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>
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>

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)
atlas>

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)
atlas>

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
 atlas>

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
atlas>

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.