How can I calculate the derivative of $\| \mathrm{d}f \|^2_h$?

Asked May 31 '21 at 12:38

Active May 31 '21 at 15:22

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The present question refers to one of the answers of this other one.

Let $h=h_{ij}dx^i \otimes dx^j$ be a Riemannian metric.

(1) Why does the inverse $(h_{ij})^{-1}=h^{ij}$ appear here: $$\| \mathrm{d}f \|^2_h = h^{-1}(\mathrm{d}f ,\mathrm{d}f) = f_{x_i}f_{x_j}h^{ij}=\langle \nabla f , \nabla f \rangle_h ?$$

(2) Putting $f(x,t) = f(x) + t g(x)$, how can I calculate $\left. \frac{d}{dt} \right|_{t=0} \| \mathrm{d}f \|^2_h$?

Thank you in advance.

edited May 31 '21 at 15:22

Didier

19,132

asked May 31 '21 at 12:38

Alice

2

(1) is just definition. For (2), what did you try? – Arctic Char May 31 '21 at 12:40
(1) Why not $h(df,df)$?, (2) $\left. \frac{d}{dt} \right|_{t=0} | \mathrm{d}f |^2_h = 2\langle df, dg \rangle_h$, but I wonder if $h^{ij}$ shouldn't play a role in it... – Alice May 31 '21 at 15:49
Are you assuming that $f : M\to \mathbb R$ or $f: M \to N$? – Arctic Char May 31 '21 at 16:41
I'm assuming $f$ and $g$ are smooth functions on M, $f,g:M \to \mathbb{R}$. – Alice May 31 '21 at 17:16
1

(1) You can't define $h$ on the cotangent space. (2) you are correct. $h$ depends only on $x$ and not on $f$. – Arctic Char May 31 '21 at 18:27

How can I calculate the derivative of $\| \mathrm{d}f \|^2_h$?

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