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Let $M$ be a Riemannian manifold with a metric $g$ and $(U,\varphi)$ a chart around a point $p\in M$.

By a Remark page 63 of Riemannian Geometry by M. Do Carmo, it seems that any open set $\mathcal{U}\subset TU$ contains an open set of the form

$$\{v_q\in TM: q\in V', ||v_q||_g<\epsilon \}$$ for $\epsilon>0$ and $V'\subset M$ an open set whose closure is compact.

I have been trying to prove that, but I am stuck at some point: By local compacity, we can take an open set $V'\subset \varphi(U)$ such that $\overline{V'}$ is compact and $\epsilon>0$ such that $$p\in V'\times B_0(r)\subset \overline{V'}\times \overline{B_0(r)}\subset T_\varphi(\mathcal{U}):=\mathcal{U}'.$$ Therefore, $$T_\varphi^{-1}(V'\times B_0(r))=\{v_q\in TM : q\in \varphi^{-1}(V'), ||v_p||_{\mathbb{R}^n}<\epsilon\}.$$ However, we do not want the norm in $\mathbb{R}^n$, but the norm from $g$! Of course, these are equivalent at each point of $M$, but here we have to deal with an open neighborhood of points. I feel there might be a compacity argument to be able to take some minimum/maximums, but I can find how. Or am I starting incorrectly ?

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(You need one more hypothesis - namely that $\mathcal{U}$ intersects the $0$ section of $TU$.) – Jason DeVito Jan 4 '13 at 18:03
up vote 4 down vote accepted

Given a coordinate chart $(U,\varphi)$ with $U \subset M$ and $\varphi : \tilde{U} \rightarrow U$, with $\tilde{U} \subseteq \mathbb{R}^n$, you get a coordinate chart on $TM$ which I'll denote by $\phi : \tilde{U} \times \mathbb{R}^n \rightarrow TU$. If the coordinates on $\mathbb{R}^n$ are denoted by $(x^1, ..., x^n)$ then the map $\phi$ is given by $$ (x^1, ..., x^n, y_1, ..., y_n) \mapsto \sum_{i = 1}^n y_i \left. \frac{\partial}{\partial x^i} \right|_{\phi(x_1, ..., x_n)}.$$

If you pullback the metric $g$ using $\varphi$ and $\phi$, you get a positive definite matrix $A = A(x^1, ..., x^n) \in C^{\infty}(\tilde{U}, M_n(\mathbb{R}))$ that depends on the coordinates $x^i$ such that $$ \left|\left| \sum_{i = 1}^n y_i \left. \frac{\partial}{\partial x^i} \right|_{\phi(x_1, ..., x_n)} \right|\right|^2_g = (y_1, \ldots, y_n) A(x^1, \ldots, x^n) (y_1, \ldots, y_n)^T.$$ If we denote by $\vec{y} = (y_1, \ldots, y_n)^T$, and by $\left< \cdot,\cdot\right>$ the usual Euclidean inner product on $\mathbb{R}^n$, then the expression above is just $\left<A(x^1, \ldots, x^n)\vec{y}, \vec{y}\right>$.

How does the (squared) norm $\left<A\vec{y},\vec{y}\right>$ compares to the regular, squared, Euclidean norm $\left<\vec{y}, \vec{y}\right>$? We have

$$ c_1 \left<\vec{y}, \vec{y}\right> \leq \left<A\vec{y},\vec{y}\right> \leq c_2 \left<\vec{y}, \vec{y}\right> $$ where we can take $c_2(x^1, \ldots, x^n) = ||A(x^1,\ldots,x^n)||$ and $c_1(x^1, \ldots, x^n) = \frac{1}{||A^{-1}(x^1,\ldots,x^n)||}$. The constants depend continuously on the point $(x^1, ..., x^n)$ because the metric and norm depends on the point, but on a compact subset $K \subset \tilde{U}$, you can take $$ C_1 = \min_{(x^1,\ldots,x^n) \in K} \frac{1}{||A^{-1}(x^1,\ldots,x^n)||}, \; C_2 = \max_{(x_1,\ldots,x^n) \in K} ||A(x^1,\ldots,x^n)|| $$ and have $$ C_1 \left<\vec{y}, \vec{y}\right> \leq \left<A(x^1,\ldots,x^n)\vec{y},\vec{y}\right> \leq C_2 \left<\vec{y}, \vec{y}\right> $$ with constants that doesn't depend on $x^i$. That is, the norms are uniformly equivalent on compact subsets of $\tilde{U}$. This is enough to finish your argument.

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Great answer, thank you! – Klaus Jan 5 '13 at 12:19

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