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\begin{document}

\author{Colin McSwiggen}

\title{The Harish-Chandra integral \\ \large An introduction with examples}

\maketitle

\abstract{This paper introduces Harish-Chandra's integral formula for compact, connected, semisimple Lie groups. It is intended for mathematicians and physicists who are familiar with the basics of Lie groups and Lie algebras but who may not be specialists in representation theory.  I present Harish-Chandra's proof of the formula in contemporary language, work out the integrals for all compact classical groups in detail, and show how to generalize the formula to compute similar integrals over arbitrary compact Lie groups.}

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\tableofcontents

\pagebreak

\section{Introduction}

This paper is an introduction to a powerful formula proved in 1957 by Harish-Chandra \cite{HC} for computing an integral over a compact, connected, semisimple Lie group $G$.  The formula reads \begin{equation} \label{eqn:hc} \Pi(h_1) \Pi(h_2) \int_G e^{\langle \mathrm{Ad}_g h_1, h_2 \rangle} dg = \frac{ [ \! [ \Pi, \Pi ] \!] }{|W|} \sum_{w \in W} \epsilon(w) e^{\langle w(h_1),h_2 \rangle}.\end{equation} To avoid overwhelming the reader with definitions at the very beginning, the full explanation of (\ref{eqn:hc}) and all the necessary notation will wait until Section \ref{sec:statement} below.  Instead, as a motivating example, we start by discussing a corollary that is more widely known than (\ref{eqn:hc}) itself.

A direct consequence of (\ref{eqn:hc}) is the Harish-Chandra-Itzykson-Zuber (HCIZ) integral formula, an important identity in random matrix theory and mathematical physics that was discovered independently by Itzykson and Zuber in 1980 \cite{IZ}.  The HCIZ integral is typically stated as
\begin{equation} \label{eqn:hciz} \int_{U(N)} e^{\mathrm{tr} (AUBU^\dagger)} dU = \left( \prod_{p=1}^{N-1}p! \right) \frac{\det(e^{a_i b_j})_{i,j = 1}^N}{\Delta(A) \Delta(B)} \end{equation}
where $U(N)$ is the group of $N$-by-$N$ unitary matrices, $A$ and $B$ are fixed $N$-by-$N$ diagonal matrices with eigenvalues $a_1 < \hdots < a_N$ and $b_1 < \hdots < b_N$ respectively, and $$\Delta(A) = \prod_{i < j} (a_j - a_i)$$ is the Vandermonde determinant.

This formula is significant for many reasons.  In random matrix theory, it appears in the expressions for the joint spectral densities of a number of matrix ensembles, including off-center Wigner matrices and Wishart matrices \cite[ch. 3]{AG}. Since Wishart matrices model sample covariance estimators, the formula is also of interest in statistics.  In physics, it arises in the partition functions for two-matrix models in quantum field theory and string theory \cite{IZ}, while in combinatorics it has the interpretation of a generating function for the monotone double Hurwitz numbers \cite{GGN}.

There are several proofs of the HCIZ formula in the literature.  The three proofs that are most frequently mentioned are a proof based on constructing the integral from the heat kernel on the space of Hermitian matrices; a representation-theoretic proof using a character expansion and the Cauchy-Binet formula; and a symplectic geometry proof in which the formula is first derived as a stationary phase approximation, which is shown to be exact by the Duistermaat-Heckman theorem \cite{DH, IZ, KJ, TT}.

All of these, however, are quite different from Harish-Chandra's 1957 proof of (\ref{eqn:hc}).  Harish-Chandra's derivation has received relatively little attention compared to these others, even though it predates them by over 20 years and obtains the HCIZ formula as a consequence of a much more general result.  In this paper I give an exposition of Harish-Chandra's proof technique in more contemporary language and demonstrate his general theorem by using it to derive HCIZ-like formulae for integrals over the special unitary, special orthogonal, orthogonal, and unitary symplectic groups.  I also prove a generalization of Harish-Chandra's formula to arbitrary compact real Lie groups that are not necessarily semisimple or connected.  The generalization is not difficult to prove with the aid of the classification of compact real Lie groups, but I do not believe that it has been recorded in the literature.

While some of the specific integral formulae derived in this paper have been published previously, for example in \cite{PEDZ}, I believe there is value in showing the calculations in detail.  Moreover, among those mathematicians and physicists who are familiar with Harish-Chandra's proof, it has acquired a reputation for being rather difficult.  My hope is that the examples and exegesis in this paper will help to make Harish-Chandra's ideas accessible to a wider range readers in contemporary mathematics and physics.

In Section \ref{sec:statement} below, I develop some necessary concepts and notation and then present Harish-Chandra's general theorem.  Section \ref{sec:un_sun} shows how to derive the HCIZ formula from the general theorem.  Section \ref{sec:background} situates the integral formula in the context of the paper \cite{HC} in which it was originally published, and Section \ref{sec:proof} contains the proof of the formula.  Concrete examples of integrals over specific groups are presented in Sections \ref{sec:so2n} through \ref{sec:uspn}, and the generalization to arbitrary compact real Lie groups is proven in Section \ref{sec:arbitrary_cpt}.

\section{Statement of the theorem}
\label{sec:statement}

Before explaining Harish-Chandra's formula, we first have to fix a substantial amount of notation.  Let $G$ be a compact, connected, semisimple, real Lie group of rank $N$ with Lie algebra $\mathfrak{g}_0$ and normalized Haar measure $dg$.  Let $\mathfrak{h}_0$ be a Cartan subalgebra of $\mathfrak{g}_0$, and $\mathfrak{g} = \mathfrak{g}_0 \otimes \mathbb{C}$, $\mathfrak{h} = \mathfrak{h}_0 \otimes \mathbb{C}$ be the complexifications of $\mathfrak{g}_0$ and $\mathfrak{h}_0$.

Let $W$ be the Weyl group of $\mathfrak{g}$ with respect to $\mathfrak{h}$.  Then $W$ acts on $\mathfrak{h}$ as a group of linear transformations generated by reflections, and for each $w \in W$ we denote by $\epsilon(w)$ the sign of $w$, that is $\epsilon(w) = (-1)^{|w|}$ where $|w|$ is the number of reflections required to generate $w$.

Let $\langle \cdot, \cdot \rangle: \mathfrak{g}_0 \times \mathfrak{g}_0 \to \mathbb{C}$ be the Killing form, defined by $$\langle x, y \rangle = \mathrm{tr}(\mathrm{ad}_x \circ \mathrm{ad}_y).$$  This is the unique (up to a scalar multiple) nondegenerate, Ad-invariant symmetric bilinear form on $\mathfrak{g}_0$.  For general finite-dimensional Lie algebras over $\R$, the Killing form is nondegenerate if and only if the algebra is semisimple, and is negative-definite if and only if the associated adjoint group is compact.  We can extend $\langle \cdot, \cdot \rangle$ linearly to a complex-valued form on $\mathfrak{g}$.  For the algebras of the compact classical groups in their fundamental representations, the Killing form is a negative multiple of the Hilbert-Schmidt inner product, $$\langle x, y \rangle = -c_N \mathrm{tr}(x^\dagger y)$$ for some $c_N > 0$.  Clearly it is no problem to leave out the factor of $-c_N$ in the exponents on both sides formulas such as (\ref{eqn:hciz}).

The bilinear form $\langle \cdot, \cdot \rangle$ induces an isomorphism $\mathfrak{g} \to \mathfrak{g}^*$ by $x \mapsto \langle x, \cdot \rangle$.  Let $n = \dim \mathfrak{g}$.  Given a basis $\{e_1, \hdots, e_n \}$ of $\mathfrak{g}$, we can define {\it coordinate functions} in the usual way by setting $x_i = \langle e_i, x \rangle$, so that we may write $x = (x_1, \hdots, x_n)$.  We can then identify $\mathfrak{g}_0$ with the real subspace of all $x \in \mathfrak{g}$ that have strictly real coordinates.

For any vector space $V$ over $\mathbb{C}$ with $\dim V = n$, the {\it symmetric algebra} $\mathrm{Sym}(V)$ is the commutative algebra of polynomials with indeterminates in $V$. We can define it formally as follows: $$\mathrm{Sym}(V) = T(V) / J$$ where $T(V) = \bigoplus_{k=0}^\infty V^{\otimes k}$ is the tensor algebra over $V$, and $J$ is the ideal of $T(V)$ generated by all elements of the form $v \otimes w - w \otimes v$.  Taking the quotient by $J$ has the effect of forcing products in the tensor algebra to become commutative, so that $$\mathrm{Sym}(V) \cong \mathbb{C}[v_1, \hdots, v_n]$$ where $\{v_1, \hdots, v_n \}$ is any basis of $V$.

The elements of the algebra $\mathrm{Sym}(\mathfrak{g})$ are {\it formal} polynomials over $\mathfrak{g}$.  These are distinct from the actual polynomial {\it functions} on $\mathfrak{g}$, which are elements of $\mathrm{Sym}(\mathfrak{g}^*)$, but we identify these two algebras via the isomorphism $\mathfrak{g} \cong \mathfrak{g}^*$ provided by the Killing form.  Furthermore, following Harish-Chandra, we will identify polynomial functions on $\mathfrak{g}$ or $\mathfrak{h}$ with their restrictions to $\mathfrak{g}_0$ or $\mathfrak{h}_0$ respectively.

Given $p(x) = \sum_\beta c_\beta x^\beta \in \mathrm{Sym}(\mathfrak{g})$ where $\beta$ ranges over multi-indices, denote by $p(\partial)$ the differential operator\footnotemark $$p(\partial) = \sum_\beta c_\beta \frac{\partial^{|\beta|}}{\partial x^\beta}.$$ We can extend $\langle \cdot, \cdot \rangle$ to a scalar product $[ \! [ \cdot, \cdot ] \!]$ on $\mathrm{Sym}(\mathfrak{g})$ by defining $$[ \! [ p, q ] \!] = p(\partial)q(x) \big |_{x=0}.$$  If $\{e_1, \hdots, e_N\}$ are an orthonormal basis of $\mathfrak{g}$ with respect to $\langle \cdot, \cdot \rangle$, then an orthonormal basis for $\mathrm{Sym}(\mathfrak{g})$ with respect to $[ \! [ \cdot , \cdot ] \!]$ is given by monomials of the form $(\prod_i e_i^{\beta_i})/\sqrt{\beta!}$, where $\beta$ is a multi-index and the multi-index factorial has the usual meaning $\beta ! = \beta_1 ! \hdots \beta_N!$.  This fact is a corollary of Lemma \ref{lem:normconst} below, which shows how to compute $[ \! [ p , q ] \!]$.  Lemma \ref{lem:normconst} also shows that $[ \! [ \cdot , \cdot ] \!]$ is symmetric and non-degenerate, and that $[ \! [ x , y ] \!] = \langle x , y \rangle$ for $x, y \in \mathfrak{g}$.

\footnotetext{Note that in \cite{HC}, this differential operator would be written as $\partial(p)$, but I've chosen to use the notation $p(\partial)$ since it's more consistent with contemporary authors and does not invite the misinterpretation that the function $p$ is being differentiated.  Additionally, Harish-Chandra writes the Killing form as $B(\cdot, \cdot)$ while using the notation $\langle p, q \rangle$ for polynomials $p$ and $q$ to indicate $p(\partial) q(x) |_{x = 0}$.}

Finally, let $\alpha_1, \hdots, \alpha_r \in \mathfrak{h}^*$ be the positive roots of $\mathfrak{g}$ in some fixed order.  The {\it discriminant} of $\mathfrak{g}$ is the homogeneous polynomial $\Pi: \mathfrak{h} \to \mathbb{C}$ given by taking the product of the positive roots: \begin{equation} \label{eqn:pidef} \Pi(x) = \prod_{i = 1}^r \alpha_i(x).\end{equation}  The discriminant has many interesting properties and plays an important role in geometric analysis on $\mathfrak{g}$.

We will make use of the fact that $\Pi$ is {\it skew} with respect to the action of $W$: for $w \in W$, $\Pi(w(h)) = \epsilon(w) \Pi(h)$.  This follows from the fact that if $\alpha$ is a simple root, the reflection through the plane $\{ \alpha = 0 \}$ sends $\alpha \mapsto -\alpha$ and permutes the other positive roots.

The sets $\mathfrak{h}'_0 = \{ h \in \mathfrak{h}_0 \ | \ \Pi(h) \not = 0 \}$ and $\mathfrak{h}' = \{ h \in \mathfrak{h} \ | \ \Pi(h) \not = 0 \}$ are the {\it regular elements} of $\mathfrak{h}_0$ and $\mathfrak{h}$ respectively.  Non-vanishing of $\Pi$ is significant because it provides a generic condition under which the adjoint orbit of a point $h \in \mathfrak{h}$ enjoys some nice geometric properties, discussed below.

We now can state Harish-Chandra's formula:

\begin{theorem} \label{thm:hc} For all $h_1, h_2 \in \mathfrak{h}$, $$ \Pi(h_1) \Pi(h_2) \int_G e^{\langle \mathrm{Ad}_g h_1, h_2 \rangle} dg = \frac{ [ \! [ \Pi, \Pi ] \!] }{|W|} \sum_{w \in W} \epsilon(w) e^{\langle w(h_1),h_2 \rangle}. $$
\end{theorem}

Let's look a little more closely at what this says.  On the left-hand side of this equation, we have an integral over the Lie group $G$.  On the right-hand side, we have a finite sum over the Weyl group $W$.  Since $G$ is compact, the action of $W$ on $\mathfrak{h}$ is represented by a finite subgroup of $G$ acting by the adjoint representation, so that (\ref{eqn:hc}) has the interpretation that the integral on the left is equal to a normalized, alternating sum of the integrand's values at finitely many points.  In other words, the integral is {\it localized} at the elements of $G$ that represent elements of $W$.

The following lemma shows how to compute the constant $[ \! [ \Pi, \Pi ] \!]$.

\begin{lemma} \label{lem:normconst}
Let $$p(x) = \sum_{\beta} p_\beta x^\beta \quad \textrm{and} \quad q(x) = \sum_{\beta} q_\beta x^\beta$$ be two polynomial functions on $\mathfrak{g}$, where $(x_1, \hdots, x_n)$ are orthogonal coordinates and $\beta$ runs over multi-indices.  Then $$p(\partial) q(x) \big |_{x=0} = \sum_{\beta} p_\beta q_\beta \beta!$$
\end{lemma}

Note that since only finitely many $p_\beta$ and $q_\beta$ are nonzero, the sum is finite.

\begin{proof}[Proof]
Expanding out terms, we have $$p(\partial) q(x) = \sum_{\alpha, \beta} p_\alpha q_\beta \frac{\partial^{|\alpha|} x^\beta}{\partial x^\alpha}.$$  If $|\alpha | \ge |\beta|$ and $\alpha \not = \beta$ then ${\partial^{|\alpha|} x^\beta}/{\partial x^\alpha} = 0$. If $|\alpha | < |\beta|$ then ${\partial^{|\alpha|} x^\beta}/{\partial x^\alpha}$ has positive degree and is killed by evaluating at $x = 0$, so that we are left only with terms where $\alpha = \beta$, and the sum becomes $$p(\partial) q(x) \big |_{x=0} = \sum_{\beta} p_\beta q_\beta \frac{\partial^{|\beta|} x^\beta}{\partial x^\beta} = \sum_{\beta} p_\beta q_\beta \beta!$$ as desired.
\end{proof}

This lemma allows us to compute $[\![ \Pi, \Pi ] \!]$ from the coefficients of $\Pi$.  Writing $$\Pi(x) = \sum_{|\beta| = r} \pi_\beta x^\beta$$ for some constants $\pi_\beta$, we have \begin{equation} \label{eqn:normconst} [\![ \Pi, \Pi ] \!] = \left( \sum_{|\beta| = r} \pi_\beta \frac{\partial^{|\beta|}}{\partial x^\beta} \right) \left( \sum_{|\beta| = r} \pi_\beta x^\beta \right) = \sum_{|\beta| = r} \pi_\beta^2 \beta ! \end{equation}

Next we study the particular cases $G = U(N)$ and $G=SU(N)$.

\section{Recovering HCIZ: the cases $G=U(N)$ and $G=SU(N)$}
\label{sec:un_sun}

Let's see how we can recover the HCIZ formula (\ref{eqn:hciz}) from (\ref{eqn:hc}).  The computation illustrates a general procedure for deriving an HCIZ-like formula for an arbitrary compact, connected, semisimple real Lie group $G$, which we will apply several times in the following sections:

\renewcommand{\labelenumi}{\arabic{enumi}.}
 \begin{enumerate}
 \item Identify the root system of the complexified Lie algebra $\mathfrak{g}$, find a Cartan subalgebra $\mathfrak{h} \subset \mathfrak{g}$, and compute the roots explicitly in a suitable basis of the dual space $\mathfrak{h}^*$.
 \item Identify the Weyl group $W$, compute $|W|$, and find a representation of $W$ as a subgroup of $G$ acting on $\mathfrak{h}$ by the adjoint action.  Use this representation to compute the sum on the right-hand side of (\ref{eqn:hc}).
 \item Use the explicit form of the roots to compute the coefficients of the discriminant $\Pi$, and use Lemma \ref{lem:normconst} to compute the constant $[\![ \Pi, \Pi ]\!]$.  Plug everything into (\ref{eqn:hc}) to arrive at the final formula.
 \end{enumerate}

Now, the astute reader may have noted that the group $U(N)$ is not semisimple!  In fact this doesn't matter: with the help of the classification of compact Lie groups, Theorem \ref{thm:hc} actually allows us to compute analogous integrals over arbitrary compact (real Lie) groups, which may be neither semisimple nor connected, by reducing to an integral over a connected semisimple group.  I will discuss this in detail below.  For now we simply note that the formula (\ref{eqn:hc}) still works as written for the unitary group.

The derivation of (\ref{eqn:hciz}) from (\ref{eqn:hc}) goes as follows.  Let $G = U(N)$ and $\langle \cdot, \cdot \rangle$ be the Hilbert-Schmidt inner product.  Then $\mathfrak{g}_0$ consists of $N$-by-$N$ skew-Hermitian matrices, and $\mathfrak{h}$ consists of $N$-by-$N$ diagonal matrices.  First we write down the discriminant $\Pi$.  The root system of $U(N)$ is $A_{N-1}$ (this is the same root system as $SU(N)$ -- more on this below).  In the orthonormal basis $\{e_i\}$ of $\mathfrak{h}$ where $e_i$ is the matrix with a single 1 in position $(i,i)$ and zeros everywhere else, we can choose as positive roots the linear functionals $\langle e_i, \cdot \rangle - \langle e_j, \cdot \rangle$ for $i < j$. Then $\Pi$ becomes the Vandermonde determinant $\Delta$.  Next we consider the Weyl group $W$, which for $U(N)$ is the symmetric group $S_N$, so $|W| = N!$.  In this case the action of $W$ on $\mathfrak{h}$ has a convenient representation as conjugation by the group of $N$-by-$N$ permutation matrices $\mathrm{Perm}(N)$.  For $w \in W = S_N$, $\epsilon(w)$ is equal to the sign of the permutation, which is the determinant of the corresponding permutation matrix.  Taking $h_1$ as $B^\dagger$ and $h_2$ as $A$, (\ref{eqn:hc}) thus becomes \begin{equation} \label{eqn:mid_un} \Delta(A) \Delta(B) \int_{U(N)} e^{\mathrm{tr} (AUBU^\dagger)} dU = \frac{[\![ \Delta, \Delta ]\!]}{N!} \sum_{P \in \mathrm{Perm}(N)} (\det P) e^{\mathrm{tr}(APBP^\dagger)}.\end{equation}
The assumption $a_1 < \hdots < a_N$, $b_1 < \hdots < b_N$ allows us to divide through on both sides by the Vandermondes, and we observe that the sum on the right-hand side is exactly the Leibniz formula for the determinant of the matrix $(e^{a_i b_j})_{i,j = 1}^N$.

It now only remains to compute $\Delta(\partial)\Delta(x) |_{x=0}$.  A straightforward induction shows that the polynomial $$\Delta(x_1, \hdots, x_N) = \prod_{i < j} (x_i - x_j) =: \sum_{|\beta| = \frac{N(N-1)}{2}} \pi_\beta x^\beta$$ consists of $N!$ nonzero monomials, each with $\beta = \sigma(0,...,N-1)$ for some $\sigma \in S_N$, and \begin{equation} \label{eqn:pibeta} \pi_\beta = \mathrm{sgn}(\sigma) = \pm 1. \end{equation}  Plugging this into (\ref{eqn:normconst}), we have \begin{equation} \label{eqn:un_norm} [\![ \Delta, \Delta ]\!] = \sum_{\sigma \in S_N} \left( \mathrm{sgn}(\sigma)^2 \prod_{p = 0}^{N-1} \sigma(p)! \right) = N! \prod_{p=1}^{N-1} p! = \prod_{p=1}^{N} p!\end{equation} which gives the correct normalization for the HCIZ formula (\ref{eqn:hciz}), and we are done.

For the case $G=SU(N)$, the formula we end up with is {\it exactly the same}: \begin{equation} \label{eqn:hciz_su} \int_{SU(N)} e^{\mathrm{tr} (AUBU^\dagger)} dU = \left( \prod_{p=1}^{N-1}p! \right) \frac{\det(e^{a_i b_j})_{i,j = 1}^N}{\Delta(A) \Delta(B)}.\end{equation} Since $SU(N)$ is semisimple unlike $U(N)$, (\ref{eqn:hciz_su}) is actually a direct specialization of (\ref{eqn:hc}) if $A$ and $B$ are taken to be {\it traceless} diagonal matrices.\footnotemark \footnotetext{This is because the complexified Lie algebra $\mathfrak{g} = \mathfrak{su}(N) \otimes \mathbb{C} = \mathfrak{sl}(N, \mathbb{C})$ consists of $N$-by-$N$ traceless matrices, among which the diagonal matrices form the Cartan subalgebra $\mathfrak{h}$.  However, it is easily checked that (\ref{eqn:hciz_su}) holds even if $A$ and $B$ are not traceless.}

The similarity of the formulas for $U(N)$ and $SU(N)$ is due to the fact that these two groups have the same root system.  The root system of $\mathfrak{sl}(N, \mathbb{C})$ is $A_{N-1}$.  The group $U(N)$ has the structure of a semidirect product $$U(N) \cong SU(N) \rtimes U(1) \cong SU(N) \rtimes SO(2),$$ and its Lie algebra is $\mathfrak{u}(N) \cong \mathfrak{su}(N) \oplus \mathfrak{so}(2)$ with complexification $\mathfrak{u}(N) \otimes \mathbb{C} = \mathfrak{gl}(N, \mathbb{C}) \cong \mathfrak{sl}(N, \mathbb{C}) \oplus \mathbb{C}$.  The factor $\C$ lies in the center of $\mathfrak{gl}(N, \mathbb{C})$, so that it contributes no roots but adds one dimension to the Cartan subalgebra (which takes us from traceless diagonal matrices in the special unitary case to arbitrary diagonal matrices in the unitary case).  Thus, just as in the unitary case, we find that $\Pi = \Delta$ and $W = S_N$.

The calculation then can proceed exactly as above for $U(N)$, although to prevent confusion I should note that if we want to insist on finding a representation of $W$ as a subgroup of $G$, then we have to observe that $S_N$ is not represented in $SU(N)$ as permutation matrices but as {\it signed} permutation matrices $\{ (\det P) P \textrm{ } | \textrm{ } P \in \mathrm{Perm}(N) \}$, so that all of the representing matrices have determinant 1.  Thus, the sum on the right-hand side of (\ref{eqn:mid_un}) becomes $$\sum_{P \in \mathrm{Perm}(N)} (\det P) e^{\mathrm{tr}(A(\det P)PB((\det P)P)^\dagger)},$$ but of course each $\det P$ squares to 1 in the exponent so that the sum is exactly the same as in the unitary case.

\section{Context of the theorem}
\label{sec:background}

To motivate the formula (\ref{eqn:hc}), it's helpful to understand what Harish-Chandra was trying to do more broadly in the paper \cite{HC}.  Although he seems to have viewed this theorem as an auxiliary result, it nicely illustrates the central theme of the paper, which addresses the relationship between the algebras of differential operators on $\mathfrak{g}$ and on $\mathfrak{h}$, and in particular the question of how much information about $G$ and $\mathfrak{g}$ we can recover just by looking at $W$ and $\mathfrak{h}$.  As the theorem demonstrates, we can often recover a surprisingly large amount.  In particular, for a function $f \in C^\infty(\mathfrak{g}_0)$, Harish-Chandra's explicit goal is to express $\overline{p(\partial) f}$ in terms of $\bar p(\partial)$ and $\bar f$, where the bar indicates restriction to $\mathfrak{h}_0$.  What he finds is that \begin{equation} \label{eqn:deriv_restr} \Pi \ \overline{p(\partial) f} = \bar p(\partial) (\Pi \bar f) \end{equation} where $\Pi$ is the discriminant defined in (\ref{eqn:pidef}).

Many of the results in the paper can be viewed as building on the Chevalley restriction theorem (Theorem \ref{thm:chev} below), which says that the $G$-invariant polynomials on $\mathfrak{g}$ are isomorphic to the $W$-invariant polynomials on $\mathfrak{h}$, and that an isomorphism is given in the $\mathfrak{g}$-to-$\mathfrak{h}$ direction simply by restriction of functions.  Harish-Chandra strengthens that result by expanding the domain of the Chevalley restriction isomorphism, showing that the map can be extended to a homomorphism from a particular algebra of $G$-invariant differential operators on $\mathfrak{g}$ to the algebra of all $W$-invariant differential operators on $\mathfrak{h}$.  He then elaborates a number of consequences including the relation (\ref{eqn:deriv_restr}) and the integral formula (\ref{eqn:hc}).

In the rest of this section we'll make these ideas concrete, starting with a formal statement of the Chevalley restriction theorem.  The adjoint group $G$ acts on functions $f: \mathfrak{g} \to \mathbb{C}$ by sending $f(x) \mapsto f(\mathrm{Ad}_{g^{-1}} x)$ for each $g \in G$.  Likewise, the Weyl group $W$ acts on functions $v: \mathfrak{h} \to \mathbb{C}$ by $v(h) \mapsto v(w^{-1}h)$ for $w \in W$.  Let $I(\mathfrak{g}) \subset \mathrm{Sym}(\mathfrak{g})$ consist of polynomials on $\mathfrak{g}$ that are invariant under the adjoint action of $G$, and let $I(\mathfrak{h}) \subset \mathrm{Sym}(\mathfrak{h})$ consist of polynomials on $\mathfrak{h}$ that are invariant under the action of $W$.  Then we have:

\begin{theorem}[Chevalley restriction theorem] \label{thm:chev}
For all $p \in I(\mathfrak{g})$, $\bar p \in I(\mathfrak{h})$.  Moreover, $p \mapsto \bar p$ is an isomorphism of $I(\mathfrak{g})$ onto $I(\mathfrak{h})$.
\end{theorem}

To understand this theorem, it is helpful to consider an example.  In the case $G = U(N)$ above, we have $\mathfrak{g} = \mathfrak{u}(N) \otimes \mathbb{C} = \mathfrak{gl}(N, \mathbb{C})$. The underlying vector space of this Lie algebra is $\mathrm{Mat}_{N \times N}^\mathbb{C}$, the space of all $N$-by-$N$ complex matrices, with the bracket given by the usual matrix commutator.  In this setting, the Chevalley restriction theorem tells us that if $p$ is a polynomial function of the entries of an $N$-by-$N$ complex matrix $M$, then we have $p(M) = p(UMU^\dagger)$ for all $U \in U(N)$ if and only if $p$ is a symmetric polynomial of the eigenvalues of $M$.

In order to explain how Harish-Chandra's results build on those of Chevalley, we have to introduce some more definitions and notation.

\subsection{Generalities on differential operators}
The first thing that we need to clarify is what exactly we mean by a ``differential operator.''  For a finite-dimensional vector space $V$ over $\mathbb{C}$ or $\mathbb{R}$, let $\partial \mathrm{Sym}(V) = \{ p(\partial) \textrm{ } | \textrm{ } p \in \mathrm{Sym}(V)\}$, and let $\mathcal{D}(V)$ denote the algebra generated by $\mathrm{Sym}(V) \cup \partial \mathrm{Sym}(V)$.  $\mathcal{D}(V)$ is the algebra of {\it polynomial differential operators} on $V$, and these are the operators that we will primarily study.  When we regard a polynomial function $p$ (without any actual derivatives) as a differential operator, it simply acts by multiplication, $f \mapsto pf$.  By applying the product rule, it is always possible to write any element of $\mathcal{D}(V)$ in the form $\sum_i p_i \cdot q_i(\partial)$, where $p_i$ and $q_i$ are polynomials.\footnotemark  \footnotetext{Here the dot ``$\cdot$'' indicates multiplication but also emphasizes that $p_i$ acts as a multiplication operator {\it following} differentiation by $q_i(\partial)$, as opposed to differentiation by $(p_i q_i)(\partial) \in \partial \mathrm{Sym}(V)$.  We will sometimes also use the notation $p(\partial) \circ q$ to indicate the composition of operators, i.e. $(p(\partial) \circ q)f = p(\partial)(qf)$.}

More generally, if $U \subseteq V$ is a nonempty open set, then we can define a {\it differential operator on $U$} to be any operator acting on $C^\infty(U)$ that has the form $\sum_{i=1}^n a_i \cdot q_i(\partial)$ where $a_i \in C^\infty(U)$ and $q_i \in \mathrm{Sym}(V)$.  For example, in this case the coefficients $a_i$ could blow up at the boundary of $U$ or could be rational functions with no poles in the interior of $U$.  When we use the term ``differential operator'' with no further qualification, we will mean an operator of this form. Such operators form an algebra $\mathfrak{D}(U)$, and there is a natural inclusion of $\mathcal{D}(V)$ as a subalgebra of $\mathfrak{D}(U)$ given by $p_i \cdot q_i(\partial) \mapsto p_i |_U \cdot q_i(\partial)$.

If $W \subset V$ is a subspace, then $\mathcal{D}(W) \subset \mathcal{D}(V)$ is a subalgebra, and each $D \in \mathcal{D}(W)$ can be thought of as a differential operator either on $W$ or on $V$.  Following Harish-Chandra, we will ignore this distinction with impunity, since for any $f \in C^\infty(V)$ we have $(Df) |_W = D(f|_W)$.  Observe also that if $V$ is a vector space over $\R$, then there is a natural correspondence between differential operators on $V$ and differential operators on the complexification $V \otimes \C$, obtained by identifying $$\frac{\partial}{\partial x_j} \leftrightarrow \frac{\partial}{\partial z_j} = \frac{\partial}{\partial x_j} - i \frac{\partial}{\partial y_j}$$ where $x_j$, $y_j$ are real coordinates on $V$ and $z_j = x_j + i y_j$ is a complex coordinate on $V \otimes \C$.

\subsection{Group actions on differential operators}
We now return to the case where the underlying vector space is one of the Lie algebras $\mathfrak{g}$ or $\mathfrak{h}$. There is a natural way in which $G$ acts on $\mathcal{D}(\mathfrak{g})$ and $W$ acts on $\mathcal{D}(\mathfrak{h})$, extending the respective actions on $\mathrm{Sym}(\mathfrak{g})$ and $\mathrm{Sym}(\mathfrak{h})$.  We have already seen how an element $g \in G$ acts on $p \in \mathrm{Sym}(\mathfrak{g})$ by sending $p \mapsto p \circ \mathrm{Ad}_{g^{-1}}$.  To extend this action to $\mathcal{D}(\mathfrak{g})$, we need to construct an analogous action on $\partial\mathrm{Sym}(\mathfrak{g})$.  In order for the actions on $\mathrm{Sym}(\mathfrak{g})$ and $\partial\mathrm{Sym}(\mathfrak{g})$  to be compatible, differential operators must transform in the opposite way to functions: under the map $x \mapsto \mathrm{Ad}_{g^{-1}} x$, we have $$\frac{\partial}{\partial x} \mapsto \frac{\partial}{\partial (\mathrm{Ad}_{g^{-1}} x)} = (\mathrm{Ad}_g x) (\partial)$$ so that $p(\partial) \mapsto (p \circ \mathrm{Ad}_g )(\partial)$. Since every element of $\mathcal{D}(\mathfrak{g})$ can be written as $\sum_i p_i q_i(\partial)$, the action of $G$ on $\mathcal{D}(\mathfrak{g})$ is fully determined by the actions on $\mathrm{Sym}(\mathfrak{g})$ and $\partial\mathrm{Sym}(\mathfrak{g})$: an element $g \in G$ sends $$p \cdot q(\partial) \mapsto (p \circ \mathrm{Ad}_{g^{-1}}) \cdot (q \circ \mathrm{Ad}_{g})(\partial)$$ and this action extends to all of $\mathcal{D}(\mathfrak{g})$ by linearity.  The action of $W$ on $\mathcal{D}(\mathfrak{h})$ is defined completely analogously.  Note that the map $\mathcal{D}(\mathfrak{g}) \to \mathcal{D}(\mathfrak{g})$ thus induced by any $g \in G$ is inverted by the map induced by $g^{-1}$, so that for each $g \in G$ or $w \in W$ we obtain an automorphism of $\mathcal{D}(\mathfrak{g})$ or $\mathcal{D}(\mathfrak{h})$ respectively.  Let $\mathcal{I}'$ denote those elements of $\mathcal{D}(\mathfrak{g})$ that are invariant under the action of $G$, and let $\mathcal{I}(\mathfrak{g})$ be the subalgebra of $\mathcal{I}'$ generated by $I(\mathfrak{g}) \cup \partial I(\mathfrak{g}) $, where $\partial I (\mathfrak{g}) = \{ p(\partial) \textrm{ } |\textrm{ } p \in I(\mathfrak{g})\}$.  Let $\mathcal{I}(\mathfrak{h})$ denote those elements of $\mathcal{D}(\mathfrak{h})$ that are invariant under the action of $W$.

The following proposition further illustrates the significance of $\Pi$ and its relationship with the invariant differential operators on $\mathfrak{h}$.
\begin{proposition} \label{prop:pi_harmonic}
$D\Pi = 0$ for all $D \in \mathcal{I}(\mathfrak{h})$ that annihilate the constants.
\end{proposition}
\begin{proof}
Since $D$ is $W$-invariant and $\Pi$ is skew, $D\Pi$ must be skew.  For each reflection $w_\alpha \in W$ we thus have $w_\alpha(D\Pi) = -D\Pi$, which implies that $D\Pi$ vanishes on the plane $\{\alpha = 0 \}$, so that $\alpha$ divides $D\Pi$.  But this implies $\Pi$ divides $D\Pi$ since the positive roots are relatively prime, and $D\Pi$ is strictly lower degree than $\Pi$, so we must have $D\Pi = 0$.
\end{proof}
We say that $\Pi$ is {\it $W$-harmonic}.  In fact, the space of {\it all} $W$-harmonic polynomials on $\mathfrak{h}$ is exactly the linear span of the partial derivatives of $\Pi$ \cite[ch. 3, Theorem 3.6]{SH}.  In particular, since the Laplacian on $\mathfrak{h}$ is $W$-invariant, $\Pi$ is harmonic in the traditional sense.

\subsection{Radial part of a differential operator}
There is one final idea that we need before we can understand Harish-Chandra's homorphism of invariant differential operators.  This is the notion of the {\it radial part} of a differential operator, which was fully developed by Helgason in the 1960s and 1970s\footnotemark \footnotetext{See Helgason's book \cite[ch. 2, sec. 3]{SH} for a detailed reference on radial parts and other geometric operations on differential operators in a general setting.} but which already plays an important role in \cite{HC}.  We say that a submanifold $M \subset \mathfrak{g}$ is {\it transverse} to the adjoint orbits in $\mathfrak{g}$ if for each $p \in M$ we have a decomposition of tangent spaces \begin{equation} \label{eqn:transverse}T_p \mathfrak{g} = T_p \mathcal{O}_p \oplus T_p M,\end{equation} where $\mathcal{O}_p = \{ \mathrm{Ad}_g p \ | \ g \in G \}$ is the adjoint orbit of $p$.  We state without proof the following theorem, which is a special case of \cite[ch.~2, Theorem 3.6]{SH}.

\begin{theorem} \label{thm:radial-part} Let $M \subset \mathfrak{g}$ be a submanifold of $\mathfrak{g}$ that is transverse to the adjoint orbits. Let $D$ be a differential operator on $\mathfrak{g}$.  Then there exists a unique differential operator $\gamma(D)$ on $M$ such that, for each function $f \in C^\infty(\mathfrak{g})$ that is locally $\mathrm{Ad}$-invariant in the sense that $f(\mathrm{Ad}_g x) = f(x)$ for $g$ in some neighborhood of $\mathrm{id}_G$, $$\overline{(Df)} = \gamma(D) \bar f,$$ the bar indicating restriction to $M$. \end{theorem}

The differential operator $\gamma(D)$ is called the {\it radial part of $D$ with transversal manifold $M$}.

Now, the importance of the regular elements $\mathfrak{h}' = \{ h \in \mathfrak{h} \ | \ \Pi(h) \not = 0 \} \subset \mathfrak{g}$ for our purposes is that they form a submanifold that is both dense in $\mathfrak{h}$ and transverse to the adjoint orbits.  To see that transversality holds, consider the root space decomposition of $\mathfrak{g}$, $$\mathfrak{g} = \mathfrak{h} \oplus \bigoplus_\alpha \mathfrak{g}_\alpha,$$ where $\alpha$ runs over the roots of $\mathfrak{g}$ with respect to $\mathfrak{h}$.  Under the usual identification $T_h \mathfrak{g} \cong \mathfrak{g}$, at $h \in \mathfrak{h}'$ we have $$T_h \mathfrak{h}' \cong \mathfrak{h}, \quad T_h \mathcal{O}_h \cong [ \mathfrak{g}, h] = \bigoplus_\alpha \mathfrak{g}_\alpha,$$ which gives the transversality property.  Thus for each $D \in \mathfrak{D}(\mathfrak{g})$ we have a well-defined operator $\gamma(D) \in \mathfrak{D}(\mathfrak{h}')$, the radial part of $D$ with transversal manifold $\mathfrak{h}'$.  Moreover, we have the following fact:

\begin{lemma} \label{lem:rp_hom}
The map $D \mapsto \gamma(D)$ is a homomorphism from $\mathcal{I}'$ to $\mathfrak{D}(\mathfrak{h}')$.
\end{lemma}

\begin{proof}
We follow the proof of  \cite[Lemma 7]{HC} but work over $\mathfrak{h}'$ instead of $\mathfrak{h}_0'$.  We first note that $\gamma$ is clearly linear from its definition.  Let $h \in \mathfrak{h}'$ and let $T \subset G$ be the maximal torus in $G$ corresponding to $\mathfrak{h}_0$.  Let $g \mapsto gT$ denote the quotient map $G \to G/T$, and set $(gT)h = \mathrm{Ad}_g h$ for $g \in G, h \in \mathfrak{h}$.  Let $U \subset \mathfrak{h}'$, $V \subset G$ be open connected neighborhoods of $h$ and $\mathrm{id}_G$ respectively, and let $VT$ be the image of $V$ under the quotient map.  Then the map $\phi: VT \times U \to \mathfrak{g}$ is regular, and since $\dim(VT \times U) = \dim \mathfrak{g}$ (as real manifolds), $N = \phi(VT \times U)$ is an open submanifold of $\mathfrak{g}$.  If $V$ and $U$ are taken to be sufficiently small, then $\phi$ is bijective and defines an analytic isomorphism of $VT \times U$ onto $N$.  For $\psi \in C^\infty(U)$, define $f_\psi \in C^\infty(N)$ by $f_\psi(\phi(gT, h)) = \psi(h)$.  Then $f_\psi$ is locally $\mathrm{Ad}$-invariant.  Let $D_1, D_2 \in \mathcal{I}'$.  We have $$\overline{D_1 D_2 f_\psi} = \gamma(D_1 D_2) \bar f_\psi = \gamma(D_1 D_2) \psi.$$ On the other hand, since $D_2$ is $\mathrm{Ad}$-invariant, $D_2 f_\psi$ must be locally $\mathrm{Ad}$-invariant, so $$\overline{D_1 D_2 f_\psi} = \gamma(D_1) \overline{(D_2 f_\psi)},$$ and $\overline{(D_2 f_\psi)} = \gamma(D_2) \bar f_\psi = \gamma(D_2) \psi,$ so that $$\gamma(D_1 D_2) \psi =  \gamma(D_1) \gamma(D_2) \psi.$$ Since $\psi \in C^\infty(U)$ was arbitrary, $\gamma$ is a homomorphism.
\end{proof}

\subsection{The $\delta$ homomorphism}
We can now state the first major theorem proved in \cite{HC}, which Harish-Chandra calls ``the central result of this paper'' and which lies in the background of most of its other results.  The theorem takes Chevalley's theorem a step further by giving a meaningful sense to the idea of ``restriction to $\mathfrak{h}$'' for invariant differential operators on $\mathfrak{g}$, rather than merely invariant polynomials.  Further, it relates the restrictions of these operators to their radial parts. Indeed, Harish-Chandra finds that Chevalley's isomorphism $I(\mathfrak{g}) \to I(\mathfrak{h})$ extends uniquely to a homomorphism $\mathcal{I}(\mathfrak{g}) \to \mathcal{I}(\mathfrak{h})$:

\begin{theorem} \label{thm:central} There exists a unique homomorphism $\delta: \mathcal{I}(\mathfrak{g}) \to \mathcal{I}(\mathfrak{h})$ such that $\delta(p) = \bar p$ and $\delta(p(\partial)) = \bar p (\partial)$ for all $p \in I(\mathfrak{g})$. Moreover, on $\mathfrak{h}'_0$, we have \begin{equation} \label{eqn:radconj} \gamma(D) = \Pi^{-1} \delta(D) \circ \Pi \end{equation} for all $D \in \mathcal{I}(\mathfrak{g})$. \end{theorem}

The bulk of section 3 of \cite{HC} is devoted to a concrete construction of the homomorphism $\delta$ and to showing the relation (\ref{eqn:radconj}) between $\delta$ and the radial part map.  The details of Harish-Chandra's construction are beyond the scope of this paper, since we don't actually need the full power of the $\delta$ homomorphism to prove the integral formula.  Instead, it will suffice to understand the relationship between $\gamma(p(\partial))$ and $\bar p(\partial)$ for $p \in I(\mathfrak{g})$, as described in the following theorem \cite[Lemma 8]{HC} \cite[ch. 2, Theorem 5.33]{SH}.

\begin{theorem} \label{thm:sh5.33}
For $p \in I(\mathfrak{g})$, $\gamma(p(\partial)) = \Pi^{-1} \bar p(\partial) \circ \Pi$.
\end{theorem}

We will sketch the proof, following \cite{SH}.  Let $\omega(x) = \langle x, x \rangle$, the {\it quadratic Casimir polynomial} on $\mathfrak{g}$. Then the Laplacian on $\mathfrak{g}$ is $\omega(\partial)$.  One first shows by a direct calculation\footnotemark \footnotetext{See \cite[ch. 3, Proposition 3.14]{SH} for details.} using the root space decomposition that $\gamma(\omega(\partial)) = \Pi^{-1} \bar \omega(\partial) \circ \Pi.$ The main idea of the proof is then to extend this result from $\omega$ to all $p \in I(\mathfrak{g})$ by way of a neat trick of taking commutators with the Laplacian, which we will show in detail.

For differential operators $D_1$ and $D_2$, we write $\{D_1, D_2 \} = D_1 \circ D_2 - D_2 \circ D_1$ for the commutator.  Consider the derivations $$\mu: D \mapsto \frac{1}{2} \{\omega(\partial), D \}$$ on $\mathfrak{D}(\mathfrak{g})$ and $$\bar \mu: d \mapsto \frac{1}{2} \{\gamma(\omega(\partial)), d \}$$ on $\mathfrak{D}(\mathfrak{h}')$.  Then $\mu$ has the following property \cite[ch. 2, Lemma 5.34]{SH}.

\begin{lemma} \label{lem:mu_powers}
If $p$ is a homogeneous polynomial on $\mathfrak{g}$ of degree $m$, then $$\mu^m(p) = m!p(\partial).$$
\end{lemma}

\begin{proof}
The proof proceeds by induction on $m$.  The base case $m=1$ follows from direct calculation.  Let $p = q_1 \hdots q_m$, where each $q_i$ is an arbitrary linear function, and make the inductive hypothesis that $$\mu^{m-1}(q_1 \hdots q_{m-1}) = (m-1)!(q_1 \hdots q_{m-1})(\partial).$$  Observing that $\mu^2(q_i) = \mu(q_i(\partial)) = 0$, by the Leibniz rule for derivations we have $$\mu^m(q_1 \hdots q_{m-1} q_m) = \mu^m(q_1 \hdots q_{m-1}) \circ q_m + m \mu^{m-1}(q_1 \hdots q_{m-1}) \circ \mu(q_m).$$  Applying the inductive hypothesis on the right-hand side, we find that the first term vanishes and the second term is exactly $m!p(\partial)$.
\end{proof}

We will also need the following commutator identity, which holds in any associative algebra.

\begin{proposition} \label{prop:comm_id_rp}
Let $A$ be an associative algebra. For $a \in A$, define the derivation $d_a: A \to A$ by $d_a(b) = \frac{1}{2}(ab - ba).$  If $c \in A$ is invertible and commutes with $b$, then $$d_{c^{-1}ac}^k(b) = c^{-1}d_a^k (b)c.$$
\end{proposition}
\begin{proof}
This follows from the observation that $$d_a^k(b) = 2^{-k} \sum_{j=0}^k {k \choose j}(-1)^j a^{k-j}ba^j,$$ which is shown by an easy induction on $k$.
\end{proof}

Now we can complete the proof of Theorem \ref{thm:sh5.33}.

\begin{proof}[Proof of Theorem \ref{thm:sh5.33}]
It suffices to assume that $p$ is homogeneous of degree $m$, because $\gamma$ is linear.  Since $\gamma$ is a homomorphism on $\mathcal{I}(\mathfrak{g})$, for $D \in \mathcal{I}(\mathfrak{g})$ we have $$\gamma(\mu(D)) = \bar \mu(\gamma(D)),$$ and thus by Lemma \ref{lem:mu_powers} we have $$m!\!\ \gamma(p(\partial)) = \gamma(\mu^m(p)) = \bar \mu^m(\gamma(p)) = \bar \mu^m(\bar p).$$  Finally we apply Proposition \ref{prop:comm_id_rp} with $A = \mathfrak{D}(\mathfrak{h}'),$ $a = \bar \omega(\partial)$, $b = \bar p$, and $c = \Pi$, observing that $c^{-1} a c = \Pi^{-1} \bar \omega(\partial) \circ \Pi = \gamma(\omega(\partial)).$  This gives $$m!\!\ \gamma(p(\partial)) = \bar \mu^m(\bar p) = (d_{c^{-1} a c})^m(\bar p) = \Pi^{-1} d_a^m(\bar p) \circ \Pi = m!\! \ \Pi^{-1} \bar p(\partial) \circ \Pi,$$ which completes the proof.
\end{proof}

\section{Proof of the integral formula}
\label{sec:proof}

We turn now to the proof of the integral formula (\ref{eqn:hc}).  While I have reorganized the presentation and simplified some steps with particular help from the invaluable reference \cite{SH}, the proofs in this section are quite close to Harish-Chandra's originals, with some additional explanation and a few modifications to notation and terminology to bring them more in line with contemporary usage.  The argument below can also be given a physical interpretation in the language of quantum integrable systems; see \cite[chs. 4 and 5]{PE} for details on this topic.

To recap definitions and assumptions, here $G$ is a compact, connected, semisimple real Lie group of rank $N$ with Lie algebra $\mathfrak{g}_0$ and normalized Haar measure $dg$, $\mathfrak{h}_0 \subset \mathfrak{g}_0$ is a Cartan subalgebra, $W$ is the corresponding Weyl group, $\mathfrak{g}$ and $\mathfrak{h}$ are the complexifications of $\mathfrak{g}_0$ and $\mathfrak{h}_0$ respectively, and $\langle \cdot, \cdot \rangle$ is the Killing form on $\mathfrak{g}$.

The proof outline goes as follows.  Before proving Theorem \ref{thm:hc} itself, we need a few preliminary lemmas.  The most important of these is a characterization of the space of analytic functions on which all operators in $\partial I(\mathfrak{h})$ can be simultaneously diagonalized with a specific choice of eigenvalues.  After these lemmas, the core argument of the proof proceeds in three steps:
\begin{enumerate}
\item Define a function $\phi_f(h_1) = \Pi(h_1) \int_G e^{\langle \mathrm{Ad}_g h_1, h_2 \rangle} dg$.  With the assumptions that $h_1 \in \mathfrak{h}_0$ and $\Pi(h_2) \not = 0$, show that $\phi_f$ is a joint eigenfunction for all operators $q(\partial) \in \partial I(\mathfrak{h})$, and use the simultaneous diagonalization lemma to write down an explicit formula for $\phi_f$ containing a number of unknown constants indexed by the Weyl group $W$.
\item Use the same explicit formula to write down expressions for $\phi_f(w(h_1))$, $w \in W$, and average these over $W$ to eliminate all but a single unknown constant.
\item Determine the value of the remaining constant by computing $\Pi(\partial) \phi_f(h) |_{h=0}$ in two different ways, and setting the resulting expressions equal to each other.  This gives the integral formula for $h_1 \in \mathfrak{h}_0$ and $\Pi(h_2) \not = 0$, and we then use the analyticity of $\phi_f$ to extend the result to all $h_1, h_2 \in \mathfrak{h}$.
\end{enumerate}

The first lemma that we need relates the algebraic structures of $\mathrm{Sym}(\mathfrak{h})$ and $I(\mathfrak{h})$.

\begin{lemma} \label{lem:sym_free_alg_over_i}  There are homogeneous elements $v_1, \hdots, v_{|W|} \in \mathrm{Sym}(\mathfrak{h})$ such that every $v \in \mathrm{Sym}(\mathfrak{h})$ can be written uniquely in the form $v = \sum_{i=1}^{|W|} u_i v_i$ where each $u_i \in I(\mathfrak{h})$. \end{lemma}

In other words, $\mathrm{Sym}(\mathfrak{h})$ is a {\it free algebra of rank $|W|$} over $I(\mathfrak{h})$.

\begin{proof}[Proof]
Let $J_+$ be the ideal in $I(\mathfrak{h})$ generated by elements of positive degree.  Since $I(\mathfrak{h})$ is a subalgebra of $\mathrm{Sym}(\mathfrak{h})$, we can consider the ideal $J_+\mathrm{Sym}(\mathfrak{h})$ generated by $J_+$ in the larger algebra $\mathrm{Sym}(\mathfrak{h})$.  A theorem by Chevalley \cite{Ch} shows that the dimension of the quotient $\mathrm{Sym}(\mathfrak{h})/J_+\mathrm{Sym}(\mathfrak{h})$ as a vector space over $\mathbb{C}$ is equal to $|W|$, so that we can choose $v_1, \hdots, v_{|W|}$ in $\mathrm{Sym}(\mathfrak{h})$ such that $$\mathrm{Sym}(\mathfrak{h}) = \mathrm{span}\{v_1, \hdots, v_{|W|}\} + J_+\mathrm{Sym}(\mathfrak{h}).$$ Then an easy induction shows that in fact for any $m \ge 1$, $$\mathrm{Sym}(\mathfrak{h}) = \mathrm{span}\{v_1, \hdots, v_{|W|}\} \cdot I(\mathfrak{h}) + J_+^m \mathrm{Sym}(\mathfrak{h}).$$  If $p \in I(\mathfrak{h})$ and $(p)_d$ is its degree $d$ homogeneous component then we must have $(p)_d \in I(\mathfrak{h})$ as well, so that we may take the elements $v_i$ to be homogeneous.  Moreover, since $\mathrm{Sym}(\mathfrak{h})$ contains no zero-divisors we can consider its field of fractions $\mathrm{Frac}(\mathrm{Sym}(\mathfrak{h}))$, and this is a field extension of degree $|W|$ over $\mathrm{Frac}(I(\mathfrak{h}))$.

Now let $v \in \mathrm{Sym}(\mathfrak{h})$ be homogeneous of degree $d$.  We will show that $v$ can be written uniquely in the form stated in the lemma.  Let $d_i = \deg v_i$ and choose $m > d$ and $u_1', \hdots, u_{|W|}' \in I(\mathfrak{h})$ such that $v - \sum_{i = 1}^{|W|} u'_i v_i \in J_+^m \mathrm{Sym}(\mathfrak{h})$.  Let $u_i = (u'_i)_{d- d_i}.$ Then, since $$\bigg( v - \sum_{i = 1}^{|W|} u'_i v_i \bigg)_d = 0$$ by construction and $v = (v)_d$ is homogeneous, we must have $v - \sum_{i = 1}^{|W|} u_i v_i = 0.$  Thus we can conclude that $$\mathrm{Sym}(\mathfrak{h}) = \mathrm{span}\{v_1, \hdots, v_{|W|}\} \cdot I(\mathfrak{h}),$$ whereby it follows that $\mathrm{Frac}(\mathrm{Sym}(\mathfrak{h}))$ is spanned over $\mathrm{Frac}(I(\mathfrak{h}))$ by the elements $v_i$.  But since $\mathrm{Frac}(\mathrm{Sym}(\mathfrak{h}))$ is a degree $|W|$ extension of $\mathrm{Frac}(I(\mathfrak{h}))$, the elements $v_i$ must therefore be linearly independent, showing that the decomposition $v = \sum_{i=1}^{|W|} u_i v_i$ is unique.
\end{proof}

The purely algebraic statement of Lemma \ref{lem:sym_free_alg_over_i} will be our main tool in proving the next lemma, which is the aforementioned simultaneous diagonalization result for $\partial I (\mathfrak{h})$. We pose an infinite system of eigenvalue problems for {\it each point} in $\mathfrak{h}'$, identify a family of analytic functions that solve all of them simultaneously, and then show that this family of solutions is exhaustive up to the assumption of analyticity.

\begin{lemma} \label{lem:diffeqs}
Let $U$ be a nonempty connected open subset of $\mathfrak{h}$.  Let $h_0 \in \mathfrak{h}$ such that $\Pi(h_0) \not = 0$.  Suppose $\phi$ is an analytic function on $U$ satisfying the system of differential equations \begin{equation} \label{eqn:diffeqs} q(\partial) \phi = q(h_0) \phi \quad \quad \forall q \in I(\mathfrak{h}). \end{equation}  Then there exist constants $c_w \in \mathbb{C}$ ($w \in W$) such that for all $h \in U$, $$\phi(h) = \sum_{w \in W} c_w e^{\langle h, w(h_0) \rangle}.$$  Moreover, given $\phi$, the constants $c_w$ are uniquely determined.
\end{lemma}

In other words, the functions $e^{\langle h, w(h_0) \rangle}$ for $w \in W$ form a basis of the complex vector space of analytic solutions to the linear system (\ref{eqn:diffeqs}).

\begin{proof}[Proof]
We first note an auxiliary result from one of Harish-Chandra's earlier papers showing that the $|W|$ points $w(h_0)$ are all distinct (see \cite[Lemma 4]{HCe}), so that the functions $$\phi_w(h) = e^{\langle h, w(h_0)\rangle}, \quad w \in W$$ are linearly independent analytic functions on $U$. (That the $\phi_w$ are linearly independent given distinct $w(h_0)$ may seem intuitively obvious, but see \cite[Lemma 41]{HCb} for a detailed proof.)

If we fix a point $h' \in \mathfrak{h}$ and identify $h'$ with the linear functional $\langle h', \cdot \rangle$ on $\mathfrak{h}_0$, then $h'(\partial) \phi_w(h) = \langle h', w(h_0) \rangle \phi_w(h)$, and so we must have $p(\partial) \phi_w = p(w(h_0))\phi_w$ for $p \in \mathrm{Sym}(\mathfrak{h})$.   For $q \in I(\mathfrak{h})$, then, we have $q(\partial) \phi_w = q(w(h_0)) \phi_w = q(h_0) \phi_w$.  In other words, each $\phi_w$ solves (\ref{eqn:diffeqs}).

Let $E$ be the vector space over $\mathbb{C}$ consisting of all analytic solutions to (\ref{eqn:diffeqs}).  We know already that $\dim E \ge |W|$ since the $\phi_w$ are linearly independent.  To show that the $\phi_w$ form a basis for $E$, it is therefore sufficient to show that if we assume $\dim E > |W|$ then we get a contradiction.

Choose a point $h_1 \in U$ and let $v_1, \hdots, v_{|W|}$ be as in the statement of Lemma \ref{lem:sym_free_alg_over_i}.  If $\dim E > |W|$, we can choose $\psi \not = 0$ in $E$ satisfying the $|W|$ linear conditions $$v_i(\partial) \psi(h) \big |_{h = h_1} = 0, \quad 1 \le i \le |W|.$$  But this is impossible: by Lemma \ref{lem:sym_free_alg_over_i}, for any $v \in \mathrm{Sym}(\mathfrak{h})$ we can write $v = \sum_{i=1}^{|W|} u_i v_i$ with $u_i \in I(\mathfrak{h})$, and since $\psi$ solves (\ref{eqn:diffeqs}), we have $$v(\partial) \psi(h) \big |_{h = h_1} = \sum_{i = 1}^{|W|} u_i(h_0) v_i(\partial) \psi(h) \big |_{h = h_1} = 0.$$  In other words all derivatives of $\psi$ vanish at $h_1$, and since $\psi$ is analytic it must therefore be identically zero, which contradicts our assumption.
\end{proof}

Now, for any $f \in C^\infty(\mathfrak{g}_0)$, define a function $\phi_f \in C^\infty(\mathfrak{h}_0)$ by $$\phi_f(h) = \Pi(h) \int_G f(\mathrm{Ad}_g h) dg.$$  We make the following observation about how $\phi_f$ transforms under the action of invariant differential operators.

\begin{lemma} \label{lem:hc15_subfromSH}
For all $p \in I(\mathfrak{g})$, $$\phi_{p(\partial) f} = \bar p(\partial) \phi_f.$$
\end{lemma}

\begin{proof}
Let $F(x) = \int_G f(\mathrm{Ad}_g x) dg.$  Then by Theorem \ref{thm:sh5.33}, for $h \in \mathfrak{h}'_0$ we have $(p(\partial)F)(h) = (\Pi^{-1} \bar p(\partial)(\Pi \bar F))(h)$, so that $$\phi_{p(\partial) f} (h) = \Pi(h) p(\partial) F(h) = \bar p(\partial) \phi_f(h).$$  Since $\phi_{p(\partial) f} (h)$ and $\bar p(\partial) \phi_f(h)$ are both continuous and $\mathfrak{h}'_0$ is dense in $\mathfrak{h}_0$, this equality must in fact hold for all $h \in \mathfrak{h}_0$.
\end{proof}

We now can give Harish-Chandra's original proof of the integral formula (\ref{eqn:hc}), following the outline of steps at the beginning of this section.

\begin{proof}[Proof (Harish-Chandra's integral formula)]
\

{\bf Step 1: Identify an appropriate joint eigenfunction of $\partial I (\mathfrak{h})$.}

Choose $h_2 \in \mathfrak{h}$ such that $\Pi(h_2) \not = 0$, and define $f: \mathfrak{g}_0 \to \mathbb{C}$ by $f(x) = e^{\langle x, h_2 \rangle}$, so that for $h_1 \in \mathfrak{h}_0$ we have \begin{equation} \label{eqn:phi_f_def} \phi_f(h_1) = \Pi(h_1) \int_G e^{\langle \mathrm{Ad}_g h_1, h_2 \rangle} dg.\end{equation}  We can see that for $y \in \mathfrak{g}$ we have $\partial_y f = \langle y, h_2 \rangle f$, so that for $q \in \mathrm{Sym}(\mathfrak{g})$ we have $q(\partial)f = q(h_2) f$.  But by Lemma \ref{lem:hc15_subfromSH}, for $p \in I(\mathfrak{g})$ we have $\phi_{p(\partial) f} = \bar p (\partial) \phi_f$, so that $$\bar p (\partial) \phi_f = \phi_{p(\partial) f} = \phi_{p(h_2) f} = p(h_2) \phi_f.$$  Since this holds for all $p \in I(\mathfrak{g})$, we may apply the isomorphism of the Chevalley restriction theorem to conclude that $$q(\partial) \phi_f = q(h_2) \phi_f \quad \quad \forall q \in I(\mathfrak{h}).$$  In other words the analytic function $\phi_f$ satisfies the system of differential equations (\ref{eqn:diffeqs}).  Therefore, by Lemma \ref{lem:diffeqs}, there is a unique choice of constants $c_w$ for $w \in W$ such that we can write $$\phi_f (h_1) = \sum_{w \in W} c_w e^{\langle w(h_1), h_2 \rangle}$$ for all $h_1 \in \mathfrak{h}_0$. \\

{\bf Step 2: Average over $W$ to eliminate all but one unknown constant.}

By the skewness of $\Pi$ and the $\mathrm{Ad}$-invariance of the integral in (\ref{eqn:phi_f_def}), we have $$\phi_f(w(h_1)) = \epsilon(w) \phi_f(h_1).$$  Multiplying this identity on both sides by $\epsilon(w)$ and taking the average over all of $W$, we get \begin{multline} \label{eqn:phi_avg} \phi_f(h_1) = |W|^{-1} \sum_{w \in W} \epsilon(w) \phi_f(w(h_1)) = |W|^{-1} c \sum_{w \in W} \epsilon(w) e^{\langle w(h_1),h_2 \rangle}\end{multline}
where $c = \sum_{w \in W} \epsilon(w) c_w$. \\

{\bf Step 3: Determine the unknown constant by computing $\Pi(\partial)\phi_f(h)|_{h=0}$ in two different ways.}

Now define, for each $w \in W$, $$\psi_w(h_1) = e^{\langle w(h_1),h_2 \rangle},$$ so that (\ref{eqn:phi_avg}) becomes $$\phi_f(h_1) = |W|^{-1} c \sum_{w \in W} \epsilon(w) \psi_w.$$Then for $q \in \mathrm{Sym}(\mathfrak{h})$, $q(\partial) \psi_w = q(w^{-1}(h_2)) \psi_w$.  In particular, $$\Pi(\partial) \psi_w = \Pi(w^{-1}(h_2)) \psi_w =  \epsilon(w^{-1}) \Pi(h_2) \psi_w = \epsilon(w) \Pi(h_2) \psi_w,$$ and therefore \begin{multline} \label{eqn:exp1} \Pi(\partial) \phi_f (h) \big |_{h=0} = \Pi(\partial) \left[ |W|^{-1} c \sum_{w \in W} \epsilon(w) \psi_w(h) \right]_{h = 0} \\ = |W|^{-1}c \sum_{w \in W} \epsilon(w)^2 \Pi(h_2) \psi_w(0) = |W|^{-1} c \textrm{ } \Pi(h_2) |W| = c \textrm{ } \Pi(h_2).\end{multline}

Now that we have one expression for the value of $\Pi(\partial) \phi_f (h) |_{h=0}$, we'll calculate it again in a different way and equate the two answers to each other.  We apply the product rule to compute \begin{multline} \label{eqn:exp2} \Pi(\partial) \phi_f (h) \big |_{h=0} = \Pi(\partial) \left( \Pi(h) \int_G f(\mathrm{Ad}_g h) dg \right) \bigg |_{h = 0} \\ = [\![\Pi,\Pi]\!] \int_G f(0) dg + \Pi(0) \cdot \Pi(\partial) \int_G f(\mathrm{Ad}_g h) dg \bigg |_{h = 0} \\ = [\![\Pi,\Pi]\!] f(0) + 0= [\![\Pi,\Pi]\!], \end{multline} where we have used the fact that the Haar measure $dg$ is taken to be normalized.

Equating (\ref{eqn:exp1}) and (\ref{eqn:exp2}), we see that $c = [\![\Pi,\Pi]\!] / \Pi(h_2)$.  Plugging this result into (\ref{eqn:phi_avg}) and multiplying both sides by $\Pi(h_2)$ gives the desired formula $$\Pi(h_1) \Pi(h_2) \int_G e^{\langle \mathrm{Ad}_g h_1, h_2 \rangle} dg = \frac{[\![\Pi,\Pi]\!]}{|W|} \sum_{w \in W} \epsilon(w) e^{\langle w(h_1),h_2 \rangle}$$ which we have established {\it with the assumptions} that $h_1 \in \mathfrak{h}_0$ and $\Pi(h_2) \not = 0$.

To extend the result to all $h_1, h_2 \in \mathfrak{h}$, we observe that the left- and right-hand sides are both holomorphic functions on $\mathfrak{h} \times \mathfrak{h}$, and that these functions agree on $\mathfrak{h}_0 \times \{h \in \mathfrak{h} \textrm{ }|\textrm{ } \Pi(h) \not = 0 \}$, so that they must agree on all of $\mathfrak{h} \times \mathfrak{h}$.  This completes the proof.
\end{proof}

\section{The cases $G = SO(2N)$ and $O(2N)$, and integrals over covering groups}
\label{sec:so2n}

In this section, we'll derive an HCIZ-like formula for the special orthogonal groups $SO(2N)$, and then use this formula to obtain analogous results for $\mathrm{Spin}(2N)$ and $O(2N)$.  The Lie algebra of $SO(2N)$ is $\mathfrak{so}(2N)$, the algebra of $2N$-by-$2N$ skew-symmetric real matrices.  Its complexification is $\mathfrak{g} = \mathfrak{so}(2N, \mathbb{C})$, the algebra of $2N$-by-$2N$ skew-symmetric complex matrices.  Let $\langle \cdot, \cdot \rangle$ be the Hilbert-Schmidt inner product on $\mathfrak{g}$.  The Cartan subalgebra $\mathfrak{h}$ of $\mathfrak{so}(2N, \mathbb{C})$ consists of skew-symmetric complex matrices in which the only nonzero entries are on the off-diagonals.  An orthonormal basis of $\mathfrak{h}$ with respect to $\langle \cdot, \cdot \rangle$ is given by \begin{equation} \label{eqn:so2n_cartan_basis} e_j = \frac{i}{2}(E_{2j-1\textrm{ }2j} - E_{2j\textrm{ }2j-1}), \quad j = 1,\hdots, N \end{equation} where $E_{jk}$ is the $2N$-by-$2N$ matrix with a 1 at the intersection of row $j$ and column $k$ and zeros everywhere else.

The root system of $SO(2N)$ is $D_N$; as positive roots we may take the $N^2 - N$ covectors $e_j^* \pm e_k^*$ for $j < k$, where $e_j^* = \langle e_j, \cdot \rangle$.  The Weyl group $W$ acts on the Cartan subalgebra by permuting the eigenvalues and changing an even number of their signs.  That is, $W$ has the form $$W \cong H_{N-1} \rtimes S_N \trianglelefteq (\mathbb{Z}/2\mathbb{Z})^N \rtimes S_N,$$ where $H_{N-1}$ is the normal subgroup of $(\mathbb{Z}/2\mathbb{Z})^N$ consisting of those elements with an even number of nonzero entries.  The semidirect product structure of $W$ means that each $w \in W$ can be written uniquely in the form $w = \eta \sigma$ with $\eta \in H_{N-1}$ and $\sigma \in S_N$, and $\epsilon(w) = \mathrm{sgn}(\sigma)$. Each $w \in W$ is represented in $SO(2N)$ by a matrix of the form $H_\eta P_\sigma$, where $P_\sigma$ represents $\sigma \in S_N$ as a block permutation matrix with 2-by-2 blocks, and $H_\eta$ represents $\eta \in H_{N-1}$ as a block-diagonal matrix with blocks equal to $$I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \quad \textrm{or} \quad Q_2 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$ where the number of $Q_2$ blocks is even.  The order of the Weyl group is $$|W| = |H_{N-1}| \cdot |S_N| = 2^{N-1} N!$$

Now suppose $A, B \in \mathfrak{h}$ are given by $A = \sum_{j=1}^N a_j e_j$, $B = \sum_{j=1}^N b_j e_j$. Let $\Delta(A) = \prod_{j < k} (a_j - a_k)$ be the Vandermonde determinant, and define \begin{equation} \label{eqn:Xi_def} \Xi(A) = \prod_{j < k} (a_j + a_k). \end{equation}  Then $\Pi(A) = \Xi(A) \Delta(A)$.  Plugging all this into (\ref{eqn:hc}), we have
\begin{multline} \label{eqn:partial_hc_so2n}
\Xi(A) \Delta(A) \Xi(B) \Delta(B) \int_{SO(2N)} e^{\mathrm{tr}(AOBO^T)} dO \\ = \frac{[\![\Xi \Delta,\Xi \Delta]\!]}{2^{N-1} N!} \sum_{\eta \in H_{N-1}} \sum_{\sigma \in S_N} \mathrm{sgn}(\sigma) e^{\mathrm{tr}(A H_\eta P_\sigma B P_\sigma^T H_\eta^T)}.
\end{multline}

There are two steps remaining: to simplify the sum over the Weyl group, and to determine the leading constant that gives the normalization.  We first turn to simplifying the sum.  Here too there is a determinantal structure under the surface, albeit a more complicated one than in the unitary case.  Since the trace of a product of matrices is invariant under cyclic permutations of the product, we have $$e^{\mathrm{tr}(A H_\eta P_\sigma B P_\sigma^T H_\eta^T)} = e^{\mathrm{tr}(P_\sigma^T H_\eta^T A H_\eta P_\sigma B)} = e^{2 \sum_{j=1}^N \eta^{-1}(j) a_{\sigma^{-1}(j)} b_j},$$ where we consider $\eta$ as a function $\eta : \{1, \hdots, N\} \to \{\pm 1\}$.  We can write $\mathrm{sgn}(\eta) = \prod_{j=1}^N \eta(j)$. Then $H_{N-1}$ consists of $\eta \in (\mathbb{Z}/2\mathbb{Z})^N$ with $\mathrm{sgn}(\eta) = 1$.

Consider the map $A \mapsto H_\eta A H_\eta^T.$  The nonzero entries of $H_\eta$ are contained in $N$ 2-by-2 blocks $L_1, \hdots, L_N$ along the diagonal, with $$L_j = \begin{cases} I_2, & \eta(j) = 1 \\ Q_2, & \eta(j) = -1. \end{cases}$$  Under conjugation by $H_\eta$, each block $L_j$ sends $a_j \mapsto \eta(j) a_j$. So we have

\begin{multline} \label{eqn:hn-1_sum} \sum_{\eta \in H_{N-1}} \sum_{\sigma \in S_N} \mathrm{sgn}(\sigma) e^{\mathrm{tr}(A H_\eta P_\sigma B P_\sigma^T H_\eta^T)} \\ = \sum_{\sigma \in S_N} \mathrm{sgn}(\sigma) \sum_{\eta \in H_{N-1}} e^{2 \sum_{j = 1}^N \eta(j) a_{\sigma(j)} b_j}.\end{multline}

Now we need to rearrange terms to make this expression more concise.  We note that $$\sum_{\eta \in H_{N-1}} e^{2 \sum_{j = 1}^N \eta(j) a_{\sigma(j)} b_j} = \frac{1}{2}\sum_{\eta \in (\mathbb{Z}/2\mathbb{Z})^N} \left[ e^{2 \sum_{j = 1}^N \eta(j) a_{\sigma(j)} b_j} + \mathrm{sgn}(\eta) e^{2 \sum_{j = 1}^N \eta(j) a_{\sigma(j)} b_j} \right]$$
so that the sum over $H_{N-1}$ can be written as an average of two sums over $(\mathbb{Z}/2\mathbb{Z})^N$, one signed and one unsigned.  Each of these sums can be factored into a product of hyperbolic sines or cosines:
\begin{multline} \label{eqn:eta_sum} \sum_{\eta \in (\mathbb{Z}/2\mathbb{Z})^N} e^{2 \sum_{j = 1}^N \eta(j) a_{\sigma(j)} b_j} = \prod_{j = 1}^N ( e^{2a_{\sigma(j)}b_j} + e^{2a_{\sigma(j)}b_j} ) \\ = 2^N \prod_{j = 1}^N \mathrm{cosh}(2a_{\sigma(j)} b_j), \end{multline}
\begin{multline} \label{eqn:eta_signed_sum} \sum_{\eta \in (\mathbb{Z}/2\mathbb{Z})^N} \mathrm{sgn}(\eta) e^{2 \sum_{j = 1}^N \eta(j) a_{\sigma(j)} b_j} = \prod_{j = 1}^N ( e^{2a_{\sigma(j)}b_j} - e^{2a_{\sigma(j)}b_j} ) \\ = 2^N \prod_{j = 1}^N \mathrm{sinh}(2a_{\sigma(j)} b_j). \end{multline}

These identities finally give $$\sum_{\eta \in H_{N-1}} \sum_{\sigma \in S_N} \mathrm{sgn}(\sigma) e^{\mathrm{tr}(A H_\eta P_\sigma B P_\sigma^T H_\eta^T)} $$$$= \sum_{\sigma \in S_N} \mathrm{sgn}(\sigma) \frac{1}{2} \left[2^N \prod_{j = 1}^N \mathrm{cosh}(2a_{\sigma(j)} b_j) + 2^N \prod_{j = 1}^N \mathrm{sinh}(2a_{\sigma(j)} b_j)\right] $$$$= 2^{N-1} \big( \det[\mathrm{cosh}(2a_j b_k)]_{j,k = 1}^N + \det[\mathrm{sinh}(2a_j b_k)]_{j,k = 1}^N \big).$$

Next we turn our attention to evaluating $[\![\Xi \Delta,\Xi \Delta]\!]$.  We will find a way of rewriting the polynomial $\Xi \Delta$ that also allows us to simplify the expression $\Xi(A) \Delta(A) \Xi(B) \Delta(B)$ that appears in (\ref{eqn:partial_hc_so2n}).  To that end, we introduce a last piece of notation.  Given $A = \sum a_i e_i \in \mathfrak{h}$, define $$A^{(2)} = \sum a_i^2 e_i \in \mathfrak{h}.$$  Note that if $\mathfrak{h}$ happens to be represented by the diagonal matrices then in the basis $e_i = E_{ii}$ we have $A^{(2)} = A^2$, but in general we must be careful to distinguish $A^{(2)}$ and $A^2$ as we may not even have $A^2 \in \mathfrak{h}$.  Similarly, given $p(x) = \sum_\beta c_\beta x^\beta \in \mathrm{Sym}(\mathfrak{h})$, denote by $p(\partial^2)$ the differential operator $$p(\partial^2) = \sum_\beta c_\beta \frac{\partial^{2|\beta|}}{\partial x^{2\beta}}.$$

Now assume $a_1 < \hdots < a_N$ and $b_1 < \hdots < b_N$ so that we can divide by the Vandermondes.  We find that we can write $$\Xi(A) = \prod_{i < j} (a_i + a_j) = \prod_{i < j} \frac{a_i^2 - a_j^2}{a_i - a_j} = \frac{\Delta(A^{(2)})}{\Delta(A)}.$$  That is, the polynomial $\Xi$ is in fact a ratio of two Vandermondes, and so we have $$\Xi(A) \Delta(A) \Xi(B) \Delta(B) = \Delta(A^{(2)}) \Delta(B^{(2)}).$$  Moreover, $$\Xi(\partial) \Delta(\partial) (\Xi \Delta)(x) \big |_{x=0} = \Delta(\partial^2) \Delta(x^{(2)}) \big |_{x = 0}$$ and this is an expression that we already know how to evaluate using Lemma \ref{lem:normconst} and our calculation (\ref{eqn:un_norm}) for the unitary case.  With $\pi_\beta$ as in (\ref{eqn:pibeta}) and $\beta_0 = (0,\hdots,N-1)$ we can write \begin{equation} \label{eqn:delta_sq} \Delta(x^{(2)}) = \sum_{\sigma \in S_N} \pi_\beta x^{2 \sigma(\beta_0)}.\end{equation}  Then Lemma \ref{lem:normconst} gives $$\Delta(\partial^2) \Delta(x^{(2)}) \big |_{x = 0} = N! \prod_{p=1}^{N-1} (2p)!$$
Plugging all of the above results into (\ref{eqn:partial_hc_so2n}), we arrive at the HCIZ-like formula for $SO(2N)$:

\begin{multline} \label{eqn:hc_so2n}
\int_{SO(2N)} e^{\mathrm{tr}(AOBO^T)} dO \\ = \left( \prod_{p = 1}^{N-1} (2p)! \right) \frac{\det[\mathrm{cosh}(2a_j b_k)]_{j,k = 1}^N + \det[\mathrm{sinh}(2a_j b_k)]_{j,k = 1}^N} {\Delta(A^{(2)}) \Delta(B^{(2)})} .
\end{multline}

We now move on to the calculations for $\mathrm{Spin}(2N)$ and $O(2N)$.  Just as for $SO(2N)$, in both of these cases we take $A$ and $B$ to be skew-symmetric complex matrices, since all three groups have the same Cartan subalgebra $\mathfrak{h}$.

The group $\mathrm{Spin}(2N)$ is a double cover of $SO(2N)$, so that it has the same Lie algebra $\mathfrak{so}(2N)$. Since the right-hand side of (\ref{eqn:hc}) depends only on the algebra, we can immediately conclude that the formula for $SO(2N)$ applies exactly as written: \begin{equation} \label{eqn:hc_spin2n} \int_{\mathrm{Spin}(2N)} e^{\mathrm{tr}(ASBS^{-1})} dS = \int_{SO(2N)} e^{\mathrm{tr}(AOBO^T)} dO. \end{equation}

In general, if $\tilde G$ is a connected compact covering group of $G$, then the integral formula for $G$ will hold as written for $\tilde G$ as well.  This equality is a consequence of the following fact, which we will use below when proving the integral formula for arbitrary compact groups in Theorem \ref{thm:hc_generalization}.  Note that if $\tilde G$ is a covering group of $G$ then their Lie algebras are isomorphic and carry an adjoint action of both groups.

\begin{proposition} \label{prop:cover_integrals} Let $G$ be a compact, connected Lie group, not necessarily semisimple, with Lie algebra $\mathfrak{g}_0$.  Let $\tilde G$ be a compact, connected covering group of $G$, and let $f: \mathfrak{g} \to \C$.  Then for $x \in \mathfrak{g}$, \begin{equation} \label{eqn:cover_int} \int_{\tilde G} f(\mathrm{Ad}_{\tilde g} x) \ d\tilde g = \int_G f(\mathrm{Ad}_g x) \ dg \end{equation} where $d\tilde g$ and $dg$ are the normalized Haar measures. \end{proposition}

\begin{proof}
Let $\pi: \tilde G \to G$ be the covering homomorphism. The exponential maps $\mathrm{exp}_{\tilde G}$ and $\mathrm{exp}_G$ are surjective since both groups are compact and connected, so we can write $\tilde g \in \tilde G$ as $\tilde g = \mathrm{exp}_{\tilde G}(y)$ for some $y \in \mathfrak{g}_0$.  We then have $\pi(\tilde g) = \mathrm{exp}_{G}(y)$, so that $$\mathrm{Ad}_{\tilde g} = \sum_{k = 0}^\infty \frac{1}{k!} \mathrm{ad}_y^k = \mathrm{Ad}_{\pi(\tilde g)},$$ and so $x \in \mathfrak{g}$ has the same orbit $\mathcal{O}_x$ under both adjoint actions.  The pushforwards of $d\tilde g$ and $dg$ to $\mathcal{O}_x$ are normalized measures that are invariant under either adjoint action, so they must be equal, and writing both sides of (\ref{eqn:cover_int}) as an integral over $\mathcal{O}_x$ gives the desired equality.
\end{proof}

In the case $G = O(2N)$, there is an additional subtlety: even though $O(2N)$ is compact and has the same Lie algebra as $SO(2N)$, it is not connected, so that we cannot use (\ref{eqn:hc}) directly.  Instead we must use a trick that allows us to reduce an integral over a non-connected group to an integral over its identity component, which we will revisit in the proof of Theorem \ref{thm:hc_generalization} below.  We have $$O(2N) = SO(2N) \sqcup O^-(2N),$$ where $O^-(2N)$ is the orientation-reversing component consisting of $2N$-by-$2N$ orthogonal matrices with determinant $-1$.  Now, notice that every matrix in $O^-(2N)$ can be written as $\tilde I O$, where $O \in SO(2N)$ and $\tilde I$ is the matrix with $\tilde I_{11} = -1$, $\tilde I_{ii} = 1$ for $i > 1$, and $\tilde I_{ij} = 0$ for $i \not = j$.  In other words, $\tilde I$ is the $2N$-by-$2N$ identity matrix but with the $1$ in the top left corner changed to a $-1$.  Thus $$\int_{O(2N)} e^{\mathrm{tr}(AOBO^T)} dO $$$$ = \int_{SO(2N)} e^{\mathrm{tr}(AOBO^T)} dO + \int_{O^-(2N)} e^{\mathrm{tr}(AOBO^T)} dO $$$$ = \int_{SO(2N)} \left[ e^{\mathrm{tr}(AOBO^T)} + e^{\mathrm{tr}(A\tilde I OBO^T \tilde I^T)} \right ] dO.$$ Here we will take the Haar measure $dO$ to be normalized over all of $O(2N)$, so that in this case $\int_{SO(2N)} dO = 1/2$ and the integration measure differs from that used in (\ref{eqn:hc_so2n}) by a factor of 2.

Next, notice that by the invariance of the trace under cyclic permutations of the product of matrices and the fact that $\tilde I^T = \tilde I$, we have $$e^{\mathrm{tr}(A\tilde I OBO^T \tilde I^T)} = e^{\mathrm{tr}(\tilde I A \tilde I OBO^T)}.$$  Conjugation of $A$ by $\tilde I$ has the effect of sending $a_1 \mapsto -a_1$.  This has the consequence that for the orientation-reversing component, the sum over the Weyl group in (\ref{eqn:hn-1_sum}) becomes a sum over $-\eta$ rather than $\eta$.  Therefore in order to integrate over both components of $O(2N)$, we should simply take the sum in (\ref{eqn:hn-1_sum}) over all of $(\mathbb{Z}/2\mathbb{Z})^N$ rather than merely over $H_{N-1}$, and then divide by 2 to account for the different normalization.  Another application of (\ref{eqn:eta_sum}) then gives the HCIZ-like formula for $O(2N)$:
\begin{equation} \label{eqn:hc_o2n}
\int_{O(2N)} e^{\mathrm{tr}(AOBO^T)} dO = \left( \prod_{p = 1}^{N-1} (2p)! \right) \frac{\det[\mathrm{cosh}(2a_j b_k)]_{j,k = 1}^N} {\Delta(A^{(2)}) \Delta(B^{(2)})}.
\end{equation}

\section{The cases $G = SO(2N+1)$ and $O(2N+1)$}
\label{sec:so2n1}

The orthogonal groups $SO(2N+1)$ and $O(2N+1)$ need to be treated separately from $SO(2N)$ and $O(2N)$, as their root system $B_N$ contains more roots than the system $D_N$ of $SO(2N)$.

We'll start with $G = SO(2N+1)$.  The Lie algebra of $SO(2N+1)$ is $\mathfrak{so}(2N+1)$, the algebra of $(2N+1)$-by-$(2N+1)$ skew-symmetric real matrices, with complexification $\mathfrak{g} = \mathfrak{so}(2N+1, \mathbb{C})$, the algebra of $(2N+1)$-by-$(2N+1)$ skew-symmetric complex matrices.  The Cartan subalgebra $\mathfrak{h}$ of $\mathfrak{so}(2N+1, \mathbb{C})$ consists of skew-symmetric complex matrices in which the only nonzero entries are on the off-diagonals, and all entries in the final row and column are zero.  In other words, just as above in (\ref{eqn:so2n_cartan_basis}), an orthonormal basis of $\mathfrak{h}$ with respect to the Hilbert-Schmidt inner product is given by $$e_j = \frac{i}{2}(E_{2j-1\textrm{ }2j} - E_{2j\textrm{ }2j-1}), \quad j = 1,\hdots, N$$ with $E_{jk}$ as defined in (\ref{eqn:so2n_cartan_basis}).

The difference from the case $G=SO(2N)$ arises because roots of $B_N$ come in two different lengths: we must consider both the $N^2 -N$ {\it long} positive roots $e_j^* \pm e_k^*$ for $j < k$, corresponding to the roots of $D_N$, and also the $N$ {\it short} positive roots $e_i^*$, $1 \le i \le N$.

Since $SO(2N+1)$ has more roots than $SO(2N)$, its Weyl group is bigger as well.  We have $$W \cong (\mathbb{Z}/2\mathbb{Z})^N \rtimes S_N,$$ so that $|W| = |(\mathbb{Z}/2\mathbb{Z})^N| \cdot |S_N| = 2^N N!$.  Like before, due to the semidirect product structure each $w \in W$ can be written uniquely in the form $w = \eta \sigma$, with $\sigma \in S_N$, but now $\eta \in (\mathbb{Z}/2\mathbb{Z})^N$ and $\epsilon(w) = \mathrm{sgn}(\eta)\mathrm{sgn}(\sigma)$. Then each element of $W$ is represented in $SO(2N+1)$ by a matrix of the form $H_\eta P_\sigma$, where $P_\sigma$ represents $\sigma \in S_N$ as a block permutation matrix with 2-by-2 blocks followed by a final $1$ in the bottom right corner, and $H_\eta$ represents $\eta \in (\mathbb{Z}/2\mathbb{Z})^N$ as a block-diagonal matrix with $N$ 2-by-2 blocks equal to $$I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \quad \textrm{or} \quad Q_2 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$ followed by a final $\pm 1$ in the bottom right corner chosen so that the matrix has unit determinant.  Because we can choose this final entry to make the determinant work out, unlike in the case of $SO(2N)$ we do not need to require the number of $Q_2$ blocks to be even.

Let $A, B \in \mathfrak{h}$ be given by $A = \sum_{j=1}^N a_j e_j$, $B = \sum_{j=1}^N b_j e_j$. Let $\Delta(A) = \prod_{j < k} (a_j - a_k)$ be the Vandermonde, and define $\Xi$ as in (\ref{eqn:Xi_def}). Additionally, define $\Phi(A) = \prod_{j=1}^N a_j$.  Plugging into (\ref{eqn:hc}), we have
\begin{multline} \label{eqn:partial_hc_so2n+1}
\Xi(A) \Delta(A) \Phi(A) \Xi(B) \Delta(B) \Phi(B) \int_{SO(2N+1)} e^{\mathrm{tr}(AOBO^T)} dO \\ = \frac{[\![\Xi \Delta \Phi,\Xi \Delta \Phi]\!]}{2^N N!} \sum_{\eta \in (\mathbb{Z}/2\mathbb{Z})^N} \sum_{\sigma \in S_N} \mathrm{sgn}(\eta)\mathrm{sgn}(\sigma) e^{\mathrm{tr}(A H_\eta P_\sigma B P_\sigma^T H_\eta^T)}.
\end{multline}

Proceeding as before, we next turn to simplifying the sum over the Weyl group.  Again we can write $$e^{\mathrm{tr}(A H_\eta P_\sigma B P_\sigma^T H_\eta^T)} = e^{\mathrm{tr}(P_\sigma^T H_\eta A H_\eta^T P_\sigma B)},$$ and the map $A \mapsto H_\eta A H_\eta^T$ acts on $\mathfrak{h}$ in a very similar way.  The nonzero entries of $H_\eta$ are contained in $N$ diagonal blocks $L_1, \hdots, L_N$, each of which is equal to one of the 2-by-2 matrices $I_2$ or $Q_2$, plus the final $\pm 1$ on the diagonal, which we ignore since the final row and column of any matrix in $\mathfrak{h}$ are all zeros.  Just like before, under conjugation by $H_\eta$, the block $L_j$ sends $a_j \mapsto \eta(j) a_j$.  We apply (\ref{eqn:eta_signed_sum}) to conclude that $$\sum_{\eta \in (\mathbb{Z}/2\mathbb{Z})^N} \sum_{\sigma \in S_N} \mathrm{sgn}(\eta) \mathrm{sgn}(\sigma) e^{\mathrm{tr}(A H_\eta P_\sigma B P_\sigma^T H_\eta^T)} = \sum_{\sigma \in S_N} \mathrm{sgn}(\eta) 2^N \prod_{j=1}^N \mathrm{sinh}(2 a_{\sigma(j)} b_j)$$$$ = 2^N \det[\mathrm{sinh}(2 a_j b_k)]_{j,k=1}^N.$$

It now remains to compute the constant $[\![\Xi \Delta \Phi,\Xi \Delta \Phi]\!]$. With reference to (\ref{eqn:delta_sq}), we can write $$\Xi(x) \Delta(x) \Phi(x) = \Delta(x^{(2)}) \Phi(x) = \sum_{\sigma \in S_N} \mathrm{sgn}(\sigma) x^{2 \sigma(\beta_0) + \vec 1}$$ where $\vec 1$ is the $N$-component multi-index $(1, 1, \hdots, 1)$.  Lemma \ref{lem:normconst} then gives $$\big [\Xi(\partial) \Delta(\partial) \Phi(\partial) \cdot \Xi(x) \Delta(x) \Phi(x) \big]_{x=0} = \sum_{\sigma \in S_N} \left( \mathrm{sgn}(\sigma)^2 \prod_{p=0}^{N-1} (2 \sigma(p) + 1)! \right)$$$$ = N! \prod_{p = 1}^{N-1} (2p + 1)!$$

We substitute this into (\ref{eqn:partial_hc_so2n+1}) under the assumptions that $a_1 < \hdots < a_N$, $b_1 < \hdots < b_N$, and all $a_i$, $b_i$ are non-zero so that we can clear the leading constant on the left-hand side.  This yields the integral formula for $SO(2N+1)$:
\begin{multline} \label{eqn:hc_so2n+1}
\int_{SO(2N+1)} e^{\mathrm{tr}(AOBO^T)} dO \\ = \left( \prod_{p = 1}^{N-1} (2p+1)! \right) \frac{\det[\mathrm{sinh}(2 a_j b_k)]_{j,k=1}^N}{\Delta(A^{(2)}) \Delta(B^{(2)}) \prod_{i=1}^N a_i b_i}.
\end{multline}

Proposition \ref{prop:cover_integrals} yields an identical formula for integrals over $\mathrm{Spin}(2N+1)$.

For the case $G = O(2N+1)$, we must compute the integral over the orientation-reversing component.  Applying the same logic as in the case $G = O(2N)$, we find $$\int_{O(2N+1)} e^{\mathrm{tr}(AOBO^T)} dO = \int_{SO(2N+1)} e^{\mathrm{tr}(AOBO^T)} dO + \int_{O^-(2N+1)} e^{\mathrm{tr}(AOBO^T)} dO$$$$ = \int_{SO(2N+1)} \left[ e^{\mathrm{tr}(AOBO^T)} +  e^{\mathrm{tr}(\tilde I A \tilde I OBO^T)} \right] dO$$ where the measure $dO$ is now taken to be normalized over the entire group $O(2N+1)$.  The same argument used in the calculation for $O(2N)$ shows that integrating over the orientation-reversing component amounts to sending $\eta \mapsto -\eta$ in the sum over the Weyl group on the right-hand side of (\ref{eqn:hc}).  But since in this case we are summing over all of $(\mathbb{Z}/2\mathbb{Z})^N$, this does not actually change the sum, so that the integrals over both connected components of $O(2N)$ are equal.  Therefore, due to the normalization of the measure, the final formula for $O(2N+1)$ is identical to the formula for $SO(2N+1)$:
\begin{multline} \label{eqn:hc_o2n+1}
\int_{O(2N+1)} e^{\mathrm{tr}(AOBO^T)} dO \\ = \left( \prod_{p = 1}^{N-1} (2p+1)! \right) \frac{\det[\mathrm{sinh}(2 a_j b_k)]_{j,k=1}^N}{\Delta(A^{(2)}) \Delta(B^{(2)}) \prod_{i=1}^N a_i b_i}.
\end{multline}

\section{The case $G = \mathrm{USp}(N)$}
\label{sec:uspn}

The unitary symplectic, or compact symplectic, group $\mathrm{USp}(N)$ is the compact real form of $\mathrm{Sp}(2N, \mathbb{C})$, the symplectic group over the complex numbers defined by $$S \in \mathrm{Sp}(2N, \mathbb{C}) \iff SJS^T = J$$ where $$J = \begin{bmatrix} 0 & I_N \\ -I_N & 0 \end{bmatrix}$$ is the standard symplectic matrix. The name ``unitary symplectic'' derives from the fact that $$\mathrm{USp}(N) \cong U(2N) \cap \mathrm{Sp}(2N, \mathbb{C}) = \left\{ \begin{bmatrix} A & - \bar B \\ B & \bar A \end{bmatrix} \in U(2N) \right\}.$$ In fact, $\mathrm{USp}(N)$ can also be defined as the quaternionic unitary (or ``hyperunitary'') group $U(N, \mathbb{H})$ of linear operators on $\mathbb{H}^N$ that preserve the standard Hermitian form $$( v, w ) = \sum_{i=1}^N \bar v_i w_i.$$  Thus we can consider $\mathrm{USp}(N)$ as a subgroup of either $GL(2N, \mathbb{C})$ or $GL(N, \mathbb{H})$.  Hereafter we will regard $\mathrm{USp}(N)$ as a group of $2N$-by-$2N$ complex matrices.  In this representation, its complexified Lie algebra $\mathfrak{g}$ is $$\mathfrak{sp}(2N, \mathbb{C}) = \left\{ \begin{bmatrix} A & B \\ C & -A^T \end{bmatrix} \in \mathfrak{gl}(2n, \mathbb{C}) \textrm{  } \bigg | \textrm{  } B^T = B, \textrm{  } C^T = C \right\}$$ with Cartan subalgebra $\mathfrak{h}$ given by the diagonal matrices in $\mathfrak{sp}(2N, \mathbb{C})$.  The Cartan subalgebra is spanned by the basis $$e_i = E_{ii} - E_{i+N \textrm{ } i+N}, \quad 1 \le i \le N.$$  $\mathrm{USp}(N)$ thus has rank $N$.  The root system in this case is $C_N$, which is similar to the root system $B_N$ but with the long and short roots reversed.  That is, in the basis $\{e_i^*\}_{i=1}^N = \{\langle e_i, \cdot \rangle\}_{i=1}^N$ of $\mathfrak{h}^*$, the $N$ long positive roots are given by $2 e_i^*$ for $1 \le i \le N$ and the $N^2 - N$ short positive roots are given by $e_j^* \pm e_k^*$ for $j < k$.  Thus for the $C_N$ root system, \begin{equation} \label{eqn:uspn_poly} \Pi(x) = 2^N \Xi(x) \Delta(x) \Phi(x) \end{equation} which differs from the $B_N$ case only by a factor of $2^N$.  Our previous calculation for the $B_N$ system therefore immediately gives $$\Pi(\partial)\Pi(x) \big |_{x=0} = 2^{2N} N! \prod_{p=1}^{N-1} (2p + 1)!$$

The Weyl group for the $C_N$ system is isomorphic to the group for the $B_N$ system: $$W \cong (\mathbb{Z}/2\mathbb{Z})^N \rtimes S_N,$$ with $|W| = 2^N N!$, and it acts on the roots in an analogous fashion.  Thus if in (\ref{eqn:hc}) we let $h_1 = A = \sum_{j=1}^N a_j e_j$ and $h_2 = B = \sum_{j=1}^N b_j e_j$, then the sum over the Weyl group on the right-hand side is equal to $$\sum_{\eta \in (\mathbb{Z}/2\mathbb{Z})^N} \sum_{\sigma \in S_N} \mathrm{sgn}(\eta) \mathrm{sgn}(\sigma) e^{\mathrm{tr}(A H_\eta P_\sigma B P_\sigma^T H_\eta^T)} = 2^N \det[\mathrm{sinh}(2 a_j b_k)]_{j,k=1}^N$$ just as in the case $G = SO(2N+1)$.  Putting this all together in (\ref{eqn:hc}) and assuming as before that $a_1 < \hdots < a_N$, $b_1 < \hdots < b_N$, and all $a_i$, $b_i$ are non-zero, we conclude
\begin{multline} \label{eqn:hc_uspn}
\int_{\mathrm{USp}(N)} e^{\mathrm{tr}(ASBS^\dagger)} dS \\ =  \left( \prod_{p = 1}^{N-1} (2p+1)! \right) \frac{\det[\mathrm{sinh}(2 a_j b_k)]_{j,k=1}^N}{\Delta(A^{(2)}) \Delta(B^{(2)}) \prod_{i=1}^N a_i b_i}.
\end{multline}

For the unitary symplectic integral, we end up with exactly the same expression as for the odd special orthogonal groups: the sum over the Weyl group in each case gives the same result, and although the form of $\Pi$ differs slightly, this only causes factors of $2^{2N}$ to appear on both sides of (\ref{eqn:hc}), which cancel each other so that the right-hand side matches that of (\ref{eqn:hc_so2n+1}).

\section{Integrals over arbitrary compact groups}
\label{sec:arbitrary_cpt}

In this section, we show a generalization of Theorem \ref{thm:hc} that holds for integrals over arbitrary compact real Lie groups that may be neither semisimple nor connected.  This result provides a general justification for using (\ref{eqn:hc}) to evaluate integrals over groups like $U(N)$ and $O(N)$ that don't actually satisfy the assumptions of Theorem \ref{thm:hc}.  The generalization is not difficult to prove using the classification of compact real Lie groups, but to my knowledge it has not previously appeared in the literature.  Our main tool will be the following theorem.\footnotemark \footnotetext{This theorem is a slightly weakened version of \cite[ch.~10, sec.~7.2, Theorem 4]{CP}.}

\begin{theorem} \label{thm:cp_classification}
Every compact, connected real Lie group is of the form $(K \times T)/Z$, where $K$ is connected, compact, and semisimple, $T = (S^1)^m$ is a torus, and $Z$ is a finite subgroup of the center of $K \times T$ satisfying $Z \cap T = \{\mathrm{id}_{K\times T}\}$.
\end{theorem}

The idea is that we will express an integral over an arbitrary compact group $G$ in terms of a sum of integrals over its identity component $G_1 = (K \times T)/Z$, each of which we can then write as an integral over the semisimple group $K$ and evaluate using (\ref{eqn:hc}).  Note that $G$, $G_1$, $K \times T$ and $K$ all have the same root system, so that the discriminant $\Pi$ and the Weyl group $W$ are the same for all of these groups.  However, the Lie algebras and Cartan subalgebras differ in some cases.  To avoid ambiguity we will use $\mathfrak{g}$ and $\mathfrak{h}$ for the complexified Lie algebra and Cartan subalgebra of $G$, $G_1$ and $K \times T$, since these are the same.  For $K$ and $T$ we will use $\mathfrak{g}_K, \mathfrak{h}_K$ and $\mathfrak{g}_T, \mathfrak{h}_T$ respectively.  Likewise we will use subscripts to distinguish between the normalized Haar measures on different groups: $dg_G$, $dg_{K\times T}$, etc.

The first fact we must show is that (\ref{eqn:hc}) is not changed by the presence of the torus factor $T$.

\begin{lemma} \label{lem:torus_factor}
For all $h_1, h_2 \in \mathfrak{h}$, $$\Pi(h_1) \Pi(h_2) \int_{K \times T} e^{\langle \mathrm{Ad}_g h_1, h_2 \rangle} dg_{K \times T} = \frac{ [ \! [ \Pi, \Pi ] \!] }{|W|} \sum_{w \in W} \epsilon(w) e^{\langle w(h_1),h_2 \rangle}.$$
\end{lemma}

\begin{proof}
We first observe that $\mathfrak{h} = \mathfrak{h}_K \oplus \mathfrak{h}_T$, and that $\mathrm{Ad}_g h = h$ for all $g \in K \times T$ and $h \in \mathfrak{h}_T$.  For $h \in \mathfrak{h}$, write $h = h^K + h^T$, with $h^K \in \mathfrak{h}_K,$ $h^T \in \mathfrak{h}_T$.  All of the roots vanish on $\mathfrak{h}_T$, so that $\Pi(h) = \Pi(h^K)$.  Since $\{1\} \times T$ lies in the center of $K \times T$, writing $g = (k, t)$ we have $\mathrm{Ad}_g h = h^T + \mathrm{Ad}_k h^K$. Because $K \times T$ is not semisimple, the Killing form is degenerate: in fact $\mathfrak{h}_T$ is the center of $\mathfrak{g}$, so that $\langle h, x \rangle = 0$ for all $x \in \mathfrak{g}$ if and only if $h \in \mathfrak{h}_T$.  We therefore have \begin{multline*} \langle Ad_g h_1, h_2 \rangle = \langle Ad_g (h_1^T + h_1^K), h_2^T + h_2^K \rangle \\\ = \langle h_1^T + Ad_g h_1^K, h_2^T + h_2^K \rangle = \langle Ad_g h_1^K, h_2^K \rangle. \end{multline*} In particular, $\langle w(h_1), h_2 \rangle = \langle w(h_1^K), h_2^K \rangle$ for $w \in W$. Additionally, writing $g = (k, t) \in K \times T$, $dg_{K \times T}$ factors as a product measure $dg_{K \times T} = dk_K \otimes dt_T$, where all of these measures are normalized Haar measures on the respective groups.  Accordingly we have
\begin{multline*} \Pi(h_1) \Pi(h_2) \int_{K \times T} e^{\langle \mathrm{Ad}_g h_1, h_2 \rangle} dg_{K \times T} \\ = \Pi(h_1^K) \Pi(h_2^K) \int_{K} \int_T e^{\langle \mathrm{Ad}_k h_1^K, h_2^K \rangle} dt_{T} dk_K \\= \Pi(h_1^K) \Pi(h_2^K) \int_{K} e^{\langle \mathrm{Ad}_k h_1^K, h_2^K \rangle} dk_K \\= \frac{ [ \! [ \Pi, \Pi ] \!] }{|W|} \sum_{w \in W} \epsilon(w) e^{\langle w(h_1^K),h_2^K \rangle} = \frac{ [ \! [ \Pi, \Pi ] \!] }{|W|} \sum_{w \in W} \epsilon(w) e^{\langle w(h_1),h_2 \rangle}\end{multline*} as desired.
\end{proof}

Because $K \times T$ is a compact, connected, $|Z|$-fold covering group of $(K \times T)/Z$, by Proposition \ref{prop:cover_integrals} we have \begin{equation} \label{eqn:torus_factor_cover_int} \int_{K \times T} e^{\langle \mathrm{Ad}_g h_1, h_2 \rangle} dg_{K \times T} = \int_{(K \times T)/Z} e^{\langle \mathrm{Ad}_g h_1, h_2 \rangle} dg_{(K \times T)/Z}.\end{equation}  We now have all the tools to generalize Theorem \ref{thm:hc} to arbitrary compact real Lie groups.

\begin{theorem}
\label{thm:hc_generalization}
Let $G$ be a compact real Lie group with $c$ connected components $G_1, \hdots, G_c$.  For $j = 1, \hdots, c$ let $g_j \in G_j$.  Then for all $h_1, h_2 \in \mathfrak{h}$, $$\Pi(h_1) \Pi(h_2) \int_G e^{\langle \mathrm{Ad}_g h_1, h_2 \rangle} dg_G = \frac{1}{c} \frac{ [ \! [ \Pi, \Pi ] \!] }{|W|} \sum_{j=1}^c \sum_{w \in W} \epsilon(w) e^{\langle w(\mathrm{Ad}_{g_j} h_1),h_2 \rangle}.$$
\end{theorem}

\begin{proof}
Let $G_1$ be the identity component of $G$. First we note that \begin{multline} \label{eqn:component_decomp} \int_G e^{\langle \mathrm{Ad}_g h_1, h_2 \rangle} dg_G = \sum_{j=1}^c \int_{G_j} e^{\langle \mathrm{Ad}_{g} h_1, h_2 \rangle} dg_G \\ = \sum_{j=1}^c \int_{G_1} e^{\langle \mathrm{Ad}_{gg_j} h_1, h_2 \rangle} dg_G = \frac{1}{c} \sum_{j=1}^c \int_{G_1} e^{\langle \mathrm{Ad}_{g}(\mathrm{Ad}_{g_j} h_1), h_2 \rangle} dg_{G_1}, \end{multline} where the factor of $1/c$ appears because $dg_G$ and $dg_{G_1}$ are both normalized.  By Theorem \ref{thm:cp_classification} we can write $G_1 = (K \times T)/Z$ as above.  Applying (\ref{eqn:torus_factor_cover_int}) and then Lemma \ref{lem:torus_factor} to the final expression in (\ref{eqn:component_decomp}) completes the proof of the theorem.
\end{proof}

\bibliographystyle{alpha}


TITLE: Why are metals worse conductors when heated?
QUESTION [2 upvotes]: When metals, (such as in circuits), are heated, their ability to conduct electric current is hampered. Why is this? Does the transition towards liquid disrupt a metal's ability to conduct, or is something else at play? I know as any material increases in temperature its atoms vibrate more, but I don't know how this affects conductivity. Thanks for your time!

REPLY [3 votes]: The hotter a material, the more its atoms or ions jiggle about and the harder it is for electrons to wriggle through, which translates into higher electrical resistance. Broadly speaking, the resistivity of most materials increases linearly with temperature (so if you increase the temperature by 10 degrees, the resistivity increases by a certain amount, and if you increase it by another 10 degrees, the resistivity rises by the same amount again). If you cool a material, you lower its resistivity—and if you cool it to an extremely low temperature, you can sometimes make the resistivity disappear altogether, in a phenomenon known as superconductivity.

REPLY [2 votes]: Electrical conductivity is for the most part determined by electron-phonon(quantized lattice vibration) scattering. As temperature increases and the phonon amplitude increases, there are an increased amount of scattering events which result in decreased conductivity.


TITLE: Confused about this "counterexample" to Chern-Gauss-Bonnet theorem (for manifold with boundary).
QUESTION [2 upvotes]: Let $\Sigma$ be a compact surface with boundary $\partial \Sigma$. Fix a Riemannian metric on $\Sigma$ and let $D \in \Omega^1(SO(\Sigma),\mathfrak{o}(2) )$ be a principal connection on the associated oriented orthonormal frame bundle $p:SO(T\Sigma) \to \Sigma$. There exists a unique $\Omega \in \Omega^2(\Sigma)$ such that $p^* \Omega = \text{Pf}(\text{Curv}_D)$, the Pfaffian of the curvature form (up to some constant involving $\pi$ which I ignore). Let $\pi_0 : S(T\Sigma) \to \Sigma $ be the unit sphere bundle with $N: \partial \Sigma \to S(T \Sigma)$ the unit normal and let $\Phi \in \Omega^1(S(T\Sigma))$ satisfy $d \Phi = \pi_0^* \Omega$. Then the Chern-Gauss-Bonnet theorem for manifolds with boundary says:
$$  \int_{\partial \Sigma}N^*\Phi = \int_{\Sigma} \Omega - \chi(\Sigma). $$
$T\Sigma$ is a trivial bundle so $SO(T\Sigma)$ is a trivial principal $SO(2)$-bundle so pick the canonical flat connection $A=\theta^L =g^{-1}dg$ on $\Sigma \times SO(2)$ (i.e. the left-invariant Maurer-Cartan form). The curvature of $A$ is $0$ so the Pfaffian of the curvature of $A$ is $0$ hence $\Omega=0$ and $\Phi=0$. Putting this into the Chern-Gauss-Bonnet theorem we conclude $\chi(\Sigma)=0$ which is certainly not true for all $\Sigma$.

REPLY [2 votes]: As I commented, when you put a flat metric on $\Sigma$, the geodesic curvature integral on $\partial\Sigma$ will definitely not vanish. (Think about classical Gauss-Bonnet for planar regions.)
Your claim that $T\Sigma$ is trivial may not be obvious to everyone. But you can put a nowhere-zero vector field on a manifold with nonempty boundary (for example, double the manifold, take a generic vector field, and then move its zeroes to "the other half"). On a (Riemannian) surface, once you have a nowhere-zero vector field, rotating it $\pi/2$ gives another, and so you've trivialized the tangent bundle.


TITLE: Distribution equation
QUESTION [2 upvotes]: I am trying to solve $xT'+T = 0$ where $T$ is a distribution
I already know (it is asked in the previous point) that $(xT)'= T + xT'$
my thoughts : $ (xT)'(f) = - (xT)(f')$ and from this point I immediately think of $T$ equals the dirac measure at $0$ since  $\int xδ(f'(x)) = 0*f'(0) = 0$, or an arbitrary constant $c$ times the dirac measure at $0$
is it correct ?

REPLY [1 votes]: $(xT)' = 0$ means $xT = c$, then let $S = pv(1/x)$,
$x  S = 1$ thus $x(T-c S) = 0$ which means $T-c S = b \delta$


TITLE: Problems regarding probability
QUESTION [0 upvotes]: A fair die is alternately thrown by two persons . The first one wins if one dot appears and the second one wins if 2-dots or 3-dots appear. The first one starts throwing the die. What is the probability that the first one ultimately wins ?

REPLY [2 votes]: if the probability of player 1 winning is P, then you can note that if both players fail to win on their first roll, then player 1 has returned to an identical position with a probability of winning of P.  The chances of each player failing on round one is
($\frac{5}{6})(\frac{4}{6}) = \frac{5}{9}$
and the chance of player 1 winning on roll 1 is $\frac{1}{6}$
P = prob(winning first roll) + P x prob(game goes to second round)
$P = \frac{1}{6} + \frac{5}{9} P$
$P = \frac{3}{8}$

REPLY [0 votes]: Let $p$ denote the probability that first one wins and let $q$ denote the probability that the first one wins if one dot does not appear by his first throw.
Then $p=\frac16+\frac56q$ and $q=\frac46p$  (do you see why?) so that $p=\frac16+\frac56\frac46p$ leading to $p=\frac38$.