Introduction to GLn Representation Theory-2

September 17 – Cauchy’s identity, flawed attempt Our goal is to show that the Schur polynomials are an orthonormal basis for h , i. Equivalently, this means we want to show Y i,j Since i,j 1−x1i yj = left side is Q P X 1 = sλ (x)sλ (y). 1 − xi y j λ Aλµ mλ (x)mµ (y), the coefficient of xα1 1 xα2 2 · · · xαnn y1β1 y2β2 · · · ynβn on the   row sum = α Aαβ = # Z≥0 - matrices . column sum = β λ,µ 20 The coefficient for this term on the right side (which we wish to prove is equal) is   T, U ∈ SSYT(λ) for some λ,  # of j’s in T is αj , # (T, U ) .   # of j’s in U is βj For instance, with the term x1 x2 y1 y2 , this says that       
1 0 0 1 1 1 # , =# , 1 2, 1 2 , 0 1 1 0 2 2 which is true. There is a useful way to think of semistandard Young tableaux using chains of partitions, which we now describe. Given an SSYT T , the position of boxes with labels ≤ j form a partition. For example, if 1 1 2 3 T = 2 6 , 5 then we have ≤5 ≤6 ≤2 ≤3 ≤4 ≤0 ≤1 . ∅ ⊆ ⊆ ⊆ ⊆ ⊆ ⊆ Note that equality is possible when a number is omitted, as 4 is in this example. Given two partitions µ and ν, we can have µ be {boxes≤ i} and ν be {boxes≤ i + 1} if and only if ν1 ≥ µ 1 ≥ ν2 ≥ µ 2 ≥ ν3 ≥ · · · We’ll say that “ν is µ plus a horizontal strip.” We’ll write µ ↑ ν and ν ↓ µ. µ1 µ3 ν4 We have that #    (T, U ) µ2 ν3 ν1 ν2  T, U ∈ SSYT(λ) for some λ, # of j’s in T is αj ,  # of j’s in U is βj    T, U ∈ SSYT(λ) for some λ,  = # (T, U ) T corresponds to ∅ = γ0 ↑ γ1 ↑ γ2 ↑ · · · ↑ λ, ∀j : αj = |γj | − |γj−1 | ,  U corresponds to ∅ = δ0 ↑ δ1 ↑ δ2 ↑ · · · ↑ λ, ∀j : βj = |δj | − |δj−1 |  so X X |γ |−|γ | |γ |−|γ | |δ |−|δ | |δ |−|δ | sλ (x)sλ (y) = x1 1 0 x2 2 1 · · · y1 1 0 y2 2 1 · · · λ ∅=γ0 ↑γ1 ↑···↑λ↓···↓δ1 ↓δ0 =∅ To evaluate the right hand side, we’ll essentially flip one by one until ∅ = γ0 ↑ γ1 ↑ · · · ↑ λ ↓ · · · ↓ δ1 ↓ δ0 = ∅ turns into ∅ = γ0 ↓ γ1 ↓ · · · ↓ λ ↑ · · · ↑ δ1 ↑ δ0 = ∅, at which point all γj , δj will be forced to be ∅. 21 ... Our goal now is to prove the lemma which will do the flipping. This was garbled in class, and we will try to present it correctly now. Lemma. Let β and γ be two partitions. Then X X 1 x|γ|−|α| y |β|−|α| x|δ|−|β| y |δ|−|γ| = 1 − xy β↓α↑γ β↑δ↓γ Example. The smallest example is β = γ = ∅. In this case, the lemma says X X 1 x|∅|−|α| y |∅|−|α| . x|δ| y |δ| = 1 − xy ∅↓α↑∅ ∅↑δ↓∅ P 1 1 The left hand side is k≥0 xk y k = 1−xy , while the right hand side is 1−xy · 1, so the lemma holds. Example. The next smallest example is β = ∅, γ = (1), where the lemma says we should have X X 1 x|δ| y |δ|−1 = x1−|α| y 0−|α| . 1 − xy ∅↑δ↓(1) ∅↓α↑(1) P x 1 x k k−1 Here the left hand side is k≥1 x y = 1−xy while the right hand side gives 1−xy · x = 1−xy , so again the lemma holds. Start of proof. Let’s outline what a proof should look like. The left hand side is a sum over the set {δ : β ↑ δ ↓ γ}. P 1 If we expand 1−xy on the right hand side as k≥0 xk y k , then the right hand side is a sum over {(α, k) : β ↓ α ↑ γ, k}. So we need a bijection between the set of partitions δ and the set of pairs (α, k). Moreover, this bijection needs to map x|δ|−|β| y |δ|−|γ| to x|γ|−|α|+k y |β|−|α|+k . In other words, we need |δ| − |β| = |γ| − |α| + k |δ| − |α| = |γ| − |β| + k. Note that these conditions are equivalent to each other, and can be written more symmetrically as |α| + |δ| = |β| + |γ| + k. (∗) At this part the proof in class was broken. We will record the wrong approach below and start over next time.  The broken proof. Let’s write down the conditions on δ and α explicitly. The condition that β ↑ δ ↓ γ means that δ1 ≥ β1 , γ1 ≥ δ2 ≥ β2 , γ2 ≥ δ3 ≥ · · · 22 and the condition that β ↓ α ↑ γ means that β1 , γ1 ≥ α1 ≥ β2 , γ2 ≥ α2 ≥ · · · . Thus, if β ↑ δ ↓ γ, and we define δ 0 = (δ2 , δ3 , . . .), then β ↓ δ 0 ↑ γ. Professor Speyer thought that we should take (α, k) = (δ 0 , δ1 − max(β1 , γ1 )). This is a bijection between {δ : β ↑ δ ↓ γ} and {(α, k) : β ↓ α ↑ γ, k}, but it does not obey condition (∗). We will see the correct bijection next time. “” September 19 – Cauchy’s identity, Corrected proof Our goal is to finish the proof of Y i,j X 1 = sλ (x)sλ (y). 1 − xi y j λ We will begin by finishing the proof of the Lemma from last class. Lemma. Let β and γ be two partitions. Then X X 1 x|δ|−|β| y |δ|−|γ| = x|γ|−|α| y |β|−|α| 1 − xy β↑δ↓γ β↓α↑γ Proof of Lemma. We can rewrite this as ∞ X X X |δ|−|β| |δ|−|γ| xk y k x|γ|−|α| y |β|−|α| x y = k=0 β↑δ↓γ β↓α↑γ or, in other words, X x|δ|−|β| y |δ|−|γ| = β↑δ↓γ X x|γ|−|α|+k y |β|−|α|+k β↓α↑γ, k≥0 We need is to find a bijection between {δ : β ↑ δ ↓ γ} and {(α, k) : β ↓ α ↑ γ, k ≥ 0} |δ| − |β| = |γ| − |α| + k and |δ| − |γ| = |γ| − |β| + k. Note that these equations are equivalent to each other, and both can be written more elegantly as |δ| + |α| − k = |β| + |γ|. Now, we translate the conditions β ↑ δ ↓ γ and β ↓ α ↑ γ into inequalities. They are equivalent to: δ1 ≥ max(β1 , γ1 ) ≥ min(β1 , γ1 ) ≥δ2 ≥ max(β2 , γ2 ) ≥ min(β2 , γ2 ) ≥ δ3 ≥ · · · min(β1 , γ1 ) ≥α1 ≥ max(β2 , γ2 ) ≥ min(β2 , γ2 ) ≥ α2 ≥ · · · Given δ = (δ1 , δ2 · · · ) with β ↑ δ ↓ γ, define (α, k) by αj = min(βj , γj ) + max(βj+1 , γj+1 ) − δj+1 and k = δ1 − max(β1 , γ1 ). Because of the inequalities, we have k ≥ 0 and β ↓ α ↑ γ. This mapping is also clearly invertible. 23 We now verify: P P |δ| + |α| − k = Pj≥1 δj + j≥1 (min(βj , γj ) + max(βj+1 , γj+1 ) − δj+1 ) − (δ1 − max(β1 , γ1 )) = Pj≥1 (min(βj , γj ) + max(βj+1 , γj+1 ) − δj + δj ) + max(β1 , γ1 ) = Pj≥1 (min(βj , γj ) + max(βj , γj )) = j≥1 (βj + γj ) = |β| + |γ|. So this bijection satisfies the conditions we needed.  We are now ready to prove Cauchy’s identity: This lemma is exemplified in the picture below, where the edges are weighted by the variable with exponent the difference of of the size of the partitions from its end to its start. To go from the top path to the bottom path, we have to multiply by the factor in the middle. δ y x 1 1−xy β y α γ x Theorem. For every n, n Y X 1 sλ (x1 , · · · , xn )sλ (y1 , · · · , yn ). = 1 − xi y j i,j=1 λ Proof. We can rewrite the right hand side as X n |−|δ n−1 | |δ n |−|δ n+1 | |δ 2n−1 |−|δ 2n | |δ 1 |−|δ 0 | . · · · x|δ yn · · · y1 x1 n ∅=δ 0 ↑δ 1 ↑···δ n ↓δ n−1 ↓···↓δ 2n =∅ This is equivalent to a path going from the empty partition to δ and back down, with arrows for each step of the way weighted with the respective xi or yi . By the lemma, we can transform this to a path that goes down to some α and then up, multiplying by the appropriate weights. That is, we can instead rewrite the right hand side as n X Y 1 |α |−|α | |α |−|α | 2n |−|α2n−1 | yn|α0 |−|α1 | · · · y1 n−1 n x1 n+1 n · · · x|α · . n 1 − x y i j 0 1 n n−1 2n i,j=1 ∅=α ↓α ↓···α ↑α ↑···↑α =∅ But since the only thing we get from the empty partition by going down is the empty partition, and the only partition that goes up to the empty partition is also the empty partition, αi = ∅ for every i, so the only possible chain is composed of all empty partitions, and therefore all the Q exponents are zero, so the sum is 1. This means we can simplify the right hand side to ni,j=1 1−x1i yj , the result we wanted.  Q By taking the limit of these identities as n → ∞, we get the general result i,j=1 1−x1i yj = P λ sλ (x)sλ (y). Corollary. The sλ (x) form an orthonormal basis for Λ. 24 Also, we can see that if `(λ) > n, there are no SSYT of shape λ filled with Q elements of {1, 2, · · · , n}, since the elements in row i of any SSYT have to be ≥ i. So ni,j=1 1−x1i yj = P `(λ)≤n sλ (x1 , · · · , xn )sλ (y1 , · · · , yn ). ( 1 λ=µ In particular, if we define h , in on Λn by hhλ , mµ i = for `(λ), `(µ) ≤ n, we 0 λ 6= µ get that {sλ |`(λ) ≤ n} are orthonormal. We can P construct a summary of the relations between the basis we have so far. We know that sλ = µ Kλµmµ , where Kλµ is the number of SSYt of shape λ and content µ. We also P know that hπ = µ Bπµ mµ , where Bπµ is the number of Z≥0 −matrices with row sum π and column sum µ. The following corollary shows how to go directly from hπ to sλ . Corollary. We have P that hhπ , sλ i is the number of SSYT of shape λ and content pi, that is, Kλπ . Thus hπ = λ Kλπsλ . Proof. We have hhπ , sλ i = hhπ , X Kλµ mµ i µ = X Kλµ hhπ , sµ i µ = Kλπ . Since the sλ are orthonormal, the inner product gives the coefficients of the decomposition of h this basis.  In the diagram below, we have labeled the combinatorial objects that are counted by each transition coefficient. The labels π  λ and λ  µ are meant to remind the reader that, if the coefficient is nonzero, then this inequality holds. hπ SSYT, shape λ, content π πλ / sλ SSYT, shape λ, content µ µλ Z≥0 -matrices with row sum π and column sum µ   mµ September 21: The Jacobi-Trudi Identity Last time, we ended with the following diagram showing how the h, s, and m bases of symmetric functions relate to each other: 25 hπ SSYT, shape λ, content π πλ / sλ SSYT, shape λ, content µ µλ Z≥0 -matrices with row sum π and column sum µ   mµ We denote the number of SSYT with shape λ and content π (with π  λ) by Kλπ , and the number of Z0 -matrices with row sum π and column sum µ by Aπµ . Then the above diagram can be interpreted as saying that X Kλπ Kλµ . Aπµ = λ That is, we have a bijection between the set of Z≥0 matrices with row sum π, column sum µ, and the set of pairs (T, U ) of SSYT of the same shape with content(T ) = π and content(U ) = µ. This is the Robinson-Schensted-Knuth (RSK) correspondence. The above equation looks like a matrix equation: A = K T K. Indeed, here are the transition matrices for degree 3 symmetric polynomials and their relation:      1 0 0 1 1 1 1 1 1 1 2 3 = 1 1 0 0 1 2 . 1 2 1 0 0 1 1 3 6 Our goal today is to begin to prove the following identity, which allows us to write the sλ in terms of the hπ , and hence gives the coefficients of the matrix K −1 . Theorem (Jacobi-Trudi). If λ = (λ1 , . . . , λ` ), then   hλ1 hλ1 +1 · · · hλ1 +(`−1)  hλ2 −1 hλ2 · · · hλ2 +(`−2)  , sλ = det  . .. ...   .. . hλ` −(`−1) ··· hλ` where we adopt the conventions h0 = 1, hk = 0 for k < 0. Example. If λ = (2, 1), we have   h2 h3 s21 = det = h21 − h3 = (m3 + 2m21 + 3m111 ) − (m3 + m21 + m111 ) = m21 + 2m111 , 1 h1 which is the decomposition of s21 into monomial symmetric functions that we have seen before. 26 Example. If λ = (5, 2, 1), we have s521   h5 h6 h7 = det h1 h2 h3  . 0 1 h1 Example. If λ = 1k = (1, . . . , 1), we have  h1 h2  1 h1  0 1 s1k = det  .  .. 0 0  · · · hk · · · hk−1   · · · hk−2  , ..  ... .  · · · h1 which is equal to ek by homework 1, problem 5. This gives the expected result that s1k = ek . Before we begin to prove the identity, we will compute explicitly what it gives for the coefficients of K −1 . We have   hλ1 hλ1 +1 · · · hλ1 +(`−1)  hλ2 −1 hλ2 · · · hλ2 +(`−2)  X = (−1)w hλ1 +w(1)−1 · · · hλ` +w(`)−` . det  . .. ..   .. . . hλ` −(`−1) ··· w∈S` hλ` The coefficient for hπ is then (K −1 )πλ = X (−1)w . w∈S` sort(λ1 +w(1)−1,...,λ` +w(`)−`)=(π1 ,...,π` ) The way we will prove Jacobi-Trudi next time is by viewing both sides as counting paths in directed graphs. This interpretation of sλ comes from the bijection constructed in homework 3, problem 4, between SSYTs of shape λ with entries from 1 to n and a set of collections of vertex disjoint directed paths connecting certain points in Z2 . For the remainder of class, we build our intuition toward relating the determinant in the Jacobi-Trudi identity with counting paths. We first note that hk (x1 , . . . , xn ) is the generating function for the weighted paths connecting the opposite corners in the following directed graph (with n − 1 columns and k + 1 rows): • 1 • 1 x1 • 1 • 1 ··· 1 ··· 1 x2 x1 xn x2 .. . • • xn .. . x1 .. . x2 1 xn 1 1 • • ··· • For example, when n = 3 and k = 2, the sum of the weights of all possible paths from the bottom left to the top right corner are h2 (x1 , x2 , x3 ) = x21 + x22 + x23 + x1 x2 + x2 x3 + x1 x3 , where the weight of a path is given by the product of the weights of its edges. 27 Next time we will put these into the context of the determinant in Jacobi-Trudi via the Lindström-Gessel-Viennot theorem. As a sneak peak, consider the expression.   h2 h3 det = h2 h1 − h3 h0 h0 h1 When n = 3, each of the hk counts weighted paths in • 1 • 1 • 1 x1 • 1 • 1 • 1 x2 x1 x3 x2 x1 • • x3 x2 • x3 1 1 • • • Namely, h2 counts paths (• → •), h1 counts paths (• → •), h3 counts paths (• → •), and h0 counts paths (• → •). The expression given by the determinant then counts all pairs of a path (• → •) and a path (• → •), then subtracts off those pairs of paths that intersect each other. September 23 – Jacobi-Trudi identity, proof conclusion Today’s goal is to finish the proof of the Jacobi-Trudi Identity, which states that   hλ1 hλ1 +1 · · · hλ1 +(`−1)  hλ2 −1 hλ2 · · · hλ2 +(`−2)  , sλ1 sλ2 . . . sλ` = det  . .. ...   .. . ··· hλ` −(`−1) hλ` where we put h0 = 1 and hk = 0 for k < 0. Example. s(5,2,1)   h5 h6 h7 = det h1 h2 h3  0 1 h1 Let Γ be a finite, acyclic directed graph with weights x(e) on each edge e. The weight of a path γ in Γ of the form e e1 e2 e3 v0 → v1 → v2 → . . . →` v` Q` is defined to be wt(γ) = i=1 x(ei ). For two vertices v, w of Γ, we define X P (v → w) = wt(γ) γ v →w 28 Therefore, if v1 , v2 , . . . , v` , w1 , w2 , . . . , w` are vertices of Γ then det[P (vi → wj )] = X (−1)σ = P (vi → wσ(i) ) i=1 σ∈S` X ` Y X (−1)σ wt(γ1 )wt(γ2 ) . . . wt(γ` ) γ1 σ∈S` v1 →w(σ(1) γ 2 v2 →w (σ(2) ... γ` v` →w(σ(`) We claim then that det[P (vi → wj )] = X σ (−1) X P (vi → wσ(i) ) i=1 σ∈S` = ` Y (−1)σ X wt(γ1 )wt(γ2 ) . . . wt(γ` ) γ1 σ∈S` v1 →wσ(1) γ 2 v2 →w σ(2) ... γ` v` →wσ(`) = X σ∈S` (−1)σ X wt(γ1 )wt(γ2 ) . . . wt(γ` ) γ1 v1 →wσ(1) γ 2 v2 →w σ(2) ... γ` v` →wσ(`) γi ∩γj =∅ ∀i6=j since all terms which involve intersecting paths cancel. Proof. Choose a total order on the vertices of Γ with i < j if i → j. If γ1 , . . . , γ` is a collection of paths which are not disjoint, choose x to be the first vertex which is in two of the paths. Then choose the first i and j such that x is on γi ad γj . Then swap the portions of γi and γj which come before the paths reach x. This procedure is a sign reversing involutions. That is, the new paths have the same weight, same point x, and opposite sign, and when we apply the map again it undos itself. Thus, everything cancels except when the paths are disjoint.  In the particular case where Γ is embedded in a disc in R2 and v1 , . . . , v` , w1 , . . . , w` are vertices in Γ such that Γ looks like v1 v2 w1 w2 .. . v` .. . w` 29 then the only way to have disjoint paths between these vertices is to have the paths be of the form vi → wi . In this case, we then have X wt(γ1 ) . . . wt(γ` ) det[P (vi → wj )] = γ 1 v1 →w 1 ... γ` v` →w` γi ∩γj =∅ We recall from last class that hk (x1 , . . . , xn ) is the generating function for directed paths in the following graph which connect the bottom left point to the top right point • 1 • 1 x1 • 1 • 1 ··· 1 ··· 1 x2 x1 xn x2 .. . • • xn .. . x1 .. . x2 xn 1 1 1 • • ··· • where each column has k or vertical arrows. To compute sλ1 sλ2 . . . sλ` using our above formulas, we use a graph as above with λ1 + (` − 1) vertical steps in each column. We place v` in the bottom left corner at height 0, and the remaining v2 , . . . , v` in above v` along the leftmost “avenue” at height ` − i. We place the wj for j = 1, . . . , ` along the rightmost avenue at height λj + (` − j) = λj + vj . Example. As as example, consider again the function s(5,2,1) . To this function we associate the following grid • 1 • 1 • 1 x1 • 1 • 1 • 1 x2 x1 1 • 1 • 1 • 1 x1 • 1 • 1 • 1 • 1 • 1 • 1 • 1 • 1 • 1 • 1 • x5 • x5 • 1 • 1 • 1 • 1 x4 • x5 x4 x3 • x5 x4 x3 x2 1 x4 x3 x2 x1 • 1 • x4 x3 x2 x1 • 1 x3 x2 • 1 x3 x2 x1 • • x5 x4 • x5 • 30 where v1 and w1 are the yellow vertices, v2 and w2 are the blue vertices, and v3 and w3 are the red vertices. We see that the statement of the Jacobi-Trudi identity in this case   h5 h6 h7 s(5,2,1) = det h1 h2 h3  0 1 h1 makes sense with the paths we see in the graph. For example, the (3, 1) entry in the matrix, corresponding to v1 and w3 , is zero, and there are no paths in the graph from v1 and w3 . In general, we note that λ1 ≥ λ2 ≥ · · · ≥ λ` implies in our construction that the height of wi is strictly greater than the height of wi+1 for all i = 1, . . . , ` − 1. Therefore, we do get a planar picture w1 .. . w` v1 .. . v` so the only collection of disjoint paths connecting the vi to the wj are the paths v1 → w1 , v2 → w2 , ... , v` → w` . Therefore, the formula X det[P (vi → wj )] = wt(γ1 ) . . . wt(γ` ) γ 1 v1 →w 1 ... γ` v` →w` γi ∩γj =∅ applies and yields the Jacobi-Trudi formula. September 28 – The ratio of alternants formula Remark. In problem 4 on the homework the goal is to have a graphical way to draw the Jacobi-Trudi identity. For example   h6 h7 h8 s(6,5,3) = det h4 h5 h6  = h(6,5,3) − h(6,6,2) − h(7,4,3) − h(7,6,1) + h(8,4,2) − h(8,5,1) . h1 h2 h3 We can draw these like this: 31 We run into an issue the way the problem was written with the example   h1 h2 h3 s(1,1,1) = det h0 h1 h2  = h(1,1,1) − h(2,1,0) − h(2,1,0) + h(3) 0 h0 h1 We’d like this to give the pictures: This problem will be rewritten for next week. We now turn to our main topic, the Ratio of Alternants Formula. Definition. A function f (x1 , . . . , xn ) ∈ Z[x1 , . . . , xn ] is called alternating if f (xσ(1) , xσ(2) , . . . , xσ(n) ) = (−1)σ f (x1 , . . . , xn ) for all σ ∈ Sn . Example. The Vandermonde determinant ∆(x1 , . . . , xn ) := Y (xi − xj ) i < j is alternating. Lemma. (From Homework) The alternating polynomials are exactly ∆·(symmetric polynomial), i.e. elements of ∆ · Λn . We are looking for an analogous basis to the symmetric polynomial bases, but for the alternating polynomials. For α1 > α2 > · · · > αn , set  α1 α2  x1 x1 . . . xα1 n xα2 1 xα2 2 . . . xαnn  X α α α = Aα = det  (−1)σ x1 σ(1) x2 σ(2) · · · xnσ(n) . . . .  ..  .. . ... . xαn1 ... ... xαnn σ∈Sn Note we want α1 > α2 > · · · > αn , if αi = Lαi+1 this is 0. Then the alternating polynomials are α1 >α2 >···>αn ZAα . So M ∆ · Λn = Z · Aα α1 >···>αn Λn = M α1 >···>αn Theorem. Z· Aα ∆ Aλ1 +n−1,λ2 +n−2,...,λn ∆ Remark. The theorem above could have been the definition of sλ , see Friday for an alternate order of the definitions we’ve seen so far. s(λ1 ,λ2 ,...,λn ) (x1 , . . . , xn ) = 32 A special case of this theorem states that A(n−1,n−2,...,2,1,0) 1 = S∅ = ∆ n−j det |x |1≤i,j≤n . = Q i i < j (xi − xj ) Q In other words, det(xn−j ) = i < j (xi − xj ), which is called Vandermonde’s identity. i A Proof. (Proof of special case of theorem.) The degree of (n−1,n−2,...,2,1,0) is ∆   n 0 + 1 + · · · + (n − 1) − , 2 which is 0, so this quantity is an integer. We can look at the coefficient of x1n−1 x2n−2 . . . xn−1 and see that it is 1.  We will define ρ = (n − 1, n − 2, . . . , 2, 1, 0), with ρj = n − j. Q We just showed Aρ = i < 0j (xi − xj ). Our next goal is to show that Aλ+ρ = sλ Aρ or in other words that Aλ+ρ = sλ · Aρ . We want to show that X λ +ρ λ +ρ λ +ρ (−1)τ x1 τ (1) τ (1) x2 τ (2) τ (2) · · · xnτ (n) τ (n) = τ ∈Sn ! ! X ρ ρ ρ (−1)σ x1σ(1) x2σ(2) · · · xnσ(n) X Kλµ mµ µ σ∈Sn or we can write the far right term on the right hand side as   X  Kλ,sort(µ) xµ1 1 xµ2 2 · · · xµnn . µ1 ,...,µn ∈Zn ≥0 On the right hand side the coefficient of x1ν1 +ρ1 x2ν2 +ρ2 · · · xnνn +ρn is X X X

(−1)σ Kλ,sort(µ) νj + ρj = ρσ(j) + µj = Kλµ (−1)σ (νj + ρj − ρσ(j) ) ∼ (µ) µ1 ≥µ2 ≥···≥µn σ where ∼ means “if the left thing is a permutation of the right thing”. Set X
Lµν := (−1)σ (νj + ρj − ρσ(j) ) ∼ (µ) σ P then the coefficient above is equal to Kλµ Lµν . Thus we want to know where K · L = Id. That is where X
(−1)δ vj + ρj − ρσ(j) ∼ µ . Lµν = σ We will compare to K −1 from the Jacobi-Trudi identity. 33 Jacobi-Trudi says  hλ1 hλ1 +1 . . . .. sλ = det  ... . ... ... ... ... X = (−1)σ hλj +j−σ(j)  ... ..  . hλn σ = X (−1)σ hλ−ρ+σ(ρ) . σ The coefficient of hµ is X
(−1)σ λ − ρ + σ(ρ) ∼ µ = Lµλ . σ By Jacobi-Trudi, X  Kνµ Lµλ = µ 1 λ=ν 0 λ= 6 ν So we get Lµν = X (−1)δ vj + ρj − ρσ(j) ∼ µ
σ as desired. September 30 – Starting from the ratio of alternants The goal for today is to follow Cauchy’s definition of the Schur polynomials to derive Jacobi-Trudi identity and Cauchy’s identity.  α As before, let ρn = (n − 1, n − 2, . . . , 2, 1, 0). Let Aα1 α2 ...αn = det xi j 1≤i,j6=n . What Aλ+ρn (x1 , x2 , · · · , xn ) ? The ratio is in Λn . It is happens if we define sλ (x1 , x2 , · · · , xn ) to be Aρn (x1 , x2 , · · · , xn ) not clear that this ratio stabilizes as n grows, but we will know that it is true once we prove the Jacobi-Trudi identity. Consider the following n by ∞ matrix M (x).   1 x1 x21 . . . xi1 · · · 1 x2 x22 · · · xi2 · · ·  M (x) =  .. . . . ..   ... ... . .. .  . 1 xn x2n · · · xin · · · Then Aλ+ρ is the n by n minor of M (x) corresponding to columns indexed by λi + n − i. For an invertible n by n matrix g, the n by n minors of gM (x) are det(g)Aλ+ρ . So ratio of minors in M is the same as ratio of minors in gM . We prove Jacobi-Trudi identity by finding an invertible g such that ratio of minors in gM gives us the desired expression in the h0 s. Suppose that x1 , x2 ,. . . , xn are distinct. Then the rows of M (x) span all solutions [a0 , a1 , · · · ] to the recursion ak+n − e1 ak+n−1 + e2 ak+n−2 · · · + (−1)n ak . 34 We can also express this in terms of generating The solutions to this recursion P functions: degree ≤ n − 1 k . The are the coefficients of generating functions ak t of the form polynomialQof(x−t j) 1 rows of M (x), specifically, are the rational functions 1−txj . polynomial Qof degree ≤ n − 1 (1−txj ) k k+1 t +h1 t +h2 tk+2 + Now, another basis for the vector space of rational functions of the form k k x x is the rational functions Q(1−tx , for 0 ≤ k ≤ n−1. We have Q(1−tx = j) j) · · · . So the row span of M (x) is the same as the row span of the matrix   1 h1 h2 h3 ··· ··· 0 1 h1 h2 ··· ···    0 0 1 h1 ··· ··· . H :=  . .  .. ... ...  .. .. . ···  0 0 0 · · · · · · 1 h1 · · · By our previous argument, the quotient of corresponding n by n minors of M is equal to the quotient of corresponding n by n minors of H. And this is the statement of Jacobi-Trudi identity because the first n by n minor of H is upper triangular with all diagonal entries equal to 1, and thus has determinant 1. Now, we provide Cauchy’s proof of Cauchy’s identity. First, we write    1 1 1 ... 1 1 x1 x21 . . .   1 x2 x22 · · ·  y1 y2 y3 . . . yn  1  2 2 2 2 = . . .. . .  1 − xi yj 1≤i,j≤n  .. .. .  y1 y2 y3 . . . yn  . .. .. .. . . . . .. 1 xn x2n · · · . . .   P 1 T So det = α1 >α2 >···>αn det M (x)α det M (y)α , where M (x)α denotes the minor 1 − xi y j corresponding to columns α1 > α2 > · · · > αn . On the right hand side, going through strictly decreasing n-tuples α1 > α2 > · · · > αn is the same as going through non-increasing n-tuples α1 − 1 ≥ α2 P − 2 ≥ · · · ≥ αn − n, which corresponds to a parition λ. So the right hand side is equal to λ Aλ+ρ (x)Aλ+ρ (y). What we need to show is that   1 Aρ (x)Aρ (y) =Q . (1 − xi yj ) 1 − xi yj 1≤i,j≤n This follows from a determinantal identity proved by Cauchy in another source, 26 years later: Q Q   1 i < j (ui − uj ) i < j (vi − vj ) Q = . ui + vj 1≤i,j≤n (ui + vj ) Q n ,v1 ,...,vn ) for some To prove this, note that the left hand side must be of the form f (u1 ,...,u (ui +vj ) polynomial f . Q The polynomial f must vanish whenever u = u and whenever vi = vj , so it i j Q is divisible by i < j (ui − uj ) i < j (vi − vj ). But also, computing degrees, the polynomial f Q Q must be homogeneous of degree n(n−1), so f is a scalar multiple of i < j (ui −uj ) i < j (vi −vj ), and a little work checks that the scalar is 1. 35 October 3-October 7 Introduction to Representation Theory of Finite Groups (IBL) We solved problems about representation theory. Here they are! Let G be a group and k a field. A representation of G is a k-vector space V and a group homomorphism ρ : G → GL(V ). In other words, a representation is an action of G on V by linear maps. We will often denote the representation as “V ”, and we will sometimes write ρV for ρ. Given two representations (V1 ,hρ1 ) and (Vi2 , ρ2 ), the direct sum representation is the action of G on V1 ⊕ V2 where g acts by ρ10(g) ρ20(g) . A representation V is called indecomposable if V 6= 0 and we cannot write V = V1 ⊕ V2 for V1 , V2 6= 0. It is easy to see that any finite dimensional representation is a direct sum of indecomposable representations. Given a representation V , a subrepresentation of V is a vector subspace U of V such that G maps U to U . A representation V is called simple if V 6= 0 and the only subrepresentations of V are V and {0}. Given two G-representations U and V , a morphism of G-representations is a linear map φ : U → V obeying φ(g ∗ u) = g ∗ φ(u) for g ∈ G, u ∈ U. Problem 1. Let U and V be G-representations and let φ : U → V be a morphism of Grepresentations. We write HomG (U, V ) for the space of morphism of G-representations from U to V . Show that Ker(φ) and Im(φ) are subrepresentations of U and V respectively. Our first goal is to prove: Theorem (Maschke’s Theorem, first version). Let k have characteristic zero and let G be a finite group. Then V is simple if and only if it is indecomposable.. Problem 2. Show that, if V is not indecomposable then V is not simple. Problem 3. Let k have characteristic zero, let G be a finite group and let V be a finite dimensional nonzero representation of G. Suppose that V is not simple, so that U ⊂ V is a nontrivial subrepresentation. Let η : V → U be a linear map with η(u) = u for u ∈ U . Define X
1 π(v) = |G| ρ(g) ◦ η ◦ ρ(g)−1 (v). g∈G (1) Show that π is a morphism of G-representations (even though η may not be). (2) Show that π has image U and that π(u) = u for u ∈ U . (3) Show that V = U ⊕ Ker(π). Deduce that V is not indecomposable. Thus we conclude: Theorem (Maschke’s Theorem, second version). Let k have characteristic zero, let G be a finite group and let V be a finite dimensional representation of G. Then V is a direct sum of simple representations. 36 We next will consider the question of whether the decomposition of V as a direct sum of simple representations is unique. Problem 4 (Schur’s Lemma, first version). Let U and V be simple representations of G and let φ : U → V be a morphism of G-representations. Show that either φ = 0 or else φ is an isomorphism. Problem 5.L Let U1 , U2 , . . . , Uk be a collection of nonisomorphic simple G-representations ⊕c and let V = kj=1 Uj j for some nonnegative integers cj . (1) Show that dimk HomG (Uj , V ) = cj dimk HomG (Uj , Uj ). L Lk ⊕c ⊕d (2) Deduce that, if kj=1 Uj j ∼ = j=1 Uj j , then (c1 , c2 , . . . , ck ) = (d1 , d2 , . . . , dk ). (Nitpicker alert: Did you use that Uj 6= 0?) We thus deduce: Theorem. Let G be a L group and let V be a finite dimensional representation of G which ⊕c can be written as V = kj=1 Uj j for a collection of nonisomorphic simple representations U1 , U2 , . . . , Uk of G. Then the Uj (up to isomorphism) and the multiplicities cj are uniquely determined by V . This theorem doesn’t require that G is finite or that k has characteristic zero, but without those hypotheses, Maschke’s theorem does not apply, so there may be many representations V which cannot be written in this way. We now see how things are better if k is algebraically closed. Problem 6. Let k be an algebraically closed field and let U be a simple G-representation. Let φ : U → U be a morphism of G-representations. Show that φ is a scalar multiple of the identity. (Hint: Let λ be an eigenvalue of φ, and apply Problem 4 to the linear map φ − λId.) Problem 7. Let k be an algebraically closed field. LetL U1 , U2 , . . . , Uk be a collection ⊕cj k for some nonnegative of nonisomorphic simple G-representations and let V = j=1 Uj integers cj . Show that dimk HomG (Uj , V ) = cj . 37 We now introduce character theory into the story. Given a representation V , the character of V is the function G → k defined by χV (g) = Tr(ρV (g)). Problem 8. Suppose that g1 and g2 are conjugate elements of G, meaning that g2 = hg1 h−1 for some h ∈ G. For any character χ of G, show that χ(g1 ) = χ(g2 ). Problem 9. Check that χV1 ⊕V2 (g) = χV1 (g) + χV2 (g). Problem 10. Let k have characteristic zero, let G be a finite group, and let V be a finitedimensional representation of G. Let V G = {v ∈ V : g(v) = v ∀g ∈ G}. Define π : V → V by π(v) = 1 |G| X g ∗ v. g∈G G G (1) Show that Im(π) = V and V = V ⊕ Ker(π). (2) Show that Tr(π) = dim V G . (3) Deduce that X 1 dim V G = |G| χV (g). g∈G Problem 11. Let k have characteristic zero, let G be a finite group, and let U and V be finite-dimensional representations of G. Let Hom(U, V ) be the vector space of k-linear maps U → V (all maps, not just the morphisms of G-representations). Define an action of G on Hom(U, V ) by
(g ∗ φ)(u) = ρV (g) ◦ φ ◦ ρU (g −1 ) (u). (1) Show that HomG (U, V ) = Hom(U, V )G for this G-action. (2) Show that X 1 dimk HomG (U, V ) = |G| χU (g −1 )χV (g). g∈G We can improve the formula above in the case that k = C. Problem 12. Let G be a finite group, and let V be a finite dimensional representation of G over the field C. Let g ∈ G. (1) Show that all the eigenvalues of ρV (g) are roots of unity. (2) Show that χV (g −1 ) = χV (g), where z is the complex conjugate of z. Thus, let G be a finite group and let p and q be C-valued functions on g. Define X 1 hp, qi := |G| p(g)q(g). g∈G This is a positive definite Hermitian inner product. Then our results above specialize to say: Theorem. With the above assumptions and notations, let U and V be two representations of G over C. Then dim HomG (U, V ) = hχU , χV i. In particular, if U is a simple representation, then hχU , χV i is the multiplicity of the summand U in V .