Example: \(SL(2,\mathbb R)\)

Let us start with the group \(G=SL(2,\mathbb C)\). For this group set \(x_b = diag(i,-i)\).

Then \((x_b)^2 =-Id \in Z(G)\). The stabilizer in \(K\) of this element is the diagonal torus

\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ K^{{\theta }_x}=\left\{ \left( \begin{array}{cc} z & 0 \\ 0 & \frac{1}{z} \end{array}\right) :z\in {\mathbb C}^{\times }\right\}\cong SO(2,\mathbb C)\)

So the real points of this group is the compact real form \(K(\mathbb R)=SO(2)\) and the real form of \(G\) with this maximal compact subgroup is \(G(\mathbb R)=SL(2,\mathbb R)\)

In this setting it is better to think of \(G(\mathbb R)\) as \(SU(1,1)\)

Then, the \(K\) orbits on \(G/B\) consist of three elements:

\(\ \ \\ \ \ \ \ \ \ \ \ \ x_b =\left( \begin{array}{cc} i&0\\ 0&-i \end{array}\right),\quad-x_b=\left(\begin{array}{cc} -i&0\\ 0&i \end{array}\right) ,\quad u=\left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right)\)

So, \(x_b\) and \(-x_b\) are all the elements of the cartan that are conjugate to \(x_b\). And there is only one other element, \(u\), up to conjugacy by \(H\), which is in the normalizer of the Cartan and is conjugate to \(x_b\).

Note that \(x_b\) and \(-x_b\) are both fixed by conjugation by \(H\) and \(H\) acts by conjugation on \(u\). Moreover, we can replace \(u\) by any element of the form

\(\left(\begin{array}{cc} 0 & z \\ -\frac{1}{z} & 0 \end{array} \right)\)

So, \(K\) acting on \(G/B\) has three elements, representatives of the \(K\) orbits on the conjugacy classes of Borel subgroups.

Observation: This is the usual action of \(Sl(2,\mathbb C)\) on the projective plane that gives three orbits, \(0\), \(\infty\) and \({\mathbb C}^{\times }\).

To obtain these elements with the software there is a command print_KGB (G):

atlas> whattype print_KGB ?
Overloaded instances of 'print_KGB'
  RealForm->void
  KGBElt->void
atlas>

atlas> set G=SL(2,R)
Variable G: RealForm
atlas> G
Value: connected split real group with Lie algebra 'sl(2,R)'
atlas>
atlas> print_KGB (G)
kgbsize: 3
Base grading: [1].
0:  0  [n]   1    2  (0)#0 e
1:  0  [n]   0    2  (1)#0 e
2:  1  [r]   2    *  (0)#1 1^e
atlas>

So \(KGB\) has three elements labeled 0, 1, 2 and the second to last column give the number of the Cartan. So the first two elements correspond to the compact Cartan and the last one to the split Cartan.

Now let us look at the block of the trivial representation of \(G\):

atlas> set B=block_of (trivial (G))
Variable B: [Param]
atlas> void: for p in B do prints(p) od
final parameter (x=0,lambda=[1]/1,nu=[0]/1)
final parameter (x=1,lambda=[1]/1,nu=[0]/1)
final parameter (x=2,lambda=[1]/1,nu=[1]/1)
atlas>

Another way to do this is to define a new function, say show as follows:

atlas> set show([Param] params)= void: for p in params do prints(p) od
Added definition [6] of show: ([Param]->)
atlas> show(B)
final parameter (x=0,lambda=[1]/1,nu=[0]/1)
final parameter (x=1,lambda=[1]/1,nu=[0]/1)
final parameter (x=2,lambda=[1]/1,nu=[1]/1)
atlas>

Then you can use the same function show for different parameter sets.

So, the first two elements parametrize the discrete series with Harish Chandra parameters \(\rho\) and \(-\rho\). More on this later.

Now as representatives of Borels we have:

\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x_b \mapsto B=\left\{ \left( \begin{array}{cc} z & w \\ 0 & \frac{1}{z} \end{array} \right) |z\in {\mathbb C}^{\times },w\in \mathbb C \right\}\)

which is the Borel that was fixed at the begining. Now, taking an element that conjugates $x_b$ to its negative we have:

\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -x_b=s_{\alpha }(x_b) \mapsto B'=s_{\alpha }(B)=\left\{ \left( \begin{array}{cc} z & 0 \\ w & 1/z \end{array} \right) |z\in {\mathbb C}^{\times },w\in \mathbb C \right\}\);

and for \(u\), the element that conjugates \(x_b\) to \(u\) is

\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g=\frac{1}{\sqrt{2}} \left( \begin{array}{cc} 1 & -1 \\ 1 & 1 \end{array} \right)\).

Then

\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ B''=gBg^{-1} =\left\{ \left(\begin{array}{cc} cosh(z) & sinh(z) \\ sinh(z) & cosh(z) \end{array} \right) + \frac{1}{2} \left(\begin{array}{cc} w & w \\ -w & w \end{array} \right) |z\in {\mathbb C}^{\times },w\in \mathbb C \right\}\).

One of the key points comes from just looking at the Cartan part of the last \(B''\):

\[\begin{split}H''=\left\{ \left(\begin{array}{cc}cosh(z)&sinh(z)\\ sinh(z)&cosh(z)\end{array}\right) |z\in {\mathbb C}^{\times} \right\}.\end{split}\]

Since we fixed the Cartan involution \({\theta }_{x_b} = diag(i,-i)\), it is acting on this Cartan by \(-1\) (i.e. by taking the inverse). It acts trivially on the diagonal Cartan.

The set of real points of this Cartan is

\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ H''(\mathbb R)=\left\{ \pm Id \left(\begin{array}{cc} cosh(x) & sinh(x) \\ sinh(x) & cosh(x) \end{array} \right) | x\in \mathbb R \right\} \cong {\mathbb R}^{\times }\)

Which is the ususal way of writing split Cartan in \(SU(1,1)\).

The point is that, the pair \((H'', {\theta }_{x_b} )\) is conjugate under \(G\) to the pair \((H, {\theta }_u )\). That is, to \(H\) and the conjugation action of this element \(u\).

In other words, the first pair is how we normally think of this Cartan in the real group: we fix a real form (determined by the Cartan involution \({\theta }_{x_b}\)) and vary the Cartans within this real group. And in this case there are two Cartans, one compact and one split.

The second pair is how atlas thinks of it. That is, it fixes the original (diagonal) Cartan and varies the Cartan involution which acts by \(-1\) on the fixed diagonal Cartan.

Moral of the Story

To summarize, we always fix:

\[H\subset B,\quad x_b ,\quad \theta = int(x_b ),\quad \text{and}\quad K=G^{\theta };\]

we vary

\[x\in \mathcal X ,\quad \text{and} \quad {\theta }_x ;\]

and we map

\[\{ (H',\theta ) \}/K \leftrightarrow \{ (H, {\theta }_{x} ) | x\in \mathcal X \}.\]

So, rather than talking about the Cartan subgroups of \(G\) with their action of the fixed \(\theta\) up to conjugacy by \(K\), we conjugate everything back to the fixed \(H\) and we vary the \({\theta }_x\).

Similarly for the Borels we have:

\[\{ (B',\theta )\}/K\leftrightarrow \{ (B,{\theta _x})|x\in \mathcal X \}\]