# Symmetrization is the unique $k$-tensor

$$\newcommand{\Sym}[1]{\operatorname{Sym}{#1}}$$

Let $$V$$ be a $$n$$-dim real vector space with dual space $$V^*$$. Let $$\alpha$$ be a covariant $$k$$-tensor, i.e., $$\alpha \in T^k(V^*) \equiv (V^*)^{\otimes k}$$. Then how would you show that the symmetrization $$\Sym{\alpha}$$ of $$\alpha$$ is the unique symmetric $$k$$-tensor such that $$\boxed{\Sym{\alpha} (v,...,v) =\alpha(v,...,v), \qquad v\in V}$$

Note the symmetrization is defined by $$\Sym{\alpha} (v_1,...,v_k) = \frac{1}{k!} \sum_{\sigma\in S_k} \alpha (v_{\sigma_1},...,v_{\sigma_k})$$ Where $$S_k$$ is the symmetric group of $$k$$ and $$T^k(V^*)$$ is identified as the space of multi-linear real functionals on $$V^k$$.

EDIT: The key seems to be proving the following fact $$\boxed{k!(v_1\cdots v_k) = \sum_{l=0}^k (-1)^l \sum_{|J|=l,J\subseteq \{1,...,k\}}\left( \sum_{i\in\{1,...,k\}-J} v_i \right)^k}$$ where the $$|J|$$ is the number of elements in set $$J$$. I used $$v_1\cdots v_k$$ to denote $$\beta(v_1,..., v_k)$$ where $$\beta$$ is a symmetric $$k$$-tensor. Similarly, I used $$v^k$$ to denote $$\beta(v,...,v)$$.

E.g. when $$k=3$$, we have $$3!(abc)=(a+b+c)^3-(a+b)^3-(a+c)^3-(b+c)^3+a^3+b^3+c^3$$ However I'm having troubles proving this identity for general $$k$$.

EDIT 2: I just learned that the formula in the first EDIT refers to the polarization formula, which can be found in this post

• Good, i think it is correct. Yes it is the polarization – Federico Fallucca Apr 12 '19 at 6:00

The idea is the following:

$$k=2$$

In this case $$V^*\otimes V^*$$ is the space of bilinear forms on $$V$$ in which you know that it is possibile identifies each symmetric bilinear forms by its quadratic form:

For each $$\beta\in V^*\otimes V^*$$ you have that

$$\beta(v,w)=\frac{1}{2}(\beta(v+w,v+w)-\beta(v-w,v-w))$$

so you have that if $$\beta$$ is a symmetric tensor such that

$$\beta(v,v)=\alpha(v,v)$$ then $$\beta=\alpha$$

• How would you extend this idea to $k\ge 3$. The algebra seems quite unwieldy. – Andrew Yuan Apr 11 '19 at 20:20
• It is difficult to generalize, I think that you can do by induction – Federico Fallucca Apr 11 '19 at 21:20