Condition for nullity of quadrilinear form

Question

I have been told the following.

Lemma

Suppose $V$ is a vector space over a field $K$, and $T:V\times V\times V\times V\to K$ is a multilinear map with the following properties holding for all $a,b,c,d\in V$:

\begin{align*} T(a,b,c,d)={}&-T(b,a,c,d) \tag{P1}\\ T(a,b,c,d)={}&-T(a,b,d,c) \tag{P2}\\ T(a,b,c,d)={}&T(c,d,a,b) \tag{P3} \end{align*} $$T(a,b,c,d)+T(b,c,a,d)+T(c,a,b,d)=0. \tag{P4}$$

Then the following are equivalent:

1. $T=0$, i.e. $T(a,b,c,d)=0$ for all $a,b,c,d\in V$;
2. $T(a,b,b,a)=0$ for all $a,b\in V$.

I tried proving it, but I got stuck. My attempt was to compute $T(v+v',w+w',w+w',v+v')$ and the same with minuses, because the teacher suggested it was a sort of ``polarization identity for $T$'', and that is an analog of what you do when proving the polarization identity. So I expanded, then I added $T(v-v',w-w',w-w',v-v')$, taking the cancellations of terms with an odd number of primed vectors and the summing of terms with an even number of primed vectors into account. And I got:

\begin{multline*} \hspace{-1cm} \underline{T(v+v',w+w',w+w',v+v')}+\underline{T(v-v',w-w',w-w',v-v')}={} \\ {}=2T(v,w,w',v')+\underline{2T(v,w,w,v)}+\underline{2T(v,w',w',v)}+2T(v,w',w,v')+{} \\ {}+\underline{2T(v',w,w,v')}+2T(v',w,w',v)+2T(v',w',w,v)+\underline{2T(v',w',w',v')}.\hspace{-1cm} \end{multline*}

And that only uses multilinearity to expand the two terms, admitting I didn't get any calculations wrong. Then I noticed how condition 2 makes the underlined terms zero, reducing the above to:

$$\underline{2T(v,w,w',v')}+\overline{2T(v,w',w,v')}+\overline{2T(v',w,w',v)}+\underline{2T(v',w',w,v)}=0.$$

Now by properties P1, P2 and P3, one deduces that:

$$T(a,b,c,d)=T(d,c,b,a) \tag{A}$$

for all $a,b,c,d\in V$. Applying that to the above, we can sum the similarly marked terms, getting, after dividing by 4, that:

$$T(v,w,w',v')+T(v,w',w,v')=0.$$

But if I apply property P4 to this, I get to nothing better. I would like to simplify this into $T(v,w,w',v')$ or the likes, but I can't manage.

Any ideas?

Answer 1

I'm sure that with enough care and effort the method you are trying would eventually work. But if you know about tensor products there's a way to shortcut some of the hard work by using the actual polarization identity.

Because of P3 we can view $T$ as a symmetric bilinear form on $V\otimes V$. That is to say $T:(V\otimes V)\times(V\otimes V)\rightarrow K$. Then by the polarization identity $T$ is zero iff it's zero for all $\alpha\in V\otimes V$. But $V\otimes V$ is spanned by simple tensors of the form $a\otimes b$. So $T$ is zero iff $T(a,b,a,b)=0$ for all $a,b\in V$. Now apply P2.