# A Tensor Problem: A skew-symmetric tensor and another tensor

## Homework Statement

If $A_{ij}$ is a skew-symmetric tensor, and $B_{ij}$ is a second-order tensor, evaluate the expression

$$(B_{ij} B_{kl} + B_{il}B_{kj})A_{ik}$$

and express the final answer in its simplest form.

## Homework Equations

For a skew-symmetric tensor, $A_{ik}=-A_{ki}$

## The Attempt at a Solution

I'm stuck and unsure what's the first step. I notice that the expression in the bracket looks similar to what happens when two Levi-Civita symbols come together to form an expression of two pairs of the Kronecker delta. Other than that I'm quite lost. Can I get a tip please?

## Answers and Replies

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You are probably summing over all the repeated indices, right? Remember then that the dummy indices are arbitrary, and you can for example swap k and i if you feel like it. Using this, maybe you can write the expression into a form where you take BijBkl as a common factor, multiplying some expression containing the tensor A.

Here's what I have so far:

$$(B_{ij} B_{kl} + B_{il}B_{kj})A_{ik}\\ =B_{ij} B_{1l}A_{i1}+B_{ij}B_{2l}A_{i2}+B_{il} B_{1l}A_{i1}+B_{il}B_{2j}A_{i2}\\ =B_{1j} B_{1l}A_{11}+B_{2j} B_{1l}A_{21}+B_{1j}B_{2l}A_{12}+B_{2j}B_{2l}A_{22}+B_{1l} B_{1l}A_{11}+B_{2l} B_{1l}A_{21}+B_{1l}B_{2j}A_{12}+B_{2l}B_{2j}A_{22}\\ =2B_{1j} B_{1l}A_{11}+2B_{2l}B_{2j}A_{22}$$

So the final answer that I can give is... $$2\sum_i B_{ij} B_{il} A_{ii}$$ or in the Einstein summation, $$2B_{mj} B_{ml} A_{nn}$$ with nn no sum.

If this is correct, is there any other way to write this without the no sum?