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I tried to google the following but couldn't find an answer that helped - so I hope I might find some here - the question is short and very basic (I guess) :

what does it mean when someone writes that a Riemannian metric $g$ on $\mathbb{R}^n$ is such that $g(x)$ and $g^{-1}(x)$ are uniformly bounded ?

From the theory that I know (unfortunately not much so far) I understand that for each $x$ $g$ is a symmetric positive definite $n \times n$ matrix (given a local coordinate system), and as such it is a linear transformation of $\mathbb{R}^n$. So, does the uniform bound mean that there exists $M > 0$ such that \begin{equation} \|g(x)\| \leq M \quad \forall x \quad ? \end{equation} (Here $\| . \|$ is understood to be some norm on $GL(n, \mathbb{R})$).

Thanks for your help!

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Basically yes. Usually whenever I make such a statement what I mean is that there exists a constant $M$ so that for all $x \in \mathbb R^n$ and all $V \in \mathbb R^n$, $V \neq 0$, $$\frac{g_x(V,V)}{V\cdot V} \leq M$$ here $V$ is an element of $\mathbb R^n$ that can be thought of as living in the tangent plane at $x$. And a similar inequality for the inverse.

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ok that helps, thanks! –  harlekin Mar 5 '12 at 23:19

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