1
$\begingroup$

I found the following definition of regular representation:

Let $V$ be a complex vector space. A representation $(\rho\colon G\to \mathbf{GL}(V), V)$ of a Linear Algebraic Group $G$ is regular if $\text{dim}(V)<\infty$ and the functions on $G$, \begin{align} G &\to \mathbb{C}\\ g&\mapsto \left<v^{\ast},\rho(g)v \right>, \end{align} which we call matrix coefficients of $\rho$, are regular for all $v\in V$ and $v^{\ast}\in V^{\ast}$.

Here, $\left<v^{\ast},v \right>:= v^{\ast}(v)$. I have several questions; first of all, is there any reason why the evaluation $v^{\ast}(v)$ is denoted as an inner product? Why do they call these functions the coefficients of the matrix? Is this the same definition of regular representation that one sees in representation theory? I mean, the representation $L\colon G\to \mathbf{GL}(\mathbb{C}G)$ defined by $$L_{g}\sum_{h\in G}c_{h}h = \sum_{h\in G}c_{h}gh = \sum_{x\in G}c_{g^{-1}x}x$$

$\endgroup$

1 Answer 1

3
$\begingroup$

The reason for denoting the evaluation as a pairing (for example, the pairing you seem to be most familiar with is the inner product) is that often $V^{**}\simeq V$, and so one wants to view $V$ and $V^*$ as being dual to each other, but not with one space being the "original" and one being "the dual of the original space." The pairing notation emphasizes this symmetry, but it is just notation.

If you choose a basis of $V$ and the corresponding dual basis of $V^*$, then the matrix coefficients are exactly the entries of the matrix $\rho(g)$. This is the reason for their name.

You last question is slightly vague; $V$ above is any finite-dimensional representation, such $V$ are certainly not all going to be isomorphic to the regular representation of $G$. In fact, none of them are! If you define the "regular representation" naively, copying the definition from finite groups, you end up with an inifinite-dimensional space with action as you've defined. However, it doesn't have any useful properties; representations of $G$ do not correspond to modules over this algebra, and as a representation of $G$, it doesn't decompose into all the simple finite-dimensional representations of $G$. For linear algebraic groups, the revelant algebra is actually the coordinate algebra $\mathbb{C}[G]$ of $G$ as a variety. The group structure of $G$ makes this algebra into a Hopf algebra, by the Yoneda lemma, and algebraic representations of $G$ are equivalent to co-modules over $\mathbb{C}[G]$. You can find this written in any introductory book on algebraic groups.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.