Reading some of my old notes, I came across a remark, I don't understand. Summarized:

Let $(M^n,g)$ be an open riemannian manifold (in that case it came with sectional curvature $K\ge0$, but I don't think this is necessary here). One way of ricciflowing this manifold is to consider the universal cover $(\tilde{M},\tilde{g})$ and find/construct a ricci flow $(\tilde{M},\tilde{g}(t))$, which is invariant under the isometry group $Iso(\tilde{M},\tilde{g}(0))$. Then this ricci flow descends to $(M^n,g)$.

How does this "descent" work and why do you need the invariance under the isometry group?

**EDIT**: I changed some notation and now want to specify my problem regarding the first part of my question:

Let $f:\tilde{M}\to M$ be the universal cover (local diffeomorphism?) of $(M,g)$. Then the metric $\tilde{g}$ on $\tilde{M}$ can be defined by $\tilde{g}_{p}(X,Y):= g_q(dfX,dfY) $ for $p\in\tilde{M}$, $q=f(p)$ and $X,Y\in T_p\tilde{M}$. Now we have of course by definition $\forall q\in M~\forall p_1,p_2\in f^{-1}(q)\forall X,Y\in T_qM:~~\tilde{g}_{p_1}(d(f^{-1})X,d(f^{-1})Y)=\tilde{g}_{p_2}(d(f^{-1})X,d(f^{-1})Y)$ and therefore $g_q$ can be obtained by choosing and lifting any $\tilde{g}_{f^{-1}(q)}$.

Given a ricci flow solution $(\tilde{M},\tilde{g}(t))$ with $\tilde{g}(0)=\tilde{g}$, the problem is, whether for $t>0$ $\tilde{g}_{p_1}(t)(d(f^{-1})X,d(f^{-1})Y)=\tilde{g}_{p_2}(t)(d(f^{-1})X,d(f^{-1})Y)$ holds true and $g_q(t)$ therefore can be obtained from any $\tilde{g}_{f^{-1}(q)}(t)$.

**SOLUTION (Thanks to Robert!)**: This might not be the best notation, but I hope one can understand it:
Since $f:(\tilde{M},\tilde{g})\to(M,g)$ is a local isometry, every deck transformation $h\in Aut(f)$ is an isometry. Given a ricci flow solution $(\tilde{M},\tilde{g}(t))$ with $\tilde{g}(0)=\tilde{g}$, which is invariant under $Iso(\tilde{M},\tilde{g})$ one has therefore $h\in Iso(\tilde{M},\tilde{g}(t))$. Furthermore it is known, that the action of $Aut(f)$ is transitive on every fiber $f^{-1}(q)$ with $q\in M$. Consequently one obtains the desired result

$\forall p_1,p_2\in f^{-1}(q)~\exists h\in Aut(f)~\forall X,Y\in T_qM~:$

$\tilde{g}_{p_1}(t)(d(f^{-1})X,d(f^{-1})Y)=\tilde{g}_{h(p_1)}(dh\cdot d(f^{-1})X,dh\cdot d(f^{-1})Y)=\tilde{g}_{p_2}(t)(d(f^{-1})X,d(f^{-1})Y)$

where the first Identity is true by $h\in Iso(\tilde{M},\tilde{g}(t))$ and the second is obtained from $f\circ h=f$.

anyRicci flow. Quoting from Hamilton [JDG, 1982], "degeneracies are there because the equation is invariant under the full diffeomorphism group. This has the interesting consequence that any isometries which exist in the metric to begin with are preserved as the metric evolves". Although he was talking about closed 3-manifolds in this context, in my understanding these observations about the evolution equation are general. – Renato G Bettiol Nov 23 '12 at 21:56