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$M \subset \mathbb{R}^{N}$ 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.

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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 $$\mathrm{d}vol_M(p)=v_1^\flat\wedge\dots\wedge v_{n-1}^\flat=\iota_\nu(\nu^\flat\wedge v_1^\flat\wedge\dots\wedge v_{n-1}^\flat)=\iota_\nu\mathrm{d}vol_{\mathbb{R}^n}.$$

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  • $\begingroup$ what is dvol, and the superscript b, and what is $\iota_\nu$? sorry.. $\endgroup$ – AkatsukiMaliki Dec 2 '18 at 11:57
  • $\begingroup$ 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. $\endgroup$ – user10354138 Dec 2 '18 at 12:00
  • $\begingroup$ 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? $\endgroup$ – AkatsukiMaliki Dec 2 '18 at 12:06
  • $\begingroup$ 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 $p$. $\endgroup$ – user10354138 Dec 2 '18 at 12:15
  • $\begingroup$ 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? $\endgroup$ – AkatsukiMaliki Dec 2 '18 at 12:29

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