Consider Ricci flow on a compact smooth Riemannian manifold $(M,g(t))$,the Ricci tensor is $Ric(t)$. Then , they meets $$ \partial_tg_{ij}=-2R_{ij} \\ \partial_tR_{ik}=\Delta R_{ik}+2g^{pr}g^{qs}R_{piqk}R_{rs}-2g^{pq}R_{pi}R_{qk} $$ If at $t=0$, the metric and Ricci tensor is diagonal, namely , $g_{ij}(0)=0 ~R_{ij}(0)=0~,~i\ne j$ , whether the metric and Ricci tensor keep diagonal under Ricci flow ?

  • $\begingroup$ The answer to this is no. I don't have a great intuitive sense for what happens in the general case. For the case of Lie algebras and Lie groups this has been investigated extensively by Jorge Lauret and Cynthia Will, where they connect the notion of a Stably Diagonal Ricci Flow with the notion of a `nice basis' for the Lie algebra. To get a feel for this, you might consider looking at the Ricci flow on three-dimensional unimodular Lie groups in a Milnor frame where the Ricci flow is stably diagonal. (I can provide references if needed). Additionally, you then might . . . $\endgroup$
    – THW
    Commented Aug 15, 2016 at 14:04
  • $\begingroup$ . . . look at the Ricci flow on four-dimensional Lie groups where one encounters Ricci tensors that are not stably diagonal. I can provide an example of this if you would like. $\endgroup$
    – THW
    Commented Aug 15, 2016 at 14:06
  • $\begingroup$ @THW I want to see your example.Could you print it as answer ? $\endgroup$
    – Enhao Lan
    Commented Aug 20, 2016 at 7:15
  • $\begingroup$ Yes, I will post the answer tomorrow. $\endgroup$
    – THW
    Commented Aug 21, 2016 at 1:56

1 Answer 1


(I apologize for the delayed response. I thought I had a counter example in an old set of notes but was unable to find it/them. As a result, this is probably not the simplest counter example.)

As indicated in the comments, the answer to the question is `no.'

Let $\mathfrak{h}$ be a four-dimensional Lie algebra with basis $X_{1}, X_{2}, X_{3},$ and $X_{4}$ and Lie algebra structure generated by the non-trivial brackets

\begin{align*} \left[X_{1}, X_{2}\right]&= (3/2)\,{\frac {3+2\,\sqrt {2}}{ \left( 1+\sqrt {2} \right) \left( 2+ \sqrt {2} \right) }} X_{2} + (1/2)\,{\frac {5+4\,\sqrt {2}}{2+\sqrt {2}}} X_{3},\\ \left[X_{1}, X_{3}\right] &= (1/2)\,{\frac {1+2\,\sqrt {2}}{2+\sqrt {2}}} X_{2} - (3/2)\,{\frac {1+\sqrt {2}}{2+\sqrt {2}}} X_{3},\\ \left[X_{1}, X_{4}\right] &= X_{4}, \end{align*} with corresponding simply-connected Lie group $H$.

Denoting the co-frame dual to the indicated basis by $\omega^{1}, \omega^{2}, \omega^{3},$ and $\omega^{4}$, suppose that $$ g = \alpha \omega^{1} \otimes \omega^{1} + \beta \omega^{2} \otimes \omega^{2} + \gamma \omega^{3} \otimes \omega^{3} + \delta \omega^{4} \otimes \omega^{4} $$ is a diagonal left-invariant metric.

Calculating the Ricci tensor of $g$ one finds that $Ric\left[g\right] =R_{ij}\omega^{i}\otimes\omega^{j}$ is not diagonal. (For simplicity, I will record only the non-diagonal terms of $Ric\left[g \right]$). The non-diagonal terms of $Ric\left[g\right]$ are $$ R_{23} = R_{32}= (7/4)\,{\frac { \left( -\beta+\gamma \right) \left( 7+5\,\sqrt {2} \right) }{ \left( 2+\sqrt {2} \right) ^{2} \left( 1+\sqrt {2} \right) \alpha}}. $$

The Ricci tensor of $g$ will thus be diagonal if $\gamma = \beta$.

Supposing now that $g$ is of the form $$ g = \alpha \omega^{1} \otimes \omega^{1} + \beta \omega^{2} \otimes \omega^{2} + \beta \omega^{3} \otimes \omega^{3} + \delta \omega^{4} \otimes \omega^{4}, $$ we find that the Ricci tensor $Ric\left[g\right]=R_{ij}\omega^{i} \otimes \omega^{j}$ is diagonal with component functions \begin{align*} R_{11}&= -\frac{11}{2}\\ R_{22}&= -3\,{\frac {\beta\, \left( 24+17\,\sqrt {2} \right) }{ \left( 1+\sqrt {2} \right) ^{2} \left( 2+\sqrt {2} \right) ^{2}\alpha}}\\ R_{33}&= 3\,{\frac {\beta\, \left( 4+3\,\sqrt {2} \right) }{ \left( 2+\sqrt {2} \right) ^{2}\alpha}}\\ R_{44}&= -{\frac {\delta\, \left( 4+3\,\sqrt {2} \right) }{ \left( 1+\sqrt {2} \right) \left( 2+\sqrt {2} \right) \alpha}}. \end{align*}

If the Ricci flow $g(t)$ were to keep both the metric diagonal and the Ricci tensor diagonal, then it would have to be of the form $$ g(t) = \alpha(t) \omega^{1}\otimes \omega^{1} + \beta(t)\omega^{2} \otimes \omega^{2} + \beta(t)\omega^{3} \otimes \omega^{3} + \delta(t)\omega^{4}\otimes\omega^{4}, $$ while satisfying the system of differential equations determined by $\frac{\partial g(t)}{\partial t} = -2Ric\left[g(t)\right]$.

The resulting system of differential equations is \begin{align*} \frac{d\alpha}{dt} &= 11\\ \frac{d\beta}{dt} &= 6\,{\frac {\beta\, \left( 24+17\,\sqrt {2} \right) }{ \left( 1+\sqrt {2} \right) ^{2} \left( 2+\sqrt {2} \right) ^{2}\alpha}}\\ \frac{d \beta}{dt} &= -6\,{\frac {\beta\, \left( 4+3\,\sqrt {2} \right) }{ \left( 2+\sqrt {2} \right) ^{2}\alpha}}\\ \frac{d\delta}{dt} &= -{\frac {\delta\, \left( 4+3\,\sqrt {2} \right) }{ \left( 1+\sqrt {2} \right) \left( 2+\sqrt {2} \right) \alpha}}, \end{align*} subject to the initial conditions determined by the metric $g$. However, the system above is only satisfied if $\beta(t) = 0$ for all $t$, which can't happen if the initial metric $g$ is diagonal with respect to the indicated basis.

Up to a calculation error--it was triple checked--this counterexample should be enough to show that the Ricci flow does not preserve the diagonalization of the metric and the Ricci tensor (although it might circumvent some more delicate analysis issues by working with left-invariant metrics). I hope that this is of some help.

  • $\begingroup$ Really thanks. I do want a example of Ricci flow. $\endgroup$
    – Enhao Lan
    Commented Aug 24, 2016 at 0:38
  • 1
    $\begingroup$ @lanse7pty Are you asking for explicit examples of the Ricci flow? I can direct you to some really nice (and readable) papers that have closed form solutions and you can analyze the long-term behavior. $\endgroup$
    – THW
    Commented Aug 24, 2016 at 1:25
  • $\begingroup$ Yes, I want it . $\endgroup$
    – Enhao Lan
    Commented Aug 24, 2016 at 1:32
  • $\begingroup$ Could you recommend a book of Lie algebra ? Although I understand the outline of your proof, but I don't know how to compute Ricci tensor by Lie structure and don't know what left-invariant is . But I think it should be right , because I can do same thing in 4 dimension vetor space. Besides, you can use Maple calculate Ricci tensor. $\endgroup$
    – Enhao Lan
    Commented Aug 24, 2016 at 3:10
  • $\begingroup$ I ask another question about the Lie structure and Ricci tensor, could you help me ? math.stackexchange.com/questions/1902015/… $\endgroup$
    – Enhao Lan
    Commented Aug 24, 2016 at 15:43

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