Example \(SL(2,R)\)ΒΆ

Recall the \(K\backslash G/B\) elements of \(SL(2,R)\):

atlas> set G=SL(2,R)
Variable G: RealForm
atlas> G
Value: connected split real group with Lie algebra 'sl(2,R)'
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

If we again look at the block of the trivial

atlas> set B=block_of (trivial (G))
Variable B: [Param]
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)

We focus on the first two elements:

atlas> B[0]
Value: final parameter (x=0,lambda=[1]/1,nu=[0]/1)
atlas> B[1]
Value: final parameter (x=1,lambda=[1]/1,nu=[0]/1)

Remark: There are two commands that give us discrete series directly. For other groups with more representations it is more helpful to list them directly. See the example of \(Sp(4,R)\) in the next section.

Recall that these are the (discrete series) representations associated to the compact Cartan. Note that they both have Harish-Chandra parameter lambda=rho. This is because the software is using a different x. Remember that we have to fix a KGB element x_b to fix a real group \(K\). Let us fix it to be x=0:

atlas> set x_b=KGB(G,0)
 Variable x_b: KGBElt
 atlas> x_b
 Value: KGB element #0

Now, in order to identify the representation associated to x=1` with a representation associated to ``x=0, we need to conjugate x=1 to x=0. This will conjugate lambda to -lambda. Then the harish chandra parameters of the discrete series with respect to the fixed element x=0 will be:

atlas> hc_parameter(B[0],x_b)
Value: [ 1 ]/1
atlas> hc_parameter(B[1],x_b)
Value: [ -1 ]/1

So, one is the holomorphic discrete series and the other is the anti-holomorphic one. But by choosing x_b =1 we get the opposite situation for the Harish Chandra parameters and holomorphic convention.