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Suppose you have a manifold $M$ with a volume form $\omega$. Let $f\in C^\infty(M)$ with a regular value $0$. Consider the codimension 1 submanifold $\Sigma=f^{-1}(\{0\})$. Intuitively, one could define the volume form $\omega_{\Sigma}=\omega\delta(f)$ on $\Sigma$. Is there a more geometric way of understanding this measure? For example, is there a natural way of writing $\omega_\Sigma=\iota_X\omega$, for some vector field $X\in\mathfrak{X}(M)$.

This sort of thing appears in the microcanonical analysis of continuous systems. For example, the measure appearing in the treatment of the classical harmonic oscillator https://physics.stackexchange.com/questions/406972/harmonic-oscillator-in-microcanonical-ensemble. In there one has $$\text{d}q\text{d}p\delta\left({\frac{p^2}{2}+\frac{q^2}{2}-E}\right)=\frac{2\text{d}q}{\sqrt{2E-q^2}},$$ where the $q$ on the right-hand side is actually the pullback of the coordinate $q$ on phase space $\mathbb{R}^2$ to the circle centered at the origin of radius $\sqrt{2E}$.

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  • $\begingroup$ Precisely: Contract with the unit normal to the hypersurface. You need a Riemannian metric compatible with your volume form. (I have no idea what $\omega\delta(f)$ is supposed to mean.) $\endgroup$ Aug 31, 2022 at 2:41
  • $\begingroup$ Thank you for your comment Prof. Shiffrin! My question is precisely in the setting where there is no Riemannian metric. In the example above, we have a symplectic manifold $\mathbb{R}^2$ with cartesian coordinates $(q,p)$ and symplectic form $\omega=\text{d}q\text{d}p$. We want to induce a volume form on the circle defined by $p^2+q^2=1$. But, instead of considering the volume form coming from the Riemannian metric on $\mathbb{R}^2$, we want to consider the measure supported on the circle given by $\text{d}q\text{d}p\delta(q^2+p^2-1)$, with $\delta$ the "Dirac delta function." $\endgroup$ Aug 31, 2022 at 5:16
  • $\begingroup$ My question is precisely on how one can make sense of this measure in a geometric way. The only way that I currently understand it is interpreting integrals with respect to this measure as iterated integrals. Fixing $q$, one can resolve the delta function as $\delta(q^2+p^2-1)=\frac{1}{2\sqrt{1-q^2}}(\delta(p-\sqrt{1-q^2})+\delta(p+\sqrt{1-q^2}))\chi_{[-1,1]}(q)$. Integrating with respect to $p$ we have $$\int_{-\infty}^\infty\text{d}q\int_{-\infty}^\infty\text{d}p\delta(q^2+p^2-1)f(q)=\int_{-1}^1\frac{\text{d}q}{\sqrt{1-q^2}}f(q).$$ $\endgroup$ Aug 31, 2022 at 5:23
  • $\begingroup$ Of course, the resulting measure coincides with the one coming from the Riemannian metric, as seen from the substitution $q=\cos(\theta)$. However, this procedure seems decidedly different from using the metric $\text{d}s^2=\text{d}q^2+\text{d}p^2$. This is particularly important in physics because this metric does not make sense in that setting. Indeed, $q$ and $p$ do not have the same units and thus this combination does not make sense. $\endgroup$ Aug 31, 2022 at 5:27

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Choose a (local) vector field $X$ such that $X(f)=1$. The interior product $\iota_X\omega$ will depend on the choice of $X$, but its pullback to $\Sigma$ will not.

This volume form is related to the induced measure on the "infinitesimally fattening of $\Sigma$" $f^{-1}([0,\epsilon))$. Each such vector field $X$ defines a diffeomorphism $\varphi:\Sigma\times[0,\epsilon)\to f^{-1}([0,\epsilon))$ where the coordinate on the second factor is exactly $f$ (the flowout of $\Sigma$ along $X$). We can define a measure $\mu$ on $\Sigma$ by $$ \mu(U)=\lim_{\delta\to 0}\frac{1}{\delta}\mu_\omega(\varphi(U\times[0,\delta))) $$ Where $U\subseteq\Sigma$ is open and $\mu_\omega$ is the measure on $M$ induced by $\omega$. It turns out this limit does not depend on the choice of $X$, and is equivalent to the measure induced by $\iota_X\omega|_{\Sigma}$.

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  • $\begingroup$ Interesting! Let me try it out in the example above to get a feeling for how it works. $\endgroup$ Aug 31, 2022 at 14:29
  • $\begingroup$ This worked wonderfully! It is so cool! Could you give me a hint towards why this works? Namely, how is it related to the delta function intuition and why is it independent of the choice of $X$? $\endgroup$ Aug 31, 2022 at 16:35
  • $\begingroup$ @IvanBurbano I've added a bit more detail. The idea is that, given $U\subseteq\Sigma$, we can "fatten" $U$ to a subset of $f^{-1}([0,\epsilon))$, and, to first order in $\epsilon$, the volume of the "fattening" of $U$ does not depend on the choice of vector field. $\endgroup$
    – Kajelad
    Aug 31, 2022 at 20:39
  • $\begingroup$ Amazing! Thank you so much @Kajelad! I'll try to work on showing the equivalence of $\mu$ and the measure induced by $\iota_X\omega|_\Sigma$. $\endgroup$ Aug 31, 2022 at 20:53

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