# How/why does the contraction of standard volume form give the canonical form.

$$M \subset \mathbb{R}^{N}$$ is a (oriented) $$n-1$$ dimensional submanifold. Suppose $$\nu \in T_{p}M^{\bot}$$, of length one (a normal unit vector on $$M$$).

How and why does the contraction $$\nu_{\neg}(dx_{1}\wedge...\wedge dx_{n})$$ (yes the contraction symbol is reversed, sorry) give the canonical volume form, vol$$_{M}$$ (or vol$$_{g}$$ ) , on $$M$$?

There is a theorem that says that the vol$$_{g}(p)$$ can be written as $$\sqrt{\det g_{ij }}dy_{1}\wedge ... \wedge dy_{n-1}$$, where $$y_i$$ would be local coordinates of $$M$$. Do I need to do something with this. Or something with the fact that vol$$_{g}(X_1,...,X_{n-1}) = 1$$ for an oriented orthonormal basis $$\{X_1,...X_{n-1}\}$$ of $$T_{p}M$$. I'm kind of getting lost in what I can and have to use.

(The contraction was defined as $$\nu_{\neg}\alpha(v_1,...,v_{n-1}) = \alpha(\nu \wedge v_1 \wedge , ... , \wedge v_{n-1})$$, where $$\alpha$$ is an n-form.

At every point $$p\in M$$, you can extend an (oriented) orthonormal basis $$v_1,\dots,v_{n-1}$$ of $$T_pM$$ to an (oriented) orthonormal basis $$\nu,v_1,\dots,v_{n-1}$$ of $$T_p\mathbb{R}^n$$. So $$\mathrm{d}vol_M(p)=v_1^\flat\wedge\dots\wedge v_{n-1}^\flat=\iota_\nu(\nu^\flat\wedge v_1^\flat\wedge\dots\wedge v_{n-1}^\flat)=\iota_\nu\mathrm{d}vol_{\mathbb{R}^n}.$$

• what is dvol, and the superscript b, and what is $\iota_\nu$? sorry.. – AkatsukiMaliki Dec 2 '18 at 11:57
• dvol is the volume form, the superscript $\flat$ is the musical isomorphism $T_p\to T^*p$ from the Riemannian metric, and $\iota_\nu$ is the interior multiplication by $\nu$. MathJax does not have MnSymbol so there isn't a real way to get the interior product symbol (either $\llcorner$ or $\lrcorner$ orientation) to work properly. – user10354138 Dec 2 '18 at 12:00
• I got even more confused. Could you perhaps explain it without the musical isomorphism? and also argue why the form you get on the submanifold is the canonical one? – AkatsukiMaliki Dec 2 '18 at 12:06
• Let me try another way. There is (at least locally) a 1-form $\theta$ (which is morally $\nu^\flat$) with $\theta_p$ annihilates $T_pM$, and for which $\theta(\nu)=1$. Then it suffices to check $\theta\wedge\mathrm{d}vol_M=\mathrm{d}vol_{\mathbb{R}^n}$, which amounts to the same statement as $\nu,v_1,\dots,v_{n-1}$ is an oriented orthonormal basis of $\mathbb{R}^n$ at point $p$. – user10354138 Dec 2 '18 at 12:15
• is this essentially the same as saying, we have local coordinates $y_{1},...,y_{n-1}$ for $M$. so the standard volume form is $\sqrt{\det g_{ij}}dy_1\wedge , ..., \wedge dy_{n-1}$. where $\nu, y_1, ...., y_{n-1}$ is an oriented orthonormal basis for $\mathbb{R}^n$. so the volume form there is $d\nu \wedge dy_{1} \wedge , ..., \wedge dy_{n-1}$ ? and basically that $\iota_{\nu} ( d\nu \wedge dy_{1} \wedge , ..., \wedge dy_{n-1}) = \sqrt{\det g_{ij}}dy_1\wedge , ..., \wedge dy_{n-1}/$ . Is this what I have to show? – AkatsukiMaliki Dec 2 '18 at 12:29