# symmetrizer

Let $V$ be a vector space over a field $F$. Let $n$ be an integer, where $n<\mathrm{char}(F)$ if $\mathrm{char}(F)\neq 0$. Let $S_{n}$ be the symmetric group on $\{1,\ldots,n\}.$ 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 $A$ is symmetric (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 $S^{n}(V)$.

Title symmetrizer Symmetrizer 2013-03-22 16:15:44 2013-03-22 16:15:44 Mathprof (13753) Mathprof (13753) 8 Mathprof (13753) Definition msc 15A04