# Totally symmetric tensor

by ehrenfest
Tags: symmetric, tensor, totally
 P: 1,996 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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P: 4,300
 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?
 P: 1,996 $$v_{a}t_{bc} + v_{c}t_{ab} + v_{b}t_{ca}= 0$$
P: 893
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

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

$$2 v_{a}t_{ab} + v_{b}t_{aa}= 0$$
P: 1,996
 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.
P: 893
 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.
P: 1,996
 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
P: 893
 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.
P: 1,996
 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.
 P: 1,996 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?

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