MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 1 down vote accepted

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 $\xi$.

share|cite|improve this answer

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.

share|cite|improve this answer
Where can I find proof of your claim? It is extremely important for me. thanks in advance. – Sepideh Bakhoda Nov 19 '13 at 4:46
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 $\langle Av,v \rangle \leq \langle Bv, v\rangle$). – Rafe Mazzeo Nov 19 '13 at 5:03
@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. – Ali Taghavi Jun 25 '14 at 14:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.