &= 0, \end{align}\] so \(j(a,b)\neq 0\) on \(\mathbb{R}_{\geq 0}^2\cap (U\setminus\{(0,0)\})\) as claimed. ◻

Proof. Suppose for a contradiction that such a closed curve \(C\) exists. By compactness of \(C\), the map \(C\rightarrow \mathbb{R}\) taking a point in \(C\subseteq \overline{S}\subseteq \mathbb{R}^2\) to its \(b\)-coordinate achieves an absolute maximum at some \((a_0,b_0)\in C\) with \(b_0 > 0\). Being an absolute maximum, the tangent line to \(C\) at \((a_0,b_0)\) must be horizontal, so the slope \(-\frac{j_a}{j_b}\) vanishes, so that \(j_a (a_0,b_0)= 0\). But we showed in the proof of Lemma 16 that any common solution \((a,b)\) of the system of equations \(j = 0\) and \(j_a = 0\) must have \(b=0\). This contradicts the fact that \(b_0 > 0\). ◻

Proof. The set \(\overline{S}\) contains three boundary arcs with equations \(b=\frac{a^2}{4}\) with \(a\in [0,1]\), \(b=0\) with \(a\in \left[0,\frac{3}{4}\right]\), and \(b=a-\frac{3}{4}\) with \(a\in\left[\frac{3}{4},1\right]\).

To begin with, if we substitute \(b= \frac{a^2}{4}\) into \(j = 0\), Sturm’s Theorem 1 (see also Remark 2) shows that the resulting function of \(a\) has no zeros on \((0,1]\).

If we substitute \(b= 0\) into \(j\), Sturm’s Theorem shows the resulting function of \(a\) has precisely one zero on \(\left(0,\frac{3}{4}\right]\) Moreover, this zero must lie in the interval \(\left(0,\frac{3}{4}\right)\) because, as it is easy to verify, the point \(\left(\frac{3}{4},0\right)\) does not belong to \(Z_j\).

Finally, if we substitute \(b=a-\frac{3}{4}\) into \(j\), Sturm’s Theorem shows the resulting function of \(a\) has precisely one zero on \(\left(\frac{3}{4},1\right]\). ◻

Proof of Proposition 21. Consider the points \(\left(\frac{19}{25}, \frac{1}{100}\right)\) and \(\left(\frac{9}{10}, \frac{3}{20}\right)\). These points lie on the curve \(b = a-\frac{3}{4}\) which lies in the boundary of \(S\). If we substitute these points into \(j\), we find that the outputs are respectively \(\frac{9274567512848}{3814697265625} > 0\) and \(-\frac{1820807829}{125000000} <0\). It follows that they lie in different components of \(\overline{S}\setminus Z_j\). By continuity, it follows that \(j\) changes sign on \(S\). Thus, there are at least two components of \(\overline{S}\setminus Z_j\). We complete the proof by arguing that there are only two components.

We showed above (see Lemma 19 and the preceding discussion) that \(Z_j\cap\overline{S}\) equals either three isolated points lying in \(\partial S\) or the \((0,0)\) together with one line segment joining the two other points. In the first possibility it is clear that \(\overline{S}\setminus Z_j\) is connected, which is a contradiction with the fact that \(\overline{S}\setminus Z_j\) has at least two components. Thus, \(\overline{S}\setminus Z_j\) equals \((0,0)\) together with one line segment joining the two other points. Now note that removing removing \((0,0)\) does not affect the number of components, so \(\overline{S}\setminus Z_j\) has the same number of components as \(\overline{S}\) with the interval with endpoints in \(\partial S\) removed. It is easy to see that the interval divides \(\overline{S}\) in two components, see for example [18]. It follows that \(\overline{S}\setminus Z_j\) has two components, and the same holds for \(S\setminus Z_j\). ◻

Proof. Let \(g_i(\ell)\) denote the Ricci flow of \(g_i\). Since \(g_1= \pi^\ast g_2\) is invariant under the action of the deck transformations group of \(\pi\), so is \(g_1(\ell)\) for all \(\ell\in[0, L_1)\). Therefore, \(g_1(\ell)\) descends to a curve of metrics \(\overline{g}_1(\ell)\) on \(M_2\) with \(\overline{g}_1(0) = g_2\). Since \(\pi\) is a local isometry, \(\overline{g}_1(\ell)\) satisfies the Ricci flow equation on \(M_2\) and by uniqueness of Ricci flow, we conclude that \(\overline{g}_1(\ell) = g_2(\ell)\) for all \(\ell\in [0,L_1)\). In particular, this implies \(L_1\leq L_2\).

Conversely, the metric \(\pi^\ast g_2(\ell)\) satisfies the Ricci flow equation on \(M_1\) with \(\pi^\ast g_2(0) = g_1\), so \(\pi^\ast g_2(\ell) = g_1(\ell)\) by uniqueness of the Ricci flow. It now follows that \(L_2\leq L_1\). ◻

Proof of Theorem 2. First of all, in the simply connected case the proof has been mostly given in the introduction. Again, from the classification of positively curved homogeneous spaces, collected and completed in [8], we know that such a space \(M\) is diffeomorphic to a compact rank one symmetric space, a Berger space \(B^7,B^{13}\), a Wallach space \(W^6,W^{12},W^{24}\), or an Aloff-Wallach space \(W_{p,q}^7\). Now, if \(M\) has a homogeneous \(\sec >0\) metric which evolves to planes with non-positive curvature, Theorem 1 implies that \(M\) cannot be a compact rank one symmetric space, and it cannot be \(B^7\) either, since \(B^7\) admits a unique homogeneous metric up to scaling and this metric is Einstein [9]. Thus \(M\) must be diffeomorphic to one of \(W^6\), \(W^{12}\), \(W^{24}\), \(W_{p,q}^7\) or \(B^{13}\). Conversely, if \(M\) is diffeomorphic to one of \(W^6\), \(W^{12}\), \(W^{24}\), \(W_{p,q}^7\) or \(B^{13}\), it follows from [6] and [7] that \(M\) has a homogeneous \(\sec >0\) metric which evolves to planes with non-positive curvature.

Now let \(M\) be a compact connected manifold admitting a homogeneous metric \(g\) with \(\sec>0\) which evolves to metrics with non-positively curved planes under the Ricci flow. In fact, the following proof works in the more general case where \(g\) is locally homogeneous. Let \((\tilde{M}, \tilde{g})\) be the universal cover of \((M, g)\). Then \((\tilde{M}, \tilde{g})\) is a homogeneous space [19]). Let \(\tilde{g}(t)\) and \(g(t)\) be the Ricci flows of \(\tilde{g}\) and \(g\), respectively. Lemma 20 implies that \(\tilde{g}(t)\) and \(g(t)\) have the same maximal forward existence time. Consequently, the homogeneous metric \(\tilde{g}\) on \(\tilde{M}\) evolves to a metric with non-positively curved planes since the universal covering maps are local isometries. It follows from the simply connected case settled above that \(\tilde{M}\) must be diffeomorphic to one of the spaces listed in the statement.

Now, we prove the converse. If \(M\) is simply connected, the result has been proven above. Therefore, we assume that \(M\) is a non-simply connected homogeneous manifold whose universal cover is one of \(W^6\), \(W^{12}\), \(W^{24}\), \(W_{p,q}^7\) or \(B^{13}\). For the proof we rely on the classification of positively curved homogeneous spaces in [8], and for simplicity we do not repeat this citation throughout the proof. In particular, \(B^{13}\) does not admit a positively curved finite quotient, so it suffices to consider spaces \(M\) whose universal cover \(\tilde{M}\) is one of \(W^6\), \(W^{12}\), \(W^{24}\) or \(W_{p,q}^7\). We will show that the universal cover has a homogeneous metric \(g_1\) of \(\sec >0\) which, on the one hand evolves to metrics with non-positively curved planes, and on the other hand descends to a homogeneous metric on the quotient \(M\) in the sense that \((\tilde{M},g_1)\to (M,g_2)\) is a Riemannian covering. By the properties of coverings and of the Ricci flow it then follows that \(g_2\) is a \(\sec >0\) metric which evolves to metrics with non-positively curved planes, as desired. We treat each case separately.

Suppose that \(\tilde{M}\) is diffeomorphic to an Aloff–Wallach space \(W^7_{p, q}\). First note that from [20], if \(M\) is diffeomorphic to the homogeneous presentation \((\mathsf{U}_{3}/\mathsf{T}^2_{p,q})/\Gamma\), for some finite subgroup \(\Gamma\subseteq \frac{{N}_{\mathsf{U}_{3}}(\mathsf{T}^2_{p, q})}{\mathsf{T}^2_{p, q}}\), we can find a finite subgroup \(\tilde{\Gamma}\subseteq \frac{N_{\mathsf{SU}_{3}}(\mathsf{S}^1_{p, q})}{\mathsf{S}^1_{p, q}}\) such that \(M\) is diffeomorphic to \((\mathsf{SU}_{3}/\mathsf{S}^1_{p,q})/\tilde{\Gamma}\). Hence without loss of generality, we can assume that \(M\) is given as a finite quotient of \(\mathsf{SU}_{3}/\mathsf{S}^1_{p,q}\).

Consider the case \((p, q)=(1, 1)\). Let \(g_1\) be a metric on \(W^7_{1, 1}\) which is \(\mathsf{U}_{2}\)-invariant and evolves under the Ricci flow to a metric with non-positively curved planes. The existence of such metrics is guaranteed by [7]. Since \(M\) is diffeomorphic to \(W^7_{1, 1}/\Gamma\), where \(\Gamma\) is a finite subgroup of \(\mathsf{U}_{2}/\mathsf{S}^1_{1,1}\), the metric \(g_1\) descends to a metric \(g_2\) on \(M\) with the desired properties.

Now, let \(p\neq q\). Then there exists a \(\mathsf{T}^2\)-invariant metric \(g_1\) on \(W^7_{p, q}\) which evolves under the Ricci flow to a metric with non-positively curved planes [7]. Similarly, since \(M\) is diffeomorphic to \(W^7_{p,q}/\Gamma\), where \(\Gamma\) is a finite subgroup of \(\mathsf{T}^{2}/\mathsf{S}^1_{p,q}\), the metric \(g_1\) descends to a metric \(g_2\) on \(M\) as in the statement of the theorem.

Suppose now that \(\tilde{M}\) is diffeomorphic to a Wallach space \(W:=W^n\), for \(n=6, 12, 24\). Since the arguments in all cases are analogous, we treat them simultaneously. Recall that, by Synge’s Theorem, any non-trivial isometric quotient of an even-dimensional \(\sec>0\) manifold must have fundamental group isomorphic to \(\mathbb{Z}_2\). In the case of \(W\), there are three \(\mathbb{Z}_2\)-quotients which give rise to non-simply connected homogeneous spaces with positive sectional curvature. These three \(\mathbb{Z}_2\) subgroups are conjugate in \(S_3=N_{G}(H)/H\), where \(G/H\) denotes the corresponding homogeneous presentation of \(W\). Hence the quotients of \(W\) by these groups are isometric [21]. Therefore, we only need to deal with one of the quotients.

Let \(g_1\) be a \(G\)-invariant right \(\mathbb{Z}_2\)-invariant metric with \(\sec>0\) which evolves under the Ricci flow to a metric with non-positively curved planes [6]. Since \(\mathbb{Z}_2\subseteq N_{G}(H)/H\), it follows that \(g_1\) descends to a metric \(g_2\) on \(M\) as desired. ◻