Question

Hello,

Given a symmetric bilinear form $f:V\times V \to K$ , where $V$ is a vectorspace and $K$ is an appropriate field, define the quadratic form $q:V \to K$ as $q(v):= f(v,v)$.

The Polarisation Formula states that $f(x,y) = 1/2\big( q(x+y) - q(x) - q(y)\big)$, which is easily proven.

This means that any symmetric bilinear form $f:V\times V \to K$ is fully determined by the values $f(v,v)$ for all $v \in V$.

I now want to prove the following theorem: Prove that any symmetric $k$-linear form $M:V\times\cdots \times V \to K$ is determined by the values $M[v]^k := M[v,...,v]$ for all $v\in V$.

How does that work?

Answer 1

The reason is Schur-Weyl duality.

The subspace $W = \langle \forall v \in V \, | \, v \otimes v \otimes v \otimes \cdots \otimes v\rangle$ forms a $GL(V)$ subrepresentation of $\otimes^kV$ if we allow $GL(V)$ to act diagonally on tensors.

If we consider the dual action, which is the symmetric group $S_k$ permuting tensor factors, we see that all of the generators of $W$ have the symmetry type of the trivial representation of $S_k$ since they are invariant under these permutations. It follows that the $GL(V)$ subrepresentation $W$ is contained within $\mbox{Sym}^kV \subset \otimes^kV$.

However, by Schur-Weyl duality, the symmetric tensors form an irreducible representation of $GL(V)$---the subrepresentation $W$ is either $0$ or all of $\mbox{Sym}^kV$.

It isn't $0$, so every symmetric tensor $s$ can be written

$$s = \alpha_0 \cdot v_0 \otimes v_0 \otimes v_0 \otimes \cdots \otimes v_0 + \alpha_1 \cdot v_1 \otimes v_1 \otimes v_1 \otimes \cdots \otimes v_1 + \cdots \alpha_l \cdot v_l \otimes v_l \otimes v_l \otimes \cdots \otimes v_l$$

for some suitable choice of $v_i$ and $\alpha_i$.

In other words, the $k^{th}$ powers of the elements of $V$ span $\mbox{Sym}^k V$. It follows that knowing a symmetric multilinear form on the $k^{th}$ powers is enough to determine the form.

Answer 2

You can be completely explicit in this matter. For $T_j$ in a commutative algebra $$ T_1T_2\dots T_k=\frac{1}{2^k k!}\sum_{\epsilon_j=\pm 1} \epsilon_1\dots\epsilon_k(\epsilon_1T_1 +\dots+\varepsilon_{k}T_{k})^k. $$ The following lemma in available in the Euclidean case.

Lemma. Let $V$ be an Euclidean finite-dimensional vector space, and $A$ a symmetric $k$-multilinear form. We have $ \sup_{\Vert T\Vert=1} \vert{A T^k}\vert =\sup_{\Vert{T_j}\Vert=1} \vert{AT_1\dots T_k}\vert. $

This lemma is a consequence of the 1928 paper by O.D. Kellogg [MR1544896]. This is not true in the non-Euclidean case where the inequality $$ \sup_{\Vert T\Vert=1} \vert{A T^k}\vert \le \sup_{\Vert{T_j}\Vert=1} \vert{AT_1\dots T_k}\vert\le \kappa_k \sup_{\Vert T\Vert=1} \vert{A T^k}\vert, $$ holds true in general with the best constant $ \kappa_{k}= k^k/k!. $

Answer 3

This Wikipedia article contains the Polarization of an algebraic form in general.