Solve Ra[bcd]=0: What Permutations Work?

[Kevin McHugh]

Given: R_a[bcd] = 0
What permutations of bcd make this expression 0? TIA for your response.

 

[Samy_A]

Given: R_a[bcd] = 0
What permutations of bcd make this expression 0? TIA for your response.

R_abcd+R_acdb+R_adbc=0

 

[Ibix]

I think this is an anti-symmetrised tensor, not a symmetrised one, isn't it?

 

[Kevin McHugh]

I think this is an anti-symmetrised tensor, not a symmetrised one, isn't it?

Yes it is an antisymmetric tensor.

 

[Kevin McHugh]

R_abcd+R_acdb+R_adbc=0

Samy, I'm afraid I'm too thick to understand your answer. Can you elaborate further please? IIRC, this expression only eliminates one expression from the 21 possible components.

 

[Ibix]

I'm pretty new to index gymnastics too, so check what I am saying with other sources.

I think Samy has written $R_{a (bcd)}$, not $R_{a [bcd]}$. The latter is $R_{abcd}- R_{acbd} +R_{acdb} -R_{adcb} +R_{adbc} -R_{abdc}=0$. Note that every ordering of the last three indices appears and that any pair of orderings that you can make by swapping two indices have opposite signs.

Both Samy's and my expressions are interpreted to mean that for any a, b, c, d, that statement must be valid. Because you are permuting b, c and d, though, there is a lot of replication in the expressions. a,b,c,d=0,1,2,3 and 0,1,3,2 yield identical expressions, for example.

 

[Kevin McHugh]

What I'm looking for is a rule to help me understand. For instance, for R_[ab][cd], R = 0 when a=b and c=d. Is there similar rule for the antisymmetric part of [bcd]?

 

[Ibix]

It's possible, for example, that Samy has used the symmetries of the Riemann tensor to simplify his expression. I'd need to think about it with a pen and paper.

Perhaps best to wait for Samy or someone else to clarify...

 

[robphy]

The latter is $R_{abcd}- R_{acbd} +R_{acdb} -R_{adcb} +R_{adbc} -R_{abdc}=0$.

Since $R_{pqdc}=-R_{pqcd}$, you can write the left-hand side as
$R_{abcd}- (-R_{acdb}) +R_{acdb} -(-R_{adbc}) +R_{adbc} -(-R_{abcd})=2(R_{abcd}+R_{acdb}+R_{adbc})$

 

[Ibix]

Ok. The Riemann tensor is anti-symmetric in its last two indices. That is, $R_{abcd}=-R_{abdc}$. So my expression (correct for a general four index tensor) simplifies to Samy's in the case of the Riemann tensor.

Edit: ...as robphy just pointed out.

Samy's expression is not $R_{a (bcd)}$. Forget I said that.

I'm going to shut up now before I confuse anything even worse.

 

[Samy_A]

My apologies, I thought R was the Riemann tensor.

 

[Kevin McHugh]

No apologies Samy, it is the Riemann tensor.

 

[Kevin McHugh]

Am I not being clear in my question? Nobody has yet answered it. What combinations of [bcd] make R_a[bcd] zero?

 

[Ibix]

No apologies Samy, it is the Riemann tensor.

Indeed. It was me who was confused, not you, @Samy_A

 

[Samy_A]

Am I not being clear in my question? Nobody has yet answered it. What combinations of [bcd] make R_a[bcd] zero?

Any combination (if we interpret the expression as I did in post ##2).

 

[Samy_A]

Am I not being clear in my question? Nobody has yet answered it. What combinations of [bcd] make R_a[bcd] zero?

Any combination (if we interpret the expression as I did in post ##2).

To elaborate on this, the first Bianchi identity R_abcd+R_acdb+R_adbc=0 is true for any combination of indices, but not all combination yield interesting information.

Let's take the case where a=b.
We then have R_aacd+R_acda+R_adac=0.
But R_aacd=0.
Also R_acda=R_daac=-R_adac, so the identity ends up telling us that 0=0 in case a=b. And similarly for a=c or a=d.

Let's take the case where b=c.
R_abbd+R_abdb+R_adbb=0.
Here R_adbb=0, and R_abbd=-R_abdb, so again the identity ends up telling us that 0=0 in case b=c.

The Bianchi identity is only interesting when the indices are all different, and by the symmetries of the Riemann tensor, it allows you to reduce the number of independent components by one.