Mixing
A question about the relation between the exterior derivative of $1$-forms and the metric
$begingroup$
Let $theta_X$ be a $1$-form. Petersen's "Riemannian Geometry" says the following on pg 24:
$dtheta_X(partial_k,partial_l)=partial_kg(X,partial_l)-partial_lg(X,partial_k)-g(X,[partial_k,partial_l])$
How is this? Can I get a reference for this fact?
riemannian-geometry
asked Jan 31 at 14:00
Anju GeorgeAnju George
1088
$endgroup$
add a comment |
$begingroup$
Let $theta_X$ be a $1$-form. Petersen's "Riemannian Geometry" says the following on pg 24:
$dtheta_X(partial_k,partial_l)=partial_kg(X,partial_l)-partial_lg(X,partial_k)-g(X,[partial_k,partial_l])$
How is this? Can I get a reference for this fact?
riemannian-geometry
asked Jan 31 at 14:00
Anju GeorgeAnju George
1088
$endgroup$
add a comment |
$begingroup$
Let $theta_X$ be a $1$-form. Petersen's "Riemannian Geometry" says the following on pg 24:
$dtheta_X(partial_k,partial_l)=partial_kg(X,partial_l)-partial_lg(X,partial_k)-g(X,[partial_k,partial_l])$
How is this? Can I get a reference for this fact?
riemannian-geometry
asked Jan 31 at 14:00
Anju GeorgeAnju George
1088
$endgroup$
Let $theta_X$ be a $1$-form. Petersen's "Riemannian Geometry" says the following on pg 24:
$dtheta_X(partial_k,partial_l)=partial_kg(X,partial_l)-partial_lg(X,partial_k)-g(X,[partial_k,partial_l])$
How is this? Can I get a reference for this fact?
riemannian-geometry
riemannian-geometry
asked Jan 31 at 14:00
Anju GeorgeAnju George
1088
asked Jan 31 at 14:00
Anju GeorgeAnju George
1088
asked Jan 31 at 14:00
Anju GeorgeAnju George
1088
asked Jan 31 at 14:00
Anju GeorgeAnju George
1088
asked Jan 31 at 14:00
Anju GeorgeAnju George
1088
1088
add a comment |
add a comment |
1 Answer
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$begingroup$
If $theta_X$ is the 1-form defined by $Ymapsto theta_X(Y)=g(X,Y)$, then this is a direct application of the formula for the exterior derivative of a 1-form $omega$:
$$domega(X,Y)=Xomega(Y)-Yomega(X)-omega([X,Y]),$$
and you can find a proof for this one in John M. Lee's Introduction to smooth manifolds, Proposition 14.29.
answered Jan 31 at 14:19
BalloonBalloon
4,710822
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1 Answer
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1 Answer
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active
oldest
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active
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active
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$begingroup$
If $theta_X$ is the 1-form defined by $Ymapsto theta_X(Y)=g(X,Y)$, then this is a direct application of the formula for the exterior derivative of a 1-form $omega$:
$$domega(X,Y)=Xomega(Y)-Yomega(X)-omega([X,Y]),$$
and you can find a proof for this one in John M. Lee's Introduction to smooth manifolds, Proposition 14.29.
answered Jan 31 at 14:19
BalloonBalloon
4,710822
$endgroup$
add a comment |
$begingroup$
If $theta_X$ is the 1-form defined by $Ymapsto theta_X(Y)=g(X,Y)$, then this is a direct application of the formula for the exterior derivative of a 1-form $omega$:
$$domega(X,Y)=Xomega(Y)-Yomega(X)-omega([X,Y]),$$
and you can find a proof for this one in John M. Lee's Introduction to smooth manifolds, Proposition 14.29.
answered Jan 31 at 14:19
BalloonBalloon
4,710822
$endgroup$
add a comment |
$begingroup$
If $theta_X$ is the 1-form defined by $Ymapsto theta_X(Y)=g(X,Y)$, then this is a direct application of the formula for the exterior derivative of a 1-form $omega$:
$$domega(X,Y)=Xomega(Y)-Yomega(X)-omega([X,Y]),$$
and you can find a proof for this one in John M. Lee's Introduction to smooth manifolds, Proposition 14.29.
answered Jan 31 at 14:19
BalloonBalloon
4,710822
$endgroup$
If $theta_X$ is the 1-form defined by $Ymapsto theta_X(Y)=g(X,Y)$, then this is a direct application of the formula for the exterior derivative of a 1-form $omega$:
$$domega(X,Y)=Xomega(Y)-Yomega(X)-omega([X,Y]),$$
and you can find a proof for this one in John M. Lee's Introduction to smooth manifolds, Proposition 14.29.
answered Jan 31 at 14:19
BalloonBalloon
4,710822
answered Jan 31 at 14:19
BalloonBalloon
4,710822
answered Jan 31 at 14:19
BalloonBalloon
4,710822
answered Jan 31 at 14:19
BalloonBalloon
4,710822
4,710822
add a comment |
add a comment |
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