I've been trying to understand the asymptotic behavior of Ricci flow, and there are two facts which I am unable to square away. I'm interested in higher dimensional manifolds, but my question is easier to state for Riemann surfaces. I suspect that the solution for surfaces will also solve the case in higher dimensions, so I'll focus on the two-dimensional case.

Short version

Given a Riemann surface which is topologically a sphere, the normalized Ricci flow is known to converge exponentially to a round metric. However, when we analyze the evolution of the scalar curvature for a small deformation of the round metric, it appears that there is a non-trivial center which seems to prevent the metric from converging exponentially. As such, I'm trying to determine the correct asymptotics for the scalar curvature, and why my calculation is going wrong.

Long version

Fact 1: In [1], Richard Hamilton proved that given a Riemann surface of positive curvature, if one evolves the metric by normalized Ricci flow \begin{equation} \frac{\partial}{\partial t} g_{i j}=(r-R) g_{i j}, \end{equation} then the metric will converge exponentially quickly to a metric of constant positive curvature (i.e., a round sphere). Here, $R$ is the scalar curvature and $r$ is the average scalar curvature, which acts to normalize the flow.

"Fact 2": The round unit sphere is a fixed point for normalized Ricci flow, so we study asymptotics of the flow for small (volume-preserving) deformations of the round metric. Under Ricci flow, the scalar curvature $R$ evolves via the reaction-diffusion equation \begin{equation} \frac{\partial R}{\partial t}=\Delta R+R^{2}-r R, \end{equation} where $\Delta$ is the Laplace-Beltrami operator with respect to $g(t)$. Since we are considering the unit sphere, the Gauss-Bonnet theorem implies that $r = 2$, so we consider the function $\phi = R-2$ and have that \begin{equation} \frac{\partial \phi}{\partial t}=\Delta \phi+R \phi. \end{equation}

If we linearize this equation at the round metric, we find that \begin{equation} \frac{\partial \phi}{\partial t}=\Delta_{\mathbb{S}^2} \phi+ 2 \phi, \end{equation}

To understand the asymptotic behavior of $\phi$, we can decompose it as $\phi = \sum_{\ell, m} a^{\ell}_m Y^m_{\ell}$, where $Y^m_{\ell}$ are spherical harmonics. Doing so, the flow under the preceding equation is simply \begin{equation} \phi(z,t) = \sum_{\ell, m} a^{\ell}_m e^{(-\ell(\ell+1)+2)t} Y^m_{\ell} \end{equation}

Here, this is using the fact that the spectrum of $\Delta_{\mathbb{S}^2}$ is $\ell (\ell+1)$. However, this seems to introduce a problem; the reaction term $2 \phi$ exactly cancels the diffusion effect on the $Y^m_{1}$-components of $\phi$. In other words, it seems like if $\phi$ is a principle eigenfunction of the Laplace-Beltrami operator (on the round sphere), then it is unchanged by Ricci flow, which seems to preclude exponential convergence of the flow.

Reintroducing the non-linearity

One might suspect that the issue is that I've linearized the flow at the round metric. However, the flow converges exponentially quickly to a round metric and a metric's principle eigenvalue is Lipschitz in the metric (which can be seen using the Rayleigh quotient property). As such, for a small deformation of the round metric where $\phi$ is close to a principal eigenfunction, we have that $ \left| \frac{\partial \phi}{\partial t}-\Delta_{\mathbb{S}^2} \phi- 2 \phi \right | < C e^{-\delta t} \phi $ for some $C$ and $\delta>0$. In other words, it seems that we can construct $\phi$ so that $$ \left| \frac{\partial \phi}{\partial t} \right | < C e^{-\delta t} \phi,$$ which again seems to preclude exponential convergence of the metric.

What's wrong with Fact 2?

Clearly, something is wrong with the analysis in the second argument. There is something that I have missed or computed incorrectly. However, I can't find the flaw in the argument, and I'd really like to know where I am going astray here.

[1] Hamilton, Richard S., The Ricci flow on surfaces, Mathematics and general relativity, Proc. AMS-IMS-SIAM Jt. Summer Res. Conf., Santa Cruz/Calif. 1986, Contemp. Math. 71, 237-262 (1988). ZBL0663.53031.

  • 3
    $\begingroup$ I believe there are issues with the gauge that are creeping in. I glanced at Hamilton's paper and it seems he proves exponential convergence of a modified flow (where there is a term that I think is supposed to correspond to the soliton potential in the limit. Of course there is no non-trivial soliton on the the two sphere so you then can recover the desired exponential convergence. $\endgroup$
    – RBega2
    Apr 26, 2021 at 19:32
  • $\begingroup$ For a more recent take on these issues in the case of the (volume normalized) Kahler-Ricci flow on Fano manifolds which admit KE metrics, in all complex dimensions (so including $S^2$) see arxiv.org/pdf/0705.4048.pdf, in particular Remark (7) at the very end of the paper: they show that if the flow converges to a KE metric smoothly modulo automorphisms(=gauge change), then it actually converges smoothly and exponentially fast (without the need for any gauge change). $\endgroup$
    – YangMills
    Apr 27, 2021 at 18:10
  • $\begingroup$ Of course, it is also known that when a KE metric exists then the volume-normalized flow (starting at any Kahler metric in the anticanonical cohomology class) does converge to some KE, see e.g. arxiv.org/pdf/1207.5441.pdf $\endgroup$
    – YangMills
    Apr 27, 2021 at 18:11
  • 1
    $\begingroup$ Small update to my first comment: it actually converges smoothly and exponentially fast to some (potentially different) KE metric. $\endgroup$
    – YangMills
    Apr 27, 2021 at 18:37

2 Answers 2


Having thought about this a little more, I think a more detailed explanation of the precise issue you are describing is that you are being too cavalier in "Fact 2" about what "linearizing" means. You are treating the scalar curvature as something you can freely vary, but that is not really true since $\phi$ is not an independent variable but is something computed in terms of the underlying metric. In other words, what you can freely vary is the metric and if you do so (within a fixed conformal class) you resolve the issue you described.

Indeed, if $g_s=e^{2u_s} g_{\mathbb{S}^2}$ is a variation of the metric (so $u_0=0$), then $$ \phi_s=2e^{-2u_s}-2e^{-2u_s} \Delta_{\mathbb{S}^2} u_s-2\frac{4\pi}{\int_{\mathbb{S}^2}e^{-2u_s} dvol_{\mathbb{S}_2}}. $$ Linearizing at $s=0$, we see that the linearized $\phi=\frac{d}{ds}|_{s=0} \phi_s$ from your Fact 2 must satisfy $$ \phi=-2 \Delta_{\mathbb{S}^2} v-4v $$ where $v=\frac{d}{ds}|_{s=0}u_s$ is the variation of the conformal factor (we used that the average scalar curvature is fixed so $v$ has average zero in order to eliminate the last term). This means $\phi$ is not arbitrary, but is in the image of the operator $\Delta_{\mathbb{S}^2} +2$. By the Fredholm alternative, this means that $\phi$ does not have any of the modes you are worried about in your original post.

You could also linearize the the Kazhdan-Warner identity Otis mentioned in his answer to get the same effect.

  • $\begingroup$ That makes sense. Intuitively, the scalar curvature being correlated with a principle eigenfunction means that one hemisphere has greater scalar curvature than the other, which doesn't seem possible geometrically. Your argument shows how to make this precise. $\endgroup$
    – Gabe K
    Apr 28, 2021 at 1:14

See Struwe, Curvature Flows on Surfaces. http://www.numdam.org/item/ASNSP_2002_5_1_2_247_0/, Section 6.2, (particularly equation (64) and surrounding text) where he uses the Kazdan-Warner identity to control the modes you are concerned about.

@RBega's answer is related to this answer, since the modes you are worried about correspond to linearized Mobius transformations (gauge) so you need a reason that they do not mess things up. In Struwe's approach, the KW identity saves you.

Note that in general, for a nonlinear parabolic equation of this form, when there are no zero eigenvalues of your linearized equation, convergence implies exponentially fast convergence (just use the linearized equation). But when there are zero eigenvalues, the story is a lot more complicated (sometimes convergence can occur polynomially, and in bad situations probably it can occur even slower).

  • 1
    $\begingroup$ Thanks for the reference and for explaining where these terms are coming from. I wish I could accept both of these answers. I accepted the other answer only because the variational approach works for the particular case I am thinking about. $\endgroup$
    – Gabe K
    Apr 28, 2021 at 17:04

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