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}=
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{