Mixing
Identity with Riemann Tensor
$begingroup$
I am currently reading some lecture notes on Riemannian geometry and it is stated that the fact that
$d|Rm|^2=2langle Rm,nabla Rmrangle$
implies that
$Delta|Rm|^2=2langle Rm,Delta Rmrangle + 2|nabla Rm|^2$.
Would someone mind clarifying why this is? To define the notation, $Delta$ is the Laplace-Beltrami operator and $Rm$ is the Riemann curvature tensor. $nabla$ is the covariant derivative.
differential-geometry riemannian-geometry
edited Jan 28 at 19:49
Neal
24k24087
asked Jan 28 at 19:22
TomTom
345111
$endgroup$
add a comment |
$begingroup$
I am currently reading some lecture notes on Riemannian geometry and it is stated that the fact that
$d|Rm|^2=2langle Rm,nabla Rmrangle$
implies that
$Delta|Rm|^2=2langle Rm,Delta Rmrangle + 2|nabla Rm|^2$.
Would someone mind clarifying why this is? To define the notation, $Delta$ is the Laplace-Beltrami operator and $Rm$ is the Riemann curvature tensor. $nabla$ is the covariant derivative.
differential-geometry riemannian-geometry
edited Jan 28 at 19:49
Neal
24k24087
asked Jan 28 at 19:22
TomTom
345111
$endgroup$
$begingroup$
Could you define the quantities in your expression? Is $Delta=nabla^munabla_mu$? What is $Rm$?
$endgroup$
– Alec B-G
Jan 28 at 19:29
$begingroup$
I've added in notation, Rm is just the Riemann tensor.
$endgroup$
– Tom
Jan 28 at 19:34
$begingroup$
With laplacian, you mean the Laplace-Beltrami operator ($Delta T=nabla^{alpha}nabla_{alpha}T$)?
$endgroup$
– Gabriele Cassese
Jan 28 at 19:38
$begingroup$
Yes, sorry, the Laplace-Beltrami operator.
$endgroup$
– Tom
Jan 28 at 19:39
add a comment |
$begingroup$
I am currently reading some lecture notes on Riemannian geometry and it is stated that the fact that
$d|Rm|^2=2langle Rm,nabla Rmrangle$
implies that
$Delta|Rm|^2=2langle Rm,Delta Rmrangle + 2|nabla Rm|^2$.
Would someone mind clarifying why this is? To define the notation, $Delta$ is the Laplace-Beltrami operator and $Rm$ is the Riemann curvature tensor. $nabla$ is the covariant derivative.
differential-geometry riemannian-geometry
edited Jan 28 at 19:49
Neal
24k24087
asked Jan 28 at 19:22
TomTom
345111
$endgroup$
I am currently reading some lecture notes on Riemannian geometry and it is stated that the fact that
$d|Rm|^2=2langle Rm,nabla Rmrangle$
implies that
$Delta|Rm|^2=2langle Rm,Delta Rmrangle + 2|nabla Rm|^2$.
Would someone mind clarifying why this is? To define the notation, $Delta$ is the Laplace-Beltrami operator and $Rm$ is the Riemann curvature tensor. $nabla$ is the covariant derivative.
differential-geometry riemannian-geometry
differential-geometry riemannian-geometry
edited Jan 28 at 19:49
Neal
24k24087
asked Jan 28 at 19:22
TomTom
345111
edited Jan 28 at 19:49
Neal
24k24087
asked Jan 28 at 19:22
TomTom
345111
edited Jan 28 at 19:49
Neal
24k24087
edited Jan 28 at 19:49
Neal
24k24087
edited Jan 28 at 19:49
Neal
24k24087
24k24087
asked Jan 28 at 19:22
TomTom
345111
asked Jan 28 at 19:22
TomTom
345111
asked Jan 28 at 19:22
TomTom
345111
345111
$begingroup$
Could you define the quantities in your expression? Is $Delta=nabla^munabla_mu$? What is $Rm$?
$endgroup$
– Alec B-G
Jan 28 at 19:29
$begingroup$
I've added in notation, Rm is just the Riemann tensor.
$endgroup$
– Tom
Jan 28 at 19:34
$begingroup$
With laplacian, you mean the Laplace-Beltrami operator ($Delta T=nabla^{alpha}nabla_{alpha}T$)?
$endgroup$
– Gabriele Cassese
Jan 28 at 19:38
$begingroup$
Yes, sorry, the Laplace-Beltrami operator.
$endgroup$
– Tom
Jan 28 at 19:39
add a comment |
$begingroup$
Could you define the quantities in your expression? Is $Delta=nabla^munabla_mu$? What is $Rm$?
$endgroup$
– Alec B-G
Jan 28 at 19:29
$begingroup$
I've added in notation, Rm is just the Riemann tensor.
$endgroup$
– Tom
Jan 28 at 19:34
$begingroup$
With laplacian, you mean the Laplace-Beltrami operator ($Delta T=nabla^{alpha}nabla_{alpha}T$)?
$endgroup$
– Gabriele Cassese
Jan 28 at 19:38
$begingroup$
Yes, sorry, the Laplace-Beltrami operator.
$endgroup$
– Tom
Jan 28 at 19:39
$begingroup$
Could you define the quantities in your expression? Is $Delta=nabla^munabla_mu$? What is $Rm$?
$endgroup$
– Alec B-G
Jan 28 at 19:29
$begingroup$
Could you define the quantities in your expression? Is $Delta=nabla^munabla_mu$? What is $Rm$?
$endgroup$
– Alec B-G
Jan 28 at 19:29
$begingroup$
I've added in notation, Rm is just the Riemann tensor.
$endgroup$
– Tom
Jan 28 at 19:34
$begingroup$
I've added in notation, Rm is just the Riemann tensor.
$endgroup$
– Tom
Jan 28 at 19:34
$begingroup$
With laplacian, you mean the Laplace-Beltrami operator ($Delta T=nabla^{alpha}nabla_{alpha}T$)?
$endgroup$
– Gabriele Cassese
Jan 28 at 19:38
$begingroup$
With laplacian, you mean the Laplace-Beltrami operator ($Delta T=nabla^{alpha}nabla_{alpha}T$)?
$endgroup$
– Gabriele Cassese
Jan 28 at 19:38
$begingroup$
Yes, sorry, the Laplace-Beltrami operator.
$endgroup$
– Tom
Jan 28 at 19:39
$begingroup$
Yes, sorry, the Laplace-Beltrami operator.
$endgroup$
– Tom
Jan 28 at 19:39
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
$$begin{align}Delta|Rm|^2&=nabla^munabla_mu|Rm|^2\
&=nabla^mu2langle Rm,nabla_mu Rmrangle\
&=2g^{munu}nabla_nulangle Rm,nabla_mu Rmrangle\
&=2g^{munu}langlenabla_nu Rm,nabla_mu Rmrangle+2langle Rm,g^{numu}nabla_nunabla_mu Rmrangle\
&=2|nabla Rm|^2+2langle Rm,Delta Rmrangle.end{align}$$
answered Jan 28 at 20:51
Amitai YuvalAmitai Yuval
15.6k11127
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$$begin{align}Delta|Rm|^2&=nabla^munabla_mu|Rm|^2\
&=nabla^mu2langle Rm,nabla_mu Rmrangle\
&=2g^{munu}nabla_nulangle Rm,nabla_mu Rmrangle\
&=2g^{munu}langlenabla_nu Rm,nabla_mu Rmrangle+2langle Rm,g^{numu}nabla_nunabla_mu Rmrangle\
&=2|nabla Rm|^2+2langle Rm,Delta Rmrangle.end{align}$$
answered Jan 28 at 20:51
Amitai YuvalAmitai Yuval
15.6k11127
$endgroup$
add a comment |
$begingroup$
$$begin{align}Delta|Rm|^2&=nabla^munabla_mu|Rm|^2\
&=nabla^mu2langle Rm,nabla_mu Rmrangle\
&=2g^{munu}nabla_nulangle Rm,nabla_mu Rmrangle\
&=2g^{munu}langlenabla_nu Rm,nabla_mu Rmrangle+2langle Rm,g^{numu}nabla_nunabla_mu Rmrangle\
&=2|nabla Rm|^2+2langle Rm,Delta Rmrangle.end{align}$$
answered Jan 28 at 20:51
Amitai YuvalAmitai Yuval
15.6k11127
$endgroup$
add a comment |
$begingroup$
$$begin{align}Delta|Rm|^2&=nabla^munabla_mu|Rm|^2\
&=nabla^mu2langle Rm,nabla_mu Rmrangle\
&=2g^{munu}nabla_nulangle Rm,nabla_mu Rmrangle\
&=2g^{munu}langlenabla_nu Rm,nabla_mu Rmrangle+2langle Rm,g^{numu}nabla_nunabla_mu Rmrangle\
&=2|nabla Rm|^2+2langle Rm,Delta Rmrangle.end{align}$$
answered Jan 28 at 20:51
Amitai YuvalAmitai Yuval
15.6k11127
$endgroup$
$$begin{align}Delta|Rm|^2&=nabla^munabla_mu|Rm|^2\
&=nabla^mu2langle Rm,nabla_mu Rmrangle\
&=2g^{munu}nabla_nulangle Rm,nabla_mu Rmrangle\
&=2g^{munu}langlenabla_nu Rm,nabla_mu Rmrangle+2langle Rm,g^{numu}nabla_nunabla_mu Rmrangle\
&=2|nabla Rm|^2+2langle Rm,Delta Rmrangle.end{align}$$
answered Jan 28 at 20:51
Amitai YuvalAmitai Yuval
15.6k11127
answered Jan 28 at 20:51
Amitai YuvalAmitai Yuval
15.6k11127
answered Jan 28 at 20:51
Amitai YuvalAmitai Yuval
15.6k11127
answered Jan 28 at 20:51
Amitai YuvalAmitai Yuval
15.6k11127
15.6k11127
add a comment |
add a comment |
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$begingroup$
Could you define the quantities in your expression? Is $Delta=nabla^munabla_mu$? What is $Rm$?
$endgroup$
– Alec B-G
Jan 28 at 19:29
$begingroup$
I've added in notation, Rm is just the Riemann tensor.
$endgroup$
– Tom
Jan 28 at 19:34
$begingroup$
With laplacian, you mean the Laplace-Beltrami operator ($Delta T=nabla^{alpha}nabla_{alpha}T$)?
$endgroup$
– Gabriele Cassese
Jan 28 at 19:38
$begingroup$
Yes, sorry, the Laplace-Beltrami operator.
$endgroup$
– Tom
Jan 28 at 19:39