symmetrizer

Let $V$ be a vector space over a field $F$. Let $n$ be an integer, where if $\mathrm{char}(F)\ne 0$. Let $S_n$ be the symmetric group on$\{1,\mathrm{\dots },n\}.$The linear operator $S:V^{\otimes n}\to V^{\otimes n}$ defined by:

$$S=\frac{1}{n!}\sum _{\sigma \in S_n}P(\sigma )$$

is called the symmetrizer.Here $P(\sigma )$ is the permutation operator.It is clear that $P(\sigma )S=SP(\sigma )=S$ for all $\sigma \in S_n$.

Let $S$ be the symmetrizer for $V^{\otimes n}$. Then an order-n tensor $A$ issymmetric (http://planetmath.org/SymmetricTensor) if and only $S(A)=A$.

Proof
If $A$ is then

$$S(A)=\frac{1}{n!}\sum _{\sigma \in S_n}P(\sigma )A=\frac{1}{n!}\sum _{\sigma \in S_n}A=A.$$

If $S(A)=A$ then

$$P(\sigma )A=P(\sigma )S(A)=P(\sigma )S(A)=S(A)=A$$

for all $\sigma \in S_n$, so $A$ is .

The theorem says that a is an eigenvector of the linear operator $S$ corresponding to the eigenvalue 1. It is easy to verify that$S^2=S$, so that $S$ is a projection onto $S^n(V)$.