More on $$K\backslash G/B$$¶

Effect on $$(\mathfrak g , K_x)$$ modules¶

Another important point is that now we will be talking about $$(\mathfrak g , K_x )$$ modules as we vary $$x$$. The $$x$$’s are all conjugate to $$x_b$$, but not literally equal.

Therefore, the $$K_x$$ are all conjugate to K but they are not equal. So, we get all these $$(\mathfrak g , K_x )$$ modules that are all equivalent to $$(\mathfrak g , {K_x}_b )$$ modules; and by using this conjugation we can conjugate them all back to a $$(\mathfrak g , {K_x}_b )$$ module.

More precisely, if $$\pi$$ (resp. $$\pi '$$) are $$(\mathfrak g , K_x )$$ (resp. $$(\mathfrak g , K_{x'} )$$ modules, then $$\pi \cong {\pi }'$$ if there is $$g \in G$$ with

$gxg^{-1} =x'`, \quad {\pi }^g \cong {\pi}$

In this way, the software is varying $$x$$, but in the end you can conjugate back to $$x_b$$.

Involutions and Conjugacy classes of Cartan subgroups¶

Recall that we are fixing $$x_b$$ and $$\mathcal X =\mathcal X (\ x_b )$$

This determines a fixed $$K$$ and $$\mathcal X$$ parametrizes:

$K\backslash G/B \leftrightarrow \mathcal X$

And in the software, this gives a finite set of parameters:

$KGB= \mathcal X = \{x_0, \ldots x_{n-1} \}$

Now the Weyl group $$W$$ acts naturally by conjugation $$\mathcal X$$. Then,

$\mathcal X /W \leftrightarrow \text{conjugacy classes of Cartan subgroups}$

This is how we associate a Cartan subgroup to an element $$x$$. Namely, via this map from $$\mathcal X$$.

Moreover

$Stab_W (x)\simeq W(K,H)\simeq W(G(\mathbb R),H(\mathbb R)),$

This is the rational Weyl group of the real form of the group with respect to the real Cartan subgroup, Which in the $$\theta$$ world we think of it as $$W(K,H)$$.

Finally, there is a map $$\rho :\mathcal X\rightarrow {\mathcal I}_W$$ (involutions in $$W$$). The map is the obvious one: $$x$$ is an element in the normalizer of $$H$$ so we take its image in the Weyl group and that is an involution. Taking the conjugacy classes of involutions in W gives a map:

${\mathcal I}_W /W\leftrightarrow \text{conjugacy classes of Cartan subgroups in quasisplit group.}$

The map $$\rho$$ is not necessarily surjective. But it is surjective if the group is quasisplit. So this $$\mathcal I$$ is telling us about Cartan subgroups of the quasisplit form.

The Algorithms paper has a picture of the $$KGB$$ space for $$Sp(4,R)$$. There are 11 elements in the space. The picture gives the fibers of elements in $$KGB$$ that go to the same conjugacy class of involutions and in turn to the same Cartan subgroup: Four elements get mapped to the identity involution which corresponds to the compact Cartan subgroup; two are mapped to the involutions from the short root reflections $$s_{\alpha _1}$$ and $$s_{\alpha _2}$$ corresponding to the intermediate Cartan subgroup isomorphic to $${\mathbb C}^{\times}$$; four are mapped to the long root reflections $$s_{\beta _1}$$, $$s_{\beta _2}$$, which correspond to the Cartan subgroup $$S^1 \times {\mathbb R}^\times$$; and one element is mapped to $$-Id$$, corresponding to the split Cartan subgroup.

In terms of representations, looking at each fiber of $$KGB$$ elements corresponding to a given Cartan subgroup, will give us representations attached to that subgroup. For example all the representations attached to the split Cartan subgroup correspond to the last element $$x_10$$ which is the fiber above the involution $$-1$$, etc.

KGB ordering¶

There is a partial order on the $$KGB$$ elements coming from the closure relations of the corresponding orbits. For example in the Hasse diagram for KGB for Sp(4,R), the vertical lines indicate closure relations.

There are four closed orbits at the bottom of the diagram which correspond to the elements $$x_0 ,x_1 ,x_3$$ and $$x_4$$, which in turn get mapped to the identity involution.

At the top of the diagram there is only one open orbit which is the element $$x_{10}$$, mapped to $$-Id$$. Below $$x_{10}$$ we have the elements corresponding to $$x_7 ,x_8$$ and $$x_9$$ and below them we have $$x_4 ,x_5$$ and $$x_6$$.

The output of the software respects this partial order. More on this later.