# Symmetric multilinear form from an homogenous form.

Let $V$ be a $n$-dimensional $\mathbb{R}$-vector space. Let $\phi:V\to\mathbb{R}$ a homogeneous form of degree $n$, i.e. $\phi(\lambda v)=\lambda^n \phi(v)$.

If we define the symmetric multinear [!see edit!] operator $\Phi:V^n\to\mathbb{R}$ by $$\Phi[v_1,\ldots,v_n]=\frac{1}{n!} \sum_{k=1}^n \sum_{1\leq j_1<\cdots<j_k\leq n} (-1)^{n-k}\phi (v_{j_1}+\cdots+v_{j_k}),$$ then we can see that for any $v\in V$, $\Phi(v,\ldots,v)=\phi(v)$ (using homogeneity and this combinatoric formula).

The question is: Is this necessarily the only symmetric multilinear form $\Phi$ satisfying $\Phi(v,\ldots,v)=\phi(v)$? If yes, why? If not, is there extra condition(s) on $\phi$ which would imply the uniqueness?

[Edit: It is actually not clear that with this setting $\Phi$ is multilinear. We probably need to add some condition on $\phi$.]

• Where do the $v_i$'s enter on the r.h.s? Are those $s_j$'s supposed to be $v$'s? – Alex R. Aug 16 '13 at 22:52
• Thanks and sorry. yes they are $v$'s. I edited it already. – Gilles Bonnet Aug 16 '13 at 22:55

The answer is almost written down in your question! You simply need to prove the identity written with $\Phi[\sum v,\sum v, \cdots, \sum v]$ instead of $\phi(\sum v)$ on the RHS must hold for any multilinear symmetric $\Phi$. This, together with the condition on the diagonal, shows this is necessarily the form of $\Phi$.
To prove this, you could take as inspiration the famous bilinear case $Q(u+v,u+v)-Q(u,u)-Q(v,v) = 2 Q(u,v)$.
Alternatively, you can prove uniqueness by deriving the expression from the above Wiki article. Consider $A=\lambda_1 v_1+\cdots+\lambda_n v_n$. Then $$\partial/\partial \lambda_1 \Phi[A,A,\cdots,A]\mid_{\lambda_1=0} = \cdots = n\Phi[v_1,A',A',\cdots,A']$$ and so on inductively allows you to express $\Phi$ at any point as a derivative of $\phi$. ($A'$ has no 1 term.)
• Nice. I just add here the complement to understand the dots in the last equality. By multilinearity and symmetry we have: $$\Phi[A,\ldots,A]=\sum_{k=0}^n \binom{n}{k} \lambda_1^k\Phi[\underbrace{v_1,\ldots,v_1}_{k},\underbrace{A',\ldots,A'}_{n-k}].$$ But $\partial/\partial\lambda_1 (\lambda_1^k\Phi[v_1,\ldots,v_1,A',\ldots,A'])|_{\lambda_1=0}=1$ for $k=1$ and $0$ otherwise. Hence $\partial/\partial\lambda_1 \Phi[A,\ldots,A] |_{\lambda_1=0}=n\Phi[v_1,A',\ldots,A'].$ – Gilles Bonnet Aug 17 '13 at 8:29
• This is an old question, but I want to say thank you for the great answer! The "alternatively" is kind of messy, actually. Directly tackling $\Phi[v_1,\ldots,v_n]=\frac{1}{n!} \sum_{k=1}^n \sum_{1\leq j_1<\cdots<j_k\leq n} (-1)^{n-k}\Phi (\sum_{m=1}^k v_{j_m}, \dots, \sum_{m=1}^k v_{j_m})$ term-wise, on the other hand, leads to a nice and concise combinatorial proof. – 4ae1e1 May 21 '14 at 5:41