Riemann tensor symmetries
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The Riemann tensor has its component expression:
$R^{mu}_{nurhosigma}=partial_{rho}Gamma^{mu}_{sigmanu}-partial_{sigma}Gamma^{mu}_{rhonu}+Gamma^{mu}_{rholambda}Gamma^{lambda}_{sigmanu}-Gamma^{mu}_{sigmalambda}Gamma^{lambda}_{rhonu}.$
It is straight forward to prove the antisymmetry of $R$ in the last two indices; but how to prove the antisymmetry in the first two ones without assuming symmetric connection/torsion-free metric?
differential-geometry general-relativity
asked Dec 12 '13 at 16:17
user115376
515
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up vote
5
down vote
favorite
The Riemann tensor has its component expression:
$R^{mu}_{nurhosigma}=partial_{rho}Gamma^{mu}_{sigmanu}-partial_{sigma}Gamma^{mu}_{rhonu}+Gamma^{mu}_{rholambda}Gamma^{lambda}_{sigmanu}-Gamma^{mu}_{sigmalambda}Gamma^{lambda}_{rhonu}.$
It is straight forward to prove the antisymmetry of $R$ in the last two indices; but how to prove the antisymmetry in the first two ones without assuming symmetric connection/torsion-free metric?
differential-geometry general-relativity
asked Dec 12 '13 at 16:17
user115376
515
6
Riemann tensor can be equivalently viewed as curvature 2-form $Omega$ with values in a Lie algebra $mathfrak g$ of group $G = SO(n)$. The antisymmetry in one pair comes from being a 2-form, the antisymmetry in the other pair comes from the antisymmetry of ${mathfrak so}(n)$. This holds even when the connection has torsion. But it won't neccesarily hold for non-metric connections (such as symplectic connections). en.wikipedia.org/wiki/Curvature_form
– Marek
Dec 12 '13 at 17:28
Doesn't this just fall from definition?
– IAmNoOne
Nov 21 at 5:24
$R(X,Y)Z=nabla_Y nabla_X Z - nabla_X nabla_Y Z+ nabla_{[X,Y]}Z.$ Switching $X,Y$ $R(Y,X)Z=nabla_X nabla_Y Z - nabla_Y nabla_X Z+ nabla_{[Y,X]}Z = -(nabla_Y nabla_X Z - nabla_X nabla_Y Z - nabla_{[X,Y]}Z) = -R(X,Y)Z$
– IAmNoOne
Nov 21 at 5:29
add a comment |
up vote
5
down vote
favorite
up vote
5
down vote
favorite
The Riemann tensor has its component expression:
$R^{mu}_{nurhosigma}=partial_{rho}Gamma^{mu}_{sigmanu}-partial_{sigma}Gamma^{mu}_{rhonu}+Gamma^{mu}_{rholambda}Gamma^{lambda}_{sigmanu}-Gamma^{mu}_{sigmalambda}Gamma^{lambda}_{rhonu}.$
It is straight forward to prove the antisymmetry of $R$ in the last two indices; but how to prove the antisymmetry in the first two ones without assuming symmetric connection/torsion-free metric?
differential-geometry general-relativity
asked Dec 12 '13 at 16:17
user115376
515
The Riemann tensor has its component expression:
$R^{mu}_{nurhosigma}=partial_{rho}Gamma^{mu}_{sigmanu}-partial_{sigma}Gamma^{mu}_{rhonu}+Gamma^{mu}_{rholambda}Gamma^{lambda}_{sigmanu}-Gamma^{mu}_{sigmalambda}Gamma^{lambda}_{rhonu}.$
It is straight forward to prove the antisymmetry of $R$ in the last two indices; but how to prove the antisymmetry in the first two ones without assuming symmetric connection/torsion-free metric?
differential-geometry general-relativity
differential-geometry general-relativity
asked Dec 12 '13 at 16:17
user115376
515
asked Dec 12 '13 at 16:17
user115376
515
asked Dec 12 '13 at 16:17
user115376
515
asked Dec 12 '13 at 16:17
user115376
515
asked Dec 12 '13 at 16:17
user115376
515
515
6
Riemann tensor can be equivalently viewed as curvature 2-form $Omega$ with values in a Lie algebra $mathfrak g$ of group $G = SO(n)$. The antisymmetry in one pair comes from being a 2-form, the antisymmetry in the other pair comes from the antisymmetry of ${mathfrak so}(n)$. This holds even when the connection has torsion. But it won't neccesarily hold for non-metric connections (such as symplectic connections). en.wikipedia.org/wiki/Curvature_form
– Marek
Dec 12 '13 at 17:28
Doesn't this just fall from definition?
– IAmNoOne
Nov 21 at 5:24
$R(X,Y)Z=nabla_Y nabla_X Z - nabla_X nabla_Y Z+ nabla_{[X,Y]}Z.$ Switching $X,Y$ $R(Y,X)Z=nabla_X nabla_Y Z - nabla_Y nabla_X Z+ nabla_{[Y,X]}Z = -(nabla_Y nabla_X Z - nabla_X nabla_Y Z - nabla_{[X,Y]}Z) = -R(X,Y)Z$
– IAmNoOne
Nov 21 at 5:29
add a comment |
6
Riemann tensor can be equivalently viewed as curvature 2-form $Omega$ with values in a Lie algebra $mathfrak g$ of group $G = SO(n)$. The antisymmetry in one pair comes from being a 2-form, the antisymmetry in the other pair comes from the antisymmetry of ${mathfrak so}(n)$. This holds even when the connection has torsion. But it won't neccesarily hold for non-metric connections (such as symplectic connections). en.wikipedia.org/wiki/Curvature_form
– Marek
Dec 12 '13 at 17:28
Doesn't this just fall from definition?
– IAmNoOne
Nov 21 at 5:24
$R(X,Y)Z=nabla_Y nabla_X Z - nabla_X nabla_Y Z+ nabla_{[X,Y]}Z.$ Switching $X,Y$ $R(Y,X)Z=nabla_X nabla_Y Z - nabla_Y nabla_X Z+ nabla_{[Y,X]}Z = -(nabla_Y nabla_X Z - nabla_X nabla_Y Z - nabla_{[X,Y]}Z) = -R(X,Y)Z$
– IAmNoOne
Nov 21 at 5:29
6
6
Riemann tensor can be equivalently viewed as curvature 2-form $Omega$ with values in a Lie algebra $mathfrak g$ of group $G = SO(n)$. The antisymmetry in one pair comes from being a 2-form, the antisymmetry in the other pair comes from the antisymmetry of ${mathfrak so}(n)$. This holds even when the connection has torsion. But it won't neccesarily hold for non-metric connections (such as symplectic connections). en.wikipedia.org/wiki/Curvature_form
– Marek
Dec 12 '13 at 17:28
Riemann tensor can be equivalently viewed as curvature 2-form $Omega$ with values in a Lie algebra $mathfrak g$ of group $G = SO(n)$. The antisymmetry in one pair comes from being a 2-form, the antisymmetry in the other pair comes from the antisymmetry of ${mathfrak so}(n)$. This holds even when the connection has torsion. But it won't neccesarily hold for non-metric connections (such as symplectic connections). en.wikipedia.org/wiki/Curvature_form
– Marek
Dec 12 '13 at 17:28
Doesn't this just fall from definition?
– IAmNoOne
Nov 21 at 5:24
Doesn't this just fall from definition?
– IAmNoOne
Nov 21 at 5:24
$R(X,Y)Z=nabla_Y nabla_X Z - nabla_X nabla_Y Z+ nabla_{[X,Y]}Z.$ Switching $X,Y$ $R(Y,X)Z=nabla_X nabla_Y Z - nabla_Y nabla_X Z+ nabla_{[Y,X]}Z = -(nabla_Y nabla_X Z - nabla_X nabla_Y Z - nabla_{[X,Y]}Z) = -R(X,Y)Z$
– IAmNoOne
Nov 21 at 5:29
$R(X,Y)Z=nabla_Y nabla_X Z - nabla_X nabla_Y Z+ nabla_{[X,Y]}Z.$ Switching $X,Y$ $R(Y,X)Z=nabla_X nabla_Y Z - nabla_Y nabla_X Z+ nabla_{[Y,X]}Z = -(nabla_Y nabla_X Z - nabla_X nabla_Y Z - nabla_{[X,Y]}Z) = -R(X,Y)Z$
– IAmNoOne
Nov 21 at 5:29
add a comment |
1 Answer
1
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0
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This simple change in definition,
$$nabla_anabla_b-nabla_bnabla_a$$
for
$$nabla_anabla_b-nabla_bnabla_a+T^d_{quad{ab}}nabla_d$$
However, that the Riemann tensor with torsion is no longer symmetric under exchange of the first pair of indices with the second.
answered Dec 13 '13 at 15:40
jimbo
1,630612
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
This simple change in definition,
$$nabla_anabla_b-nabla_bnabla_a$$
for
$$nabla_anabla_b-nabla_bnabla_a+T^d_{quad{ab}}nabla_d$$
However, that the Riemann tensor with torsion is no longer symmetric under exchange of the first pair of indices with the second.
answered Dec 13 '13 at 15:40
jimbo
1,630612
add a comment |
up vote
0
down vote
This simple change in definition,
$$nabla_anabla_b-nabla_bnabla_a$$
for
$$nabla_anabla_b-nabla_bnabla_a+T^d_{quad{ab}}nabla_d$$
However, that the Riemann tensor with torsion is no longer symmetric under exchange of the first pair of indices with the second.
answered Dec 13 '13 at 15:40
jimbo
1,630612
add a comment |
up vote
0
down vote
up vote
0
down vote
This simple change in definition,
$$nabla_anabla_b-nabla_bnabla_a$$
for
$$nabla_anabla_b-nabla_bnabla_a+T^d_{quad{ab}}nabla_d$$
However, that the Riemann tensor with torsion is no longer symmetric under exchange of the first pair of indices with the second.
answered Dec 13 '13 at 15:40
jimbo
1,630612
This simple change in definition,
$$nabla_anabla_b-nabla_bnabla_a$$
for
$$nabla_anabla_b-nabla_bnabla_a+T^d_{quad{ab}}nabla_d$$
However, that the Riemann tensor with torsion is no longer symmetric under exchange of the first pair of indices with the second.
answered Dec 13 '13 at 15:40
jimbo
1,630612
answered Dec 13 '13 at 15:40
jimbo
1,630612
answered Dec 13 '13 at 15:40
jimbo
1,630612
answered Dec 13 '13 at 15:40
jimbo
1,630612
1,630612
add a comment |
add a comment |
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6
Riemann tensor can be equivalently viewed as curvature 2-form $Omega$ with values in a Lie algebra $mathfrak g$ of group $G = SO(n)$. The antisymmetry in one pair comes from being a 2-form, the antisymmetry in the other pair comes from the antisymmetry of ${mathfrak so}(n)$. This holds even when the connection has torsion. But it won't neccesarily hold for non-metric connections (such as symplectic connections). en.wikipedia.org/wiki/Curvature_form
– Marek
Dec 12 '13 at 17:28
Doesn't this just fall from definition?
– IAmNoOne
Nov 21 at 5:24
$R(X,Y)Z=nabla_Y nabla_X Z - nabla_X nabla_Y Z+ nabla_{[X,Y]}Z.$ Switching $X,Y$ $R(Y,X)Z=nabla_X nabla_Y Z - nabla_Y nabla_X Z+ nabla_{[Y,X]}Z = -(nabla_Y nabla_X Z - nabla_X nabla_Y Z - nabla_{[X,Y]}Z) = -R(X,Y)Z$
– IAmNoOne
Nov 21 at 5:29