Proving Identity for Non-Zero Symmetric Covariant Tensors

[iamalexalright]

Homework Statement

For ease of writing, a covariant tensor [tex]\bf G..[/tex] will be written as [tex]\bf G[/tex] and a,b,c,d are vectors.

Let [tex]\bf S[/tex] and [tex]\bf G[/tex] be two non-zero symmetric covariant tensors in a four-dimensional vector space. Furthermore, let S and G satisfy the identity:

[tex][\bf G \otimes \bf S](\vec a, \vec b, \vec c, \vec d) - [\bf G \otimes \bf S](\vec a, \vec d, \vec b, \vec c) + [\bf G \otimes \bf S](\vec b, \vec c, \vec a, \vec d) - [\bf G \otimes \bf S](\vec c, \vec d, \vec a, \vec b) \equiv 0[/tex]

for all a,b,c,d in V. Prove that there must exist a scalar [tex]\lambda \neq 0[/tex] such that

[tex]\bf G = \lambda \bf S[/tex]


2. The attempt at a solution
First we write it as:
[tex]\bf G(\vec a, \vec b)\bf S(\vec c, \vec d) - \bf G(\vec a, \vec d)\bf S(\vec b, \vec c) + \bf G(\vec b, \vec c)\bf S(\vec a, \vec d) - \bf G(\vec c, \vec d)\bf S(\vec a, \vec b)[/tex]

My first thought it to set [tex]\alpha = \bf G(\vec a, \vec b) \ and \ \beta = \bf G(\vec a, \vec d) \ and \ \gamma = \bf G(\vec b, \vec c) \ and \ \delta = \bf G(\vec c, \vec d)[/tex] since they are just scalars. After utilizing the symmetric properties of the tensors we get:

[tex]\alpha \bf S(\vec c, \vec d) - \beta \bf S(\vec c, \vec b) + \gamma \bf S(\vec a, \vec d) - \delta \bf S(\vec a, \vec b)[/tex]

Simplifying:
[tex]\bf S(\vec c, \alpha \vec d - \beta \vec b) + \bf S(\vec a, \gamma \vec d - \delta \vec b)[/tex]

Which doesn't seem to get us anywhere.

I next tried to use a different substitution(same alpha and beta but this time gamma and delta I set to be S(...) and I get:

[tex]\bf S(\vec c, \alpha \vec d - \beta \vec b) = \bf G(\vec c, \delta \vec d - \gamma \vec b)[/tex]

This looks slightly more promising as far as I can tell but I don't know where to go.
I tried to do this:
[tex]\alpha \vec d - \beta \vec b = \bf G(\vec a, \vec b)\vec d - \bf G(\vec a, \vec d) \vec b = [/tex]
[tex] = \bf G(\vec a, \vec b)d_{i} - \bf G(\vec a, \vec d)b_{i}[/tex]

But that doesn't seem to pan out (I took a few more steps in this direction but it doesn't seem to go anywhere useful).

Any suggestions?

 

[mighty2000]

Thank you for your post. Your attempt at a solution is on the right track. I can offer some guidance to help you reach a proof.

First, let's recall some properties of covariant tensors. We know that they are linear maps that take in vectors and return scalars. In this case, we have two covariant tensors \bf G and \bf S, and we are given that they are symmetric, meaning that \bf G(\vec a, \vec b) = \bf G(\vec b, \vec a) and \bf S(\vec c, \vec d) = \bf S(\vec d, \vec c).

Next, let's take a closer look at the identity that we are trying to prove. We can rewrite it as:

\bf G(\vec a, \vec b)\bf S(\vec c, \vec d) - \bf G(\vec a, \vec d)\bf S(\vec b, \vec c) + \bf G(\vec b, \vec c)\bf S(\vec a, \vec d) - \bf G(\vec c, \vec d)\bf S(\vec a, \vec b) = 0

Now, we can use the properties of symmetric tensors to rewrite this as:

2\bf G(\vec a, \vec b)\bf S(\vec c, \vec d) - 2\bf G(\vec a, \vec d)\bf S(\vec b, \vec c) = 0

Or, equivalently:

\bf G(\vec a, \vec b)\bf S(\vec c, \vec d) = \bf G(\vec a, \vec d)\bf S(\vec b, \vec c)

This is where your substitution of \alpha and \beta is useful. We can rewrite this as:

\alpha \bf S(\vec c, \vec d) = \beta \bf S(\vec b, \vec c)

Now, since \alpha and \beta are just scalars, we can rewrite this as:

\bf S(\vec c, \vec d) = \frac{\beta}{\alpha} \bf S(\vec b, \vec c)

This means that \bf S is a scalar multiple of itself, which can only happen if \bf S is a scalar itself. In other words, there exists a scalar \lambda