Representation Theory - First-Year Interest Group Seminar | F A 1, Study notes of Art

Download Representation Theory - First-Year Interest Group Seminar | F A 1 and more Study notes Art in PDF only on Docsity! Representation Theory (Fall 2004) Lecture 15-16 Fernando Rodriguez-Villegas November 9 and 11 2004 1 Tue 11/9 Recall definitions: G = SL2(Fp), B = {( ∗ ∗ 0 ∗ ) ∈ G } , N = {( 1 ∗ 0 1 ) ∈ G } ' Fp, and if ϕ : F×p → C× is a character, Vϕ = IndGBϕ. Also recall that: • Vϕ is irreducible if ϕ2 6= 1. • V1 = trivial⊕ Steinberg We are working on the case where ϕ =: τ is a quadratic character (τ2 = 1). We claim that Vτ = V +τ ⊕ V −τ and V ±τ are each of dimension (p+ 1)/2. Recall that Vτ = {f : G→ C with f(nag) = τ(a)f(g)}, where n ∈ N and a = ( a 0 0 a−1 ) , a ∈ F×p . Recall the definition of U : (Uf)(g) = ∑ n∈N f(w −1ng). U : Vϕ→̃Vϕ−1 is G-linear so if τ = τ−1 (so U : Vτ → Vτ ) then V ±τ are the eigenspaces of U . Let’s verify that U is G-linear: (g1(Uf))(g) = (Uf)(gg1) = ∑ n∈N f(w−1ngg1) = (U(g1f))(g) because (g1f)(w−1ng) = f(w−1gg1). Uf ∈ Vτ ; does (Uf)(n1ag) = τ(a)(Uf)(g)? (Uf)(n1ag) = ∑ n∈N f(w −1nn1ag) = ∑ n2∈N f(w −1n2ag); if n2 = ( 1 x0 1 ) then n2a = a ( 1 x/a2 0 1 ) , which we call n3. So ... = ∑ n3∈N f(w −1an3g) = ∑ n3∈N f(a −1w−1n3g), (since w−1a = a−1w−1) = τ(a−1) ∑ n3∈N f(w −1n3g) = τ(a)(Uf)(g). Notes by Kyle Schalm ; edited by Nick Leger 1 Claim: U is an isomorphism. Idea: Restrict U to a subgroup and diagonalize U . Consider Vτ as a representation of N . The characters of N are ( 1 x 0 1 ) 7→ ψ(x) where ψ(x) is an additive character of Fp. Fix a nonzero ψ; then any other character of Fp is of the form ψt(x) := ψ(tx), for some t ∈ Fp (if t = 0, ψ is the trivial character). We will show Vτ = ⊕p−1 t=0 V t τ where V t τ is the ψt-component of Vτ , meaning that if ft ∈ V tτ , ft 6= 0, then ft(gn) = ψt(n)ft(g). Also, ft(n1agn2) = τ(a)ψt(n2)ft(g). First, we claim that ft is determined by its value at 1, w. Let’s use the Bruhat decomposition G = B t BwN : either g ∈ B, in which case g = na and ft(g) = ft(na) = τ(a)ft(1), or g ∈ BwN , in which case g = n1awn2 and ft(g) = ft(n1awn2) = τ(a)ψt(n2)ft(w). Case: t 6= 0. ft(1) = ft(n · 1) = ft(1 · n) = ψt(n)ft(1), since ψt is not trivial, there is an n for which ψt(n) 6= 1 so that ft(1) = 0. Hence ft is completely determined by its value at w. Normalize by letting ft(w) = 1. Case: t = 0. Let f0 = { 1 7→ 0 w 7→ 1 f∞ = { 1 7→ 1 w 7→ 0. These span V 0τ (which is therefore 2-dimensional). To show V tτ are stable by U : If t 6= 0: Uft = λtft for some λt ∈ C. To compute this λt, (Uft)(w) = λtft(w) = λt and (Uft)(w) = ∑ n∈N ft(w −1nw), where w−1nw = ( 0 −1 1 0 ) ( 1 x 0 1 ) ( 0 1 −1 0 ) . If x 6= 0, this is ( 0 −x−1 1 0 ) ( x−1 0 0 x ) ( 0 1 −1 0 ) ( 1 −x−1 0 1 ) , so ft(w−1nw) = ψt(−x−1)τ(x−1)ft(w) = ψt(−x−1)τ(x−1), (since w ∈ N). Hence λt = ∑ x∈F×p ψt(−x−1)τ(x−1) = ∑ x∈F×p τ(−x)ψ(tx) (x 7→ −x−1) = τ(−t) ∑ x∈F×p τ(x)ψ(x) (tx 7→ x) =: g(τ), this last sum is a Gauss sum and equal to p, so g(τ)2 = τ(−1)p and g(τ) = ± √ τ(−1)p. (Gauss actually proved g(τ) = {√ p p ≡ 1 mod 4 i √ p p ≡ 3 mod 4 ). If t = 0 (continuing our computation of λt): Using the fact that V 0τ is spanned by f0, f∞ and these are determined by their values at 1 and w, 2 and Φ = ( 0 B −B 0 ) . This is a non-degenerate skew symmetric form on L× L. Ψ((l1, l2), (l′1, l ′ 2)) = l1Bl ′ 2 and Ψ((l1, 0), (0, l2)) = B(l1, l2). ψ = e2πi/pΨ, ϕ = e2πi/pΦ, b = e2πiB. The whole point is the connection G = SL2(Fp) ↪→ {Sp(U) : ( α β γ δ ) → (u 7→ uσ)}, where U = L × L, Sp(U) = {µ : U → U preserves Φ}. u = (l1, l2), uσ = (l1, l2) ( α β γ δ ) = (αl1 + γl2, βl1 + δl2). (l1 and l2 are themselves vectors of dimension 2.) σ 7→ ( αI2 βI2 γI2 δI2 ) . σ preserves Φ, i.e.( αI2 γI2 βI2 δI2 ) ( 0 B −B 0 ) ( αI2 βI2 γI2 δI2 ) = Φ using det(σ) = 1. We can embed Sp(U) in Aut0(H) (these preserve the center). σ 7→ ((ζ, u) 7→ (ζVσ(u), uσ)) where Vσ(u) = (ψ(uσ,uσ)ψ(u,u) ) 1/2. We have the Schrödinger representation of H, ρ : H → GL(V ) where V = {f : L→ C}, defined by ρ(ζ, 0) 7→ ζ. The center acts by multiplication; ζ ∈ µp acts by sending f to ζf . • (1, (l1, 0)) acts by l 7→ f(l + l1) • (1, (0, l2)) acts by l 7→ f(l)Ψ(l, l2) = f(l)b(l, l2). If σ ∈ Aut0(H), then the composition ρσ := ρ ◦ σ acts on the center in the same way. By Stone-von Neumann, R(σ)−1ρR(σ) = ρσ for some R(σ) ∈ GL(V ). Pick such an R(σ) for each σ. It’s well defined up to a scalar by Schur. It follows that c(σ1, σ2)R(σ1σ2) = R(σ1)R(σ2) for some c(σ1, σ2) ∈ C×. R : G → PGL(V ) is a projective representation. Can we lift it to GL(V )? Plan: let’s try to understand R for some σ ∈ N,T and σ = w, and then extend with Bruhat. Case (i) σ = ( a 0 0 a−1 ) let h = (1, u) ∈ H for notation. Subcase (1): u = (l1, 0) so uσ = (al1, 0). For f ∈ V , (hσf)(l) = f(l + al1), (hf)(l) = f(l + l1). Subcase (2): u = (0, l2), uσ = (0, a−1l2). (hσf)(l) = b(a−1l2, l)f(l), (hf)(l) = b(l2, l)f(l). Define (R(σ)f)(l) = f(al). We claim that this works (satisfies R(σ)−1ρR(σ) = ρσ). Proof: (R(σ)−1hR(σ))f)(l) = (R(σ)−1hf)(al) = R(σ)−1f(a(l + l1)) = f(l + al1) = (hσf)(l). Case (ii): σ = w = ( 0 1 −1 0 ) 5 Subcase (1): u = (l1, 0), uσ = (0, l1). (hσf)(l) = b(l, l1)f(l), (hf)(l) = f(l + l1). Subcase (2): u = (0, l2), uσ = (−l2, 0). (hσf)(l) = f(l − l2), (hf)(l) = b(l, l2)f(l). To find R, let’s define a “Fourier transform”: Ff(l) = |L|−1/2 ∑ l′∈L f(l′)b(l,−l′) and define f1(l) = f(l + l1). Then Ff1(l) = |L|−1/2 ∑ l′∈L f1(l′)b(l,−l′) = |L|−1/2 ∑ l′∈L f1(l′ + l)b(l,−l′) = |L|−1/2 ∑ l′′∈L f1(l′′)b(l,−l′′ + l1) = b(l, l1)Ff(l). Now Fh = hσF =⇒ FhF−1 = hσ =⇒ R( ( 0 1 −1 0 ) ) = F−1. Case (iii): σ = ( 1 x 0 1 ) x ∈ Fp. Subcase (1): u = (l1, 0), hσ = (νσ(u), (l1, xl1)), uσ = (l1, xl1); Ψ(uσ, uσ) = Ψ((l1, xl1), (l1, xl1)) = B(l1, xl1), Ψ(u, u) = Ψ((l1, 0), (l1, 0)) = 0, so νσ(u) = b(l1, l1)x/2; (hσf)(l) = b(l1, l1)x/2b(l, xl1)f(l + l1) = b(l1, l1)x/2b(l, l1)xf(l + l1); (hf)(l) = f(l + l1). Claim (R(σ)f)(l) := b(l, l)x/2f(l) = eπixB(l,l)/pf(l) works: (R(σ)−1hR(σ)f)(l) = (R(σ)−1h)(f(l)b(l, l)x/2) = R(σ)−1f(l + l1)b(l + l1, l + l1)x/2 = b(l, l)−x/2b(l + l1, l + l1)x/2f(l + l1) = b(l, l1)xb(l1, l1)x/2f(l + l1) = (hσf)(l) Now use Bruhat to define R(σ) for arbitrary σ (γ 6= 0): If σ = ( α β γ δ ) = ( 1 αγ−1 0 1 ) ( 0 −γ−1 γ 0 ) ( 1 δγ−1 0 1 ) , define R(σ) = R(...)R(...)R(...). Verify: if σ1, σ2, σ1σ2 ∈ G \B, then c(σ1, σ2)R(σ1σ2) = R(σ1)R(σ2) with c(σ1, σ2) = { +1 (I) → done → principal series −1 (II) → replace R by − R on G \ B → cuspidal series (Something missing here....) Consider characters of the circle group λ : K = {γ ∈ F× q2 : Nγ = 1} → C×. If λ2 6= 1, Vλ is irreducible of dimension p− 1. If λ2 = 1, Vλ splits into V +λ , V − λ each of dimension p−1 2 . These give the missing representations. Case (I) recovers what we did before. 6