Complete Homogenous Symmetric Functions

LECTURE 3: COMPLETE HOMOGENOUS SYMMETRIC FUNCTIONS Define the complete homogenous symmetric functions X hk = xi1 xi2 . . . xik 1≤i1 ≤i2 ≤...≤ik Example 1. h2 = X xi xj = 1≤i≤j X x2i + i≥1 X xi xj = m2 + m11 1≤i< j For a partition λ = (λ1 , λ2 , . . . , λr ), define hλ = hλ1 hλ2 . . . hλr Example 2. X X X h21 = h2 h1 = ( x2i + xi xj )( xk ) i≥1 = X x3i +2 i≥1 1≤i< j X x2i xj +3 k≥1 X xi xj xk 1≤i< j< k i6=j = m3 + 2m21 + 3m111 For future reference, h3 = m3 + m21 h21 = m3 + 2m21 h111 = m3 + 3m21 So we have a transformation matrix between h’s just computed that the matrix is:   1 1 1 1 2 3 1 3 6 We have two main results today: + m111 + 3m111 + 6m111 and m’s. For example, in degree 3, we Theorem 3. The h’s form a basis of Λ Theorem 4. The transformation matrix between h’s and m’s is symmetric. 1. The h’s are a basis Since the h polynomials are the monomials in h1 , h2 , h3 , . . . , the claim is that Λ = Z[h1 , h2 , h3 , . . .]. We first show that the natural ring map Z[h1 , h2 , . . .] → Λ is surjective. Indeed it is enough to show every ek is a polynomial in h’s. Consider two identities Y X (1 + xi t) = ek tk i≥1 k≥0 1 Y i≥1 X 1 = hk tk 1 − xi t k≥0 Hence, 1= Y Y (1 − xi t) i≥0 i≥0 X X 1 = ( (−1)k ek tk )( hk tk ) (1 − xi t) k≥0 k≥0 Comparing the coefficients on both sides, one gets the relation: hk − e1 hk−1 + e2 hk−2 − e3 hk−3 + . . . + (−1)k ek = 0 (∗) The dimension of degree d parts of Z[h1 , h2 , . . .] and Λ are equal, which proves the injectivity.  So Z[h1 , h2 . . .] ∼ = Z[e1 , e2 . . .] ∼ = Λ. 1.1. What about Λn ? We have Λn ∼ = Z[e1 , e2 , . . .]/I where I is the ideal generated by ∼ ei = 0 for i ≥ n + 1. If one thinks Λ ∼ = Z[h1 , h2 . . .], then P Λn = iZ[h1 , h2 . . .]/J, where J is k the coefficient of t in the power series expansion of 1/( i≥0 hi t ) in t for k ≥ n + 1. The above proof can also show that Λn = Z[h1 , h2 , . . . , hn ] so {hλ : l(λT ) ≤ n} is a basis for Λn . (In the lecture on Wednesday, we will show that {hλ : l(λ ≤ n)} is also a basis for Λn ). 1.2. The map ω. (Section largely added by editor). Noticing the symmetric between e and h, define a map ω : Λ → Λ by hk 7→ ek and thus hλ 7→ eλ . By applying ω to the equation (*), one can get ω(ek ) = hk , and hence ω(eλ ) = hλ . From our perspective, ω is pretty mysterious. There are lots of applications of the ring of symmetric functions. In some of those other applications, ω is more motivated. For example, if you use Λ to study the cohomology of the Grassmannian G(d, n), then ω is the isomorphism G(d, n) ∼ = G(n − d, n) which sends a d-plane to ints orthogonal complement. If you use Λ to study the representation theory of Sn , then ω tensors with the sign representation. There is not a similarly elegant answer for GLn representation theory. Let’s see what an answer would look like. First of all, what is the representation theory meaning of ek and hk ? Remember that we go from a representation of GLn to a symmetric polynomial by taking the trace of the action of a diagonal matrix. Let V be the standard n-dimensional representation of GLn . In V , a diagonal matrix acts by itself, and thus has Vk trace x1 + x2 + · · · + xn =
e1 = h1 . Let’s look at V . If V has basis e1 , e2 , . . . , en , then Vk n a basis for V is the k elements ei1 ∧ ei2 ∧ · · · ∧ eik . A diagonal matrix of GLn acts diagonally in this basis, with diagonal elements the products ei1 ei2 · · · eik , and hence with trace ek . Similarly, Symk V will correspond V to hk . So we want an operation which switches k V and Symk V . Such a thing occurs in physics and is called the “boson-fermion correspondence”; I don’t know much about it. Such a thing also occurs Vkin the theory of super-groups where, if V is the standard representation of GL−n , then V computed in the category of super-vector spaces is what we k would normally call Sym V . I also don’t know much about this, and we won’t talk about it. In a month, we will talk about Schur-Weyl duality, which is about the relation between GLn and Sm representation theory. Since ω has a simple meaning on the symmetric group 2 side, we can try to use that to extract an interpretation for ω on the GL side; let’s remember to think about that. Question from the floor: Are you saying there is no functor GL − rep → GL − rep which realizes ω? Answer: No, I am not willing to make such a specific claim. I am saying it is not any familiar or elegant operation. 2. The matrix is symmetric Define Aλµ = coefficient of mµ in hλ . Theorem 5. Aλµ = Aµλ . Proof. To get output xµ1 1 xµ2 2 . . . xµr r the hλj term must contribute xαj = P P (k) are vectors in Z≥0 with j αj = µ and k αj = λj for all j. Thus, ( ) X (k) X Aλµ = # (α1 , . . . , αr )| αj = λj , αj = µ Q α (k) j k xk , where αj ’s j k Which is equivalent to say Aλµ = number of nonnegative integer matrices with row sum µ and column sum λ Clearly, nonnegative integer matrices with row sum µ and column sum λ is in bijection with those with row sum λ and column sum µ. Therefore, Aλµ = Aµλ .  Example 6. Λ = (2, 1), µ = (1, 1, 1), all possible such matrices are       0 1 1 0 1 0 1 0  0 1  1 0  1 0 1 0 0 1 This shows that the coefficient of m111 in h21 should be 3. We can write this proof using generating function identities. By definition, we have X X mλ (x)hλ (y) Aλµ mλ (x)mµ (y) = λ λ,µ and the above proof showed that X Y Aλµ mλ (x)mµ (y) = 1 . 1 − x y i j i,j≥1 λ,µ The right hand side is sometimes known as the Cauchy product. Similarly, X X Y 1 Aµλ mλ (x)mµ (y) = hµ (x)mµ (y) = 1 − y i xj µ i,j≥1 µ,λ The right hand sides are equal, P so so are the left handQsides, showing Aλµ = Aµλ . Let’s explain the identity λ,µ Aλµ mλ (x)mµ (y) = i,j≥1 1−x1i yj more slowly. Expand the geometric series in the Cauchy product Q 1 = (1 + x1 y1 + x21 y12 + . . .)(1 + x1 y2 + x21 y22 + . . .) · · · i,j≥1 1−xi yj (1 + x2 y1 + x22 y12 + . . .)(1 + x2 y2 + x22 y22 + . . .) · · · ··· 3 The coefficient of x71 x32 y15 y25 is the number of ways in which we pick (xi yj )k such that the powers of xi ’s is (7, 3), and powers of yj ’s is (5, 5). This is in bijection with the integer matrices with nonnegative entries with row sum (7, 3) and column sum (5, 5). For example (x1 y1 )5 (x1 y2 )2 (x2 y2 )3 corresponds to the matrix ( 50 23 ). 2.1. The Hall inner product. The symmetry of the matrix Aλµ can be used in the following way: Define aPbilinear product hhλ , mµ i = δλµ , where δλµ is the Kronecker delta, and then hhλ , hµ i = hhλ , ν Aλν mν i = Aλµ . Since Aλµ = Aµλ , this bilinear form is symmetric. Here is a useful fact about this bilinear form, which we’ll prove next time. Proposition 7. Let {fλ } and {gµ } be two P homogenous bases Q of1 Λ. Then f and g are dual basis (i.e. hfλ , gµ i = δλµ ) if and only if λ fλ (x)gλ (y) = i,j 1−xi yj . The homogeneity is just to make sure there are no issues about formal convergence of the sum; we could replace it with any condition that made the sum formally convergent. 4