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Let $H(s,t)$ be a variation of a curve $c: [a,b]\to M$. Let $V(s,t):= \frac{\partial H}{\partial t}(s,t)$, be the vector field along the map $H$. Furthermore let $\overline{D}_{\frac{\partial}{\partial t}}V$ be the covariant derivate of $V(s,t)$ in direction $t \mapsto H(s,t)$. Why do we have in this case: $$\langle\overline{D}_{\frac{\partial}{\partial t}}V,\overline{D}_{\frac{\partial}{\partial t}}V\rangle =-\langle \overline{D}^2_{\frac{\partial}{\partial t}}V,V \rangle $$

($\langle,\rangle$, denotes here a Riemann metric)

I thought, this could follow from the metric property, but then (when I did no mistake) I should have that $\frac{\partial }{\partial t}\langle V,\overline{D}_{\frac{\partial}{\partial t}}V\rangle =0$, But I don't see, why this should be the case..

Many thanks for some help!

Didier
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Frederick Manfred
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    The expression looks like it comes out taking the derivative of the norm of the vector through dot product deftn twice – tryst with freedom Jun 20 '22 at 12:12
  • I removed the abstract-algebra tag (it is unrelevant) and edited your <,> with \langle , \rangle to produce the right typesetting ($\langle\cdot , \cdot \rangle$ instead of $<\cdot ,\cdot>$). Regarding your question: this is not true for an arbitrary $V$, you have to assume something like $|V|\equiv c$ or something, and it comes from applying $\partial/\partial t$ twice to $|V|^2 = \text{cste}$ – Didier Jun 20 '22 at 12:34

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