Introduction to GLn Representation Theory-4

November 4 – Constructing the irreps of GLn with matrix minors Today was an impromptu worksheet day! David was sick, so he sent in problems to solve, We found (or nearly found) a representation with character sλ . We are specifically concerned with the right action of GLn on polynomials on n2 variables zij , where we think of these as entries in an n × n matrix Z. So A acts by replacing Z with ZA−1 . Note that the span of each row of Z is an n-dimensional sub-representation, since multiplication on the right sends zi∗ , ie row i, to zi∗ A−1 . In other words, A acts on each row 58 seperately. Because this is a subrep, we can compute the character by setting A to Pthe diagonal matrix with x−1 in the i, ith entry. This sends z to x z , so the character is xj . ij j ij i Problem 19. Fix 1 ≤ i ≤ n and a nonnegative integer d. Show that the vector space of degree d homogeneous polynomials in {zi1 , zi2 , . . . , zin } is a GLn -representation and, as such, is isomorphic to Symd Cn . As discussed before, the right action sends row i to row i linearly. We act on degree d polynomials by replacing each variable by a linear combination of the others, which is again a degree d polynomial. The torus (our diagonal matrix of x−1 i ) acts by rescaling each variable. The monomials are a basis, and we send each degree d monomial zij1 · · · zijd to xj1 · · · xjd zij1 · · · zijd . So the character is X xj1 · · · xjd = hd (x). j1 ≤j2 ≤···≤jd Since characters determine our representations, our representation must be isomorphic to Symd , whose character is hd . We can also see the isomorphism directly by identifying the product zij1 · · · zijd with ej1 · · · ejd ∈ Symd . Either way, this concludes the Problem. Let J = {j1 , . . . , jk } be a k-element subset of {1, . . . , n}. Define the k × k minor ∆J as the determinant of the first k rows and J columns of Z. Problem 20. For 1 ≤ k ≤ n, check that the subspace of R spanned by the ∆J (where V |J| = k) is a GLn -subrepresentation, and that this subrepresentation is isomorphic to k Cn as a GLn -representation. j Let Z j denote the jth column of the jth column to the first P ofi Z, and Z |k the restriction j k rows. Recall that (ZA) = i Z Aij . So ∆J = det(Z j1 |k , . . . , Z jk |k ) and acting by A gives us: n X j1 jk det((AZ) |k , . . . , (AZ) |k ) = Ai1 j1 det(Z i1 |k , (AZ)j2 |k , . . . , (AZ)jk ) i1 =1 X k Y I={i1 ,i2 ,...,ik } m=1 =
Aim jm ∆I which is in the span of ∆J . So this is a subrep. One may also decompose A in various ways to get nicer formulas, or notice some cancelation in the sum above. P The character of this representation is i1 < ... < ik xi1 · · · xik = ek (x), the same as the charV acter for k Cn . So they must be isomorphic. We can find a particular isomorphism by identifying ∆J with ej1 ∧ ej2 ∧ . . . ∧ ejk . This concludes the Problem. For λ a partition, define Uλ to be the subspace of R spanned by ∆J1 ∆J2 · · · ∆Jm where |Ji | = λTi . Recall the representations Eλ and Hλ , with characters eλ and hλ , from the previous classes. Problem 21. Define a surjection Eλ → Uλ . Let Ji = {j1i , j2i · · · }. Send ej11 ∧ ej21 ∧ · · · ∧ ejλT 1 ⊗ ej12 ∧ ej22 ∧ · · · ∧ ejλT 2 ⊗ · · · 1 2 59 V to ∆J1 ∆J2 · · · . Essentially the multiplication map after we take our identification with k Cn from the previous problem. This is surjective, we can get any particular product of ∆J ’s. Problem 22. Define an injection Uλ → Hλ . Note that the variables z1j , those in the first row of Z, appear with degree 1 in every ∆J (provided J is not empty). Likewise the zij appear with degree 1 in every ∆J with |J| ≥ i. Thus in every monomial of ∆J1 ∆J2 · · · ∆Jm ∈ Uλ can be written as a product of λi = #{k|λTk ≥ i} of the zkj s, ie grouping the variables from the same rows. We’ve already identified monomials of degree d with elements of Symd Cn in the first problem of today. For example, we would send the monomial: z1a z2b z3c z1d z2e z1f z1g 7→ ea ed ef eg ⊗ eb ee ⊗ ec Since this map is an injection on monomials, it is an injection on Uλ . And now we are done - we have found a nonzero map from Eλ → Hλ . Since the only common subrep of Eλ and Hλ is Vλ , we conclude that Uλ must have character sλ , and is thus an irrep. November 7 – Some examples of GLn representations We first review the material from the previous IBL class:   x11 · · · x1n ..  , ... R = C  ... . xn1 · · · xnn with GLn (C) acting via right multiplication. For J = {j1 , j2 , · · · , jk } ⊆ {1, · · · , n}, define   x1j1 · · · x1jk ..  . .. ∆J = det  ... . . xkj1 · · · xkjk Vk n Then we showed span∆J ∼ C as a GLn (C)-rep. If λ = (λ1 , λ2 , · · · , λn ) is a partition = |J|=k then span (∆J1 ∆J2 · · · ∆Jn ) |Jk |=λT k is an irreducible representation of GLn (C) with character sλ . For example, take n = 3 and λ = (2, 1) so that sλ (x1 , x2 , x3 ) = x21 x2 + x1 x22 + x21 x3 + 2x1 x2 x3 + x22 x3 + x1 x23 + x2 x23 . Let Vλ := span (∆J1 ∆J2 · · · ∆Jn ). |Jk |=λT k We have dim(Vλ ) = χVλ (Id3 ) = sλ (1, 1, 1) = 8, 60 despite Vλ being the span of 9 polynomials:  ∆12 ∆1   Vλ = span  ∆13 ∆1  ∆23 ∆1 ∆12 ∆2 ∆13 ∆2 ∆23 ∆2 ∆12 ∆3    ∆13 ∆3  .  ∆23 ∆3 There must therefore be a linear relationship between these polynomials. Indeed ∆12 ∆3 − ∆12 ∆2 + ∆23 ∆1 = 0. One way to find this relationship is that x11 x12 x21 x22  x11 = det x11 x21 ∆12 ∆3 − ∆12 ∆2 + ∆23 ∆1 = x13 − x11 x13 x x x + 12 13 x11 x21 x23 12 x22 x23  x12 x13 x12 x13  = 0. x22 x23 Another way to look at isito decompose Vλ into T -eigenspaces, where T ⊆ GL3 (C) is h xthis 1 x2 the torus. The matrix ∈ T acts on span(∆ab ∆c ) through multiplication by xa xb xc . x3 This is because it takes zij to xj zij , so it maps ∆ab ∆c = (z1a z2b − z2a z1b )z1c 7→ ((xa z1a )(xb z2b ) − (xa z2a )(xb z1b ))xc z1c = (xa xb xc )∆ab ∆c . Therefore this matrix will act on span(∆23 ∆1 , ∆13 ∆2 , ∆12 ∆3 ) via multiplication by x1 x2 x3 . Since the coefficient of the x1 x2 x3 term in sλ (x1 , x2 , x3 ) (the character of Vλ ) is 2, the dimension of span(∆23 ∆1 , ∆13 ∆2 , ∆12 ∆3 ) must be 2 or less, meaning the linear dependence must be among these three terms. In other words, if we associate the polynomial ∆ab ∆c with the young tableau ab c , then the coefficient for x1 x2 x3 in sλ (x1 , x2 , x3 ) is already covered by ∆13 ∆2 and ∆12 ∆3 since they’re associated with a semi standard young tableau, whereas ∆23 ∆1 ↔ 23 1 is not. Therefore we can safely discard ∆23 ∆1 without impacting the span. The T -eigenspaces of this representation can therefore be drawn as follows: 1 1 2 1 1 3 ∆13 ∆1 1 3 3 1 2 2 ∆12 ∆1 1 2 3 ∆13 ∆2 1 3 2 ∆12 ∆3 ∆13 ∆2 ∆12 ∆2 2 2 3 2 3 3 ∆23 ∆3 ∆23 ∆2 61 Recall from previous lectures the decomposition of algebraic representations of C ∗ = GL1 . Namely, if C ∗ acts on V then V can be broken up as the direct sum ∞ M V = Vk , k=−∞ k where  [t] actson Vk by  t . On homework, we showed that algebraic representations of P 1 u uk N k : u ∈ C are exp(uN ) = ∞ for some nilpotent N . Also shown on C+ ∼ = k=0 k! 0 1    t u homework was that if G = : u ∈ C, t ∈ C∗ then the nilpotent N takes Vk to Vk+1 0 1 where [ 0t 01 ] acts on Vk by multiplication by tk , i.e. N N N N V = · · · V−2 ⊕ V−1 ⊕ V0 ⊕ V1 ⊕ V2 · · · The key relationship here is that       t 0 1 u 1 tu t 0 = . 0 1 0 1 0 1 0 1 If v ∈ Vk , this means that       t2 u2 N 2 u2 N 2 v t 0 + · · · = I + tuN + + · · · tk v. ρ v + uN v + 0 1 2 2 The linear term says that   t 0 (N v) = tk+1 N v, ρ 0 1 so N v ∈ Vk+1 . More generally, if GLn acts on V then V can be written as a direct sum M V = Vµ , µ ht i ... where ∈ T ⊆ GLn acts on Vµ by multiplication by tµ1 1 · · · tµnn . Rather than just tn looking at the subgroup T of diagonal matrices, we can put a nonzero entry in the ith row and jth column, e.g. j   ∗  i ∗ ∗ ∼   = T n C+   ∗ ∗ 1 j  We’ll have i    1 1 u 1    acting by exp(uNij ) where Nij : Vµ → Vµ+ei −ej . For 1 example, with λ = (2, 1), the matrix h 1 u 1 i 1 will take ∆12 7→ ∆12 since a column operation 62 won’t change the determinant of the minor. It will map ∆1 7→ ∆1 + u∆2 . Therefore ∞ k X uk N21 ∆12 ∆1 = exp(uN21 )(∆12 ∆1 ) = ∆12 ∆1 + u∆12 ∆2 . k! k=0 Looking at the linear term tells us that N21 will take elements in span(∆12 ∆1 ) to span(∆12 ∆1 ). Similarly the matrix takes ∆13 7→ ∆13 + u∆23 and ∆2 7→ ∆2 so N21 takes ∆13 ∆2 to ∆23 ∆2 . As a final example ∆13 ∆1 7→ (∆13 + u∆23 )(∆1 + u∆2 ) = ∆13 ∆1 + u(∆13 ∆2 + ∆23 ∆1 ) + u2 ∆23 ∆2. Again looking at the linear term, we find that N21 takes ∆13 ∆1 into the span(∆13 ∆2 , ∆23 ∆1 ) = span(∆13 ∆2 , ∆12 ∆3 ). Completing this process for every basis element gives us the following arrows for N21 1 1 2 1 1 3 ∆13 ∆1 1 3 3 The transpose matrix h 1 u 1 1 2 2 ∆12 ∆1 1 2 3 ∆13 ∆2 1 3 2 ∆12 ∆3 ∆13 ∆2 i ∆12 ∆2 2 2 3 2 3 3 ∆23 ∆2 ∆23 ∆3 will have corresponding nilpotent N12 which reverses the direch i 1 1 tion of these arrows. The matrix will have corresponding nilpotent N32 which follows 1 u 1 63 different arrows: 1 1 2 1 1 3 ∆13 ∆1 1 3 3 1 2 2 ∆12 ∆1 1 2 3 ∆13 ∆2 1 3 2 ∆12 ∆3 2 2 3 2 3 3 ∆13 ∆2 ∆12 ∆2 ∆23 ∆2 ∆23 ∆3 Again N23 will just reverse the direction of these arrows. November 9 – Weight spaces and the maps between them We begin by discussing the language of weight spaces. Consider the torus T ∼ = (C∗ )n ⊂   t1 .. . If V is an algebraic representation of GLn which consists of matrices of the form . T , then V = ⊕α∈Zn Vα where 0   t1 ... Vα = v ∈ V | tn  ~v = tα1 1 . . . tαnn ~v  . tn We call (α1 , . . . , αn ) a weight. The space Vα is the α-weight space, and a vector v ∈ V is an α-weight vector. We now discuss how to prove the decomposition of V given above. When ρ : T → GLn is algebraic, this implies the entries of ρij (t1 , . . . , tn ) are Laurent polynomials in the ti . That  t1  P P ± α ± ... is, they are of the form Pij,α t ∈ C[t1 , . . . , tn ]. Thus ρ = α∈Zn Pα tα1 1 . . . tαnn . tn Since ρ is a homomorphism,  t1   u1   t1   u1  .. .. ... ... ρ =ρ ρ . . un un tn tn
P
P P P γ γ α β α β Therefore, γ Pγ t u = = α,β Pα Pβ t u . Matching coefficients α Pα t β Pβ u on both sides yields ( Pα α = β Pα Pβ = 0 α 6= β P Further, ρ(Idn ) = Idn , so α Pα = Idn . It follows that V breaks up as described above. When n = 2, we see that V = Sym2 C2 = Ce1 e1 ⊕ Ce1 e2 ⊕ Ce2 e2 64 and 2 ^ V = C2 = C(e1 ∧ e2 )   t1 0 where Vλ has character sλ . Since acts on Ce1 e1 by t21 , the weight is (2, 0). Similarly, 0 t2 the weight on Ce1 e2 and on C(e1 ∧ e2 ) is (1, 1), and the weight on Ce2 e2 is (0, 2). Example. In the homework, we particularly thought about the case where |λ| = n and we look at the weight (1, 1, . . . , 1). We called the corresponding space Wλ . Here are the values for n = 2 and n = 3: W = Ce1 e2 and W and S2 acts trivially on W V V V W W W = C(e1 ∧ e2 ) and by sign on W . In the n = 3 case we have = Sym3 C3 = Ce31 ⊕ · · · ⊕ Ce1 e2 e3 ⊕ · · · ⊕ Ce33 = C(∆12 ∆1 ) ⊕ · · · ⊕ C(∆23 ∆3 ) ⊕ C{∆12 ∆3 , ∆13 ∆2 } = C(e1 ∧ e2 ∧ e3 ) = C(e1 e2 e3 ) = C = (∆12 ∆3 ) ⊕ C(∆13 ∆2 ) = Span(∆12 ∆3 , ∆13 ∆2 , ∆23 ∆1 ) = C(e1 ∧ e2 ∧ e3 ) The W ’s are called Specht modules, and they are irreps of Sn . Let 1 ≤ i 6= j ≤ n. Let Eij be the matrix with 1 in the (i, j) entry and 0 elsewhere. P uk k Then Id + uEij ∼ N = = C forms a copy of the additive group and acts on V by k! ij   t1 ..   form a semidirect product, and Nij maps exp(uNij ). Matrices of the form . u tn Vα to Vα+ei −ej . We now begin a discussion on Lie algebras. Let G be a smooth Lie group and g = TId G. We will think of having G = GLn C and g = Matn×n C. Let ρ : G → GLn be a representation and g ∈ g be an n × n matrix. Then ρ(Id + g) is ρ(Id) + r(g) plus higher terms in  where r : g → Matn×n C is r = Dρ, a map of tangent spaces. Proposition. Let G = GLn . Then r determines ρ. Proof. Suppose we have two representations ρ1 , ρ2 : G → GLn with the same derivative r. For any g ∈ g with M >> 0, we have ρ1 ((1 + Mg )M ) = (ρ1 (1 + Mg ))M = (Id + r(g) + O( M12 )M . M 65 P gk g n Then exp(g) = ∞ k=0 k! = limn→∞ (1 + n ) , so ρ(exp(g)) = exp(r(g)). Thus, ρ1 = ρ2 on {exp(g) | g ∈ g}. There is an open set Ω containing Idn such that every matrix in Ω is an exponential. (Intuition: log series converegs near the identity.) Let X = {g ∈ GLn | ρ1 (g) = ρ2 (g)}. Clearly, X is closed in G. If g ∈ X, u ∈ Ω, then ρ1 (gu) = ρ1 (g)ρ1 (u) = ρ2 (g)ρ2 (u) = ρ2 (gu), which implies gu ∈ X. So, for any g ∈ G, gΩ is an open neighborhood of g in X. It follows that X is both open and closed. Since it is in a connected space, we conclude X = G.  If  t1 ..   ρ .  1 tα1 ...  =   tn then  s1 ..   r .     tαn N   α1 · s  ...  =   sn αN · s Consider T ⊂ GLn , t ⊂ g where g is the set of n × is the set  of diagonal  and t  n∗ matrices s1 0 ... .. and r acts by matrices. Weight spaces are also eigenspaces of r . 0 sn ∗ α1 · s1 + · · · + αn · sn on Vα . We also have   1 . = exp(uN ) ρ .. u 1 Taking the derivative at u = 0, we see  0 .. r .  1 =N 0 November 11 – The Lie Algebra Let G be a smooth Lie group and let g be Te G, the tangent space to G at the identity. Let ρ : G → GLn be a representation, . Then r : g → MatN ×N is the derivative De (ρ). Concretely, if G is GLn , then g is Matn×n , and we have ρ(Id + g) = IdN + r(g) + O(2 ). For G connected, the representation ρ is determined by r: ρ(exp(g)) = exp(r(g)). Here is a sketch of a proof for G = GLn : We have exp(g) = limM →∞ (1 + g/M )M . So   ρ(exp(g)) = ρ lim (1 + g/M )M = lim ρ((1 + g/M )M ) = lim ρ(1 + g/M )M M →∞ M →∞ M →∞ = lim (1 + r(g)/M + O(1/M 2 ))M = exp(r(g)). M →∞ 66 Next we discussed in the GLn case what happens with commutators. Let g, h ∈ Matn×n , and δ,  ∈ R>0 . Then we have (1 + gδ)(1 + h)(1 + gδ)−1 (1 + h)−1 =(1 + gδ)(1 + h)(1 − gδ + g 2 δ 2 − . . . )(1 − h + h2 2 − . . . ) =1 + (gh − gh − hg + gh)δ + O(δ + )3 =1 + [g, h]δ + O(δ + )3 where [g, h] := gh − hg. In general we get [, ] : g × g → g such that if α(t) and β(u) are paths, then α(0) =Id α0 (0) =g β(0) =Id β 0 (0) =h and α(t)β(u)α(t)−1 β(u)−1 = Id + (tu)[g, h] + . . . . So (1 + r(g)δ + . . . )(1 + r(h) + . . . )(1 + r(g)δ + . . . )−1 (1 + r(h) + . . . )−1 = 1 + r([g, h])δ + . . . which implies (1) [r(g), r(h)] = r([g, h]). A map r : g → MatN ×N obeying [r(g), r(h)] = r([g, h]) is called a Lie algebra homomorphism. Example. What happens in GL2 ? We can split gl2 = {[ ∗∗ ∗∗ ]} = C [ 00 10 ] ⊕ [ ∗0 0∗ ] ⊕C [ 00 01 ] . |{z} |{z} |{z} E t F Then we can compute  t1 0 0 t2   , [ 00 10 ] = [ 00 t01 ] − [ 00 t02 ] = (t1 − t2 )E  t1 0  0 0  0 0 0 0 0 t2 , [ 1 0 ] = [ t2 0 ] − [ t1 0 ] = (t2 − t1 )F 0 [[ 00 10 ] , [ 01 00 ]] = [ 10 00 ] − [ 00 01 ] = [ 10 −1 ]. Continuing with this example we have ρ : GL2 → GL(V ) with V = ⊕(a1 ,a2 )∈Z2 Va1 a2 . We can view ρ as a a  
T1 1 T2 2 ρ T01 T02 : Va1 a2 −− −−→ Va1 a2 then r  t1 0 0 t2 
a t +a t 1 1 2 2 : Va1 a2 −− −−− → Va1 a2 . Note that r(E) : Va1 a2 → V(a1 +1)(a2 −1) 67 and (1 + t1 )a1 (1 + t2 )a2 = 1 + (a1 t1 + a2 t2 ) + . . . . We see that   t1 r 0 0 t2   , r(E) = (t1 − t2 )r(E) so r  t1 0 0 t2  t1   r(E) = −r(E)r  t1 0 0 t2  = (t1 − t2 )r(E). If v ∈ Va1 a2 we get r 0 0 t2 r(E)v − r(E)(t1 a1 + t2 a2 )V = (t1 − t2 )r(E)v or   r t01 t02 r(E)v = (t1 (a1 + 1) + t2 (a2 − 1))r(E)v. So we get this diagram (possibly with more diagonals): 0 On va1 ,a2 , we have the relation EF − F E = (a1 − a2 )Id, since [E, F ] = [ 10 −1 ], and we 1 0 know that [ 0 −1 ] acts on the (a1 , a2 ) weight space by a1 − a2 . Let’s see how this works for the irreps of GL2 .. We know GL2 irreps correspond to (λ ∈ Z2 with λ1 ≥ λ2 , so we can write (k + `, `) with k ≥ 0. They are of the form V12, λ22) ⊗l ( C ) ⊗ Symk C2 . So the character is k (x1 x2 )l (xk1 + xk−1 1 x2 + · · · + x2 ). The weight diagram looks like (`, k + `) E i F ) (` + 1, k + ` − 1) E g F ' ... g E ' F (k + ` − 1, ` + 1) E i F ) (k + `, `) Let’s compute the actual maps. They are maps between one dimensional vector spaces, but we need to choose a particular vector in each vector space in order to write down specific 68 maps. Let e1 , e2 be a basis for C2V . Then, for 0 ≤ a ≤ k, the vector (e1 ∧ e2 )` ea1 ek−a is in 2 2 2 ⊗l k 2 (a + `, k + ` − a) weight space of ( C ) ⊗ Sym C . The subgroup [ 10 u1 ] preserves e1 ∧ e2 . The action on ea1 ek−a is 2 k−a−1 + ··· . + u(k − a)ea+1 ) = ea1 (e2 + ue1 )k−a = ea1 ek−a ρ ([ 10 u1 ]) (ea1 ek−a 1 e2 2 2 So k−a−1 = (k − a)ea+1 r ([ 00 10 ]) ea1 ek−a 1 e2 2 and the maps are (`, k + `) k i 1 ) (` + 1, k + ` − 1) k−1 g ' 2 .. . g k−2 ' k−1 (k + ` − 1, ` + 1) 2 i k ) (k + `, `) To repeat, every algebraic GL2 rep breaks up into strings like this, which we call GL2 -strings. November 14 - gl(2) Strings Recall from last time that every gl(2) representation breaks up into irreducible representations as follows: There is some ordered pair (a, b) of integers, with a ≤ b, such that the weights of the representation are (a, b), (a + 1, b − 1), (a + 2, b − 2), . . . , (b, a). The actions of the Lie algebra elements E = [ 00 10 ] and F = [ 01 00 ] are as shown below: (a, b) b−a 1 b−a−1 2 .. . 1 b−a (b, a) Let’s look in detail at the representation V(2,1) of gl(3). It is 8 dimensional h i and hspanned i 0 0 0 0 1 0 by matrix minors. We have a copy of gl(2) sitting inside generated by 1 0 0 and 0 1 0 0 0 0 0 0 0 We can obtain the action of these elements on this representation by looking at the coffeicient i h i h 1 u 0 1 0 0 of u in the action of u 1 0 and 0 1 0 0 0 1 0 0 1 h i 1 0 0 ρ( u 1 0 )(∆12 ∆1 ) = ∆12 (∆1 + u∆2 ) = ∆12 ∆1 + u∆12 ∆2 0 0 1 Thus we have the following string 69 ∆12 ∆2 ∆12 ∆1 Similarly we get the following string ∆13 ∆3 ∆12 ∆1 h i 1 0 0 ρ( u 1 0 )(∆13 ∆1 ) = (∆13 + u∆23 )(∆1 + u∆2 ) = ∆13 ∆1 + u(∆13 ∆2 + ∆23 ∆1 ) + u2 ∆23 ∆2 h 0 0 i1 1 0 0 ρ( u 1 0 )(∆13 ∆2 +∆23 ∆1 ) = (∆13 +u∆23 )(∆2 )+∆23 (∆1 +u∆2 ) = ∆13 ∆2 +∆23 ∆1 +u(2∆23 ∆2 0 0 1 So we can normalize our basis vectors and get the following string 2 ∆23 ∆2 2 ∆23 ∆1 +∆13 ∆2 2 ∆13 ∆1 ∆13 ∆2 − ∆23 ∆1 Putting it all together we get ∆12 ∆2 2 ∆23 ∆2 1 ∆12 ∆1 1 (∆23 ∆1 2 + ∆13 ∆2 ) 1 (∆13 ∆2 2 − ∆23 ∆1 ) ∆23 ∆3 We have a similar diagram for the action of h 0 0 0 1 0 0 0 0 0 i 1 2 ∆13 ∆3 h i 0 1 0 and 0 1 0 0 0 0 ∆13 ∆1 70 ∆12 ∆2 ∆12 ∆1 2 1 1 (∆13 ∆2 2 + ∆12 ∆3 ) 1 (∆13 ∆2 2 − ∆12 ∆3 ) ∆23 ∆2 ∆13 ∆1 1 2 ∆23 ∆3 ∆13 ∆3 We now consider the general gl(n). We define ek as the matrix with 1 in the position (k, k + 1) and zero everywhere else. Similarly we define fk is defined as the matrix with 1 in (k + 1, k). These are called the Chevally Generators and generate a copy of gl(2) inside gl(n). Proposition. If |i − j| > 1, then ej , fj map (ei , fi ) strings to strings of the same length or to 0 Proof. This follows from the fact that [ei , fj ] = [fi , ej ] = [ei , ej ] = [fi , fj ] = 0 Thus we have the following diagram where the the solid arrows denote (ei , fi ) strings and the dotted lines denote the action of ej and fj Now if a terminal vertex is non-zero but it maps above or below to 0 on going right we have a composite map which is zero. But if we first go right and then go up or down we get a composite map which is non zero hence contradiction.  For i, i + 1 we have the following relation called Serre’s Relation [ei , [ei , ei+1 ]] = [ei+1 , [ei , ei+1 ]] = [fi , [fi , fi+1 ]] = [fi+1 , [fi , fi+1 ]] = 0. We also still have [ei , fi+1 ] = [ei+1 , fi ] = 0. The corresponding result for strings is Proposition. ei+1 , fi+1 map length ` (ei , fi ) strings to length ` ± 1 strings. Proof. This follows from translating Serre’s relation into the commutator for the relation. We work it out for ei , ei+1 r(ei )2 r(ei+1 ) − 2r(ei )r(ei+1 )r(ei ) + r(ei+1 )r(ei )2 = 0 71 Indeed, suppose that we had an (ei , fi )-string of length ` mapping to an (ei , fi ) string of length ≥ ` + 2. We’ll draw ei mapping rightward with solid arrows and ei+1 slanting down and left with dashed arrows. Then, at the left hand end of the shorter string, we would have ··· ··· In the above picture, start at the white square node. We see that r(ei )2 r(ei+1 ) is non zero but r(ei )r(ei+1 )r(ei ) and r(ei+1 )r(ei )2 map to 0 and thus we have a contradiction to Serre’s relation. Thus the result follows.  Remark. This was the argument presented in class, but it now seems easier to start at the black square node and use the relation [ei , fi+1 ] = 0. November 16 – High weight vectors Our goal for this class is to study High Weight Vectors (HWVs) for a representation V of GLn . From here on we have ρ : GLn → GL(V ) assigns matrices to elements of the group and r : g`n → L End(V ) is the corresponding derivative map for the Lie Algebra. We also have that V = α∈Zn Vα is a decomposition of V into weight spaces.   t1   t2  ∈ T we have that t acts on Vα by For any element of the torus t =  .   .. tn  · · · 1 ·  is the matrix with a single tα1 1 tα2 2 · · · tαnn . And for any i 6= j, we have Eij =   ·  · non-zero entry in row i and column j), such that r(Eij ) : Vα → Vα+ei −ej . Note that, by applying r(Eij ) with i < j, the weight of the result keeps increasing in lexicographical order so that eventually we get to a point where Vα 6= 0, but r(Eij )(Vα ) = 0. Definition. We say v ∈ V is a high weight vector (HWV) if r(Eij )(v) = 0 for all 1 ≤ i < j ≤ n. Examples for GL3 : For each of this examples, → represents the E12 action, ← the E21 one, % the E23 one, and . the E32 one. The high weight vector is in green. (1) Representation on C3 . e1 e2 e3 (2) Representation on V2 C3 . 72 e1 ∧ e2 e1 ∧ e3 (3) Representation defined by n = 3, λ = e2 ∧ e3 . 2∆12 ∆1 2∆12 ∆2 (∆13 ∆2 + ∆23 ∆1 ) ∆13 ∆1 ∆23 ∆2 (∆12 ∆3 − ∆13 ∆2 ) ∆13 ∆3 ∆23 ∆3 Remark. The following are equivalent: (1) v is a high weight vector;   1  1  u   ; 1 (2) v is fixed by all   ...    1   1 ∗ .  . (3) v is fixed by N+ = { . } (upper triangular matrices with 1s in the diagonal). 1 Theorem. Let Vλ be a GLn −irrep. Then the vector space of high weight vectors for Vλ has dimension 1. This proof was presented badly. Read the presentation in the following class instead. November 18 – High weight vectors, and the easy part of the PBW theorem We’ll start with redoing the proof from last class, this time also giving context from general Lie algebras. Here’s our setting: we had a GLn -representation V . Inside GLn , we have the subgroups N+ of upper triangular matrices with diagonal entries 1, N− of lower diagonal matrices with diagonal entries all 1, and the torus subgroup T of (invertible) diagonal matrices. The Lie algebras corresponding to these matrices are: gl consisting of all matrices, n+ with strictly upper triangular matrices, n− with strictly lower triangular 73 matrices, and t of diagonal matrices. We have a representation ρ : GLn → GL(V ) and a corresponding Lie algebra representation r : g → End(V ). V decomposes into a direct sum L α Vα , where each Vα is an eigenspace for T and t. For eij ∈ n+ with i < j, r(eij ) maps between weight spaces, with r(eij ) : Vα → Vα+ei −ej , with α + ei − ej >lex α. (Thus we called these eij the ‘raising operators’). So there exists v ∈ V, v 6= 0, such that n+ v = 0. Equivalently, ρ(u)v = v for all u ∈ N+ . Definition. Vhigh weight = {v ∈ V |ev = 0 for all e ∈ n+ }. This is also {v ∈ V |u · v = v for all u ∈ N+ . Our claim was that Vhigh weight is one dimensional. Here’s general context from Lie algebras: Suppose g is a finite dimensional Lie algebra, with g1 , g2 , . . . , gm forming a basis for g. Let r : g → End(V ) be a Lie algebra representation. This means r([g, h]) = [r(g), r(h)] for all g, h. Let v ∈ V . The following subspaces of V are equal: (1) Span(r(gi1 )r(gi2 ) . . . r(gil )v), for l ≥ 0 and 1 ≤ i1 , i2 , . . . , il ≤ m. (2) Span(r(gi1 )r(gi2 ) . . . r(gil )v), for l ≥ 0 and 1 ≤ i1 ≤ i2 ≤ · · · ≤ il ≤ m. Proof. The inclusion of the latter subspace into the former is clear. We need the opposite inclusion. We’ll show by induction on k, that any r(gi1 )r(gi2 ) . . . r(gil )v with l ≤ k is in the span of r(gji )r(gj2 ) . . . r(gjl0 )v for l0 < k and 1 ≤ j1 ≤ j2 · · · ≤ jl0 ≤ m. First note that we can use our induction hypothesis and write r(gi2 )r(gi3 ) . . . r(gil )v in the span of r(gj1 )r(gj2 ) . . . r(gl0 )v with j2 ≤ j3 · · · ≤ jl0 , l0 ≤ k. So it’s okay to deal with vectors of the form r(gi1 )r(gj2 )r(gj3 ) . . . r(gjl0 )v with j2 ≤ j3 ≤ · · · ≤ jl0 . Suppose j2 ≤ j3 ≤ · · · ≤ ja ≤ i1 ≤ ja+1 · · · ≤ jl0 . We can write r(gi1 )r(gj2 ) . . . r(gjl0 )v = [r(gi1 ), r(gj2 )]r(gj3 ) . . . r(gjl0 )v + r(gj2 )[r(gi1 ), r(gj3 )]r(gj4 ) . . . r(gjl0 )v + ... + r(gj2 )r(gj3 ) . . . [r(gi1 ), r(gja )] . . . v + r(gj2 )r(gj3 ) . . . r(gja )r(gi1 )r(gja+1 ) . . . v Note that [r(gi1 ), r(gjb )] = r([gi1 , gjb ]). Here, all but the last term is gotten by letting atmost k − 1 elements on v, and so by induction these should be in the span of the vector space we’re looking at. The last term is one of our spanning elements, since i2 ≤ i3 ≤ · · · ≤ ja ≤ i1 ≤ ja+1 . . . .  Now we return to g = gln . This is spanned by n2 basis elements g1 , g2 , . . . gn2 , and we can pick them in such a way that the first few of them form a basis for n− , the next few for t, and the rest for n+ . If v is a high weight vector, then n+ v = 0, and tv ∈ Span(v). By our theorem, if we want to look at the subrepresentation generated by v, we can look at the span of r(gj1 )r(gj2 ) . . . r(gjl )v, with the first few gjk in n− , the next few in t and the last few in n+ . If there are a non-zero number of gjk ’s from n+ , then the vector becomes 0, and so we can disregard these. The action of the elements from t gives us a scalar multiple of v. So we are left with the span of r(gj1 )r(gj2 ) . . . r(gjl )v with all jk ∈ n− . So the subspace of V spanned by r(gj1 )r(gj2 ) . . . r(gja )v, for jb ∈ n− forms a gln -subrepresentation of V . So if V 74 is an irreducible representation, and v a high weight vector, then the n− -subrepresentation generated by v is V itself. Let v ∈ Vα . Then, if Vβ 6= 0, we must have β < lex α and Vα = C·v. Corollary. The high weight space is 1-dimensional Proof. Suppose v ∈ Vα and w ∈ Vβ are both linearly independent high weight vectors. WLOG assume α ≤ lex β. Vβ 6= 0 implies β ≤ lex α and so we must have α = β. But then w ∈ Vβ = Vα = C · v. This gives a contradiction.  In terms of constructing Vλ as products of matrix minors, ∆(λT )1 ∆(λT )2 . . . ∆(λT )l is a high 1 1 ... 1 2 ... 2 .. weight vector. Its associated SSYT is: 3 . . . . Abstractly, Vλ is characterised as the unique GLn -irrep whose high weight vector space is in weight λ. Here, λ1h ≥ λ2 ≥ · · · ≥ λn . i L L Recall that R = C x11 . . . x1n ...xn1 . . . xnn = Vλ∨ ⊗Vλ . We have N+ R = λ N+ Vλ∨ ⊗ L L Vλ = ( high weight )⊗Vλ ∼ Vλ as GLn -reps. So what we are learning is that R = C[∆I ]. = L Theorem. The ring of functions on n × n matrices invariant for left N+ action is Vλ as a representation for the right GLn action, and is generated by the minors ∆I . Vλ ⊗ Vλ∨ 1 1 ... 1 2 ... 2 is the GLn subrep generated by ∆( 3 . . . ). Remark. Question from the floor: Given the high weight space, is it clear that we could construct the irrep? If v is a high weight vector, then Vλ = Span(r(gj1 ) . . . r(gjl )v) for g1 , g2 , . . . , gl ∈ n− , j1 ≤ j2 ≤ · · · ≤ jl . We can multiply with an r(h) using the equation: X +r(gj1 ) . . . r(gjl )r(h)v. r(h)r(gj1 ) . . . r(gl )v = r(gj1 ) . . . [r(gja ), r(h)]r(gja+1 ) . . . Abstractly, we can obtain Vλ as a quotient of an infinite dimensional vector space, called the Verma module. The λ-Verma module is an infinite dimensional vector space with g-action, whose basis we think of as (gj1 , . . . , gjl ), j1 ≤ · · · ≤ jl with ja ∈ n− . Vλ turns out to be its unique finite dimensional g-invariant quotient. Remark. So, what is this about the PBW theorem? For any Lie algebra g, there is a non-commutative ring U (g), called the universal enveloping algebra of g, such that every grepresentation is an U (g)-module. In this language, Spanr(gi1 )r(gi2 ) . . . r(gil )v) is the U (g)submodule generated by v, and each individual product r(gi1 )r(gi2 ) . . . r(gil )v is the product of v with an element gi1 gi2 · · · gil in U (g). The theorem we have proved is equivalent to showing that the monomials gi1 gi2 · · · gil with 1 ≤ i1 ≤ i2 ≤ · · · ≤ il ≤ m span U (g). The full PBW theorem states that these monomials are a basis of U (g). The linear independence is significantly harder than the spanning; fortunately, we don’t need it.