Kirillov's classification

Kirillov's classification

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Department of Mathematics
University of Wisconsin, Madison
Madison, WI 53706 USA

Last updates: 23 May 2010

Kirillov's classification

Let G be a simply connected nilpotent Lie group. The coadjoint representation of G is the action of G on 𝔤*=Hom 𝔤 given by gφ x =φ Ad g -1 x ,for all  gG   and  φ 𝔤* . A coadjoint orbit is a set Gφ,φ 𝔤* . Then coadjoint orbits irreducible unitary representations of   G .

Let Ω be a coadjoint orbit Ω 𝔤* . Let φΩ so that Ω=Gφ. A Lie subalgebra 𝔥𝔤 is subordinate to φ if φ | 𝔥𝔥 =0,or, equivalently,φ xy =0for all  x,y𝔥. A subalgebra 𝔥 is subordinate to 𝔤 iff H eX e 2πiφ X defines a representation of H=exp 𝔥 .

Choose a maximal dimensional subalgebra 𝔥𝔤 which i subordinate to φ. (Choosing a maximal subalgebra 𝔥 with respect to inclusion turns out to be the same thing.) Let H=exp 𝔥 and define a one dimensional representation of H by 1φ : H * eX e 2πiφ X for x𝔥. Then VΩ = Ind GH 1φ is the irreducible representation of G associated with Ω. This says that every irreducible unitary representation of G is obtained by inducing from a one dimensional representation of some subgroup H. We will want to show that

  1. V is an irreducible representation of G,
  2. V does not depend on the choice of φ and 𝔥 (only on the orbit Ω ).
and answer the questions
  1. Why is it sufficient to define Wφ bu its values on eX for X𝔥 ?
  2. What is Ind HG W ?

Let 𝔥 𝔥~ be Lie subalgebras of 𝔤 and let φ 𝔤* be such that 𝔥 and 𝔥~ are both subordinate to φ. Then, since H H~ , there are fewer functions in Ind H~ G Wφ than in Ind HG Wφ . The G -actions on the two spaces are defined in the same way so that Ind H~ G Wφ is a submodule of Ind HG Wφ . So VΩ is not irreducible unless 𝔥 is maximal.

Suppose that Ω is a coadjoint orbit, φΩ is not irreducible unless 𝔥 is maximal.

Suppose that Ω is a coadjoint orbit and that φΩ and 𝔥 is a maximal dimensional subalgebra of 𝔤 subordinate to φ. Suppose that φ~ is another element of Ω. Then there is a gG such that φ~ =gφ- Ad g* φ. Let 𝔥~ =g𝔥= Ad g 𝔥. Then, for gXg𝔥, φ~ gX = gφ gX =φ g -1 gX =φ X , and since g XY = Ad g XY = Rg XY R g -1 = Rg XY-YX R g -1 = Rg X R g -1 Rg Y R g -1 - Rg Y R g -1 Rg X R g -1 = gX gY , it follows that 𝔥~ is subordinate to φ~ iff 𝔥 is subordinate to φ. If H=exp 𝔥 and H~ =exp 𝔥~ then H` =gH g -1 , since Ad g is the differential of Int g . So we get one dimensional representations Wφ : H h e 2πiφ X and W φ~ gH g -1 gh g -1 e 2πi φ~ gX = e 2πiφ X if h= eX . Define a map Ind HG Wφ Φ Ind H~ G W φ~ f f~ by   f~ t = f g -1 t ^~. Then f~ gh g -1 t = f h g -1 t ^~= h f g -1 t ^~= h~ f~ t and xf ^~ t = x f g -1 t ^~= f g -1 tx ^~= f~ tx = x f~ t . So Φ is a well defined G -module isomorphism.

In general, how do we construct maximal subordinate subalgebras?

(Vergne's lemma) Let V be a vector space and let , be a skew symmetric bilinear form on V. f= 0V V2 Vn =V ,dim Vi =i. Let W f , = i=0 n Vi Vi . Then

  1. W f , is an isotropic subspace of V,
  2. If xy = φ xy for some φ 𝔤* then 𝔥 is a subalgebra of 𝔤.

Let 𝔤 be a nilpotent Lie algebra such that dim Z 𝔤 =1. Then 𝔤=xyzW, where z=Z 𝔤 , xy =z, and yw =0 for wW.

Let y𝔤 such that the image of y in 𝔤/Z 𝔤 is a nonzero element of Z 𝔤/Z 𝔤 . Then ad y : 𝔤 Z 𝔤 t yt , has im ad y =Z 𝔤 , and ker ad y = Z𝔤 y . Thus dim ker ad y =dim 𝔤 -1 and ker ad y is an ideal of 𝔤 since ad y xt = y xt =- t yx - x ty =0,for  tker ad y   and  x𝔤. So Z𝔤 y =yzW,where   yw =0  for  wW, is an ideal of 𝔤. Let x𝔤 such that xy 0 and then normalise x so that xy =z.

Note that, as in the previous lemma, x+y+z is a subalgebra isomorphic to a Heisenberg algebra: 𝔤= 0 x z 0 y 0 | x,y,z and let x= 0 1 0 0 0 0 ,y= 0 0 0 0 1 0 ,z= 0 0 1 0 0 0 . Then 𝔤=x+y+z, xy =z, and z=Z 𝔤 . So, if G is a nilpotent Lie group with one dimensional center then G0 =exp Z𝔤 y ,where   Z𝔤 y =yzW, and G0 is normal since e tx eh e -tx = e t xh + , where t hx + Z𝔤 y since only involves brackets of x and h and Z𝔤 y is an ideal. So G= g0 e tx | g0 G0 ,t . Since Z𝔤 y is an ideal of 𝔤, G0 =exp Z𝔤 y is a normal subgroup of  G. In fact, every element of G can be written uniquely in the form g0 e tx , g0 G0 ,t . Since G0 is normal in G, the group G acts on G0 by automorphisms. Let V0 be an irreducible representation of G0 . Define g: V0 to be the G0 -module with the same vector space g: V0 =V and with G0 action g0 .v = g -1 g0 g v,for   g0 G0   and  v V0 . Then g: V0 is a new G0 representation and g: V0 is irreducible iff V0 is (since, if P V0 is a submodule of V0 the g: V0 then g:P is a submodule of g: V0 ). So G acts on the irreducible representations of G0 .

Why consider Lie algebras with one dimensional center? Suppose we are trying to prove that nilpotent Lie groups are monomial:
Let M be an irreducible representation. Then Z G acts on M by scalars. Let z1 ,, zk be a basis of Z 𝔤 . Let φ:Z 𝔤 be a map such that e 2πiφ z =ρ ez ,for  zZ 𝔤 ,  and let   Z0 = zZ 𝔤 | φ z =0 . So z Z0 if i=1 k ci φ bzi =0. So dim Z0 dimZ-1 and exp Z0 acts trivially on M. Let ρ:GGL M has a nontrivial kernel unless dimZ 𝔤 =1. In the case when dimZ 𝔤 =1 we can use Kirillov's lemma to obtain the normal subgroup G0 =exp Z𝔤 y .

References [PLACEHOLDER]

[BG] A. Braverman and D. Gaitsgory, Crystals via the affine Grassmanian, Duke Math. J. 107 no. 3, (2001), 561-575; arXiv:math/9909077v2, MR1828302 (2002e:20083)

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