Doing a little bit of calculations I found that the metric:


satisfies $\frac{dg(t)}{dt}=-2Ric(t)$


$$\frac{dg(t)}{dt}=\frac{4e^{-4t}(dx^2+dy^2)}{(e^{-4t}-x^2-y^2)^2}\ {\rm and} \ Ric(t)=\frac{-2e^{-4t}( dx^2+dy^2)}{(e^{-4t}-x^2-y^2)^2}$$

this should be a solution to the Ricci flow on $R^2$.

Furthermore, with some other calculation, I found that this metric $g(t)$ is the only solution within the family: $g_{\lambda}(t)=\frac{dx^2+dy^2}{e^{-4\lambda t}-x^{2 \lambda}-y^{2 \lambda}}$, in fact, the only solution to the Ricci flow is only for $\lambda=1$.

Said this, my analysis has been to see the behavior in time for $g(t)$ and $Ric(t)$, then $t$ tends to $\infty$ and $- \infty$, but I want that the metric $g(t)$  remain positive.

a) Then for $t \rightarrow \infty$, to ensure that $g(t)$ is positive, I fixed $x=0$ and $y=0$ and I get $g(t)$ tends to $\infty$ and $Ric(t)$ tends to $-2$

b) While for $t \rightarrow - \infty$ I get $g(t)$ tends to $0$ and $Ric(t)$ tends again to $-2$.

For $t=0$,  

I get $g_0= \frac{dx^2+dy^2}{1-x^2-y^2}$ and with a simple calculations, I found that the 1-parameter family of diffeomorfism is $\phi_t=(e^{2t}x, e^{2t}y)$.

Considering $r^2=(x^2+y^2)$, the Scalar curvature, for $t=0$, is $R=\frac{-4}{1-r^2}$.

Now If I make an analysis of singularities (always for $(1-r^2)>0$), I found that if $r$ tends to $1-$, $g_0$ tends to $\infty$ and $R$ tends $- \infty$

My questions are:

1) (Admit that my calculations are correct) Is this a Steady Soliton?

2) Is this, for $t=0$, the Poincaré disk?

3) What else can I say? Which other analysis can I do?

Thank you in advance!


  • $\begingroup$ ...no idea? surely I realized that the answer to 2) is a "NO", because the curvature is not constant. $\endgroup$ Oct 27, 2016 at 9:13
  • $\begingroup$ No hurry , I try it ,although I know little about Ricci flow. $\endgroup$
    – lanse7pty
    Nov 2, 2016 at 4:47
  • $\begingroup$ Thanks "lanse7pty"! reflecting believe that the answer to 1 is a Yes...this is "in some sense" a steady soliton... I think it lacks only the 3 ... I do not know how to continue the analysis $\endgroup$ Nov 2, 2016 at 16:08


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