1 Introduction↩︎

The Ricci flow on a smooth Manifold is defined as the maximal solution to the partial differential equation \[\frac{dg}{dt}=-2\mathrm{Ric}(g),\] where \(g(t)\) denotes a time-dependent Riemannian metric and \(\mathrm{Ric}(g)\) is its Ricci curvature. A solution is said to be immortal if it exists for all positive times \(t\).

On homogeneous spaces, the Ricci flow evolves within the finite-dimensional space of homogeneous metrics. This restriction simplifies the computations and has made the homogeneous setting a fruitful ground for investigating long term behavior of the flow. In this context, an alternative formulation known as the bracket flow [Lauret, [1]] has proven particularly effective. Using this framework, it was shown by Lafuente [2] that if the universal cover of a homogeneous space is diffeomorphic to \(\mathbb{R}^n\), then the Ricci flow is immortal.

The converse implication was conjectured by Böhm and Lafuente :

It is in fact the "dynamical" version of the well known Alekseevski conjecture stating that a simply-connected homogeneous negative-Einstein space is diffeomorphic to \(\mathbb{R}^n\). In that version, we replace a static condition \(Ric(g)=\lambda g\) with a dynamic condition \(\frac{dg}{dt}=-2\mathrm{Ric} (g)\). The two conjectures are clearly equivalent for Einstein spaces, on which the Ricci flow is given by \(g(t)=(1-2\lambda t)g_0\).

An immediate corollary is an "all or nothing" law for homogeneous flows: either the space has a universal cover diffeomorphic to \(\mathbb{R}^n\) and all the homogeneous flows are immortal, or all the homogeneous flows have a finite time extinction.

Up to this date, the dynamical Alekseevski conjecture has been verified in several settings, including symmetric spaces, compact homogeneous spaces [Böhm, [4]], and cases where the isometry group \(G\) has a compact normal semisimple subgroup [Araujo, [5]]. However, a full understanding of the long-time behaviour of homogeneous Ricci flows remains incomplete. In dimensions \(\leq 4\), the conjecture has been fully addressed [5]. Dimension \(5\) is the first setting where simply connected, non Euclidean homogeneous spaces exist that are not covered by previous results.

The main result of this paper is the following theorem:

Our strategy is to classify systematically all five-dimensional homogeneous spaces and to identify the exceptional cases that fall outside known criteria (Part 2). We will isolate two such exceptional spaces: \(SO(3)\ltimes \mathbb{R}^3/SO(2)\) and \(SL(2,\mathbb{C})/U(1)\). For both cases, we will derive explicit Ricci flow equations and show that the flow develops a singularity in finite time, i.e, the flow is not immortal (Parts 3 and 4).

As a consequence, we will easily obtain a result on the topology of the space of positively curved metrics on \(5-\)dimensional homogeneous spaces (Part 5):

Acknowledgments. I’m grateful to my supervisor, Ramiro Lafuente, for his continuous support and helpful suggestions.

2 Classification of 5-dimensional homogeneous spaces↩︎

The Ricci flow is immortal on \(M\) if and only if it is immortal on \(\tilde{M}\). So we will directly suppose that \(M\) is simply connected. We write \(M=G/H\) and we will assume some properties on this writing (that we can always assume without loss of generality):

• \(G\) doesn’t contain a proper Lie subgroup acting transitively by isometries on \(M\);

• One Levi decomposition \(G=G_{ss} \ltimes R\) is such that \(H \subset G_{ss}\);

• \(G_{ss}\) acts effectively on \(G_{ss}/H\);

• \(H\) is compact.

On the Lie algebra level, we have \(\mathfrak{g} = \mathfrak{g}_{ss}\ltimes \mathfrak{r}\) and \(\mathfrak{h} \subset \mathfrak{g}_{ss}\) with \(codim_{\mathfrak{g}_{ss}}(\mathfrak{h}) + dim(\mathfrak{r}) = 5\). Notice that \[M \cong G_{ss}/H \times R\] and simply connected solvable Lie groups are Euclidean. Therefore, if \(G_{ss}/H\) is diffeomorphic to \(\mathbb{R}^k\) then \(M\) is diffeomorphic to \(\mathbb{R}^5\).

We will go through the cases according to the dimension of \(r\).

2.1 Dim(r)=5↩︎

In that case, \(H=G_{ss}\) so by iii, \(dim(G_{ss})=0\) and \(M=G=R\) so \(M\cong \mathbb{R}^5\).

2.2 Dim(r)=4↩︎

In that case, \(codim_{\mathfrak{g}_{ss}}(\mathfrak{h}) = 1\). It is impossible as shown by the following lemma :

2.3 Dim(r)=3↩︎

In that case, \(codim_{G_{ss}}(H) = 2\) so \(G_{ss}/H\) is a two-dimensional Riemannian homogeneous space i.e it is homothetic to \(\mathbb{S}^2\), \(\mathbb{R}^2\) or \(\mathbb{H}^2\) with their canonical metrics.

• If it’s \(\mathbb{R}^2\) or \(\mathbb{H}^2\) then \(M\cong \mathbb{R}^5\).

• If it’s \(\mathbb{S}^2\), then by iii \(\mathfrak{g}_{ss} \hookrightarrow Lie(Isom(\mathbb{S}^2)) = \mathfrak{so}(3)\) is injective. Therefore \(g_{ss}\) is a semisimple subalgebra of \(\mathfrak{so}(3)\), so it has to be \(\mathfrak{so}(3)\) because of the dimensions. In that case, \(\mathfrak{h}=\mathfrak{so}(2)\) is the whole isotropy group. So \(\mathfrak{g} = \mathfrak{so}(3) \ltimes \mathfrak{r}\) with a specific semidirect product, \(\mathfrak{h}=\mathfrak{so}(2)\subset \mathfrak{so}(3)\) and \(M = SO(3)\ltimes R/SO(2)\).

In the first case, \(G\) contains a normal compact semisimple subgroup. In the second case, \(M=SO(3)\ltimes \mathbb{R}^3/SO(2)\) with the standard semidirect product.

2.4 Dim(r)=2↩︎

In that case, \(codim_{\mathfrak{g}_{ss}}(\mathfrak{h})=3\). By iii, \(\mathfrak{g}_{ss}\hookrightarrow Lie(Isom(G_{ss}/H))\) is injective. We also know that \(dim(Isom(G_{ss}/H)) \leq \frac{3\cdot 4}{2} = 6\). Moreover, either \(G_{ss}/H\) has a constant sectional curvature (and therefore is isometric to \(\mathbb{R}^3, \mathbb{H}^3\) or \(\mathbb{S}^3\) up to homothety) or \(dim(Isom(G_{ss}/H)) \leq \frac{3\cdot 2}{2}+1 = 4\). By looking at the semisimple Lie algebras of dimension less than \(6\) we have the following cases :

• If \(\mathfrak{g}_{ss} = \mathfrak{so}(3)\) then \(\mathfrak{h}\) is trivial and \(M = SU(2) \ltimes R\) with a homogeneous metric. The only representation of \(\mathfrak{so}(3)\) of dimension less than \(3\) is the trivial representation, so \(M=SU(2)\times R\) and \(G\) has a normal semisimple compact subgroup.

• If \(\mathfrak{g}_{ss} = \mathfrak{sl}(2,\mathbb{R})\) then \(\mathfrak{h}\) is trivial and \(M=\tilde{SL(2,R)} \ltimes R\) so \(M\cong \mathbb{R}^5\).

• If not, \(dim(\mathfrak{g}_{ss}) = 6\) so we are in the equality case and \(G_{ss}/H\) is isometric to \(\mathbb{R}^3, \mathbb{H}^3\) or \(\mathbb{S}^3\) with their natural metrics (up to homothety). In the two first cases, \(M\cong \mathbb{R}^5\). In the last case, \(\mathfrak{g}_{ss}\hookrightarrow Lie(Isom(G_{ss}/H))\) is one-to-one so \(g_{ss} = \mathfrak{so}(4)=\mathfrak{so}(3)+\mathfrak{so}(3)\). Same as before, the semidirect product is a direct product. But then \(G\) has a normal compact semisimple subgroup \(SO(4)\).

2.5 Dim(r)=1↩︎

In that case, \(codim_{\mathfrak{g}_{ss}} (\mathfrak{h}) = 4\) and \(\mathfrak{r} = \mathbb{R}\). The only \(1\)-dimensional representation of a semisimple Lie algebra is the trivial one (because the kernel of the representations contains \([g_{ss},g_{ss}]=g_{ss}\)), so \(G = G_{ss}\times R\). Just like before, \(g_{ss}\hookrightarrow Lie(Isom(G_{ss}/H))\) is injective and \(dim(Isom(G_{ss}/H)) \leq \frac{4\cdot 5}{2} = 10\). Moreover, either \(G_{ss}/H\) has a constant sectional curvature (and therefore is isometric to \(\mathbb{R}^4, \mathbb{H}^4\) or \(\mathbb{S}^4\)) or \(dim(Isom(G_{ss}/H)) \leq 9\).

• If \(dim(G_{ss})=10\) then \(G_{ss}/H\) is either \(\mathbb{R}^4, \mathbb{H}^4\) or \(\mathbb{S}^4\). In the two first cases, \(M\) is diffeomorphic to \(\mathbb{R}^5\). In the last case, \(G_{ss}=SO(5)\) so \(G\) has a compact normal subgroup .

• If \(dim(G_{ss}) = 9\) then \(\mathfrak{g}_{ss}\) is a sum of \(\mathfrak{so}(3)\), \(\mathfrak{sl}(2,R)\), \(\mathfrak{sl}(2,C)\) and \(dim(\mathfrak{h})=5\). But the dimension of a maximal compact subgroup of \(so(3)\) or \(sl(2,R)\) is \(1\), and the dimension of a maximal compact subgroup of \(\mathfrak{sl}(2,\mathbb{C})\) is \(3\). So \(\mathfrak{h}\) cannot be \(5-\)dimensional.

• If \(dim(G_{ss}) = 8\) then \(\mathfrak{g}_{ss}=\mathfrak{su}(2,1)\) or \(\mathfrak{sl}(3,R)\) and \(dim(\mathfrak{h})=4\). But the dimension of a maximal compact subgroup of \(\mathfrak{sl}(3,R)\) is \(3\), so this case is not possible. Moreover, the maximal compact subgroup of \(SU(2,1)\) is \(S(U(2)\times U(1))\) and has dimension \(4\). So in that case \(M\) is Euclidean.

• The remaining possibilities are \(\mathfrak{g}_{ss} = \mathfrak{so}(3)+\mathfrak{so}(3)\), \(\mathfrak{so}(3)+\mathfrak{sl}(2,R)\), \(\mathfrak{sl}(2,R)+\mathfrak{sl}(2,R)\), \(\mathfrak{sl}(2,C)\). If \(\mathfrak{so}(3)\) is involved we are in the case where \(G_{ss}\) has a compact normal semisimple subgroup. If \(g_{ss} = sl(2,R)+sl(2,R)\) then \(H\) is a maximal compact subgroup of \(G_{ss}\) so \(M\) is Euclidean. If \(g_{ss} = sl(2,C)\) then we can suppose \(\mathfrak{h}\subset \mathfrak{su}(2)\) (maximal compact subgroup of \(SL(2,C)\)). But \(\mathfrak{su}(2)\) is simple so it cannot have a codimension\(-1\) compact subalgebra (lemme 2.1). Contradiction.

2.6 Dim(r)=0↩︎

In that case, \(G=G_{ss}\) and \(M=G_{ss}/H\).

• If \(G_{ss}\) has a compact factor in its decomposition in simple groups, then \(G\) has a compact normal semisimple subgroup.

• If \(M\) is a Riemannian product of two homogeneous spaces, then the proposition is true.

• If \(M\) is symmetric, then the proposition is true.

• From the work of Arroyo and Lafuente [[6], Table 1], we get the remaining cases \(M=SL(2,C)/U(1)\), \(M=SL(2,R)\times SL(2,R)/\Delta_{p,q}SO(2)\) and \(M=SU(2,1)/SU(2)\). For the last one, \(SU(2)\) has codimension \(1\) in a maximal compact subgroup of \(SU(2,1)\) so \(M\cong \mathbb{R}^5\). For \(SL(2,R)\times SL(2,R)/\Delta_{p,q}SO(2)\), the space is clearly diffeomorphic to \(\mathbb{R}^5\) because the maximal compact subgroup is \(SO(2)\times SO(2)\) and \(SO(2)\times SO(2)/\Delta_{p,q}SO(2)\) has a universal cover diffeomorphic to \(\mathbb{R}\). The last interesting case is \(M=SL(2,C)/U(1)\) with a homogeneous metric.

3 First special case : \(SO(3)\ltimes R^3/SO(2)\)↩︎

3.1 Lie algebra structure↩︎

We use the unique \(3\)-dimensional representation of \(\mathfrak{ so}(3)\), with basis : \[E = \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \quad F = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix} \quad G = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{pmatrix}\] We note \(c_1, c_2, c_3\) the canonical basis of \(\mathbb{R}^3\). Then the Lie algebra \(g\) is generated by \((E,F,G,c_1, c_2, c_3)\) with bracket specified in the following table.

In this setting we can assume that the isotropy subalgebra is \(\mathfrak{so}(2) = Vect(E)\). Moreover, \(p = Vect(F,G,c_1, c_2, c_3)\) is an \(ad(E)\)-invariant complement of \(Vect(E)\) in \(g\).
It is easy to show that the decomposition of \(p\) in \(\mathfrak {so}(2)\) irreducible representations is \(p = Vect(c_3) + Vect(c_1,c_2) + Vect(F,G)\). The problem making this case hard to solve is that the two \(2\)-dimensional representations are isomorphic, so cross-terms will appear in the metric and the Ricci tensor.
It will be useful to have the killing form on \(p\) :

Finally, we see that \(tr \circ ad = 0\) so \(G\) is unimodular.

3.2 Homogeneous metrics↩︎

We want to characterize all the the homogeneous metrics on \(SO(3)\ltimes \mathbb{R}^3/SO(2)\). Such a metric correspond to an \(\mathrm{Ad}(SO(2))-\)invariant scalar product on \(p\). Using this invariance and the irreducible decomposition of \(p\), we get that the scalar product in the basis \((c_3, c_1, c_2, F,G)\) has the following form : \[g = \begin{pmatrix} \alpha & 0 & 0 & 0 & 0 \\ 0 & \beta & 0 & \mu & \nu \\ 0 & 0 & \beta & -\nu & \mu \\ 0 & \mu & -\nu & \gamma & 0 \\ 0 & \nu & \mu & 0 & \gamma \\ \end{pmatrix},\] with \(\alpha, \beta, \gamma > 0\). As a matter of fact, we can get rid of the parameter \(\mu\).

We now have to find an orthonormal basis for \(g\). We introduce \(\tau = \mu^2 + \nu^2\). Let’s suppose that \(\tau > 0\). We define \[\lambda_{\pm} = \frac{\beta+\gamma}{2} \pm \frac{1}{2} \sqrt{4\tau + (\beta - \gamma)^2}\] These are the roots of the equation \((\lambda - \beta)(\lambda - \gamma) = \tau\) and are distinct from \(\beta, \gamma\) by assumption.
Let’s introduce \[H^{\pm} = \begin{pmatrix} 0\\ \lambda_{\pm} - \gamma \\ 0\\ \mu \\ \nu \end{pmatrix} \quad B^{\pm} = \begin{pmatrix} 0\\ 0 \\ \lambda_{\pm} - \gamma\\ -\nu \\ \mu \end{pmatrix}\] Then \((H^{\pm}, B^{\pm})\) are eigenvectors of \(g\) associated to eigenvalues \(\lambda_{\pm}\) (simple computation). It is then straightforward to show that \((\frac{c_3}{\sqrt{\alpha}},\frac{H^{\pm}}{f_{\pm}}, \frac{B^{\pm}}{f_{\pm}})\) is an orthonormal basis for \(g\), where \[f_{\pm} = \sqrt{\lambda_{\pm} \cdot (\tau + (\lambda_{\pm} - \gamma)^2)}\]

3.3 Ricci tensor↩︎

We know from [[7], Corollary 7.38] that if \((X_i)\) is an orthonormal basis of \(p\) for \(g\) then the Ricci tensor in the unimodular case is given by : \[Ric(X,Y) = -\frac{1}{2}B(X,Y) - \frac{1}{2}\sum_i g([X,X_i],[Y,X_i]) + \frac{1}{4}\sum_{i,j}g([X_i, X_j],X)g([X_i,X_j],Y)\] With that formula and simplifications with the \(\lambda_{\pm}\), we are able to express the full Ricci tensor (which has the same form as the metric \(g\)): \[Ric(c_3, c_3) = \frac{\alpha^2 - \beta^2}{\beta\gamma - \tau} + \frac{2\alpha^2 \nu^2}{(\beta\gamma - \tau)^2}\] \[Ric(c_1, c_1) = Ric(c_2, c_2) = \frac{\beta}{2\alpha}\frac{\beta^2 - \alpha^2}{\beta\gamma - \tau}\] \[Ric(F,F) = Ric(G,G) = 2 - \frac{\beta}{\alpha} + \frac{\gamma}{2\alpha}\frac{\beta^2 - \alpha^2}{\beta\gamma - \tau}\] \[Ric(c_1, F) = Ric(c_2, G) = \frac{\mu}{2\alpha (\beta\gamma - \tau)} (\beta^2 - \alpha^2)\] \[Ric(c_1, G) = -Ric(c_2, F) = \frac{\nu }{2\alpha (\beta\gamma - \tau)} (\alpha^2 + \beta^2)\]

3.4 Extinction time of the Ricci flow↩︎

The Ricci flow equation is \(\frac{dg}{dt} = -2Ric(g)\) i.e \[\begin{cases} \displaystyle \dot{\alpha} = 2\frac{\beta^2 - \alpha^2}{\beta\gamma - \tau} - \frac{4\alpha^2 \nu^2}{(\beta\gamma - \tau)^2} \\ \displaystyle \dot{\beta} = \frac{\beta}{\alpha}\frac{\alpha^2 - \beta^2}{\beta\gamma - \tau} \\ \displaystyle \dot{\gamma} = -4 + 2\frac{\beta}{\alpha} + \frac{\gamma}{\alpha} \frac{\alpha^2 - \beta^2}{\beta\gamma - \tau} \\ \displaystyle \dot{\mu} = \frac{\mu}{\alpha (\beta\gamma - \tau)}(\alpha^2 - \beta^2) \\ \displaystyle \dot{\nu} = -\frac{\nu}{\alpha(\beta\gamma - \tau)}(\alpha^2 + \beta^2) \end{cases}\] We will now prove the main theorem of this section.

First thing to notice is that if \(\mu = 0\) at \(t=0\) then \(\mu=0\) at any time due to the differential equation. Combined with the observations in Section \(2\), we can therefore assume \(\mu (t)=0\) and \(\tau = \nu^2\).
Let’s suppose that the Ricci flow above is immortal. We should arrive to some contradiction. There are two relevant ratios to analyze, which are \(x = \frac{\beta}{\alpha}\) and \(\epsilon = \frac{\tau}{\beta \gamma}\). The constraints on these variables are \(x>0\) and \(\epsilon \in [0,1)\).

Now we can assume \(\tau(t)>0\).

A short computation shows that \[\dot{(\frac{\gamma}{\beta})} = \frac{2}{\beta} (\frac{\beta}{\alpha} - 2)\] If \(x_{\infty} \leq 1\) and \(x<1\) after a certain time, then \(\alpha\) decreases, \(\beta\) increases and \(\beta < \alpha\) so \(\alpha, \beta\) converge to strictly positive constants. Therefore, \(\dot{(\frac{\gamma}{\beta})} \rightarrow \frac{2}{\beta_{\infty}}(x_{\infty} - 2 <0)\) so \(\frac{\gamma}{\beta}\rightarrow -\infty\) which is a contradiction.
If \(x_{\infty}\leq 1\) and \(x>1\) after a certain time then \(x_{\infty}=1\) and \[\dot{\gamma} \leq -4 + 2\frac{\beta}{\alpha} + \frac{\gamma}{\alpha} \frac{\alpha^2 - \beta^2}{\beta \gamma} = -4 + x + \frac{1}{x}\] so after a while \(\dot{\gamma} < -1\) and \(\gamma \rightarrow -\infty\), contradiction.
If \(1<x_{\infty}<2+\sqrt{3}\) then after a certain time, \(\beta > \alpha\) and \[\dot{\gamma} \leq -4 + 2\frac{\beta}{\alpha} + \frac{\gamma}{\alpha} \frac{\alpha^2 - \beta^2}{\beta \gamma} = -4 + x + \frac{1}{x}\] The last term is null for \(x = 2+\sqrt{3}\) and strictly negative for \(1<x<2+\sqrt{3}\) so after a certain time, \(\dot{\gamma} \leq -cste < 0\), and we would have \(\gamma \rightarrow -\infty\), which is a contradiction.
The last case is \(x_{\infty} \geq 2 + \sqrt{3}\). In that case, after a while, \(\beta > \alpha\) so \(\beta\) is decreasing. Therefore \(\dot{(\frac{\gamma}{\beta})} \geq cste > 0\) and \(\frac{\gamma}{\beta} \rightarrow +\infty\).
We compute the logarithmic derivative of \(\frac{\tau}{\beta^2}\) : \[\frac{\dot{(\frac{\tau}{\beta^2})}}{\frac{\tau}{\beta^2}} = - \frac{4\alpha}{\beta \gamma - \tau} < 0\] So we get that \(\epsilon = \frac{\tau}{\beta^2} \frac{\beta}{\gamma} \rightarrow 0\). Therefore the ratio between the two terms constituting \(\dot{\alpha}\) is \[\frac{(\beta \gamma - \tau)(\beta^2 - \alpha^2)}{2\alpha^2 \tau} = \frac{(\frac{1}{\epsilon} - 1)(x^2-1)}{2}>1\] after a certain time. So \(\alpha\) is increasing and \(\beta\) is decreasing, with \(\beta > \alpha\). They are both converging respectively to \(\beta_{\infty} \geq \alpha_{\infty} > 0\). So \(\frac{\gamma}{\beta} \sim \frac{2}{\beta_{\infty}} (x_{\infty} - 2)t\) and \(\gamma \sim 2(x_{\infty} - 2)t\). But \(\dot{\gamma} \leq -4 + x + \frac{1}{x}\) so we must have \(x_{\infty} + \frac{1}{x_{\infty}} - 4 \geq 2x_{\infty} - 4\) so \(x_{\infty}\leq 1\), contradiction. Thus, we’ve proved the theorem. \(\blacksquare\).

4 Second special case : \(SL(2,\mathbb{C})/U(1)\)↩︎

4.1 Lie algebra structure↩︎

The real Lie algebra \(g = sl(2,\mathbb{C})\) is generated by the following matrices : \[X = \begin{pmatrix} i & 0\\ 0 & -i \end{pmatrix} \quad A = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix} \quad B = \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix}\] \[C = \begin{pmatrix} 0 & i\\ 0 & 0 \end{pmatrix} \quad D = \begin{pmatrix} 0 & 0\\ 1 & 0 \end{pmatrix} \quad E = \begin{pmatrix} 0 & 0\\ i & 0 \end{pmatrix}\] Of course, \(u(1)\) is the subalgebra generated by \(X\). The bracket structure is specified in the following table :

So \(p = Vect(A,B,C,D,E)\) is an \(Ad(U(1))-\)invariant complement of \(u(1)\) in \(g\). Moreover, it is easy to show the decomposition of \(p\) in \(\mathfrak{so}(2)\) irreducible representations is \(p = Vect(A) + Vect(B,C) + Vect(D,E)\). The problem making this case hard to solve is that the two \(2\)-dimensional representations are isomorphic, so cross-terms will appear in the metric and the Ricci tensor.
It will be useful to have the killing form on \(p\) :

Finally, we see that \(tr \circ ad = 0\) so \(G\) is unimodular.

4.2 Homogeneous metrics↩︎

We want to characterize all the the homogeneous metrics on \(SL(2,\mathbb{C})/U(1)\). Such a metric correspond to an \(\mathrm{Ad}(U(1))-\)invariant scalar product on \(p\). Using this invariance and the irreducible decomposition of \(p\), we get that the scalar product in the basis \((A,B,C,D,E)\) has the following form. \[g = \begin{pmatrix} \alpha & 0 & 0 & 0 & 0 \\ 0 & \beta & 0 & \mu & \nu \\ 0 & 0 & \beta & -\nu & \mu \\ 0 & \mu & -\nu & \gamma & 0 \\ 0 & \nu & \mu & 0 & \gamma \\ \end{pmatrix}\] With \(\alpha, \beta, \gamma > 0\). As a matter of fact, we can get rid of one parameter.

We now have to find an orthonormal basis for \(g\). We introduce \(\tau = \mu^2 + \nu^2\). Let’s suppose for the moment that \(\tau > 0\). We define \[\lambda_{\pm} = \frac{\beta+\gamma}{2} \pm \frac{1}{2} \sqrt{4\tau + (\beta - \gamma)^2}\] These are the roots of the equation \((\lambda - \beta)(\lambda - \gamma) = \tau\) and are distinct from \(\beta, \gamma\) by assumption.
Let’s introduce \[H^{\pm} = \begin{pmatrix} 0\\ \lambda_{\pm} - \gamma \\ 0\\ \mu \\ \nu \end{pmatrix} \quad B^{\pm} = \begin{pmatrix} 0\\ 0 \\ \lambda_{\pm} - \gamma\\ -\nu \\ \mu \end{pmatrix}\] Then \((H^{\pm}, B^{\pm})\) are eigenvectors of \(g\) associated to eigenvalues \(\lambda_{\pm}\) (simple computation). It is then straightforward to show that \((\frac{H^{\pm}}{f_{\pm}}, \frac{B^{\pm}}{f_{\pm}})\) is an orthonormal basis for \(g\), where \[f_{\pm} = \sqrt{\lambda_{\pm} \cdot (\tau + (\lambda_{\pm} - \gamma)^2)}\]

4.3 Ricci tensor↩︎

We know from [[7], Corollary 7.38] that if \((X_i)\) is an orthonormal basis of \(p\) for \(g\) then the Ricci tensor in the unimodular case is given by : \[Ric(X,Y) = -\frac{1}{2}B(X,Y) - \frac{1}{2}\sum_i g([X,X_i],[Y,X_i]) + \frac{1}{4}\sum_{i,j}g([X_i, X_j],X)g([X_i,X_j],Y)\] With that formula and simplifications with the \(\lambda_{\pm}\), we are able to express the full Ricci tensor (which by the same argument has the same form as the metric \(g\)): \[Ric(A,A) = \frac{\alpha^2 - 16\beta \gamma}{\beta \gamma - \tau}\] \[Ric(B,B) = Ric(C,C) = \frac{\beta (16\tau - \alpha^2)}{2\alpha (\beta \gamma - \tau)}\] \[Ric(D,D) = Ric(E,E) = \frac{\gamma (16\tau - \alpha^2)}{2\alpha (\beta \gamma - \tau)}\] \[Ric(B,D) = Ric(C,E) = -4 + \frac{\mu}{2\alpha (\beta \gamma - \tau)}(-\alpha^2 + 16\beta \gamma)\] \[Ric(B,E) = -Ric(C,D) = \frac{\nu}{2\alpha (\beta \gamma - \tau)}(-\alpha^2 + 16\beta \gamma)\]

4.4 Extinction of the Ricci flow↩︎

The Ricci flow equation is \(\frac{dg}{dt} = -2Ric(g)\) i.e \[\begin{cases} \displaystyle \dot{\alpha} = 2\frac{16\beta \gamma-\alpha^2}{\beta \gamma - \tau} \\ \displaystyle \dot{\beta} = -\frac{\beta (16\tau - \alpha^2)}{\alpha (\beta \gamma - \tau)} \\ \displaystyle \dot{\gamma} = -\frac{\gamma (16\tau - \alpha^2)}{\alpha (\beta \gamma - \tau)} \\ \displaystyle \dot{\mu} = 8 - \frac{\mu}{\alpha (\beta \gamma - \tau)}(-\alpha^2 + 16\beta \gamma) \\ \displaystyle \dot{\nu} = -\frac{\nu}{\alpha (\beta \gamma - \tau)}(-\alpha^2 + 16\beta \gamma) \end{cases}\] We will now prove the main theorem of this section.

First thing to notice is that if \(\beta = \gamma\) at \(t=0\) then \(\beta = \gamma\) at any time. Combined with the observations in section \(2\), we can therefore assume \(\beta (t) = \gamma (t)\).
Let’s suppose that the Ricci flow above is immortal. We should arrive to some contradiction. There are two relevant ratios to analyze, which are \(x = \frac{\alpha}{\beta}\) and \(\epsilon = \frac{\tau}{\beta \gamma}\).
Let’s compute their derivatives : \[\frac{\dot{x}}{x} = \frac{\beta^2}{\alpha (1-\epsilon)} (16\epsilon + 32 - 3x^2)\] \[\frac{\dot{\epsilon}}{\epsilon} = \frac{16}{\sqrt{\tau}}(\frac{\mu}{\sqrt{\tau}}-\frac{2\sqrt{\epsilon}}{x})\] Notice that if \(\mu=0\) then \(\dot{\mu}=8\). So just before, \(\mu <0\) and just after \(\mu>0\). Therefore, after a certain time, \(\tau>0\) and \(\mu \neq 0\).

If \(x_{\infty}>2\) then after a while \(\frac{\mu}{\sqrt{\tau}}-\frac{2\sqrt{\epsilon}}{x} \geq \kappa >0\). So \[\frac{\dot{\epsilon}}{\epsilon} \geq \frac{16\kappa}{\sqrt{\tau}} \geq \frac{cste}{cste + 8\sqrt{\delta}t}\] By integrating this expression, we find that \(\log{\epsilon} \to +\infty\), contradiction.
If \(x_{\infty}< 2\), then \(\epsilon_{\infty}\neq 1\) because if not, \(\epsilon\) should increase so \(\frac{x\mu}{2\sqrt{\tau}}\geq \sqrt{\epsilon}\) and therefore we would have \(x_{\infty} \geq 2\). Thus, \[\alpha \sim 2\frac{16-x_{\infty}^2}{1-\epsilon_{\infty}}t\] \[\beta \sim \frac{2}{x_{\infty}}\frac{16-x_{\infty}^2}{1-\epsilon_{\infty}}t\] \[\dot{\beta} \to \frac{1}{x_{\infty}}\frac{x_{\infty}^2 - 16\epsilon_{\infty}}{1-\epsilon_{\infty}}\] So by equalizing the growth rates : \[3x_{\infty}^2 = 32 + 16\epsilon_{\infty}\] Finally, \(x_{\infty}\geq \frac{4\sqrt{2}}{\sqrt{3}}>2\), contradiction.
If \(x_{\infty} = 2\) and \(\epsilon_{\infty} \neq 1\), then the previous argument also works. The last case is \(x_{\infty}=2\) and \(\epsilon_{\infty}=1\). By a straightforward argument \(\alpha\) increases and \(\beta\) decreases, both converge to non zero constants. But then \(\frac{\dot{x}}{x} \sim \frac{cste}{1-\epsilon} \to +\infty\) so \(x\to +\infty\), contradiction. This finishes the proof of the theorem. \(\blacksquare\)

5 Contractibility of the space of positively curved metrics↩︎

Here, we give a proof of the Corollary 1.3, stating that the space of all positively curved metrics \(M^G_{\mathrm{Scal} >0}\) is either empty or contractible for any five-dimensional homogeneous space \(M=G/H\). In fact, we will prove this stronger statement:

References↩︎