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Convergence of a sequence of sections of a bundle is defined as follows:

Definition: Let $E$ be a vector bundle over a manifold $M$, and let metrics $g$ and connections $∇$ be given on $E$ and on $TM$. Let $Ω ⊂ M$ be an open set with compact closure $\bar{Ω}$ in $M$, and let $(ξ_k)$ be a sequence of sections of $E$. For any $p ≥ 0$ we say that $ξ_k$ converges in $C^p$ to $ξ_∞ ∈ Γ(E\big|_{\bar{\Omega}})$ if for every $ε > 0$ there exists $k_0 = k_0(ε)$ such that $$\sup‎‎_{0\leq |\alpha | \leq p}‎‎\sup‎‎_{x\in \bar{\Omega}}‎|\nabla‎^{\alpha}‎‎(\xi_k -\xi_\infty)‎|_{‎g}‎‎<‎\varepsilon‎‎$$

whenever $k > k_0$. $\nabla^\alpha$ is the covariant derivative corresponding to the multi-index $α$.

Question: In the book "The Ricci Flow in Riemannian Geometry" by Ben Andrews and Christopher Hopper, is written: Note that since we are working on a compact set, the choice of metric and connection on $E$ and $TM$ have no affect on the convergence.

I can't understand why the sentence is true. Can someone help me? Thanks in advance.

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2 Answers 2

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This is just to supply some details to what Rafe Mazzeo wrote. Let $g_{i}$ be metrics and $^{\left( i\right) }\nabla$ be connections on $E$ and on $M$ for $i=1,2$. Since $\bar{\Omega}$ is compact, the uniform equivalence of norms reduces to a local coordinate chart $(U,\{x^{i}\})$ over which the bundle $E$ is trivialized. In the following, the constant $C$ may change from line to line. Since $C^{-1}g_{1}\leq g_{2}\leq Cg_{1}$ (uniform equivalence) on $E$ (fiberwise) and on $M$ for some $C$, for any $\xi\in\Gamma(E\otimes\bigotimes^{k}T^{\ast}M)$ we have $|\xi|_{g_{1}}\leq C|\xi|_{g_{2}}$ (same for $1$ and $2$ switched). Let $\alpha=(\alpha_{1},\ldots,\alpha_{n})$, so that $\nabla^{\alpha}=\nabla _{1}^{\alpha_{1}}\cdots\nabla_{n}^{\alpha_{n}}$ (up to uniform equivalence of norms, we may order it this way since commutators yield curvature and its derivative terms, which are bounded). Let $\lesssim$ denote $\leq C\cdot$. Now for $\xi\in\Gamma(E)$, $$ |^{\left( 1\right) }\nabla^{\alpha}\xi|_{g_{1}}\lesssim|^{\left( 1\right) }\nabla^{\alpha}\xi|_{g_{2}}\lesssim|^{\left( 2\right) }\nabla^{\alpha} \xi|_{g_{2}}+|\sum_{k=0}^{\left\vert \alpha\right\vert -1}{}^{\left( 1\right) }\nabla^{\ast k}\circ(^{\left( 1\right) }\nabla-{}^{\left( 2\right) }\nabla)\circ^{\left( 2\right) }\nabla_{g_{2}}^{\ast(\left\vert \alpha\right\vert -k-1)}|, $$ where the sum is comprised of linear combinations of $\ell$-th order covariant derivatives $^{\left( i\right) }\nabla^{\ast\ell}$. Since the sum has only covariant derivatives of lower order together with (bounded) derivative of the difference of connections terms, by induction on $p$ we obtain $\sum _{\left\vert \alpha\right\vert \leq p}|^{\left( 1\right) }\nabla^{\alpha} \xi|_{g_{1}}\lesssim\sum_{\left\vert \alpha\right\vert \leq p}|^{\left( 2\right) }\nabla^{\alpha}\xi|_{g_{2}}$ over $U$, independent of $\xi$.

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The reason is simply that over a compact set, all the choices (metrics, connections, etc.) are uniformly (or in whatever C^k topology you want) equivalent to one another.

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  • $\begingroup$ Where can I find proof of your claim? It is extremely important for me. thanks in advance. $\endgroup$ Nov 19, 2013 at 4:46
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    $\begingroup$ Once one chooses trivializations of bundles, coordinates, etc., then this reduces to the standard real analysis exercise that if $f_1$ and $f_2$ are two strictly positive functions on a compact set $K$, then there exist positive constants $C_1$, $C_2$ such that $C_1 f_1 \leq f_2 \leq C_2 f_1$. Similarly, if $A_1$ and $A_2$ are two positive definite matrices depending smoothly on a compact set $K$, then $C_1 A_1 \leq A_2 \leq C_2 A_1$ (where $A \leq B$ means $\langle Av,v \rangle \leq \langle Bv, v\rangle$). $\endgroup$ Nov 19, 2013 at 5:03
  • $\begingroup$ @RafeMazzeo I think that you do not mean the following:"If a compact topological space admit two compatible metrics $d_{1}, d_{2}$, then two metrics are equivalent, that is $\frac{d_{1}}{d_{2}}$ is bounded from below and above" There is a counter example for this statement. $\endgroup$ Jun 25, 2014 at 14:33

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