# How to get Haar measure on a compact Lie group, given the complexification?


Let $\fg$ be a complex semisimple Lie algebra. I have two objects on my mind:

-- the algebraic group over $\Cplx$ obtained by exponentiating $\fg$; and

-- the compact Lie group obtained by exponentiating the compact real form of $\fg$. Wikipedia

To be honest, what I'm really interested in the matrix coefficient ring of the latter, but according to what I've swotted up on, this is naturally isomorphic in a fairly precise sense to the coordinate ring of the former, in the sense of affine algebraic varieties. Let me denote this ring by $R$.

Anyway, if I think of $R$ as an algebra of smooth functions on the compact group, then there is a pairing $R\times R \to \Cplx$ given by $(f,g)\mapsto \int f(p)g(p)\,dp$ where $dp$ denotes a fixed choice of Haar measure on the compact group. My somewhat vague and naive question is:

Q1. What is the right way to "see" this pairing when I think of $R$ as the coordinate ring of an affine algebraic variety over $\Cplx$?

Here, by "see", I mean that I want some way of getting it in principle from the algebro-geometric object $R$, which is not just a concatenation of big existence results.

This may or may not require an answer to another simple-minded question:

Q2. What is the natural/canonical way of "seeing" the compact group, up to suitable isomorphism, if we are just given $R$ + Hopf algebra structure?

-
To be precise, $R$ is the coordinate ring of the simply connected semisimple group $G$ with a given semisimple Lie algebra $\mathfrak{g}$ over a field $k$ of char. 0. Section 3 in Hochschild's 1970 paper in vol. 34 of the Illinois Math Journal provides a commutative Hopf algebra structure on the direct limit of the duals $(U(\mathfrak{g})/J)^{\ast}$ for 2-sided ideals $J$ of finite codimension in the universal enveloping algebra, and sections 2 and 5 of his 1959 paper in the same journal show that this limit is $R$ (e.g., it really is a domain finitely generated over $k$). –  user22479 Sep 1 '12 at 6:02

I'll write $K$ for the compact group. A correction to start with: You write $\int_K f g$, but I assume you mean $\int_K f \bar{g}$.

Complex conjugation We need to identify the map $\sigma: R \to R$ which has the property that $\sigma(f)|_K = \overline{\sigma(f)}$. Note that this map will NOT be complex conjugation of functions on $G$; all of the elements of $R$ are holomorphic as functions on $G$. For example, if $K=U(n)$ and $G=GL_n$, then $\sigma(f)(g) = \overline{f}(\overline{g}^{-T})$.

We cannot define $\sigma$ completely canonically, because choosing a different $K$ will change $\sigma$. However, if you have a subalgebra $\mathfrak{k}$ of $\mathfrak{g}$ which you want to be the algebra of $K$, then I think that $\sigma$ is the unique $\mathbb{C}$-antilinear Hopf automorphism (commutes with both multiplication and co-multiplication) whose induced action on $\mathfrak{g}$ fixes $\mathfrak{k}$. And, of course, an $\mathbb{R}$-subalgera $L$ of $\mathfrak{g}$ is a possible candidate for $L$ iff $\mathfrak{g} = \mathfrak{k} \otimes \mathbb{C}$ and the Killing form is positive definite (or maybe its negative definite) on $L$.

Integration Integration is a special case of the Reynolds operator. In general, if $G$ is a reductive group, and $V$ is an algebraic representation of $G$, than $\rho: V \to V$ is the unique $G$-equivariant idempotent whose image is the $G$-invariants. I'm not sure of a really clean way to define it. One fairly algebraic way is the following: Let $v \in V$, and let $U \subset V$ be the sub-rep spanned by the $G$-orbit of $v$. Note that $U$ is finite dimensional. Let $\Omega$ denote the Casimir; let the characteristic polynomial of $\Omega$ acting on $U$ be $x^k g(x)$ with $g(0) \neq 0$; then $\rho|_U = g(\Omega)/g(0)$. Conceptually, $\rho = \lim_{t \to \infty} e^{- t \Omega}$ (the heat equation) but that's not algebraic.

In your case, you want the Reynolds operator for $G$ acting on $R$. Since I think it is easiest to define $\rho$ from the Cassimir, you need to know how to algebraically write down the action of $\mathfrak{g}$ on $R$. I think it is as follows: Let $\epsilon: R \to \mathbb{C}$ be evaluation at the identity (normally called counit). Let $\mathfrak{m}$ be the kernel of $\epsilon$, so $\mathfrak{m}/\mathfrak{m}^2 = \mathfrak{g}^{\vee}$. Then $\mathfrak{g}$ maps $R \to \mathbb{C}$ by $g \cdot f = \langle g, f-\epsilon(f)+\mathfrak{m}^2 \rangle$. This is the standard description of how to see elements of the Zariski tangent space as derivations.

We want to extend from a single vector to an entire left invariant vector field. I think the recipe is $R \stackrel{\Delta}{\longrightarrow} R \otimes R \stackrel{g \cdot \otimes 1}{\longrightarrow} \mathbb{C} \otimes R \cong R$, where $\Delta$ is comultiplication.

Hope that helps!

-
Thanks David - sorry for taking a while to respond; I will need some time to digest this. I think that "Reynolds operator" may have been the two word answer to my underlying question, but the extra detail is of course most welcome! –  Yemon Choi Sep 4 '12 at 2:36
Small comment: actually, for what I want to ask next, I think I really do want the C-bilinear map R\times R \to C and not the sesquilinear one -- I have something landing in the dual space of R and I want to show that it arises from something landing in R –  Yemon Choi Sep 4 '12 at 2:38