# 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 $A$ 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 |
---|---|

Canonical name | Symmetrizer |

Date of creation | 2013-03-22 16:15:44 |

Last modified on | 2013-03-22 16:15:44 |

Owner | Mathprof (13753) |

Last modified by | Mathprof (13753) |

Numerical id | 8 |

Author | Mathprof (13753) |

Entry type | Definition |

Classification | msc 15A04 |