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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?

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

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See also http://en.wikipedia.org/wiki/Polarization_of_an_algebraic_form

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.)

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