# Contraction of an asymmetric tensor?

1. Nov 11, 2014

Hey guys,

So in my notes ive got this statement written:

If tensor with no symmetry properties, $A^{\mu\nu}$, contracts to $a_{\mu\nu}$, we can write this as $A^{\mu\nu}a_{\mu\nu}=\frac{1}{2}a_{\mu\nu}(A^{\mu\nu}-A^{\nu\mu})$ as $a_{\mu\nu} (A^{\mu\nu}+A^{\nu\mu}) = 0$. So I dont see how the symmetric part contracts to 0.

*Note* I do also have written that $a^{\mu\nu}=-a^{\nu\mu}$ but Im not sure if this is relevant.

I understand that you can decompose the tensor $A^{\mu\nu}$ into the sum of symmetric and anti-symmetric parts, but i dont see why the symmetric part vanishes under contraction.

If someone could explain I'd be very grateful - thank you!

Last edited: Nov 11, 2014
2. Nov 11, 2014

### my2cts

Your notes are inaccurate. What you tried to note down is probably:
If $a^{\mu\nu}=-a^{\nu\mu}$, then $A^{\mu\nu}a_{\mu\nu}=\frac{1}{2}a_{\mu\nu}(A^{\mu\nu}-A^{\nu\mu})$,
for any $A^{\mu\nu}$.

3. Nov 11, 2014

Whoops that B was meant to be an A -- error fixed! but what do you mean by inaccurate exactly? what part is wrong?

4. Nov 12, 2014

### my2cts

The part where you wrote B instead of A ?

5. Nov 12, 2014