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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 $f\circ h=f$.

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@malik: Regarding the invariance under the isometry group, since both the Ricci tensor and the metric are invariant under isometries (and these constitute the equation for the Ricci flow), any isometries of the original metric $g_0$ are also isometries of $g_t$, regardless of any curvature conditions. Of course, $g_t$ has to exist and be a smooth metric for this statement to make sense, but as long as it exists the isometry group is preserved. –  Renato G Bettiol Nov 22 '12 at 17:17
    
Thank you, somehow I overlooked that fact. So there is no need for the last part about Kotschwar's result and I will delete it. –  malik Nov 22 '12 at 17:33
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Renato, don't you need a uniqueness theorem for this to be true? –  Deane Yang Nov 22 '12 at 17:43
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@malik: If the isometries are preserved, then in particular the deck transformations remain isometries. Thus it doesn't matter which lift you pick and you get a well defined metric on the quotient. –  Robert Haslhofer Nov 23 '12 at 14:14
    
@Deane: I actually had closed manifolds in mind (and just realized the OP talks about open manifolds instead), but it seems to me that isometries are preserved under any Ricci 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
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1 Answer

I was going to post this as a comment, but it got too long. I'm not exactly sure how to answer the question as written, but the following might be enlightening:

A crucial piece of information for your Ricci flow is that you start with bounded curvature, i.e. $$ \sup_M |Riem|_{g(0)} < \infty. $$ Then, according to Shi (Deforming the metric on complete Riemannian manifold, JDG 1989) there exists a Ricci flow $g(t)$ with bounded curvature, $g(t)$, defined for $t \in [0,T)$ where for $t \in [0,T)$ $$ \sup_M |Riem|_{g(t)} < \infty. $$ Shi makes no claims about uniqueness.

Later, it was proved by Chen-Zhu (Uniqueness of the Ricci Flow on Complete Noncompact Manifolds) that if $g_1(t), g_2(t)$ are solutions to Ricci flow for $t \in [0,T)$ and they both have bounded curvature in $[0,T)$, and $g_1(0) = g_2(0)$, then $g_1(t) = g_2(t)$

In particular, a corollary of Chen-Zhu's result is:

If $\phi\in Isom(M,g(0))$ is an isometry of $(M,g(0))$ where $g(0)$ has bounded curvature, then the Shi solution to Ricci flow with initial data $g(0)$ still has $\phi \in Isom(M,g(t))$.

A "fancy" way of saying this is $Isom(M,g(0)) \subseteq Isom(M,g(t))$.

To prove this, if $\phi \in Isom(M,g(0))$, let $g(t)$ be the Shi solution to Ricci flow starting at $g(0)$. Clearly $\phi^*g(t)$ is still a solution to Ricci flow, but $\phi^*g(0) = g(0)$ by assumption that it is an isometry of $g(0)$. So $\phi^*g(t)$ is a solution to Ricci flow (obviously having bounded curvature) with initial data $g(0)$, so it must be $g(t)$!

I assume that you were originally referring to Kotschwar (Backwards Uniqueness of Ricci Flow). This result says the opposite of Chen-Zhu, and is not really what you want here. In particular, it says that if $g_1(t), g_2(t)$ are both solutions to Ricci flow on $t\in[0,T]$ with uniformly bounded curvature ($T<\infty$), if $g_1(T) = g_2(T)$ then $g_1(t) = g_2(t)$ for all $t \in [0,T]$. In particular, this complements the Chen-Zhu corollary:

Under the assumptions of uniformly bounded curvature, $Isom(M,g(t)) \subseteq Isom(M,g(0))$.

Finally, I'll remark on the assumption of bounded curvature. In Chen (Strong Uniqueness of the Ricci flow, JDG 2009), it is shown that one does not always need to assume bounded curvature in dimension $3$. Instead, Chen shows:

If $(M,g(0))$ is complete, has bounded, nonnegative sectional curvature, $0 \leq Rm \leq K$, then any two $g_1(t),g_2(t)$ Ricci flows (which are complete for all $t\in[0,T)$) starting from $g(0)$ must agree.


None of this exactly answers your question as it stands, but I don't think that being simply connected makes things any easier (but I could be wrong). On the other hand, as long as your initial manifold has bounded curvature, by Shi-Chen-Zhu, there is a unique (short time) solution to Ricci flow with bounded curvature.

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@Otis Thank you for the patience to summarize these results. So given a manifold of bounded curvature I could use any Ricci flow solution $(\tilde{M},\tilde{g}(t))$ with $\tilde{g}(0)=\tilde{g}$ on the universal cover $(\tilde{M},\tilde{g})$ to obtain a solution on my original manifold $(M,g)$. (The question, how this is done, still remains.) –  malik Nov 23 '12 at 9:35
    
Sorry, I don't understand "the question, how this is done still remains." Could you rephrase your exact question? –  Otis Chodosh Nov 23 '12 at 17:30
    
Also just to be precise, you could use any ricci flow with bounded curvature although I do not see any point in passing to the universal cover. –  Otis Chodosh Nov 23 '12 at 17:32
    
I specified it in my post above (see "EDIT"). In short my question was, why $g_p(t)(X,Y) := \tilde{g}_{f^{-1}(p)}(t)(df^{-1}X,df^{-1}Y)$ is well defined for $t>0$. But this was answered by Robert, so I updated my post. –  malik Nov 23 '12 at 18:19
    
Oh, sorry I missed the update! –  Otis Chodosh Nov 23 '12 at 19:40
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