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In wikipedia, it has been written that an important 2-dimensional example of Ricci flow over $M=\mathbb{R}^2$ is given by $g((x,y),t)=\frac{dx^2+dy^2}{e^{4t}+x^2+y^2} \;\;\; (\star) $

Here are my questions;

I. Why the family $(\star)$ satisfies the Ricci flow equation, i.e. $g'(t)=-2\text{Ric}(t)$ for some $t \in I=[0,T),$ where $\text{Ric}$ is the Ricci tensor?

What I have found:

Using the traditional notations, we have $\partial_tg_{11}=\frac{-4e^{4t}}{ (e^{4t}+x^2+y^2)^2}$ which should be equal to $-2R_{11}.$ We know that, $R_{11}:=R_{1212}+R_{1111}=R_{1212}.$ So, we’re left to show that

$$R_{1212}= \frac{2e^{4t}}{ (e^{4t}+x^2+y^2)^2}$$

By definition, $R_{1212}=<R(X_1,X_2)X_1,X_2>_{g(t)}=\frac{ <R(X_1,X_2)X_1,X_2>_{\delta}}{(e^{4t}+x^2+y^2)^2}$ where $X_1:=\frac{\partial}{\partial x}$ and $X_2:=\frac{ \partial}{\partial y}$ and $<.,.>_{\delta}$ is the Euclidean metric. I cannot proceed from here!

II. Does there exist a family of smooth diffeomorphisms $\phi_t : \mathbb{R}^2 \to \mathbb{R}^2$ s.t. for any $t \in I,$ we have $g(t)=\phi^{\star}_t(g(0))$ i.e. for any $t\geq 0$ the Riemannian manifold $(\mathbb{R}^2, g(t))$ is isometric to $(\mathbb{R}^2, g(0))?$

What I have found:

We have that $g(o)=\frac{dx^2+dy^2}{1+x^2+y^2}. $ Now, $\phi^{\star}_t(g(0))(u,v)=\frac{<d \phi_t (u), d \phi_t (v)>_{\delta}}{1+x^2+y^2}$ and it must coincide with $g(t)(u,v)=\frac{<u,v>_{\delta}}{e^{4t}+x^2+y^2}$ for $u,v \in T_{(x,y)} \mathbb{R}^2.$ How can I find $\phi_t$ explicitly from here?

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1 Answer 1

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I. You can use the Christoffel symbols of the metric to compute the Ricci tensor. The Christoffel symbols $\Gamma^\alpha_{\phantom{\alpha}\beta\gamma}$ are defined by $$\Gamma^\alpha_{\phantom{\alpha}\beta\gamma} = \frac{1}{2}g^{\alpha\rho}(\partial_\gamma g_{\rho\beta} + \partial_\beta g_{\rho\gamma} - \partial_\rho g_{\beta\gamma}),$$ where we are using the Einstein summation convention. You can compute that the Christoffel symbols of $g(t)$ are $$\Gamma^x_{\phantom{x}xx} = \Gamma^y_{\phantom{y}xy} = \Gamma^y_{\phantom{y}{yx}} = - \frac{x}{e^{4t} + x^2 + y^2},$$ $$\Gamma^y_{\phantom{y}yy} = \Gamma^x_{\phantom{x}xy} = \Gamma^x_{\phantom{x}{yx}} = - \frac{y}{e^{4t} + x^2 + y^2},$$ $$\Gamma^x_{\phantom{x}yy} = \frac{x}{e^{4t} + x^2 + y^2},$$ and $$\Gamma^y_{\phantom{y}xx} = \frac{y}{e^{4t} + x^2 + y^2}.$$

In terms of the Christoffel symbols, the Ricci tensor has components $$\mathrm{Ric}_{\alpha\beta} = \partial_\rho \Gamma^\rho_{\phantom{\rho}\beta\alpha} - \partial_\beta \Gamma^\rho_{\phantom{\rho}\rho\alpha} + \Gamma^\rho_{\phantom{\rho}\rho\lambda} \Gamma^\lambda_{\phantom{\lambda}\beta\alpha} - \Gamma^\rho_{\phantom{\rho}\beta\lambda} \Gamma^\lambda_{\phantom{\lambda}\rho\alpha}.$$ Using this and the Christoffel symbols for $g(t)$ above, we have that the Ricci tensor for $g(t)$ is $$\mathrm{Ric}(t) = \frac{2e^{4t}}{(e^{4t} + x^2 + y^2)^2} (dx^2 + dy^2).$$ On the other hand, $$g'((x,y),t) = - \frac{4e^{4t}}{(e^{4t} + x^2 + y^2)^2} (dx^2 + dy^2).$$ Hence $g(t)$ satisfies the Ricci flow equation $$g'(t) = -2 \mathrm{Ric}(t),$$ as desired.

II. Yes, the flow $$\varphi_t: \mathbb{R}^2 \longrightarrow \mathbb{R}^2,$$ $$\varphi_t(x,y) = (e^{-2t}x, e^{-2t}y)$$ satisfies $$g(t) = \varphi_t^\ast g(0).$$

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    $\begingroup$ Do you have any references to recommend? $\endgroup$
    – Neal
    Commented Apr 10, 2012 at 19:32
  • $\begingroup$ @Henry T. Horton: Thank you for your solution. Could you show me, how you came up with the flow for the second part, explicitly? $\endgroup$ Commented Apr 10, 2012 at 23:23
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    $\begingroup$ @Neal: For more background on the calculations I outlined, see chapter 3 of Jost's Riemannian Geometry and Geometric Analysis, for example. Physics books are also nice for details on the computational side of Riemannian Geometry: see section 3.4 of Wald's General Relativity for information on various techniques for computing curvature. $\endgroup$ Commented Apr 11, 2012 at 2:32
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    $\begingroup$ @ehsanmo: I didn't use any special techniques. Simply note that $g((e^{-2t}x,e^{-2t}y),t) = g((x,y),0)$, so $\varphi_t(x,y) = (e^{-2t}x,e^{-2t}y)$ is the desired isometry. $\endgroup$ Commented Apr 11, 2012 at 2:33

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