Generalization of the polarisation formula for symmetric bilinear forms to symmetric multilinear forms

Question

Given a symmetric bilinear form $f:V\times V \to K$ , where $V$ is a vector space 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

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 2

The following link contains an explicit formula for bilinear and trilinear forms:

http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.53.3481&rep=rep1&type=ps

It is a ps file by Erik G.F. Thomas "A polarization identity for multilinear maps".

A general formula can be constructed in Mathematica by means of the shift operator mentioned in that paper.

I hope it will be useful. It is not so easy to find an explicit formula in the net!

Answer 3

The reason is Schur–Weyl duality. $\newcommand{\GL}{\operatorname{GL}}$

The subspace $W = \langle \forall v \in V \, \mid \, v \otimes v \otimes \cdots \otimes v\rangle$ forms a $\GL(V)$ subrepresentation of $\bigotimes^kV$ 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 $\mbox{Sym}^kV \subset \bigotimes^kV$.

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

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

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

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

Answer 4

A later version of the paper referenced above now appears on ArXiv https://arxiv.org/abs/1309.1275 edited by TH Koornwinder after Thomas's death. The appendix by Koornwinter attributes the result originally to

S. Mazur and W. Orlicz, Grundlegende Eigenschaften der polynomischen Operationen. Studia Math. 5 (1934) 50–68.

with

J. Bochnak and J. Siciak, Polynomials and multilinear mappings in topological vector spaces. Studia Math. 39 (1971), 59–76.

giving a slightly more general formula. They agree with Bazin's formula above but with 0 and 1 instead of +1 and -1 (and hence without the power of 2).

$$M(v_1,...,v_k) = \frac{1}{k!} \sum\limits_{\epsilon_1,...,\epsilon_k=0}^1 (-1) ^{n- \epsilon_1-\cdots-\epsilon_k} M(\epsilon_1 v_1+ \cdots \epsilon_k v_k)^k $$

Answer 5

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