Mixing
Derivative of Norm Squared of the Riemann Tensor
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I am currently reading some lecture notes on Ricci flow and am not sure how the following identity is derived:
$frac{partial}{partial t} bigg( g^{ij}g^{kl}g^{ab}g^{cd}R_{ikac}R_{jlbd} bigg) = 2bigg(R^{ij}g^{kl}g^{ab}g^{cd} + g^{ij}R^{kl}g^{ab}g^{cd}+g^{ij}g^{kl}R^{ab}g^{cd} +g^{ij}g^{kl}g^{ab}R^{cd}bigg)R_{ikac}R_{jlbd} :+:2<Rm,frac{partial}{partial t}Rm> $
I know there are identities for the time derivative of the Riemann tensor in general, but I don't quite follow how the author has gone from the time derivative to the right-hand side, could someone explain?
differential-geometry ricci-flow
asked Jan 28 at 0:13
TomTom
345111
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add a comment |
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I am currently reading some lecture notes on Ricci flow and am not sure how the following identity is derived:
$frac{partial}{partial t} bigg( g^{ij}g^{kl}g^{ab}g^{cd}R_{ikac}R_{jlbd} bigg) = 2bigg(R^{ij}g^{kl}g^{ab}g^{cd} + g^{ij}R^{kl}g^{ab}g^{cd}+g^{ij}g^{kl}R^{ab}g^{cd} +g^{ij}g^{kl}g^{ab}R^{cd}bigg)R_{ikac}R_{jlbd} :+:2<Rm,frac{partial}{partial t}Rm> $
I know there are identities for the time derivative of the Riemann tensor in general, but I don't quite follow how the author has gone from the time derivative to the right-hand side, could someone explain?
differential-geometry ricci-flow
asked Jan 28 at 0:13
TomTom
345111
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$begingroup$
I think you have one excess $g^{ij}$ on the LHS and the first summand of the RHS - else sum convention doesn't work.
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– 0x539
Jan 28 at 0:40
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Yes, my mistake, I've corrected this. Are you now able to see where the identity comes from?
$endgroup$
– Tom
Jan 28 at 16:19
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Hang on, I've realised it now, you just apply the product rule repeatedly and the final term is due to the metric being symmetric.
$endgroup$
– Tom
Jan 28 at 17:21
add a comment |
$begingroup$
I am currently reading some lecture notes on Ricci flow and am not sure how the following identity is derived:
$frac{partial}{partial t} bigg( g^{ij}g^{kl}g^{ab}g^{cd}R_{ikac}R_{jlbd} bigg) = 2bigg(R^{ij}g^{kl}g^{ab}g^{cd} + g^{ij}R^{kl}g^{ab}g^{cd}+g^{ij}g^{kl}R^{ab}g^{cd} +g^{ij}g^{kl}g^{ab}R^{cd}bigg)R_{ikac}R_{jlbd} :+:2<Rm,frac{partial}{partial t}Rm> $
I know there are identities for the time derivative of the Riemann tensor in general, but I don't quite follow how the author has gone from the time derivative to the right-hand side, could someone explain?
differential-geometry ricci-flow
asked Jan 28 at 0:13
TomTom
345111
$endgroup$
I am currently reading some lecture notes on Ricci flow and am not sure how the following identity is derived:
$frac{partial}{partial t} bigg( g^{ij}g^{kl}g^{ab}g^{cd}R_{ikac}R_{jlbd} bigg) = 2bigg(R^{ij}g^{kl}g^{ab}g^{cd} + g^{ij}R^{kl}g^{ab}g^{cd}+g^{ij}g^{kl}R^{ab}g^{cd} +g^{ij}g^{kl}g^{ab}R^{cd}bigg)R_{ikac}R_{jlbd} :+:2<Rm,frac{partial}{partial t}Rm> $
I know there are identities for the time derivative of the Riemann tensor in general, but I don't quite follow how the author has gone from the time derivative to the right-hand side, could someone explain?
differential-geometry ricci-flow
differential-geometry ricci-flow
asked Jan 28 at 0:13
TomTom
345111
asked Jan 28 at 0:13
TomTom
345111
edited Jan 28 at 16:18
Tom
asked Jan 28 at 0:13
TomTom
345111
asked Jan 28 at 0:13
TomTom
345111
asked Jan 28 at 0:13
TomTom
345111
345111
$begingroup$
I think you have one excess $g^{ij}$ on the LHS and the first summand of the RHS - else sum convention doesn't work.
$endgroup$
– 0x539
Jan 28 at 0:40
$begingroup$
Yes, my mistake, I've corrected this. Are you now able to see where the identity comes from?
$endgroup$
– Tom
Jan 28 at 16:19
$begingroup$
Hang on, I've realised it now, you just apply the product rule repeatedly and the final term is due to the metric being symmetric.
$endgroup$
– Tom
Jan 28 at 17:21
add a comment |
$begingroup$
I think you have one excess $g^{ij}$ on the LHS and the first summand of the RHS - else sum convention doesn't work.
$endgroup$
– 0x539
Jan 28 at 0:40
$begingroup$
Yes, my mistake, I've corrected this. Are you now able to see where the identity comes from?
$endgroup$
– Tom
Jan 28 at 16:19
$begingroup$
Hang on, I've realised it now, you just apply the product rule repeatedly and the final term is due to the metric being symmetric.
$endgroup$
– Tom
Jan 28 at 17:21
$begingroup$
I think you have one excess $g^{ij}$ on the LHS and the first summand of the RHS - else sum convention doesn't work.
$endgroup$
– 0x539
Jan 28 at 0:40
$begingroup$
I think you have one excess $g^{ij}$ on the LHS and the first summand of the RHS - else sum convention doesn't work.
$endgroup$
– 0x539
Jan 28 at 0:40
$begingroup$
Yes, my mistake, I've corrected this. Are you now able to see where the identity comes from?
$endgroup$
– Tom
Jan 28 at 16:19
$begingroup$
Yes, my mistake, I've corrected this. Are you now able to see where the identity comes from?
$endgroup$
– Tom
Jan 28 at 16:19
$begingroup$
Hang on, I've realised it now, you just apply the product rule repeatedly and the final term is due to the metric being symmetric.
$endgroup$
– Tom
Jan 28 at 17:21
$begingroup$
Hang on, I've realised it now, you just apply the product rule repeatedly and the final term is due to the metric being symmetric.
$endgroup$
– Tom
Jan 28 at 17:21
add a comment |
1 Answer
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Apply the product rule repeatedly and use the fact that $frac{partial}{partial t}g^{ij}=2R^{ij}$. The final term arises from the fact that the metric is a symmetric form.
answered Jan 28 at 17:33
TomTom
345111
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Apply the product rule repeatedly and use the fact that $frac{partial}{partial t}g^{ij}=2R^{ij}$. The final term arises from the fact that the metric is a symmetric form.
answered Jan 28 at 17:33
TomTom
345111
$endgroup$
add a comment |
$begingroup$
Apply the product rule repeatedly and use the fact that $frac{partial}{partial t}g^{ij}=2R^{ij}$. The final term arises from the fact that the metric is a symmetric form.
answered Jan 28 at 17:33
TomTom
345111
$endgroup$
add a comment |
$begingroup$
Apply the product rule repeatedly and use the fact that $frac{partial}{partial t}g^{ij}=2R^{ij}$. The final term arises from the fact that the metric is a symmetric form.
answered Jan 28 at 17:33
TomTom
345111
$endgroup$
Apply the product rule repeatedly and use the fact that $frac{partial}{partial t}g^{ij}=2R^{ij}$. The final term arises from the fact that the metric is a symmetric form.
answered Jan 28 at 17:33
TomTom
345111
answered Jan 28 at 17:33
TomTom
345111
answered Jan 28 at 17:33
TomTom
345111
answered Jan 28 at 17:33
TomTom
345111
345111
add a comment |
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$begingroup$
I think you have one excess $g^{ij}$ on the LHS and the first summand of the RHS - else sum convention doesn't work.
$endgroup$
– 0x539
Jan 28 at 0:40
$begingroup$
Yes, my mistake, I've corrected this. Are you now able to see where the identity comes from?
$endgroup$
– Tom
Jan 28 at 16:19
$begingroup$
Hang on, I've realised it now, you just apply the product rule repeatedly and the final term is due to the metric being symmetric.
$endgroup$
– Tom
Jan 28 at 17:21