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Given is an analytic function from $M$ to $N$, both equipped with conformal Riemannian metric, say $g$ and $h$ resp.

What might the $h$ norm of the derivative of the function at a point mean?

$\|f'(x)\| $ with respect to the $h$ metric.

Thanks

ADD:

It was used in this context:

d(x,y) < d(f(x),f(y)), both defined by the same metric, x,y,f(x),f(y) are in the same space. And the conclusion is that f is expanding, or $\|f'(x)\| $ > 1 with respect to this metric

Walter
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  • $f'(x)$ is a linear map from one normed space to another. Surely, its norm must be the operator norm? But you could also use the Hilbert–Schmidt norm, I suppose. – Harald Hanche-Olsen Jun 24 '14 at 07:34
  • Thanks for response, but in the paper I read, it did not talk about operator norm D: – Walter Jun 24 '14 at 08:01

1 Answers1

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The context given makes it clear they are talking about the operator norm. In this context, this can be defined as $$ \|f'(x)\|=\sup\sqrt{\frac{h(f'(x)u,f'(x)u)}{g(u,u)}}, $$ the supremum taken over all nonzero tangent vectors $u$ at $x$.

Harald Hanche-Olsen
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