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I'm going to asking a question about the polarization formula of an $n$-form.

Given an $n$-homogeneous form $q:V\rightarrow \mathbb{R}$ on a vector space $V,$ i.e., $\forall v\in V, r\in \mathbb{R},$ we have $q(rv) = r^nq(v),$ the polarization formula tells us that we can have an $n$-multilinear map $f:V^n\rightarrow\mathbb{R},$ $(v_1,...,v_n)\mapsto \frac{1}{n!}(\partial_{\alpha_1}...\partial_{\alpha_n} q(\alpha_1v_1+...+\alpha_nv_n))|_{(\alpha_1,...,\alpha_n)=(0,...,0)}$ such that $\forall v\in V,$ $f(v,...,v) = q(v).$

The equality $f(v,...,v) = q(v)$ is obvious, while my problem is that the $n$-multilinearity of this $f.$ This may be related to the linearity of the differential operators, but I cannot solve it by following the definition to prove $$\frac{1}{n!}(\partial_{\alpha_1}...\partial_{\alpha_n} q(\alpha_1(av_1+v)+...+\alpha_nv_n))|_{\alpha=(0,...,0)}=\\a\cdot\frac{1}{n!}(\partial_{\alpha_1}...\partial_{\alpha_n} q(\alpha_1v_1+...+\alpha_nv_n))|_{\alpha=(0,...,0)}+\frac{1}{n!}(\partial_{\alpha_1}...\partial_{\alpha_n} q(\alpha_1v+...+\alpha_nv_n))|_{\alpha=(0,...,0)}.$$ Or may my work be not feasible?

Thanks for any ideas in advance!

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For any $v\in V$, write $\alpha\mapsto q(\alpha v)$ as the composition $q\circ g_v$ of the maps $q:V\to\mathbb{R}$ and $g_v:\mathbb{R}\to V, g_v(\alpha)=\alpha v$. Now $g_v'(0) = v$ and by the chain rule $$\partial_\alpha q(\alpha v)\vert_{\alpha=0} = dq(g_v(0))g_v'(0) = dq(0)v.$$ Since the derivative $dq(0):V\to \mathbb{R}$ of $q$ at 0 is a linear map, $$\partial_\alpha q(\alpha (av+w))\vert_{\alpha=0} = dq(0)(av+w) = a\cdot dq(0)v+dq(0)w = a\cdot\partial_\alpha q(\alpha v)\vert_{\alpha=0}+\partial_\alpha q(\alpha w)\vert_{\alpha=0}.$$

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  • $\begingroup$ I got your ideas, thsnks a lot! $\endgroup$ – tommy xu3 Mar 28 '17 at 8:59

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