# representation ring vs burnside ring

Let $G$ be a finite group^{} and let $k$ be any field. If $X$ is a $G$-set, then we may consider the vector space ${V}_{k}(X)$ over $k$ which has $X$ as a basis. In this manner ${V}_{k}(X)$ becomes a representation of $G$ via action induced from $X$ and linearly extended to ${V}_{k}(X)$. It can be shown that ${V}_{k}(X)$ only depends on the isomorphism class of $X$, so we have a well-defined mapping:

$$[X]\mapsto [{V}_{k}(X)]$$ |

which can be easily extended to the function

$$\beta :\mathrm{\Omega}(G)\to {R}_{k}(G);$$ |

$$\beta ([X])=[{V}_{k}(X)]$$ |

where on the left side we have the Burnside ring and on the right side the representation ring^{}. It can be shown, that $\beta $ is actually a ring homomorphism^{}, but in most cases it neither injective^{} nor surjective^{}. But the following theorem^{} due to Segal gives us some properties of $\beta $:

Theorem (Segal). Let $\beta :\mathrm{\Omega}(G)\to {R}_{\mathbb{Q}}(G)$ be defined as above with rationals as the underlying field. If $G$ is a $p$-group for some prime number $p$, then $\beta $ is surjective. Furthermore $\beta $ is an isomorphism^{} if and only if $G$ is cyclic.

Title | representation ring vs burnside ring |
---|---|

Canonical name | RepresentationRingVsBurnsideRing |

Date of creation | 2013-03-22 19:19:05 |

Last modified on | 2013-03-22 19:19:05 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 4 |

Author | joking (16130) |

Entry type | Theorem |

Classification | msc 20C99 |