\classheader{2013-03-15}

This lecture is really an apology for not having done any examples.

\subsection*{Classical groups over $\R$ and $\C$, and their maximal compact subgroups}

There's a paper by Andre Weil, from around 1955, titled \textit{Algebras with Involution}, which says what we'll do today but over all fields. Another good reference is Helgason's \textit{Differential Geometry and Symmetric Spaces}.

Let $V$ be a finite-dimensional vector space over $K$, where $K=\R$, $K=\C$, or $K=\H$ (the quaternions). In each case, we have an involution $K\to K$ denoted by $z\mapsto \overline{z}$, which is additive and reverses multiplication, i.e.
\[\overline{zw}=\overline{w}\,\overline{z}.\]
Given a (left) finite-dimensional vector space $V$ over $K$, we say that $B:V\times V\to K$ is sesquilinear if it is additive in each variable separately, and
\[B(\lambda v,\mu w)=\lambda B(v,w)\overline{\mu}\]
for all $\lambda,\mu\in K$ and $v,w\in V$.

We say that a sesquilinear form $B$ is positive definite when for all non-zero $v\in V$, we have that $B(v,v)\in\R$, and that $B(v,v)>0$.

This $B$ is essentially unique, as we will now see. If $e_1,\ldots,e_n$ is a $K$-basis for $V$, then a standard example of such a $B$ is
\[B\left(\sum \lambda_ie_i,\sum \mu_j e_j\right)=\sum_{i=1}^n \lambda_i\overline{\mu}_i.\]
An easy exercise is that $\GL_K(V)$ acts transitively on the space of sesquilinear, positive definite forms $B:V\times V\to K$.

We won't prove this, but any compact group $G$ has a left-invariant measure which is unique up to scaling. Thus, there is a unique left-invariant measure $\mu$ such that $\mu(G)=1$. This lets us treat $G$ as if it were finite.

\begin{corollary-N}
If $\rho:G\to\GL(V)$ is a representation of a compact group $G$, then for all $v\in V$, we can define
\[Pv=\int_G\rho(g)v\,d\mu(g),\]
and the map $P$ is a projection of $V$ onto the subspace
\[V^G=\{v\in V\mid \rho(g)v=v\text{ for all }g\in G\}.\]
In other words, $Pv\in V^G$ for all $v\in V$, and if $v\in V^G$, then $Pv=v$.
\end{corollary-N}

\begin{corollary-N}
If $\rho:G\to\GL(V)$ is a representation of a compact group $G$, then there is a unique positive definite sesquilinear form $B:V\times V\to K$ such that
\[B(\rho(g)v,\rho(g)w)=B(v,w)\]
for all $g\in G$ and $v,w\in V$.
\end{corollary-N}
\begin{proof}
Let $B$ be any positive definite sesquilinear form on $V$. Define $\widetilde{B}$ by
\[\widetilde{B}(v,w)=\int_G B(\rho(g)v,\rho(g)w)\,d\mu(g).\]
It is clear that $\widetilde{B}$ has the desired properties.
\end{proof}

Let $A$ be a finite-dimensional associative algebra over $\R$. The units of $A$, denoted $A^\times$, form an open subset of $A$. $A^\times$ is a Lie group, with Lie algebra $\Lie(A^\times)=A$, and for all $X,Y\in\Lie(A^\times)$, we have $[X,Y]=XY-YX$.

Now assume that $A$ is simple, which forces that $A=\End_K(V)$ where $V$ is some finite-dimensional $K$-vector space, where $K=\R$, $\C$, or $\H$. We have that $A^\times=\GL_K(V)$.

Observe that, given a positive definite, sesquilinear form $B$ on $V$, the set
\[M=\{g\in\GL_K(V)\mid B(gv,gw)=B(v,w)\text{ for all }v,w\in V\}.\]
is a compact group; indeed, every maximal compact subgroup of $\GL_n(\R)$, $\GL_n(\C)$, $\GL_n(\H)$ is a conjugate of $\mathrm{O}_n(\R)$, $\mathrm{U}_n(\C)$, or $\mathrm{Sp}_n(\H)$ respectively.

\begin{theorem}
Let $G$ be a Lie group with finitely many connected components. Then there exist maximal compact subgroups of $G$, and they are all conjugate to each other.
\end{theorem}

\begin{definition}
Given an associative ring $A$, an involution on $A$ is a map $i:A\to A$ that is additive, $i(ab)=i(b)i(a)$ for all $a,b\in A$, and $i(i(a))=a$ for all $a\in A$.
\end{definition}

Let
\[G=\{a\in A\mid ai(a)=1=i(a)a\},\]
i.e. the subset consisting of $a\in A$ such that $i(a)=a^{-1}$. Note that
\[i(ab)=i(b)i(a)=b^{-1}a^{-1}=(ab)^{-1}=i(ab),\]
so that $G$ is a subgroup. As an exercise, you can show using the implicit function theorem on the map $A\to A^+=\{a\in A\mid a=i(a)\}$ defined by $a\mapsto ai(a)$, that $G$ is a Lie group and that $\Lie(G)=A^{-}$ ($A^+$ and $A^-$ are the eigenspaces of $i$ for $+1$ and $-1$, respectively).

Let $B:V\times V\to\R$ be a non-degenerate bilinear form. When we require that $B$ is symmetric, we write $\mathrm{O}(B)$ for the stabilizer of $B$ in $\GL_\R(V)$. When we require that $B$ is antisymmetric, we write $\mathrm{Sp}(B)$ for the stabilizer of $B$ in $\GL_\R(V)$.

Let's suppose that $B$ is symmetric for now. Then there is a basis $e_1,\ldots,e_n$ of $V$ such that $B(e_i,e_j)=0$ for $i\neq j$, and for some $p+q=n$, we have
\[B(e_i,e_i)=\begin{cases}
1 & \text{ if } i=1,\ldots,p,\\
-1 & \text{ if }i=p+1,\ldots,n.
\end{cases}\]
By Sylvester's law of inertia, $p$ depends only on $B$. We write $\mathrm{O}(B)=\mathrm{O}(p,q)$.

Let $H$ be a compact subgroup of $\mathrm{O}(B)$, where $B$ is a non-degenerate symmetric bilinear form on $V$. We know that $H$ preserves a positive definite form $\langle\,\cdot\,,\,\cdot\,\rangle$. It follows that the map $T:V\to V$ defined by $B(x,y)=\langle Tx,y\rangle$ for all $x,y\in V$ satisfies $T=T^*$, and because $H$ preserves $B$, we have that $T$ commutes with the $H$ action. Because $B$ is non-degenerate, we have that $T$ is invertible.

Let $V=\bigoplus_{\lambda\in\R} V_\lambda$ be the eigenspace decomposition of $T:V\to V$ (note that the $V_\lambda$'s are $\langle\,\cdot\,,\,\cdot\,\rangle$ orthogonal). We have that $H(V_\lambda)\subseteq V_\lambda$ for all $\lambda$.

Let $V^+=\bigoplus_{\lambda>0}V_\lambda$ and $V^-=\bigoplus_{\lambda<0}V_\lambda$, so that $V=V^+\oplus V^-$. We see that $B|_{V^+}$ is positive definite, and $-B|_{V^-}$ is positive definite, and $B(v,w)=0$ for all $v\in V^+$ and $w\in V^-$.

Thus, we have
\begin{center}
\begin{tikzcd}
\underbrace{H}_{\text{compact}} \ar[hook]{r} & \mathrm{O}(V^+,B|_{V^+})\times \mathrm{O}(V^-,B|_{V^-})\ar[hook]{r} & \mathrm{O}(V,B)
\end{tikzcd}
\end{center}
For any $W\subseteq V$ such that $B|_W$ is positive definite and $B|_{W^\perp}$ is negative definite, we have that
\[\{g\in \mathrm{O}(V,B)\mid gW=W\}\]
is a maximal compact subgroup. Thus, maximal compact subgroups are in bijection with
\[\{W\subseteq V\mid B|_W\text{ is positive definite, and }B|_{W^\perp}\text{ is negative definite}\}.\]
This completes the symmetric case. Now suppose that $B$ is skew-symmetric, and let $G=\mathrm{Sp}(V,B)$. Let $V=W$, where $W$ is a $k$-dimensional $\C$-vector space, and $n=2k$. Let $H:W\times W\to\C$ be a positive definite Hermitian form, so that $\mathrm{U}(H)=\mathrm{U}(k)$ is a compact subgroup of $\GL_\C(W)$. If $g\in \mathrm{U}(H)$, then $\Im(H(gv,gw))=\Im(H(v,w))$. Let $B=\Im H$. We conclude that $\mathrm{U}(H)$ is a subgroup of $\mathrm{Sp}(V,B)$.


Let $A=\End_\R(V)$. Here is how to pass from the general description, involving the arbitrary involution $i$, to the specific case of symmetric / skew-symmetric forms. By Wedderburn theory, the map $L$ in this diagram must be an isomorphism of $\R$-algebras:
\begin{center}
\begin{tikzcd}
\End_\R(V) \ar{r}{i} \ar{d}[swap]{\mathsf{T}} & \End_\R(V)\\
\End_\R(V^*) \ar{ru}[swap]{L}
\end{tikzcd}
\end{center}
Thus, Wedderburn etc. gives an isomorphism $h:V\to V^*$ such that $i(T)=h^{-1}\transpose{T}h$, and thus
\[i(i(T))=h^{-1}\transpose{i(T)}h=h^{-1}\transpose{h}T(\cdots)^{-1}.\]
There is a $\lambda\in\R$ such that $\transpose{h}=\lambda h$ and $h=\lambda \transpose{h}$, so that $\lambda^2=1$.

Now suppose that $A=\End_\C(V)$. First, we let $i:\C\to\C$ be the identity, and we get $G=\mathrm{O}(V,B)$ when $B$ is symmetric (in which case $B=\mathrm{Id}_n$), and $\mathrm{Sp}(V,B)$ when $B$ is anti-symmetric. 

Second, let us take $i:\C\to\C$ be $i(z)=\overline{z}$. We get that $H:V\times V\to\C$ is a non-degenerate Hermitian form, and $G=\mathrm{U}(V,H)$. Choosing a basis, we have that for some $p+q=n$, the matrix of $H$ is
\[\begin{bmatrix}
\mathrm{Id}_p & 0\\ 0 & -\mathrm{Id}_q
\end{bmatrix}\]
The compact subgroups are similar to the $\mathrm{O}(p,q)$ case.

\begin{remark}
$\mathrm{U}(p,q)/\mathrm{U}(p)\times\mathrm{U}(q)$ is a Hermitian symmetric space.
\end{remark}

Lastly, we come to the case $K=\H$. Let $B:V\times V\to \H$ be a non-degenerate sesquilinear form. We have two cases, where $\overline{B(v,w)}=B(w,v)$ or $\overline{B(v,w)}=-B(w,v)$. In the first case, the matrix is again of the form
\[\begin{bmatrix}
\mathrm{Id}_p & 0\\ 0 & -\mathrm{Id}_q
\end{bmatrix}\]
and in the second case, the matrix is
\[\begin{bmatrix}
i & \cdots & 0\\
\vdots & \ddots & \vdots\\
0 & \cdots & i
\end{bmatrix}\]






