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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 $\tilde{g}_{f^{-1}(q)}(t)$.

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 $f\circ h=f$.

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?

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 $\tilde{g}_{f^{-1}(q)}(t)$.

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 $(\tilde{M},g)$ and find/construct a ricci flow $(\tilde{M},g(t))$, which is invariant under the isometry group $Iso(\tilde{M},g)$. 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?

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?