# representation ring

Let $G$ be a group and $k$ a field. Consider the class

$$\mathcal{R}=\{X|X\text{is a representation of}G\text{over}k\}$$ |

and its subclass ${\mathcal{R}}_{f}$ consisting of those representations^{} which are finite-dimensional as vector spaces^{}. We consider a special representation

$$\mathcal{F}=(V,\cdot )$$ |

where $V$ is a fixed vector space with a basis $\mathcal{B}$ which is in bijective^{} correspondence with $G$. If $f:\mathcal{B}\to G$ is a required bijection, then we define ,,$\cdot $” on basis $\mathcal{B}$ by

$$g\cdot b=gf(b)$$ |

where on the right side we have a multiplication in $G$. It can be shown that this gives us a well-defined representation and further more, if $X\in {\mathcal{R}}_{f}$, then there exists an epimorphism^{} of representations

$$e:{\mathcal{F}}^{n}\to X$$ |

for some $n\in \mathbb{N}$ ($\mathcal{F}$ is a ,,free” representation). In particular every finite-dimensional representation is a quotient of a direct sum^{} of copies of $\mathcal{F}$. This fact shows that a maximal subclass $\mathcal{X}\subset {\mathcal{R}}_{f}$ consisting of pairwise nonisomorphic representations is actually a set (note that $\mathcal{X}$ is never unique). Fix such a set.

Definition. The representation semiring^{} $\overline{{R}_{k}(G)}$ of $G$ is defined as a triple $(\mathcal{X},+,\cdot )$, where $\mathcal{X}$ is a maximal set of pairwise nonisomorphic representations taken from ${\mathcal{R}}_{f}$. Addition and multiplication are given by

$$X+Y=Z$$ |

where $Z$ is a representation in $\mathcal{X}$ isomorphic to the direct sum $X\oplus Y$ and

$$X\cdot Y={Z}^{\prime}$$ |

where ${Z}^{\prime}$ is a representation in $\mathcal{X}$ isomorphic to the tensor product^{} $X\otimes Y$. Note that $\overline{{R}_{k}(G)}$ is not a ring, because there are no additive inverses.

The representation ring^{} ${R}_{k}(G)$ is defined as the Grothendieck ring (http://planetmath.org/GrothendieckGroup) induced from $\overline{{R}_{k}(G)}$. It can be shown that the definition does not depend on the choice of $\mathcal{X}$ (in the sense that it always gives us naturally isomorphic rings).

It is convenient to forget about formal definition which includes the choice of $\mathcal{X}$ and simply write elements of $\overline{{R}_{k}(G)}$ as isomorphism classes of representations $[X]$. Thus every element in ${R}_{k}(G)$ can be written as a formal difference^{} $[X]-[Y]$. And we can write

$$[X]+[Y]=[X\oplus Y];$$ |

$$[X][Y]=[X\otimes Y].$$ |

Title | representation ring |
---|---|

Canonical name | RepresentationRing |

Date of creation | 2013-03-22 19:19:02 |

Last modified on | 2013-03-22 19:19:02 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 7 |

Author | joking (16130) |

Entry type | Definition |

Classification | msc 20C99 |