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1 SCIENCE FICTION 208

2 PRELIMINARIES Let X n = {x,...,x n } and Y n = {y,...,y n }. we define the scalar product For hp,qi = P (@ x y )Q(x; y) For P (x; y),q(x; y) 2 Q[X n ; Y n ] x,y=0 2 S n, and P (x; y) 2 Q[X n ; Y n ] we define the diagonal action P (x; y) =P (x,x 2,...,x n ; y,y 2,...,y n ) Notice that since we have h P,Qi = hp, Qi The orthogonal complement of an invariant subspace is also invariant. An alternant under the diagonal action ,2(x; y) = The Frobenius map F = s [X] The Frobenius Characteristic of an S n invariant module M = M r,s H r,s (M) FM = X r,s t r q s Fch H r,s (M)

3 n(µ) = l(µ) X (i )µ i i= NOTATION T µ = t n(µ) q n(µ0 ) µ = {(a,b ), (a 2,b 2 ),...,(a n,b n ) (a i,b i )they, x coordinates B µ (q, t) = nx t a i q b i µ (q, t) = ny ( t a i q b i ) i= i=2 w µ (q, t) = Y c2µ t l(µ) q a(µ)+ q a(µ) t l(µ)+ l(c) c a(c) example B µ (q, t) =+q + q 2 + t + tq µ (q, t) =( q)( q 2 )( t)( qt)

4 BASIC FACTS For a partition µ ` n with cells (a,b ), (a 2,b 2 ),, (a n,b n )set µ(x, y) =det x a j i y b j i n i,j= and define M µ = L[@ p x@ q y µ] Theorem (Conjectured in 990 proved in 2000) The Frobenius characteristic of M µ is the modified Macdonald H e µ (X; q, t) If p(x, y) 2 M µ set flip µ p(x, y) =p(@ y ) µ(x, y) Theorem The map flip µ p(x, y) =p(@ y ) µ(x, y) is non singular for p(x, y) 2 M µ Proof If p(x, y) 2 M µ then p(x, y) =A(@ y ) µ(x, y) p(@ y )p(x, y) =A(@ y )p(@ y ) µ(x, y) =0 thus Q.E.D.

5 BASIC FACTS The definition of the map flip µ p(x, y) can be extended to any S n invariant submodule by acting with flip µ on every element of a basis Define for any symmetric F (X; q, t) # F (X; q, t) =!F (X;/q, /t) Theorem If M has Frobenius Characteristic M, (in symbols FM= M ) then F flip µ M = T µ # M Corollary T µ # e H µ (X; q, t) = e H µ (X; q, t)

6 SCIENCE FICTION Let µ ` n + be a partition with at least m removable corners. Suppose that A = {, 2,..., m } is an m-subset of the predecessors of µ Conjecture I m The polynomial A(X; q, t) = mx i= eh i (X; q, t) my j=;j6=i T i /T j is the Frobenius characteristic of the space M \ M 2 \ \ M m Notice that for m =2thisreducesto Conjecture I 2 If and are any two predecessors of a partition µ, thepolynomial is the Frobenius characteristic of the space M \ M

7 OPEN PROBLEMS M M flip M \ M? M \ M? flip M \ M For example if A = {, M = M \ M F M \ M = e H T /T + e H, } then flip (M \ M Why divided di erences? = T H e T H e T /T T F M T F(M \ M ) T F(M \ M ) \ M \ M = T = A More generally if A = {, 2,..., m } then A = T m F M \ M 2 \ \ M m T F M 2 \ M 2 \ \ M m T m T T T

8 DIAGONAL HARMONICS AND NABLA Let X n = {x,...,x n } and Y n = {y,...,y n } and set DH n = P (x, y) 2 Q[X n,y n ]: P n x s y i P (x, y) =0 8 apple r + s apple n In 994 Mark Haiman (after exposure to Procesi) was able to predict that F DH n = X µ`n T µ Hµ e [X; q, t]( t)( q)b µ (q, t) µ (q, t) w µ (q, t) In the q, t-catalan and Lagrange inversion paper this formula appeared together with the expansion e n = X eh µ [X; q, t]( t)( q)b µ (q, t) µ (q, t) w µ (q, t) µ`n Francois noticed that we could write F DH n = re n by the definition rh e µ [X; q, t] =T µ Hµ e [X; q, t] Tex t Thereafter we applied r to everything in sight with astonishing findings!!! SCIENCE FICTION WAS ENRICHED BEYOND BELIEF!!!

9 Theorem For A = {, 2,..., m } Proof my r=;r6=s = = Corollary and mx i= mx i= SCIENCE FICTION AND NABLA eh s (X; q, t) = r T r my r=;r6=s my r=;r6=s eh s (X; q, t) = A(X; q, t) = X m k=0 my r=;r6=s r T r eh i (X; q, t) T i T r eh i (X; q, t) r T r my j=;j6=i my j=;j6=i A(X; q, t) T i /T j T i /T j ( ) k r k A(X; q, t)e k T + + T m T s L{ e H,..., e H m } = L{ A, r A,...,r m A} (for all apple s apple m) = e H s (X; q, t) Q.E.D.

10 MIRACLES OF SCIENCE FICTION Let m be the collection of words m in 0,. Define for m 2 m 2 m A = ( m ) A Q m j=0 T j with (k) A =( r)m k A. Corollary For each apple s apple m we have the expansion We derived that eh s = X 2 m 2 m 2 m ( s = ) This can be rewritten as Convert subsets of {, 2...,m} into words in m Q.E.D.

11 MIRACLES OF SCIENCE FICTION For all 2 m 2 m and for each apple i apple m we have T i # i i+ m A = i i+ m A where for convenience we let j = For m = 2 M j M FchM 0 = Fchflip M \ M, = Fch(M ) FchM 0 = Fchflip M \ M M 0 M M 0 M = M M 0 M = M M 0 Flipping does not change dimension and Schur positivity (By SF and the n! Theorem!) If ` n then dim M 0 =dimm =dimm 0 = n! 2 How do we split a left regular representation?

12 Define Theorem Proof Flipping Frobenius Characteristics flip µ (X; q, t) =T µ # (X; q, t) T H flip e T H e H e H e = T T T T T H flip e T H e T T H e T T H e = T T T T T Notice: H e = T H e T H e T H + T e T H e T T T Notice: = T µ! (X;/q, /t) T H e H e = T T T Flipping a Frobenius Characteristic preserves positivity Recall that each e H [X; q, t] is the Frobenius of a left regular representation. lim t,q! eh [X;, ] = e n This given, what are your guesses as to the evaluation of the following limits? T T H e T H e = t,q! T T T H e T H e = h lim T T 2 e n 2 T Q.E.D. Thus e 2 e n 2 In other words, what is the most natural way to split e n in exactly /2?

13 SCIENCE FICTION 208 Recall that if µ ` n + and A = {, 2,... m } is any subset of the immediate predecessors of µ then the symmetric functions ( r) k A = mx Y ( T i ) k H i e m i= j=j6=i T i /T j (for 0 apple k apple m ) are all Schur positive. Guoce Xin proved that Set m k A =( r) k A Where m A k t=q= = e n 2m+2 m,k m,0 = m e m [mx] h i! h i+ h i

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18 THE CASE OF THREE PREDECESSORS M = M \ M \ M M M M 0 =(M \ M ) \ M? M 0 =(M \ M ) \ M? M 0 =(M \ M ) \ M? M 00 =(M 0 [ M [ M 0 ) \? M M 00 =(M 0 [ M [ M 0 ) \? M M 00 =(M 0 [ M [ M 0 ) \? M M 0 M 00 M 00 M M 0 M 0 M 00 FM \ M FM \ M \ M = 0 A FM 2 3 = 2 3 M flip = 00 flip = 00 flip = 00 dim M =dimm 00 = n!/3 dim M =dimm 00 = n!/3 dim M =dimm 00 = n!/3

19 For example For example FM = X r,s t r q s Fch H r,s (M)

20 eh µ [X; q, t] = P s [X] e K,µ (q, t) the marginal, modified Hall-Littlewood case $q=0 This branch of Algebraic Combinatorics resulted by my introduction In the early 90 s I discovered a recursive method for proving g the marginal, modified Hall-Littlewood case e H µ [X;0,t] THE $\bf n!\over 2$ CONJECTURE the marginal (Hall-Littlewood) case q = 0 eh µ [X; q, t] = P s [X] e K,µ (q, t)

21 In 990, working jointly with Haiman, we where led to the discovery In 990, working jointly with Haiman, we were led to the discovery must be the desired bi-graded module the coordinates of the cells of µ and the Conjecture that mx my A(X; q, t) = eh i (X; q, t) i= j=;j6=i T i /T j the coordinates of the cells of µ and the Conjecture that the desired module

22 my r=;r6=s r T r A(X; q, t) = A(X; q, t) = mx i= eh i (X; q, t) my j=;j6=i T i /T j ] my r=;r6=s r T r A(X; q, t) =

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