# Product of a symmetric and antisymmetric tensor

1. Aug 11, 2015

### stedwards

It seems there should be a list of tensor identities on the internet that answers the following, but I can't find one.

For tensors in $R^4$,

$S = S_\mu{}^\nu = S_{(\mu}{}^{\nu)}$ is a symmetric tensor.
$A = A_{\nu\rho\sigma}= A_{[\nu\rho\sigma]}$ is an antisymmetric tensor in all indices.

The basis vectors are suppressed for simplicity.

Parentheses $(\ \ )$ enclose symmetric indices; the tensor remains the same upon an exchange of any two indices.
Brackets $[\ \ ]$ enclose antisymmetric indices; the tensor changes sign on an exchange of any two indices.

Is the inner product $SA=S_\mu{}^\nu A_{\nu\rho\sigma}$ antisymmetric in all indices?

(In a related aside, $A_{\nu\rho\sigma}$ can be expressed as a scalar times a "unit valued" antiysmmetric tensor. Call it $\epsilon^3{}_{\mu\nu\rho}$ in $R^4$, or the 3 index Levi-Civita tensor in 4 dimensions. But I don't see any internet reference to this animal either. Are my keywords lacking?)

2. Aug 11, 2015

### Staff: Mentor

While I don't have the answer to your question, I did find this article on Penrose Graphical Diagrams and Tensors that seems to use some identities:

https://en.wikipedia.org/wiki/Penrose_graphical_notation

Probably not too helpful in this context though. Save it for future reference. :-)

Perhaps @HallsofIvy knows of some resource for Tensor identities.

3. Aug 12, 2015

### Ben Niehoff

I don't think the contraction will have any specific symmetry between the index $\mu$ and the other indices.

Suppose you were just talking about $n \times n$ matrices. If you multiplied a symmetric matrix S with an antisymmetric matrix A, would you expect the result to have any sort of symmetry?

4. Aug 12, 2015

### stedwards

Well...I am hoping, not expecting.

As it turns out, $S=\delta$, or ${S_\mu}^\nu = \delta_\mu^\nu$ obtains an antisymmetric tensor for the $SA$ product.

In general, in the orthonormal coordinate system where S is diagonal, $S = ({S_0}^0,\ {S_1}^1,\ {S_2}^2,\ {S_3}^3)$. The $SA$ product has 4 corresponding, independent values.

I had thought this was sufficient to supply the exception to disprove the hypothesis, but on second glance, no.

Last edited: Aug 12, 2015
5. Aug 12, 2015

### stedwards

The result has multiple interesting antisymmetric properties but not, in general, is the product antisymmetric.

In orthonormal coordinates the tensor $\epsilon_{\mu\nu\rho}$ is equal to it's symbol. Under a change of coordinates, it remains antisymmetric. Symmetric tensors likewise remain symmetric.

We can construct a product space $\hat{q}_\delta = \hat{\theta}_\alpha \wedge \hat{\theta}_\beta \wedge \hat{\theta}_\gamma$. Bases vectors are identified with hats. $\hat{\theta}_\alpha$ are the contravariant bases of the orthonormal 4-space. $\hat{\theta}_\alpha$ are the covariant basis of the orthonormal 4-space.

There are for independent bases in $q$. One choice of basis is:
$\hat{q}_0 = \hat{\theta}_1 \wedge \hat{\theta}_2 \wedge \hat{\theta}_3$
$\hat{q}_1 = \hat{\theta}_2 \wedge \hat{\theta}_3 \wedge \hat{\theta}_0$
$\hat{q}_2 = \hat{\theta}_3 \wedge \hat{\theta}_0 \wedge \hat{\theta}_1$
$\hat{q}_3 = \hat{\theta}_0 \wedge \hat{\theta}_1 \wedge \hat{\theta}_2$​

In one choice of orthonormal coordinates, S is a diagonal matrix. Define the elements as,
$a = {S_0}^0$
$b = {S_1}^1$
$c = {S_2}^2$
$d = {S_3}^3$.​

$A= k\epsilon$. All 3 index antisymmetric tensors are $k$ times the corresponding 3 index Levi-Civita in $R^4$.

Putting it all together, with more abuse of notation (leaving out the basis vectors on the left side),

${S_0}^0 A_{0\mu\nu} = ak(\hat{q}_0 - \hat{q}_1 + \hat{q}_2)$
${S_1}^1 A_{1\mu\nu} = bk(\hat{q}_1 - \hat{q}_2 + \hat{q}_3)$
${S_2}^2 A_{2\mu\nu} = ck(\hat{q}_2 - \hat{q}_3 + \hat{q}_0)$
${S_3}^3 A_{3\mu\nu} = dk(\hat{q}_3 - \hat{q}_0 + \hat{q}_1)$.

${S_\mu}^\nu A_{\nu\rho\sigma}$ is a linear combination of 3 out of 4 of 3-index Levi-Civita tensors of $R^4$ in orthonormal coordinates.

How this variegates under a general coordinate transformation, I don't know yet. I could be more literate in tensor densities.

Last edited: Aug 12, 2015