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I read substantially the following sentence in Frankel's "Geometry of physics":

Look now at the level set $$V_{E}=\left\{(p,q)\in T^{*}M:H(p,q)=E\right\}$$ where $T^{*}M$ is the cotangent space, $p$ and $q$ respectively the momenta and the positions.

If $dH\neq 0$ on $V_{E}$, then we know that $V_{E}$ is a $2n-1$ dimensional submanifold of $T^{*}M$, called the hypersurface of constant energy $E$.

Now, I do not understand how $dH$ can be different from $0$ on $V_{E}$. What am I missing? I would say that the restriction of $dH$ to $V_{E}$ is zero by construction.

Thanks for any answer... I am sure I am missing something really stupid here...

Slz2718
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  • It's the same principle as $f(0)=0 \not\Rightarrow f'(0)=0 $: $dH$ is defined by all variations, not just those in $V_E$. – Chappers Apr 04 '16 at 18:09

1 Answers1

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I'm not sure what you mean by "the restriction of $\mathrm dH$ to $V_E$", but it sounds like you mean $\mathrm d(\text{the restriction of $H$ to $V_E$})$, which is indeed zero but isn't what is meant by "$\mathrm dH\ne0$ on $V_E$", which is the differential of $H$ as defined in all of $T^*M$, evaluated on $V_E$; this includes variations in directions away from $V_E$.

joriki
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