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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?

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I'll rephrase my question: Why is any symmetric k-linear form M:Vx...xV -> determined by the values of M[v]^k := M[v,...,v] for all v in V? I am looking for a PROOF. –  Felix Wutschke Jun 3 '12 at 15:41
This looks like homework. Voting to close until some context is given –  Igor Rivin Jun 3 '12 at 16:37
This is not my homework. I have just completed Linear Algebra 1, which is more basic than the question I posted. I stumbled upon this in a German Analysis book in my free time! I need to understand this because the book uses multilinear forms to define higher order differentials of functions f:R^n -> R. I have never heard of multilinear forms before today and have a hard time dealing with them. For M(x1,---xk) I tried to substitute linear combinations of a basis v1,..,vn but I couldn't simplify it further. Please help me get this right, I am really trying to understand it. Thank you, Feli –  Felix Wutschke Jun 3 '12 at 19:54
Answers to (and the general context of) mathoverflow.net/questions/85213/… might also be somewhat enlightening. –  Vladimir Dotsenko May 29 '13 at 9:01

4 Answers 4

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!. $

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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.

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This could be considered overkill. You can write down a simple formula for $M(x_1,\dots ,x_k)$ as a linear combination of terms of the form $M(x,\dots ,x)$. And it does look like homework. –  Tom Goodwillie Jun 3 '12 at 17:08
This is the most general version of polarization that I know. I leave my answer in hopes that it helps a future google searcher. –  John Wiltshire-Gordon Jun 3 '12 at 17:21
Most general? The proof by explicit formulas works over any ring in which k factorial is invertible, and shows that symmetric multilinear functors of $k$ variables correspond bijectively with homogeneous degree $k$ polynomials. –  Tom Goodwillie Jun 3 '12 at 20:29
@Tom Goodwillie By "most general" I meant that any symmetry type is accommodated. Of course, symmetric and exterior powers are familiar, and so the polarization coefficients can be obtained through a little experimentation. But there are other symmetry types corresponding to other irreps of S_k and the corresponding polarizations would be much harder to guess. Continued... –  John Wiltshire-Gordon Jun 4 '12 at 0:09
For example, if I have a linear map from the degree 4 portion of the free Lie algebra on V and I have already specified its values on brackets of the form [v,[w,[v,w]]], is the map determined? If so, how can I write a general [x,[y,[z,t]]] as a linear combination of brackets of this unusual form? –  John Wiltshire-Gordon Jun 4 '12 at 0:09

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

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


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!

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