# monoid bialgebra

Let $G$ be a monoid and $k$ a field. Consider the vector space $kG$ over $k$ with basis $G$. More precisely,

$$kG=\{f:G\to k|f(g)=0\text{for almost all}g\in G\}.$$ |

We identify $g\in G$ with a function ${f}_{g}:G\to k$ such that ${f}_{g}(g)=1$ and ${f}_{g}(h)=0$ for $h\ne g$. Thus, every element in $kG$ is of the form

$$\sum _{g\in G}{\lambda}_{g}g,$$ |

for ${\lambda}_{g}\in k$. The vector space $kG$ can be turned into a $k$-algebra, if we define multiplication as follows:

$$g\cdot h=gh,$$ |

where on the right side we have a multiplication in the monoid $G$. This definition extends linearly to entire $kG$ and defines an algebra structure on $kG$, where neutral element^{} of $G$ is the identity^{} in $kG$.

Furthermore, we can turn $kG$ into a coalgebra as follows: comultiplication $\mathrm{\Delta}:kG\to kG\otimes kG$ is defined by $\mathrm{\Delta}(g)=g\otimes g$ and counit $\epsilon :kG\to k$ is defined by $\epsilon (g)=1$. One can easily check that this defines coalgebra structure on $kG$.

The vector space $kG$ is a bialgebra^{} with with these algebra and coalgebra structures and it is called a monoid bialgebra.

Title | monoid bialgebra |
---|---|

Canonical name | MonoidBialgebra |

Date of creation | 2013-03-22 18:58:48 |

Last modified on | 2013-03-22 18:58:48 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 4 |

Author | joking (16130) |

Entry type | Example |

Classification | msc 16W30 |