David Vogan Introduction Langlands classification A Representations of (g; K )-modules R(h; L)-mod reductive groups Langlands classification B Cartan subgroups Langlands David Vogan classification C Your friend K (R) RTNCG August-September 2021 Outline David Vogan Introduction What are these talks about? Langlands classification A (g; K )-modules Langlands classification: big picture R(h; L)-mod Langlands Introduction to Harish-Chandra modules classification B Cartan subgroups (h; L)-modules as ring modules Langlands classification C Your friend K ( ) Langlands classification: some details R Cartan subgroups of real reductive groups Langlands classification: getting explicit Representations of K (R) What real reductive groups? David Vogan Introduction Old days: assumed G(R) connected semisimple. Langlands Problem is that G( ) is studied using Levi classification A R (g; K )-modules subgroups; these aren’t connected even if G is. R(h; L)-mod Here are some possible assumptions for us: Langlands 1. Narrowest: G complex connected reductive algebraic classification B Cartan subgroups defined over R, G(R) = real points. Langlands 2. Somewhat weaker: G(R) is transpose-stable subgp classification C of GL(n; ) with G( )=G( )0 finite. R R R Your friend K (R) 3. Still weaker: G(R) is finite cover of a group as in (2). General notation: g(R) = Lie(G(R)), g = g(R) ⊗R C. Everything I say holds exactly under (1); lots is still true under the (strictly weaker) (2); most things work under (3). Structure of G(R) David Vogan Introduction G(R) ,! GL(n; R); stable by transpose; G(R)=G(R)0 finite: Langlands classification A t −1 Cartan involution of GL(n; R) is automorphism θ(g) = g . (g; K )-modules R(h; L)-mod Recall polar decomposition: Langlands GL(n; R) = O(n) × exp(symmetric matrices): classification B θ −θ Cartan subgroups = GL(n; R) × exp(gl(n; R) ) Langlands Inherited by G(R) as Cartan decomposition for G(R): classification C θ K (R) = G(R) = O(n) \ G(R); Your friend K (R) −θ s(R) = g(R) = symm matrices in g(R) S(R) = exp(s(R)) = pos def symm matrices in G(R); G(R) = K (R) × S(R) ' K (R) × s(R): Nice structures on G(R) come from nice structures on K (R) by solving differential equations along S(R). What representations (A)? David Vogan As good analytic C∗-algebra people, you understand Introduction Definition. Unitary representation of G(R) on Hilbert Langlands classification A space Hπ is weakly continuous homomorphism (g; K )-modules π : ! (H ): G U π R(h; L)-mod Irreducible if Hπ has exactly two closed G(R)-invt subspaces. Langlands classification B Chevalley told Harish-Chandra to weaken this definition. Cartan subgroups Langlands Definition. Representation of reductive G(R) on loc cvx classification C complete V is weakly continuous group homomorphism π Your friend K (R) π : G ! GL(Vπ) 1 1 Get a new loc cvx complete Vπ ⊂ Vπ on which π differentiates to action of U(g). Define Z(g) = U(g)Ad(G(R)). Schur’s lemma suggests that 1 Z(g) should act by scalars on Vπ for irreducible π. Always true for π unitary (Segal), fails sometimes for nonunitary π on any noncompact G(R) (Soergel). What representations (B)? David Vogan π ( ) Definition (Harish-Chandra) Rep of G R on complete Introduction loc cvx V is quasisimple if Z(g) acts by scalars on V 1. π π Langlands classification A [ You know to care about G(R)u = unitary equivalence (g; K )-modules classes of irr unitary representations. R(h; L)-mod Langlands HC says to care about larger G[(R)= infinitesimal classification B equivalence classes of irr quasisimple π. Cartan subgroups Defining infinitesimal equivalence is a bit complicated; soon... Langlands classification C Your friend K (R) To see the value of this, helpful to introduce G[(R)h = infl equiv classes of irr quasisimple π with nonzero (maybe indefinite) invariant Hermitian form. G[(R)u ⊂ G[(R)h ⊂ G[(R): You know that the left term is interesting. I claim that it’s best understood by understanding the right term and the two inclusions. What representations (C)? David Vogan Introduction Langlands classification A G[(R)u ⊂ G[(R)h ⊂ G[(R) (g; K )-modules unitary ⊂ hermitian ⊂ quasisimple R(h; L)-mod Langlands desirable ⊂ acceptable ⊂ available classification B Cartan subgroups [ Langlands classification beautifully describes G(R) as Langlands complex algebraic variety. classification C Your friend K (R) Knapp-Zuckerman describe G[(R)h as real points of this alg variety: fixed points of simple complex conjugation. G[(R)u is cut out inside G[(R)h by real algebraic inequalities, more or less computed by Adams, van Leeuwen, Trapa, V. What can we ask about representations? David Vogan Introduction Start with a reasonable category of representations. Langlands Example: cplx reductive g ⊃ b = h + n; BGG category O classification A consists of U(g)-modules V subject to (g; K )-modules R(h; L)-mod 1. fin gen: 9V0 ⊂ V ; dim V0 < 1; U(g)V0 = V . Langlands 2. b-locally finite: 8v 2 V ; dim U(b)v < 1. classification B P 3. h-semisimple: V = γ2h∗ Vγ . Cartan subgroups Want precise information about reps in the category. Langlands Example: V in category O classification C Your friend K (R) 1. dim Vγ is almost polynomial as function of γ. hP λi 2. V has a formal character λ2h∗ aV (λ)e =∆. Want construction/classification of reps in the category. ∗ Example: λ 2 h I(λ) =def U(g) ⊗U(b) Cλ = Verma module. n 1. (STRUCTURE THM): I(λ) has highest weight Cλ ,! I(λ) . 2. (QUOTIENT THM): I(λ) has unique irr quo J(λ). 3. (CLASSIF THM): Each irr in O is J(λ), unique λ 2 h∗. How do you do that? David Vogan ∗ + g ⊃ b = h + n, ∆ = ∆(g; h) ⊂ h roots, ∆ roots in n. Introduction ∗ Langlands partial order on h : classification A µ0 ≤ µ () µ0 2 µ − ∆+ N (g; K )-modules 0 X () µ = µ − nαα; (nα 2 N) R(h; L)-mod α2∆+ Langlands Proposition. Suppose V 2 O. classification B ∗ Cartan subgroups 1. If V 6= 0; 9 maximal µ 2 h subject to Vµ 6= 0. ∗ n Langlands 2. If µ 2 h is maxl subj to Vµ 6= 0, then Vµ ⊂ V . classification C n 3. If V 6= 0; 9µ with 0 6= Vµ ⊂ V . Your friend K (R) ∗ n 4. 8λ 2 h , Homg(I(λ); V ) ' Homh(Cλ; V ). Parts (1)–(3) guarantee existence of “highest weights;” based on formal calculations with lattices in vector P spaces, and n · Vµ0 ⊂ α2∆+ Vµ0+α. Sketch of proof of (4): n HomU(g)(U(g) ⊗U(b) Cλ; V ) ' HomU(b)(Cλ; V ) = HomU(h)(Cλ; V ): First isom: “change of rings.” Second: n · Cλ =def 0. Moral of the story David Vogan Introduction For category O, three key ingredients: Langlands classification A 1. Change of rings U(g) ⊗U(b) · Verma mods I(λ). n 2. Universality: Homg(I(λ); V ) ' Homh(Cλ; V ). (g; K )-modules 3. Highest weight exists: J irr =) Jn 6= 0. R(h; L)-mod ∗ Langlands #2 is homological alg, #3 is comb/geom in h . classification B ∗ n Cartan subgroups Irrs J in O ! λ 2 h characterized by Cλ ⊂ J(λ) . Langlands Same three ideas apply to G(R) representations. classification C Your friend K (R) Technical problem: change of rings isn’t projective, so ⊗ Tor. n 0 p Parallel problem: J = H (n; J) derived functors H (n; J). Conclusion will be: irr G(R)-reps J ! γ 2 H[(R), s some Cartan H(R) ⊂ G(R); char by Cγ ⊂ H (n; J). Next topic: Harish-Chandra’s algebraization of rep theory, making possible the program outlined above. Principal series for SL(2; R) (skip this!) David Vogan To understand how Harish-Chandra studied reductive group representations, need a serious example. Introduction Langlands But there isn’t time; so look at these slides on your own! classification A (g; K )-modules Use principal series repns for SL(2; R) =def G(R). R(h; L)-mod 2 2 G(R) y R , so get rep of G(R) on functions on R : Langlands classification B [ρ(g)f ](v) = f (g−1 · v): Cartan subgroups Lie algs easier than Lie gps write sl(2; ) action, basis Langlands R classification C 1 0 0 1 0 0 D = ; E = ; F = ; 0 −1 0 0 1 0 Your friend K (R) Action on functions on R2 is by vector fields: @f @f @f @f ρ(D)f = −x1 + x2 ; ρ(E) = −x2 ; ρ(F) = −x1 : @x1 @x2 @x1 @x2 General principle: representations on function spaces are reducible ! exist G(R)-invt differential operators. @ @ Euler deg operator E = x1 + x2 commutes with G( ). @x1 @x2 R Conclusion: interesting reps of G(R) on eigenspaces of E. Principal series for SL(2; R) (also skip) David Vogan Previous slide: expect interesting reps of G(R) = SL(2; R) on homogeneous functions on R2. Introduction Langlands For ν 2 C, 2 Z=2Z, define classification A ν; 2 −ν−1 W = ff :(R − 0) ! C j f (tx) = jtj sgn(t) f (x)g; (g; K )-modules functions on the plane homog of degree −(ν + 1; ). R(h; L)-mod Langlands ν ν + 1 simplifies MANY things later. classification B ν, Study W by restriction to circle f(cos θ; sin θ)g: Cartan subgroups W ν, ' fw : S1 ! j w(−s) = (−1)w(s)g; f (r; θ) = r −ν−1w(θ): Langlands C classification C Compute Lie algebra action in polar coords using Your friend K (R) @ @ @ @ @ @ = −x2 + x1 ; = x1 + x2 ; @x1 @θ @r @x2 @θ @r @ = −ν − 1; x = cos θ; x = sin θ: @r 1 2 Plug into formulas on preceding slide: get @ ρν,(D) = 2 sin θ cos θ + (− cos2 θ + sin2 θ)(ν + 1); @θ @ ρν,(E) = sin2 θ + (− cos θ sin θ)(ν + 1); @θ @ ρν,(F) = − cos2 θ + (− cos θ sin θ)(ν + 1): @θ A more suitable basis (skip this too!) David Vogan Have family ρν, of reps of SL(2; R) defined on functions on S1 of homogeneity (or parity) : Introduction @ Langlands ρν,(D) = 2 sin θ cos θ + (− cos2 θ + sin2 θ)(ν + 1); classification A @θ @ (g; K )-modules ρν,(E) = sin2 θ + (− cos θ sin θ)(ν + 1); @θ @ R(h; L)-mod ρν,(F) = − cos2 θ + (− cos θ sin θ)(ν + 1): @θ Langlands classification B Hard to make sense of.