# Steinberg group

Given an associative ring $R$ with identity^{}, the Steinberg group $St(R)$ describes the minimal amount of relations between elementary matrices^{} in $R$.

For $n\ge 3$, define $S{t}_{n}(R)$ to be the free abelian group^{} on symbols ${x}_{ij}(r)$ for $i,j$ distinct integers between $1$ and $n$, and $r\in R$, subject to the following relations:

$${x}_{ij}(r){x}_{ij}(s)={x}_{ij}(r+s)$$ |

$$[{x}_{ij},{x}_{kl}]=\{\begin{array}{cc}1\hfill & \text{if}j\ne k\text{and}i\ne l\hfill \\ {x}_{il}(rs)\hfill & \text{if}j=k\text{and}i\ne l\hfill \\ {x}_{kj}(-sr)\hfill & \text{if}j\ne k\text{and}i=l.\hfill \end{array}$$ |

Note that if ${e}_{ij}(r)$ denotes the elementary matrix with one along the diagonal, and $r$ in the $(i,j)$ entry, then the ${e}_{ij}(r)$ also satisfy the above relations, giving a well defined morphism $S{t}_{n}(R)\to {E}_{n}(R)$, where the latter is the group of elementary matrices.

Taking a colimit over $n$ gives the Steinberg group $St(R)$. The importance of the Steinberg group is that the kernel of the map $St(R)\to E(R)$ is the second algebraic $K$-group of the ring $R$, ${K}_{2}(R)$. This also coincides with the kernel of the Steinberg group. One can also show that the Steinberg group is the universal central extension of the group $E(R)$.

Title | Steinberg group |
---|---|

Canonical name | SteinbergGroup |

Date of creation | 2013-03-22 16:44:38 |

Last modified on | 2013-03-22 16:44:38 |

Owner | dublisk (96) |

Last modified by | dublisk (96) |

Numerical id | 4 |

Author | dublisk (96) |

Entry type | Definition |

Classification | msc 19C09 |