Mixing
Covariant derivative of a symmetric tensor
$begingroup$
Assume that a symmetric $(0,2)$ satisfies
$$nabla_iT_{jk}+nabla_jT_{ik}+nabla_kT_{ji}=0$$
where $T=T_i^i$ is constant and $nabla_jT_{ik}ne 0$.
What are the values of the constants $a,b,c$ such that
$$anabla_iT_{jk}+bnabla_jT_{ik}+cnabla_kT_{ji}=0$$
Is there any difference if the tensor $T$ is the Ricci tensor.
I think $(a,b,c)$ where $a=b=c$ are the only solution set to it however I failed to prove it nor to find the solution.
Thanks in advance.
differential-geometry riemannian-geometry tensor-products tensors
asked Jan 31 at 20:49
SemsemSemsem
6,52031534
$endgroup$
add a comment |
$begingroup$
Assume that a symmetric $(0,2)$ satisfies
$$nabla_iT_{jk}+nabla_jT_{ik}+nabla_kT_{ji}=0$$
where $T=T_i^i$ is constant and $nabla_jT_{ik}ne 0$.
What are the values of the constants $a,b,c$ such that
$$anabla_iT_{jk}+bnabla_jT_{ik}+cnabla_kT_{ji}=0$$
Is there any difference if the tensor $T$ is the Ricci tensor.
I think $(a,b,c)$ where $a=b=c$ are the only solution set to it however I failed to prove it nor to find the solution.
Thanks in advance.
differential-geometry riemannian-geometry tensor-products tensors
asked Jan 31 at 20:49
SemsemSemsem
6,52031534
$endgroup$
add a comment |
$begingroup$
Assume that a symmetric $(0,2)$ satisfies
$$nabla_iT_{jk}+nabla_jT_{ik}+nabla_kT_{ji}=0$$
where $T=T_i^i$ is constant and $nabla_jT_{ik}ne 0$.
What are the values of the constants $a,b,c$ such that
$$anabla_iT_{jk}+bnabla_jT_{ik}+cnabla_kT_{ji}=0$$
Is there any difference if the tensor $T$ is the Ricci tensor.
I think $(a,b,c)$ where $a=b=c$ are the only solution set to it however I failed to prove it nor to find the solution.
Thanks in advance.
differential-geometry riemannian-geometry tensor-products tensors
asked Jan 31 at 20:49
SemsemSemsem
6,52031534
$endgroup$
Assume that a symmetric $(0,2)$ satisfies
$$nabla_iT_{jk}+nabla_jT_{ik}+nabla_kT_{ji}=0$$
where $T=T_i^i$ is constant and $nabla_jT_{ik}ne 0$.
What are the values of the constants $a,b,c$ such that
$$anabla_iT_{jk}+bnabla_jT_{ik}+cnabla_kT_{ji}=0$$
Is there any difference if the tensor $T$ is the Ricci tensor.
I think $(a,b,c)$ where $a=b=c$ are the only solution set to it however I failed to prove it nor to find the solution.
Thanks in advance.
differential-geometry riemannian-geometry tensor-products tensors
differential-geometry riemannian-geometry tensor-products tensors
asked Jan 31 at 20:49
SemsemSemsem
6,52031534
asked Jan 31 at 20:49
SemsemSemsem
6,52031534
edited Feb 5 at 7:21
Semsem
asked Jan 31 at 20:49
SemsemSemsem
6,52031534
asked Jan 31 at 20:49
SemsemSemsem
6,52031534
asked Jan 31 at 20:49
SemsemSemsem
6,52031534
6,52031534
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
When $T_{ijk}:=nabla_{i}T_{jk}$, then $$ T_{ijk} +T_{jki}+ T_{kij}
=0 (A)$$
$$ aT_{ijk} +b T_{jki}+ cT_{kij}
=0 (B)$$
When $a=b=0$, then $T_{ijk}=0$.
When $a=0, b, cneq 0$, then
$T_{ijk}=Ccdot T_{jki}, Cneq 0$. Then $$T_{ijk}{1+C+C^2}=0$$ from
$(A)$ so that $T_{ijk}=0$.
When $a, b, cneq 0$, then $(A), (B)$ implies that $(1-b/a)T_{jki} + (1-c/a)T_{kij}=0$. Hence $a=b=c$ or
$T_{ijk}=0$.
answered Feb 6 at 2:59
HK LeeHK Lee
14.1k52361
$endgroup$
$begingroup$
I got it. Thank you.
$endgroup$
– Semsem
Feb 6 at 9:42
add a comment |
$begingroup$
I am afraid not. Take $T_{jk}=g_{jk}$, where $g_{jk}$ denotes the metric tensor with which the connection $nabla$ is compatible. In this case, $a$, $b$, and $c$ could be arbitrarily valued, and $g_{jk}$ also differs from the Ricci tensor.
answered Feb 4 at 15:56
hypernovahypernova
4,979514
$endgroup$
$begingroup$
@Semsem: May I thus confirm your question again? "Is it a must that $a=b=c$ in the second equation, provided that $T$, a tensor of type (0,2), satisfies the first equation?" - is this what you were asking or somewhat else?
$endgroup$
– hypernova
Feb 5 at 0:40
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
When $T_{ijk}:=nabla_{i}T_{jk}$, then $$ T_{ijk} +T_{jki}+ T_{kij}
=0 (A)$$
$$ aT_{ijk} +b T_{jki}+ cT_{kij}
=0 (B)$$
When $a=b=0$, then $T_{ijk}=0$.
When $a=0, b, cneq 0$, then
$T_{ijk}=Ccdot T_{jki}, Cneq 0$. Then $$T_{ijk}{1+C+C^2}=0$$ from
$(A)$ so that $T_{ijk}=0$.
When $a, b, cneq 0$, then $(A), (B)$ implies that $(1-b/a)T_{jki} + (1-c/a)T_{kij}=0$. Hence $a=b=c$ or
$T_{ijk}=0$.
answered Feb 6 at 2:59
HK LeeHK Lee
14.1k52361
$endgroup$
$begingroup$
I got it. Thank you.
$endgroup$
– Semsem
Feb 6 at 9:42
add a comment |
$begingroup$
When $T_{ijk}:=nabla_{i}T_{jk}$, then $$ T_{ijk} +T_{jki}+ T_{kij}
=0 (A)$$
$$ aT_{ijk} +b T_{jki}+ cT_{kij}
=0 (B)$$
When $a=b=0$, then $T_{ijk}=0$.
When $a=0, b, cneq 0$, then
$T_{ijk}=Ccdot T_{jki}, Cneq 0$. Then $$T_{ijk}{1+C+C^2}=0$$ from
$(A)$ so that $T_{ijk}=0$.
When $a, b, cneq 0$, then $(A), (B)$ implies that $(1-b/a)T_{jki} + (1-c/a)T_{kij}=0$. Hence $a=b=c$ or
$T_{ijk}=0$.
answered Feb 6 at 2:59
HK LeeHK Lee
14.1k52361
$endgroup$
$begingroup$
I got it. Thank you.
$endgroup$
– Semsem
Feb 6 at 9:42
add a comment |
$begingroup$
When $T_{ijk}:=nabla_{i}T_{jk}$, then $$ T_{ijk} +T_{jki}+ T_{kij}
=0 (A)$$
$$ aT_{ijk} +b T_{jki}+ cT_{kij}
=0 (B)$$
When $a=b=0$, then $T_{ijk}=0$.
When $a=0, b, cneq 0$, then
$T_{ijk}=Ccdot T_{jki}, Cneq 0$. Then $$T_{ijk}{1+C+C^2}=0$$ from
$(A)$ so that $T_{ijk}=0$.
When $a, b, cneq 0$, then $(A), (B)$ implies that $(1-b/a)T_{jki} + (1-c/a)T_{kij}=0$. Hence $a=b=c$ or
$T_{ijk}=0$.
answered Feb 6 at 2:59
HK LeeHK Lee
14.1k52361
$endgroup$
When $T_{ijk}:=nabla_{i}T_{jk}$, then $$ T_{ijk} +T_{jki}+ T_{kij}
=0 (A)$$
$$ aT_{ijk} +b T_{jki}+ cT_{kij}
=0 (B)$$
When $a=b=0$, then $T_{ijk}=0$.
When $a=0, b, cneq 0$, then
$T_{ijk}=Ccdot T_{jki}, Cneq 0$. Then $$T_{ijk}{1+C+C^2}=0$$ from
$(A)$ so that $T_{ijk}=0$.
When $a, b, cneq 0$, then $(A), (B)$ implies that $(1-b/a)T_{jki} + (1-c/a)T_{kij}=0$. Hence $a=b=c$ or
$T_{ijk}=0$.
answered Feb 6 at 2:59
HK LeeHK Lee
14.1k52361
edited Feb 6 at 9:19
answered Feb 6 at 2:59
HK LeeHK Lee
14.1k52361
answered Feb 6 at 2:59
HK LeeHK Lee
14.1k52361
answered Feb 6 at 2:59
HK LeeHK Lee
14.1k52361
14.1k52361
$begingroup$
I got it. Thank you.
$endgroup$
– Semsem
Feb 6 at 9:42
add a comment |
$begingroup$
I got it. Thank you.
$endgroup$
– Semsem
Feb 6 at 9:42
$begingroup$
I got it. Thank you.
$endgroup$
– Semsem
Feb 6 at 9:42
$begingroup$
I got it. Thank you.
$endgroup$
– Semsem
Feb 6 at 9:42
add a comment |
$begingroup$
I am afraid not. Take $T_{jk}=g_{jk}$, where $g_{jk}$ denotes the metric tensor with which the connection $nabla$ is compatible. In this case, $a$, $b$, and $c$ could be arbitrarily valued, and $g_{jk}$ also differs from the Ricci tensor.
answered Feb 4 at 15:56
hypernovahypernova
4,979514
$endgroup$
$begingroup$
@Semsem: May I thus confirm your question again? "Is it a must that $a=b=c$ in the second equation, provided that $T$, a tensor of type (0,2), satisfies the first equation?" - is this what you were asking or somewhat else?
$endgroup$
– hypernova
Feb 5 at 0:40
add a comment |
$begingroup$
I am afraid not. Take $T_{jk}=g_{jk}$, where $g_{jk}$ denotes the metric tensor with which the connection $nabla$ is compatible. In this case, $a$, $b$, and $c$ could be arbitrarily valued, and $g_{jk}$ also differs from the Ricci tensor.
answered Feb 4 at 15:56
hypernovahypernova
4,979514
$endgroup$
$begingroup$
@Semsem: May I thus confirm your question again? "Is it a must that $a=b=c$ in the second equation, provided that $T$, a tensor of type (0,2), satisfies the first equation?" - is this what you were asking or somewhat else?
$endgroup$
– hypernova
Feb 5 at 0:40
add a comment |
$begingroup$
I am afraid not. Take $T_{jk}=g_{jk}$, where $g_{jk}$ denotes the metric tensor with which the connection $nabla$ is compatible. In this case, $a$, $b$, and $c$ could be arbitrarily valued, and $g_{jk}$ also differs from the Ricci tensor.
answered Feb 4 at 15:56
hypernovahypernova
4,979514
$endgroup$
I am afraid not. Take $T_{jk}=g_{jk}$, where $g_{jk}$ denotes the metric tensor with which the connection $nabla$ is compatible. In this case, $a$, $b$, and $c$ could be arbitrarily valued, and $g_{jk}$ also differs from the Ricci tensor.
answered Feb 4 at 15:56
hypernovahypernova
4,979514
answered Feb 4 at 15:56
hypernovahypernova
4,979514
answered Feb 4 at 15:56
hypernovahypernova
4,979514
answered Feb 4 at 15:56
hypernovahypernova
4,979514
4,979514
$begingroup$
@Semsem: May I thus confirm your question again? "Is it a must that $a=b=c$ in the second equation, provided that $T$, a tensor of type (0,2), satisfies the first equation?" - is this what you were asking or somewhat else?
$endgroup$
– hypernova
Feb 5 at 0:40
add a comment |
$begingroup$
@Semsem: May I thus confirm your question again? "Is it a must that $a=b=c$ in the second equation, provided that $T$, a tensor of type (0,2), satisfies the first equation?" - is this what you were asking or somewhat else?
$endgroup$
– hypernova
Feb 5 at 0:40
$begingroup$
@Semsem: May I thus confirm your question again? "Is it a must that $a=b=c$ in the second equation, provided that $T$, a tensor of type (0,2), satisfies the first equation?" - is this what you were asking or somewhat else?
$endgroup$
– hypernova
Feb 5 at 0:40
$begingroup$
@Semsem: May I thus confirm your question again? "Is it a must that $a=b=c$ in the second equation, provided that $T$, a tensor of type (0,2), satisfies the first equation?" - is this what you were asking or somewhat else?
$endgroup$
– hypernova
Feb 5 at 0:40
add a comment |
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