Degrees of freedom of Riemann curvature tensor
$begingroup$
I know the argument that uses the symmetries
- $R_{a b c d} = -R_{b a c d} = R_{c d a b}$
- $R_{a b c d} + R_{b c a d} + R_{c a b d} = 0$
of the Riemann curvature tensor $R$ of an $n$-dimensional Riemannian manifold to show that it (the curvature tensor) has at most $frac1{12} n^2 (n^2 - 1)$ degrees of freedom (see for example here, section 7) at a given point.
As far as I know $frac1{12} n^2 (n^2 - 1)$ is the actual number of degrees of freedom, so my question is how does one get the lower bound?
EDIT: To clarify: The argument I linked to shows that using the mentioned symmetries $frac1{12}n^2 (n^2 - 1)$ of the $n^4$ entries of $R$ already determine it. However there could be other constraints like symmetries or inequalities that make the actual number of degrees of freedom even lower. As far as I know, this is not the case and my but I would like a proof of this.
(The only wat to prove this I can think of would be constructing a family of manifolds parametrized by $mathbb{R}^{n^2 (n^2 - 1)/12}$ such that for a given $p in mathbb{R}^{n^2 (n^2 - 1)/12}$ the corresponding manifold has the curvature tensor corresponding to $p$ at some point.)
differential-geometry riemannian-geometry tensors curvature general-relativity
asked Dec 30 '18 at 0:17
0x5390x539
1,445518
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show 2 more comments
$begingroup$
I know the argument that uses the symmetries
- $R_{a b c d} = -R_{b a c d} = R_{c d a b}$
- $R_{a b c d} + R_{b c a d} + R_{c a b d} = 0$
of the Riemann curvature tensor $R$ of an $n$-dimensional Riemannian manifold to show that it (the curvature tensor) has at most $frac1{12} n^2 (n^2 - 1)$ degrees of freedom (see for example here, section 7) at a given point.
As far as I know $frac1{12} n^2 (n^2 - 1)$ is the actual number of degrees of freedom, so my question is how does one get the lower bound?
EDIT: To clarify: The argument I linked to shows that using the mentioned symmetries $frac1{12}n^2 (n^2 - 1)$ of the $n^4$ entries of $R$ already determine it. However there could be other constraints like symmetries or inequalities that make the actual number of degrees of freedom even lower. As far as I know, this is not the case and my but I would like a proof of this.
(The only wat to prove this I can think of would be constructing a family of manifolds parametrized by $mathbb{R}^{n^2 (n^2 - 1)/12}$ such that for a given $p in mathbb{R}^{n^2 (n^2 - 1)/12}$ the corresponding manifold has the curvature tensor corresponding to $p$ at some point.)
differential-geometry riemannian-geometry tensors curvature general-relativity
asked Dec 30 '18 at 0:17
0x5390x539
1,445518
$endgroup$
$begingroup$
lower bound on what ?
$endgroup$
– Amr
Dec 30 '18 at 2:11
$begingroup$
@Amr on the degrees of freedom.
$endgroup$
– 0x539
Dec 30 '18 at 2:25
$begingroup$
Wouldn't $frac{1}{12}n^2(n^2-1)$ count as a lower bound already ?
$endgroup$
– Amr
Dec 30 '18 at 3:08
$begingroup$
@Amr Yes it is, but i want to know why.
$endgroup$
– 0x539
Dec 30 '18 at 3:19
2
$begingroup$
Duplicate of math.stackexchange.com/questions/2235253/… (no vote to close due to open bounty)
$endgroup$
– Dap
Jan 16 at 11:12
|
show 2 more comments
$begingroup$
I know the argument that uses the symmetries
- $R_{a b c d} = -R_{b a c d} = R_{c d a b}$
- $R_{a b c d} + R_{b c a d} + R_{c a b d} = 0$
of the Riemann curvature tensor $R$ of an $n$-dimensional Riemannian manifold to show that it (the curvature tensor) has at most $frac1{12} n^2 (n^2 - 1)$ degrees of freedom (see for example here, section 7) at a given point.
As far as I know $frac1{12} n^2 (n^2 - 1)$ is the actual number of degrees of freedom, so my question is how does one get the lower bound?
EDIT: To clarify: The argument I linked to shows that using the mentioned symmetries $frac1{12}n^2 (n^2 - 1)$ of the $n^4$ entries of $R$ already determine it. However there could be other constraints like symmetries or inequalities that make the actual number of degrees of freedom even lower. As far as I know, this is not the case and my but I would like a proof of this.
(The only wat to prove this I can think of would be constructing a family of manifolds parametrized by $mathbb{R}^{n^2 (n^2 - 1)/12}$ such that for a given $p in mathbb{R}^{n^2 (n^2 - 1)/12}$ the corresponding manifold has the curvature tensor corresponding to $p$ at some point.)
differential-geometry riemannian-geometry tensors curvature general-relativity
asked Dec 30 '18 at 0:17
0x5390x539
1,445518
$endgroup$
I know the argument that uses the symmetries
- $R_{a b c d} = -R_{b a c d} = R_{c d a b}$
- $R_{a b c d} + R_{b c a d} + R_{c a b d} = 0$
of the Riemann curvature tensor $R$ of an $n$-dimensional Riemannian manifold to show that it (the curvature tensor) has at most $frac1{12} n^2 (n^2 - 1)$ degrees of freedom (see for example here, section 7) at a given point.
As far as I know $frac1{12} n^2 (n^2 - 1)$ is the actual number of degrees of freedom, so my question is how does one get the lower bound?
EDIT: To clarify: The argument I linked to shows that using the mentioned symmetries $frac1{12}n^2 (n^2 - 1)$ of the $n^4$ entries of $R$ already determine it. However there could be other constraints like symmetries or inequalities that make the actual number of degrees of freedom even lower. As far as I know, this is not the case and my but I would like a proof of this.
(The only wat to prove this I can think of would be constructing a family of manifolds parametrized by $mathbb{R}^{n^2 (n^2 - 1)/12}$ such that for a given $p in mathbb{R}^{n^2 (n^2 - 1)/12}$ the corresponding manifold has the curvature tensor corresponding to $p$ at some point.)
differential-geometry riemannian-geometry tensors curvature general-relativity
differential-geometry riemannian-geometry tensors curvature general-relativity
asked Dec 30 '18 at 0:17
0x5390x539
1,445518
asked Dec 30 '18 at 0:17
0x5390x539
1,445518
edited Jan 12 at 11:11
0x539
asked Dec 30 '18 at 0:17
0x5390x539
1,445518
asked Dec 30 '18 at 0:17
0x5390x539
1,445518
asked Dec 30 '18 at 0:17
0x5390x539
1,445518
1,445518
$begingroup$
lower bound on what ?
$endgroup$
– Amr
Dec 30 '18 at 2:11
$begingroup$
@Amr on the degrees of freedom.
$endgroup$
– 0x539
Dec 30 '18 at 2:25
$begingroup$
Wouldn't $frac{1}{12}n^2(n^2-1)$ count as a lower bound already ?
$endgroup$
– Amr
Dec 30 '18 at 3:08
$begingroup$
@Amr Yes it is, but i want to know why.
$endgroup$
– 0x539
Dec 30 '18 at 3:19
2
$begingroup$
Duplicate of math.stackexchange.com/questions/2235253/… (no vote to close due to open bounty)
$endgroup$
– Dap
Jan 16 at 11:12
|
show 2 more comments
$begingroup$
lower bound on what ?
$endgroup$
– Amr
Dec 30 '18 at 2:11
$begingroup$
@Amr on the degrees of freedom.
$endgroup$
– 0x539
Dec 30 '18 at 2:25
$begingroup$
Wouldn't $frac{1}{12}n^2(n^2-1)$ count as a lower bound already ?
$endgroup$
– Amr
Dec 30 '18 at 3:08
$begingroup$
@Amr Yes it is, but i want to know why.
$endgroup$
– 0x539
Dec 30 '18 at 3:19
2
$begingroup$
Duplicate of math.stackexchange.com/questions/2235253/… (no vote to close due to open bounty)
$endgroup$
– Dap
Jan 16 at 11:12
$begingroup$
lower bound on what ?
$endgroup$
– Amr
Dec 30 '18 at 2:11
$begingroup$
lower bound on what ?
$endgroup$
– Amr
Dec 30 '18 at 2:11
$begingroup$
@Amr on the degrees of freedom.
$endgroup$
– 0x539
Dec 30 '18 at 2:25
$begingroup$
@Amr on the degrees of freedom.
$endgroup$
– 0x539
Dec 30 '18 at 2:25
$begingroup$
Wouldn't $frac{1}{12}n^2(n^2-1)$ count as a lower bound already ?
$endgroup$
– Amr
Dec 30 '18 at 3:08
$begingroup$
Wouldn't $frac{1}{12}n^2(n^2-1)$ count as a lower bound already ?
$endgroup$
– Amr
Dec 30 '18 at 3:08
$begingroup$
@Amr Yes it is, but i want to know why.
$endgroup$
– 0x539
Dec 30 '18 at 3:19
$begingroup$
@Amr Yes it is, but i want to know why.
$endgroup$
– 0x539
Dec 30 '18 at 3:19
2
2
$begingroup$
Duplicate of math.stackexchange.com/questions/2235253/… (no vote to close due to open bounty)
$endgroup$
– Dap
Jan 16 at 11:12
$begingroup$
Duplicate of math.stackexchange.com/questions/2235253/… (no vote to close due to open bounty)
$endgroup$
– Dap
Jan 16 at 11:12
|
show 2 more comments
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$begingroup$
lower bound on what ?
$endgroup$
– Amr
Dec 30 '18 at 2:11
$begingroup$
@Amr on the degrees of freedom.
$endgroup$
– 0x539
Dec 30 '18 at 2:25
$begingroup$
Wouldn't $frac{1}{12}n^2(n^2-1)$ count as a lower bound already ?
$endgroup$
– Amr
Dec 30 '18 at 3:08
$begingroup$
@Amr Yes it is, but i want to know why.
$endgroup$
– 0x539
Dec 30 '18 at 3:19
2
$begingroup$
Duplicate of math.stackexchange.com/questions/2235253/… (no vote to close due to open bounty)
$endgroup$
– Dap
Jan 16 at 11:12