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A homogenious polynomial of degree $k\in{\Bbb N}$ on a finite dimensional vector space $X$ over $\Bbb R$ is a polynomial $f:X\to{\Bbb R}$ satisfying the identity $$ f(\lambda\cdot x)=\lambda^k\cdot f(x),\qquad x\in X,\quad \lambda\in{\Bbb R} $$ Qiaochu Yuan explained me a week ago that the formulas $$ f(x)=\alpha(x,...,x),\qquad x\in X, $$ $$ \alpha(x_1,...,x_k)=\frac{1}{k!}\cdot\frac{\partial}{\partial \lambda_k}... \frac{\partial}{\partial \lambda_1} f(\lambda_1\cdot x_1+...+\lambda_k\cdot x_k)\Big|_{\lambda_i=0},\qquad x_i\in X, $$ establish a bijection between homogenious polynomials $f:X\to{\Bbb R}$ of degree $k$ on $X$ and symmetric polylinear forms $\alpha:X^k\to{\Bbb R}$. (As far as I understand, the last formula here is called polarization.)

When $k=2$ we can, first, define homogenious polynomial as a quadratic form, i.e. a polynomial $f:X\to{\Bbb R}$ satisfying the parallelogram identity, $$ \frac{f(x+y)+f(x-y)}{2}=f(x)+f(y),\qquad x,y\in X, $$ and, second, replace the (analytic) polarisation formula by the purely algebraic formula $$ \alpha(x,y)=\frac{f(x+y)-f(x)-f(y)}{2},\qquad x,y\in X. $$

My question:

What are the analogs of the last two identities in higher degrees (when $k>2$)?

In other words, which identities characterize homogenious polynomials $f$ of degrees 3, 4,..., instead of the homogenity condition, and which (algebraic) identities can be used for expressing $\alpha$ instead of the (analytic) polarization formula?

  • There is nothing analytic about the polarization formula. Taking derivatives of polynomials is a purely algebraic operation, and e.g. the formula makes sense and is true over any field of the right characteristic ($0$ or larger than $k$). – Qiaochu Yuan Apr 09 '15 at 17:30
  • Yes... differentiation can be defined inductively on polynomials... Qiaochu, anyway this formula (of polarization) is absurd, I don't like it. What is it called when a variable is not free and at the same time not bound in a formula? It seems to me I saw an explanation somewhere that this is not good... – Sergei Akbarov Apr 09 '15 at 20:15
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    Have you seen this post? https://math.stackexchange.com/questions/481167/polarization-formula – mr_e_man Jun 11 '20 at 00:04

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