Mixing
Characterizations of Riemannian Volume Form
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I'm trying to understand how some characterizations of the Riemannian volume form $dV$ are equivalent on an oriented Riemannian manifold of dimension $n$. I'm a bit new to Riemannian geometry (and self-studying it), and I can tell that it should just come down to a straightforward calculation, but I'm stuck for some reason.
Characterization 1: If $(omega^1,...,omega^n)$ is a local oriented orthonormal coframe for the cotangent bundle, then $dV=omega^1wedge...wedgeomega^n$.
Characterization 2: If $(y^1,...,y^n)$ are oriented local coordinates, then $dV=sqrt{det(g_{ij})}dy^1wedge...wedge dy^n$ where $g_{ij}$ is the representation of the Riemannian metric in local coordinates.
I'm guessing that the square root of the determinant factor shows up from something involving $det(A)=sqrt{det(A A^t)}$ for a suitable matrix $A$. I have a hunch that using uniqueness of the form (which follows for e.g. characterization 1 from the fact that you can cover the manifold with charts that have local oriented orthonormal coframes) might be important? Any help is very much appreciated!
differential-geometry riemannian-geometry differential-forms
asked Feb 2 at 15:52
BenBen
776
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add a comment |
$begingroup$
I'm trying to understand how some characterizations of the Riemannian volume form $dV$ are equivalent on an oriented Riemannian manifold of dimension $n$. I'm a bit new to Riemannian geometry (and self-studying it), and I can tell that it should just come down to a straightforward calculation, but I'm stuck for some reason.
Characterization 1: If $(omega^1,...,omega^n)$ is a local oriented orthonormal coframe for the cotangent bundle, then $dV=omega^1wedge...wedgeomega^n$.
Characterization 2: If $(y^1,...,y^n)$ are oriented local coordinates, then $dV=sqrt{det(g_{ij})}dy^1wedge...wedge dy^n$ where $g_{ij}$ is the representation of the Riemannian metric in local coordinates.
I'm guessing that the square root of the determinant factor shows up from something involving $det(A)=sqrt{det(A A^t)}$ for a suitable matrix $A$. I have a hunch that using uniqueness of the form (which follows for e.g. characterization 1 from the fact that you can cover the manifold with charts that have local oriented orthonormal coframes) might be important? Any help is very much appreciated!
differential-geometry riemannian-geometry differential-forms
asked Feb 2 at 15:52
BenBen
776
$endgroup$
$begingroup$
Hint: If your $partial_{y_i}$ are orthogonal then the metric is represented by a diagonal matrix with the lengths of the $partial_{y_i}$ on the diagonal. What is the determinant, then?
$endgroup$
– Angelo Brillante Romeo
Feb 2 at 16:18
$begingroup$
You can only guarantee the existence of an orthonormal coordinate frame when the metric is flat though, right?
$endgroup$
– Ben
Feb 2 at 16:22
$begingroup$
Locally (i.e. in a small enough open neighborhood of a point) you can always find such frame!
$endgroup$
– Angelo Brillante Romeo
Feb 2 at 16:38
2
$begingroup$
@AngeloBrillanteRomeo OP is referring to an orthonormal coordinate frame.
$endgroup$
– Amitai Yuval
Feb 2 at 16:43
$begingroup$
@AmitaiYuval Oh, sorry!
$endgroup$
– Angelo Brillante Romeo
Feb 2 at 16:56
add a comment |
$begingroup$
I'm trying to understand how some characterizations of the Riemannian volume form $dV$ are equivalent on an oriented Riemannian manifold of dimension $n$. I'm a bit new to Riemannian geometry (and self-studying it), and I can tell that it should just come down to a straightforward calculation, but I'm stuck for some reason.
Characterization 1: If $(omega^1,...,omega^n)$ is a local oriented orthonormal coframe for the cotangent bundle, then $dV=omega^1wedge...wedgeomega^n$.
Characterization 2: If $(y^1,...,y^n)$ are oriented local coordinates, then $dV=sqrt{det(g_{ij})}dy^1wedge...wedge dy^n$ where $g_{ij}$ is the representation of the Riemannian metric in local coordinates.
I'm guessing that the square root of the determinant factor shows up from something involving $det(A)=sqrt{det(A A^t)}$ for a suitable matrix $A$. I have a hunch that using uniqueness of the form (which follows for e.g. characterization 1 from the fact that you can cover the manifold with charts that have local oriented orthonormal coframes) might be important? Any help is very much appreciated!
differential-geometry riemannian-geometry differential-forms
asked Feb 2 at 15:52
BenBen
776
$endgroup$
I'm trying to understand how some characterizations of the Riemannian volume form $dV$ are equivalent on an oriented Riemannian manifold of dimension $n$. I'm a bit new to Riemannian geometry (and self-studying it), and I can tell that it should just come down to a straightforward calculation, but I'm stuck for some reason.
Characterization 1: If $(omega^1,...,omega^n)$ is a local oriented orthonormal coframe for the cotangent bundle, then $dV=omega^1wedge...wedgeomega^n$.
Characterization 2: If $(y^1,...,y^n)$ are oriented local coordinates, then $dV=sqrt{det(g_{ij})}dy^1wedge...wedge dy^n$ where $g_{ij}$ is the representation of the Riemannian metric in local coordinates.
I'm guessing that the square root of the determinant factor shows up from something involving $det(A)=sqrt{det(A A^t)}$ for a suitable matrix $A$. I have a hunch that using uniqueness of the form (which follows for e.g. characterization 1 from the fact that you can cover the manifold with charts that have local oriented orthonormal coframes) might be important? Any help is very much appreciated!
differential-geometry riemannian-geometry differential-forms
differential-geometry riemannian-geometry differential-forms
asked Feb 2 at 15:52
BenBen
776
asked Feb 2 at 15:52
BenBen
776
asked Feb 2 at 15:52
BenBen
776
asked Feb 2 at 15:52
BenBen
776
asked Feb 2 at 15:52
BenBen
776
776
$begingroup$
Hint: If your $partial_{y_i}$ are orthogonal then the metric is represented by a diagonal matrix with the lengths of the $partial_{y_i}$ on the diagonal. What is the determinant, then?
$endgroup$
– Angelo Brillante Romeo
Feb 2 at 16:18
$begingroup$
You can only guarantee the existence of an orthonormal coordinate frame when the metric is flat though, right?
$endgroup$
– Ben
Feb 2 at 16:22
$begingroup$
Locally (i.e. in a small enough open neighborhood of a point) you can always find such frame!
$endgroup$
– Angelo Brillante Romeo
Feb 2 at 16:38
2
$begingroup$
@AngeloBrillanteRomeo OP is referring to an orthonormal coordinate frame.
$endgroup$
– Amitai Yuval
Feb 2 at 16:43
$begingroup$
@AmitaiYuval Oh, sorry!
$endgroup$
– Angelo Brillante Romeo
Feb 2 at 16:56
add a comment |
$begingroup$
Hint: If your $partial_{y_i}$ are orthogonal then the metric is represented by a diagonal matrix with the lengths of the $partial_{y_i}$ on the diagonal. What is the determinant, then?
$endgroup$
– Angelo Brillante Romeo
Feb 2 at 16:18
$begingroup$
You can only guarantee the existence of an orthonormal coordinate frame when the metric is flat though, right?
$endgroup$
– Ben
Feb 2 at 16:22
$begingroup$
Locally (i.e. in a small enough open neighborhood of a point) you can always find such frame!
$endgroup$
– Angelo Brillante Romeo
Feb 2 at 16:38
2
$begingroup$
@AngeloBrillanteRomeo OP is referring to an orthonormal coordinate frame.
$endgroup$
– Amitai Yuval
Feb 2 at 16:43
$begingroup$
@AmitaiYuval Oh, sorry!
$endgroup$
– Angelo Brillante Romeo
Feb 2 at 16:56
$begingroup$
Hint: If your $partial_{y_i}$ are orthogonal then the metric is represented by a diagonal matrix with the lengths of the $partial_{y_i}$ on the diagonal. What is the determinant, then?
$endgroup$
– Angelo Brillante Romeo
Feb 2 at 16:18
$begingroup$
Hint: If your $partial_{y_i}$ are orthogonal then the metric is represented by a diagonal matrix with the lengths of the $partial_{y_i}$ on the diagonal. What is the determinant, then?
$endgroup$
– Angelo Brillante Romeo
Feb 2 at 16:18
$begingroup$
You can only guarantee the existence of an orthonormal coordinate frame when the metric is flat though, right?
$endgroup$
– Ben
Feb 2 at 16:22
$begingroup$
You can only guarantee the existence of an orthonormal coordinate frame when the metric is flat though, right?
$endgroup$
– Ben
Feb 2 at 16:22
$begingroup$
Locally (i.e. in a small enough open neighborhood of a point) you can always find such frame!
$endgroup$
– Angelo Brillante Romeo
Feb 2 at 16:38
$begingroup$
Locally (i.e. in a small enough open neighborhood of a point) you can always find such frame!
$endgroup$
– Angelo Brillante Romeo
Feb 2 at 16:38
2
2
$begingroup$
@AngeloBrillanteRomeo OP is referring to an orthonormal coordinate frame.
$endgroup$
– Amitai Yuval
Feb 2 at 16:43
$begingroup$
@AngeloBrillanteRomeo OP is referring to an orthonormal coordinate frame.
$endgroup$
– Amitai Yuval
Feb 2 at 16:43
$begingroup$
@AmitaiYuval Oh, sorry!
$endgroup$
– Angelo Brillante Romeo
Feb 2 at 16:56
$begingroup$
@AmitaiYuval Oh, sorry!
$endgroup$
– Angelo Brillante Romeo
Feb 2 at 16:56
add a comment |
1 Answer
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Let $dV$ denote the Riemannian volume form given by your first characterization, and let $y^1,ldots,y^n$ be oriented local coordinates. Let $frac{partial}{partial y^1},ldots,frac{partial}{partial y^n}$ denote the induced local frame of the tangent bundle, and let $e_1,ldots,e_n$ be a local orthonormal frame of the tangent bundle. By definition, the matrix representation of the metric $g$ with respect to $e_1,ldots,e_n$ at any point is the identity matrix. The matrix representation of $g$ with respect to $frac{partial}{partial y^1},ldots,frac{partial}{partial y^n}$ is $(g_{ij})$.
Now, for a point $p$, let $A_p$ denote the transition matrix of the two above mentioned frames of the tangent space at $p$. Then it follows from characterization $1$ that $$dVleft(frac{partial}{partial y^1},ldots,frac{partial}{partial y^n}right)_p=det A_pcdot dV(e_1,ldots,e_n)=det A_p.$$On the other hand, by construction we have $$(g_{ij})_p=A_p^Tcdot Icdot A_p=A_p^TA_p,$$and hence, $$det A_p=sqrt{det(g_{ij})},$$ and it follows that $dV$ coincides with the volume form of your second characterization.
answered Feb 2 at 16:42
Amitai YuvalAmitai Yuval
15.6k11127
$endgroup$
$begingroup$
Thanks! This was exactly what I was looking for.
$endgroup$
– Ben
Feb 2 at 21:59
add a comment |
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Let $dV$ denote the Riemannian volume form given by your first characterization, and let $y^1,ldots,y^n$ be oriented local coordinates. Let $frac{partial}{partial y^1},ldots,frac{partial}{partial y^n}$ denote the induced local frame of the tangent bundle, and let $e_1,ldots,e_n$ be a local orthonormal frame of the tangent bundle. By definition, the matrix representation of the metric $g$ with respect to $e_1,ldots,e_n$ at any point is the identity matrix. The matrix representation of $g$ with respect to $frac{partial}{partial y^1},ldots,frac{partial}{partial y^n}$ is $(g_{ij})$.
Now, for a point $p$, let $A_p$ denote the transition matrix of the two above mentioned frames of the tangent space at $p$. Then it follows from characterization $1$ that $$dVleft(frac{partial}{partial y^1},ldots,frac{partial}{partial y^n}right)_p=det A_pcdot dV(e_1,ldots,e_n)=det A_p.$$On the other hand, by construction we have $$(g_{ij})_p=A_p^Tcdot Icdot A_p=A_p^TA_p,$$and hence, $$det A_p=sqrt{det(g_{ij})},$$ and it follows that $dV$ coincides with the volume form of your second characterization.
answered Feb 2 at 16:42
Amitai YuvalAmitai Yuval
15.6k11127
$endgroup$
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Thanks! This was exactly what I was looking for.
$endgroup$
– Ben
Feb 2 at 21:59
add a comment |
$begingroup$
Let $dV$ denote the Riemannian volume form given by your first characterization, and let $y^1,ldots,y^n$ be oriented local coordinates. Let $frac{partial}{partial y^1},ldots,frac{partial}{partial y^n}$ denote the induced local frame of the tangent bundle, and let $e_1,ldots,e_n$ be a local orthonormal frame of the tangent bundle. By definition, the matrix representation of the metric $g$ with respect to $e_1,ldots,e_n$ at any point is the identity matrix. The matrix representation of $g$ with respect to $frac{partial}{partial y^1},ldots,frac{partial}{partial y^n}$ is $(g_{ij})$.
Now, for a point $p$, let $A_p$ denote the transition matrix of the two above mentioned frames of the tangent space at $p$. Then it follows from characterization $1$ that $$dVleft(frac{partial}{partial y^1},ldots,frac{partial}{partial y^n}right)_p=det A_pcdot dV(e_1,ldots,e_n)=det A_p.$$On the other hand, by construction we have $$(g_{ij})_p=A_p^Tcdot Icdot A_p=A_p^TA_p,$$and hence, $$det A_p=sqrt{det(g_{ij})},$$ and it follows that $dV$ coincides with the volume form of your second characterization.
answered Feb 2 at 16:42
Amitai YuvalAmitai Yuval
15.6k11127
$endgroup$
$begingroup$
Thanks! This was exactly what I was looking for.
$endgroup$
– Ben
Feb 2 at 21:59
add a comment |
$begingroup$
Let $dV$ denote the Riemannian volume form given by your first characterization, and let $y^1,ldots,y^n$ be oriented local coordinates. Let $frac{partial}{partial y^1},ldots,frac{partial}{partial y^n}$ denote the induced local frame of the tangent bundle, and let $e_1,ldots,e_n$ be a local orthonormal frame of the tangent bundle. By definition, the matrix representation of the metric $g$ with respect to $e_1,ldots,e_n$ at any point is the identity matrix. The matrix representation of $g$ with respect to $frac{partial}{partial y^1},ldots,frac{partial}{partial y^n}$ is $(g_{ij})$.
Now, for a point $p$, let $A_p$ denote the transition matrix of the two above mentioned frames of the tangent space at $p$. Then it follows from characterization $1$ that $$dVleft(frac{partial}{partial y^1},ldots,frac{partial}{partial y^n}right)_p=det A_pcdot dV(e_1,ldots,e_n)=det A_p.$$On the other hand, by construction we have $$(g_{ij})_p=A_p^Tcdot Icdot A_p=A_p^TA_p,$$and hence, $$det A_p=sqrt{det(g_{ij})},$$ and it follows that $dV$ coincides with the volume form of your second characterization.
answered Feb 2 at 16:42
Amitai YuvalAmitai Yuval
15.6k11127
$endgroup$
Let $dV$ denote the Riemannian volume form given by your first characterization, and let $y^1,ldots,y^n$ be oriented local coordinates. Let $frac{partial}{partial y^1},ldots,frac{partial}{partial y^n}$ denote the induced local frame of the tangent bundle, and let $e_1,ldots,e_n$ be a local orthonormal frame of the tangent bundle. By definition, the matrix representation of the metric $g$ with respect to $e_1,ldots,e_n$ at any point is the identity matrix. The matrix representation of $g$ with respect to $frac{partial}{partial y^1},ldots,frac{partial}{partial y^n}$ is $(g_{ij})$.
Now, for a point $p$, let $A_p$ denote the transition matrix of the two above mentioned frames of the tangent space at $p$. Then it follows from characterization $1$ that $$dVleft(frac{partial}{partial y^1},ldots,frac{partial}{partial y^n}right)_p=det A_pcdot dV(e_1,ldots,e_n)=det A_p.$$On the other hand, by construction we have $$(g_{ij})_p=A_p^Tcdot Icdot A_p=A_p^TA_p,$$and hence, $$det A_p=sqrt{det(g_{ij})},$$ and it follows that $dV$ coincides with the volume form of your second characterization.
answered Feb 2 at 16:42
Amitai YuvalAmitai Yuval
15.6k11127
answered Feb 2 at 16:42
Amitai YuvalAmitai Yuval
15.6k11127
answered Feb 2 at 16:42
Amitai YuvalAmitai Yuval
15.6k11127
answered Feb 2 at 16:42
Amitai YuvalAmitai Yuval
15.6k11127
15.6k11127
$begingroup$
Thanks! This was exactly what I was looking for.
$endgroup$
– Ben
Feb 2 at 21:59
add a comment |
$begingroup$
Thanks! This was exactly what I was looking for.
$endgroup$
– Ben
Feb 2 at 21:59
$begingroup$
Thanks! This was exactly what I was looking for.
$endgroup$
– Ben
Feb 2 at 21:59
$begingroup$
Thanks! This was exactly what I was looking for.
$endgroup$
– Ben
Feb 2 at 21:59
add a comment |
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Hint: If your $partial_{y_i}$ are orthogonal then the metric is represented by a diagonal matrix with the lengths of the $partial_{y_i}$ on the diagonal. What is the determinant, then?
$endgroup$
– Angelo Brillante Romeo
Feb 2 at 16:18
$begingroup$
You can only guarantee the existence of an orthonormal coordinate frame when the metric is flat though, right?
$endgroup$
– Ben
Feb 2 at 16:22
$begingroup$
Locally (i.e. in a small enough open neighborhood of a point) you can always find such frame!
$endgroup$
– Angelo Brillante Romeo
Feb 2 at 16:38
2
$begingroup$
@AngeloBrillanteRomeo OP is referring to an orthonormal coordinate frame.
$endgroup$
– Amitai Yuval
Feb 2 at 16:43
$begingroup$
@AmitaiYuval Oh, sorry!
$endgroup$
– Angelo Brillante Romeo
Feb 2 at 16:56