How/why does the contraction of standard volume form give the canonical form.
$begingroup$
$M subset mathbb{R}^{N}$ 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.
differential-geometry riemannian-geometry differential-forms orientation tangent-spaces
asked Dec 2 '18 at 11:20
AkatsukiMalikiAkatsukiMaliki
308110
$endgroup$
add a comment |
$begingroup$
$M subset mathbb{R}^{N}$ 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.
differential-geometry riemannian-geometry differential-forms orientation tangent-spaces
asked Dec 2 '18 at 11:20
AkatsukiMalikiAkatsukiMaliki
308110
$endgroup$
add a comment |
$begingroup$
$M subset mathbb{R}^{N}$ 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.
differential-geometry riemannian-geometry differential-forms orientation tangent-spaces
asked Dec 2 '18 at 11:20
AkatsukiMalikiAkatsukiMaliki
308110
$endgroup$
$M subset mathbb{R}^{N}$ 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.
differential-geometry riemannian-geometry differential-forms orientation tangent-spaces
differential-geometry riemannian-geometry differential-forms orientation tangent-spaces
asked Dec 2 '18 at 11:20
AkatsukiMalikiAkatsukiMaliki
308110
asked Dec 2 '18 at 11:20
AkatsukiMalikiAkatsukiMaliki
308110
edited Dec 2 '18 at 11:31
AkatsukiMaliki
asked Dec 2 '18 at 11:20
AkatsukiMalikiAkatsukiMaliki
308110
asked Dec 2 '18 at 11:20
AkatsukiMalikiAkatsukiMaliki
308110
asked Dec 2 '18 at 11:20
AkatsukiMalikiAkatsukiMaliki
308110
308110
add a comment |
add a comment |
1 Answer
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$begingroup$
At every point $pin M$, 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 $T_pmathbb{R}^n$. So $$mathrm{d}vol_M(p)=v_1^flatwedgedotswedge v_{n-1}^flat=iota_nu(nu^flatwedge v_1^flatwedgedotswedge v_{n-1}^flat)=iota_numathrm{d}vol_{mathbb{R}^n}.$$
answered Dec 2 '18 at 11:48
user10354138user10354138
7,3772925
$endgroup$
$begingroup$
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 $T_pto T^*p$ 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 $thetawedgemathrm{d}vol_M=mathrm{d}vol_{mathbb{R}^n}$, 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 $sqrt{det g_{ij}}dy_1wedge , ..., wedge dy_{n-1}$. where $nu, y_1, ...., y_{n-1}$ is an oriented orthonormal basis for $mathbb{R}^n$. so the volume form there is $dnu wedge dy_{1} wedge , ..., wedge dy_{n-1}$ ? and basically that $iota_{nu} ( dnu wedge dy_{1} wedge , ..., wedge dy_{n-1}) = sqrt{det g_{ij}}dy_1wedge , ..., wedge dy_{n-1}/$ . Is this what I have to show?
$endgroup$
– AkatsukiMaliki
Dec 2 '18 at 12:29
add a comment |
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$begingroup$
At every point $pin M$, 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 $T_pmathbb{R}^n$. So $$mathrm{d}vol_M(p)=v_1^flatwedgedotswedge v_{n-1}^flat=iota_nu(nu^flatwedge v_1^flatwedgedotswedge v_{n-1}^flat)=iota_numathrm{d}vol_{mathbb{R}^n}.$$
answered Dec 2 '18 at 11:48
user10354138user10354138
7,3772925
$endgroup$
$begingroup$
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 $T_pto T^*p$ 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 $thetawedgemathrm{d}vol_M=mathrm{d}vol_{mathbb{R}^n}$, 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 $sqrt{det g_{ij}}dy_1wedge , ..., wedge dy_{n-1}$. where $nu, y_1, ...., y_{n-1}$ is an oriented orthonormal basis for $mathbb{R}^n$. so the volume form there is $dnu wedge dy_{1} wedge , ..., wedge dy_{n-1}$ ? and basically that $iota_{nu} ( dnu wedge dy_{1} wedge , ..., wedge dy_{n-1}) = sqrt{det g_{ij}}dy_1wedge , ..., wedge dy_{n-1}/$ . Is this what I have to show?
$endgroup$
– AkatsukiMaliki
Dec 2 '18 at 12:29
add a comment |
$begingroup$
At every point $pin M$, 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 $T_pmathbb{R}^n$. So $$mathrm{d}vol_M(p)=v_1^flatwedgedotswedge v_{n-1}^flat=iota_nu(nu^flatwedge v_1^flatwedgedotswedge v_{n-1}^flat)=iota_numathrm{d}vol_{mathbb{R}^n}.$$
answered Dec 2 '18 at 11:48
user10354138user10354138
7,3772925
$endgroup$
$begingroup$
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 $T_pto T^*p$ 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 $thetawedgemathrm{d}vol_M=mathrm{d}vol_{mathbb{R}^n}$, 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 $sqrt{det g_{ij}}dy_1wedge , ..., wedge dy_{n-1}$. where $nu, y_1, ...., y_{n-1}$ is an oriented orthonormal basis for $mathbb{R}^n$. so the volume form there is $dnu wedge dy_{1} wedge , ..., wedge dy_{n-1}$ ? and basically that $iota_{nu} ( dnu wedge dy_{1} wedge , ..., wedge dy_{n-1}) = sqrt{det g_{ij}}dy_1wedge , ..., wedge dy_{n-1}/$ . Is this what I have to show?
$endgroup$
– AkatsukiMaliki
Dec 2 '18 at 12:29
add a comment |
$begingroup$
At every point $pin M$, 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 $T_pmathbb{R}^n$. So $$mathrm{d}vol_M(p)=v_1^flatwedgedotswedge v_{n-1}^flat=iota_nu(nu^flatwedge v_1^flatwedgedotswedge v_{n-1}^flat)=iota_numathrm{d}vol_{mathbb{R}^n}.$$
answered Dec 2 '18 at 11:48
user10354138user10354138
7,3772925
$endgroup$
At every point $pin M$, 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 $T_pmathbb{R}^n$. So $$mathrm{d}vol_M(p)=v_1^flatwedgedotswedge v_{n-1}^flat=iota_nu(nu^flatwedge v_1^flatwedgedotswedge v_{n-1}^flat)=iota_numathrm{d}vol_{mathbb{R}^n}.$$
answered Dec 2 '18 at 11:48
user10354138user10354138
7,3772925
answered Dec 2 '18 at 11:48
user10354138user10354138
7,3772925
answered Dec 2 '18 at 11:48
user10354138user10354138
7,3772925
answered Dec 2 '18 at 11:48
user10354138user10354138
7,3772925
7,3772925
$begingroup$
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 $T_pto T^*p$ 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 $thetawedgemathrm{d}vol_M=mathrm{d}vol_{mathbb{R}^n}$, 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 $sqrt{det g_{ij}}dy_1wedge , ..., wedge dy_{n-1}$. where $nu, y_1, ...., y_{n-1}$ is an oriented orthonormal basis for $mathbb{R}^n$. so the volume form there is $dnu wedge dy_{1} wedge , ..., wedge dy_{n-1}$ ? and basically that $iota_{nu} ( dnu wedge dy_{1} wedge , ..., wedge dy_{n-1}) = sqrt{det g_{ij}}dy_1wedge , ..., wedge dy_{n-1}/$ . Is this what I have to show?
$endgroup$
– AkatsukiMaliki
Dec 2 '18 at 12:29
add a comment |
$begingroup$
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 $T_pto T^*p$ 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 $thetawedgemathrm{d}vol_M=mathrm{d}vol_{mathbb{R}^n}$, 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 $sqrt{det g_{ij}}dy_1wedge , ..., wedge dy_{n-1}$. where $nu, y_1, ...., y_{n-1}$ is an oriented orthonormal basis for $mathbb{R}^n$. so the volume form there is $dnu wedge dy_{1} wedge , ..., wedge dy_{n-1}$ ? and basically that $iota_{nu} ( dnu wedge dy_{1} wedge , ..., wedge dy_{n-1}) = sqrt{det g_{ij}}dy_1wedge , ..., wedge dy_{n-1}/$ . Is this what I have to show?
$endgroup$
– AkatsukiMaliki
Dec 2 '18 at 12:29
$begingroup$
what is dvol, and the superscript b, and what is $iota_nu$? sorry..
$endgroup$
– AkatsukiMaliki
Dec 2 '18 at 11:57
$begingroup$
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 $T_pto T^*p$ 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
$begingroup$
dvol is the volume form, the superscript $flat$ is the musical isomorphism $T_pto T^*p$ 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
$begingroup$
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 $thetawedgemathrm{d}vol_M=mathrm{d}vol_{mathbb{R}^n}$, 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
$begingroup$
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 $thetawedgemathrm{d}vol_M=mathrm{d}vol_{mathbb{R}^n}$, 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 $sqrt{det g_{ij}}dy_1wedge , ..., wedge dy_{n-1}$. where $nu, y_1, ...., y_{n-1}$ is an oriented orthonormal basis for $mathbb{R}^n$. so the volume form there is $dnu wedge dy_{1} wedge , ..., wedge dy_{n-1}$ ? and basically that $iota_{nu} ( dnu wedge dy_{1} wedge , ..., wedge dy_{n-1}) = sqrt{det g_{ij}}dy_1wedge , ..., wedge dy_{n-1}/$ . Is this what I have to show?
$endgroup$
– AkatsukiMaliki
Dec 2 '18 at 12:29
$begingroup$
is this essentially the same as saying, we have local coordinates $y_{1},...,y_{n-1}$ for $M$. so the standard volume form is $sqrt{det g_{ij}}dy_1wedge , ..., wedge dy_{n-1}$. where $nu, y_1, ...., y_{n-1}$ is an oriented orthonormal basis for $mathbb{R}^n$. so the volume form there is $dnu wedge dy_{1} wedge , ..., wedge dy_{n-1}$ ? and basically that $iota_{nu} ( dnu wedge dy_{1} wedge , ..., wedge dy_{n-1}) = sqrt{det g_{ij}}dy_1wedge , ..., wedge dy_{n-1}/$ . Is this what I have to show?
$endgroup$
– AkatsukiMaliki
Dec 2 '18 at 12:29
add a comment |
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