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Let $(M, g)$ be a Riemannian manifold. I know the following result: For each $x \in M$ there exists a chart $V$ and a constant $C \geq 1$ such that for all $y \in V$ and $\eta \in T_y \mathbb{R}^n$ $$ \frac{1}{C}|\eta|_{g_{\mathbb{R}^n}} \leq |\eta|_g \leq C|\eta|_{g_{\mathbb{R}^n}}. $$ Now, my professor says, that it is not hard to prove, that for an arbitrary constant $C > 1$ one can find a chart s.t. the above inequality is satisfied. I cant figure out, why it's the case.

Metalhead
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  • Can you think of any possible ways to show this? Or some facts you know that you think might be useful in answering this? – Deane Jun 26 '24 at 02:53
  • If a vector space is finite dimensional, all norms are equivalent: https://math.stackexchange.com/questions/57686/understanding-of-the-theorem-that-all-norms-are-equivalent-in-finite-dimensional – hwood87 Jul 16 '24 at 01:28
  • I got a solution, I will post an answer to my question when I have time. – Metalhead Jul 20 '24 at 23:06

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Let $U \ni p$ be a chart with local coordinates $x^1 , \dots, x^n$. Because of symmetry and positive definiteness of $g^x(p)$ find a non- singular matrix $A$ such that $A^T g^x(p)A = I_n$. Now define a new coordinate system $y_1, \dots, y_n$ by

$$\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix} = A \cdot \begin{bmatrix}y_1\\\vdots\\y_n\end{bmatrix}$$.

Thus, the change of coordinates at $p$ from $x$ to $y$ at $p$ is given by $A$. So, $$g^y(p) = A^Tg^x(p)A = I_n$$ By continuity and shrinking $U$, and using $y_1, \dots, y_n$ as a coordinate system, we get the result.

Metalhead
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