## totally symmetric tensor

1. The problem statement, all variables and given/known data
If t_{ab} are the components of a symmetric tensor and v_a are the components of a vector, show that if:

$$v_{(a}t_{bc)} = 0$$

then either the symmetric tensor or the vector = 0.

Let me know if you are not familiar with the totally symmetric notation.

2. Relevant equations

3. The attempt at a solution

You can write out the 6 terms in the equation above and then reduce it to 3 terms through the symmetry of the tensor but I am not sure where to go from there.

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 Quote by ehrenfest You can write out the 6 terms in the equation above and then reduce it to 3 terms through the symmetry of the tensor
I don't really feel like doing that now... can you post what that got you?

 $$v_{a}t_{bc} + v_{c}t_{ab} + v_{b}t_{ca}= 0$$

## totally symmetric tensor

 Quote by ehrenfest $$v_{a}t_{bc} + v_{c}t_{ab} + v_{b}t_{ca}= 0$$
I think you can do it if you start with the special cases

$$v_{a}t_{aa} = 0$$

$$2 v_{a}t_{ab} + v_{b}t_{aa}= 0$$

 Quote by Daverz I think you can do it if you start with the special cases $$v_{a}t_{aa} = 0$$ $$2 v_{a}t_{ab} + v_{b}t_{aa}= 0$$
I don't know--

You're first equation only implies that either v_a or t_aa is 0 for a given a. Using the first one with the second one doesn't really imply anything because both v_a and t_aa could be 0 so then there would be no restrictions on t_ab or v_b.

 Quote by ehrenfest I don't know-- You're first equation only implies that either v_a or t_aa is 0 for a given a. Using the first one with the second one doesn't really imply anything because both v_a and t_aa could be 0 so then there would be no restrictions on t_ab or v_b.
v_a is arbitrary, so if v is not identically zero, t_aa must be zero for all a (think of v_1=1, the rest 0, then the same for a=2, 3, ...). Actually, I think that does it, because there's no way t_aa=0 for all a unless t is identically zero.

Otherwise, for v not identically zero, the second equation then gives v_a t_ab = 0, and, again, since v_a is arbitrary, we must have t_ab = 0.

 Quote by Daverz v_a is arbitrary, so if v is not identically zero, t_aa must be zero for all a (think of v_1=1, the rest 0, then the same for a=2, 3, ...). Actually, I think that does it, because there's no way t_aa=0 for all a unless t is identically zero.
First of all, v_a is not arbitrary--it is a vector that is part of the problem statement.
Second, why could you not have something like:

v_a = (1,0,1,0,1,1,1,0,...)
t_aa = (0,1,0,1,0,0,0,1,...)

This is a case where v is not identially zero and t_aa is NOT zero for all a

 Quote by ehrenfest First of all, v_a is not arbitrary--
Perhaps it's easier to think of the basis for the vector space as being arbitrary.

 it is a vector that is part of the problem statement. Second, why could you not have something like: v_a = (1,0,1,0,1,1,1,0,...) t_aa = (0,1,0,1,0,0,0,1,...) This is a case where v is not identially zero and t_aa is NOT zero for all a
But the equation has to hold for any basis for the vector space, not just this one.

 Quote by Daverz But the equation has to hold for any basis for the vector space, not just this one.
That is not stated in the problem and I think it is not safe to assume that. The problem just gives us the components of a tensor and a vector, presumably in a given basis.

I think we should be able to prove if it holds in just one basis.

 Do other people agree with Daverz? I thought when the problem said that "If t_{ab} are the components of a symmetric tensor", that meant it was symmetric in some given basis?