Is it true that $Alt(Alt(T)\bigotimes S) = Alt(T \bigotimes S)$ where $T$ and $S$ are k-tensors?

Question

I have $Alt(T) = \frac{1}{k!} \sum_{\sigma} \text{sign}(\sigma)T(v_{\sigma_1},...,v_{\sigma_k})$.

Where the above is just summing over all permutations $\sigma$ of $\{1,..,k\}$

Therefore

$Alt(T)\bigotimes S = \frac{1}{k!} \sum_{\sigma} \text{sign}(\sigma)T(v_{\sigma_1},...,v_{\sigma_k})S(v_{k+1},...,v_{2k})$

The problem is I don't know how to take ALT of this - I'm very lost here and confused, so forgive me if it's obvious that the theorem isn't even correct.

Answer 1

Following Spivak, first show that if $\text{Alt}(S)=0$ and $T$ is any multilinear operator, then $\text{Alt}(S\otimes T)=0$. To do this, show that the sum defining $\text{Alt}(S\otimes T)$ has subsums that cancel (page 80 of his book Calculus on Manifolds, or Theorem 4-4).

To show that $\text{Alt}(\text{Alt}(S)\otimes T) = \text{Alt}(S\otimes T)$, note that $\text{Alt}(S)- S$ maps to zero under $\text{Alt}$ since $\text{Alt}$ is linear and $\text{Alt}^2=\text{Alt}$, and then by the above for any $T$,

$$\text{Alt}(\text{Alt}(S)\otimes T - S\otimes T) = 0$$

which is what you want.