# Restricting a measure to a codimension 1 submanifold

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 $$\sqrt{2E}$$.

• 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.) Aug 31, 2022 at 2:41
• 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." Aug 31, 2022 at 5:16
• 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 $$\int_{-\infty}^\infty\text{d}q\int_{-\infty}^\infty\text{d}p\delta(q^2+p^2-1)f(q)=\int_{-1}^1\frac{\text{d}q}{\sqrt{1-q^2}}f(q).$$ Aug 31, 2022 at 5:23
• 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. Aug 31, 2022 at 5:27

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 $$\iota_X\omega|_{\Sigma}$$.
• 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 $X$? Aug 31, 2022 at 16:35
• @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. Aug 31, 2022 at 20:39
• Amazing! Thank you so much @Kajelad! I'll try to work on showing the equivalence of $\mu$ and the measure induced by $\iota_X\omega|_\Sigma$. Aug 31, 2022 at 20:53