Spherical Schubert Varieties Mahir Bilen Can September 29, 2020 PART 1 MUKASHIBANASHI Then R is an associative commutative graded ring w.r.t. the product χV · χW = indSn χV × χW Sk ×Sn−k V where χ is the character of the Sk -representation V . Likewise, W χ is the character of the Sn−k -representation W . ! Sk : the stabilizer of {1,...,k} in Sn Sn−k : the stabilizer of {k + 1,...,n} in Sn n R : the free Z-module on the irreducible characters of Sn n 0 R : the direct sum of all R , n ∈ N (where R = Z) n R : the free Z-module on the irreducible characters of Sn n 0 R : the direct sum of all R , n ∈ N (where R = Z) Then R is an associative commutative graded ring w.r.t. the product χV · χW = indSn χV × χW Sk ×Sn−k V where χ is the character of the Sk -representation V . Likewise, W χ is the character of the Sn−k -representation W . ! Sk : the stabilizer of {1,...,k} in Sn Sn−k : the stabilizer of {k + 1,...,n} in Sn For λ ` k, µ ` n − k, let λ λ χ : the character of the Specht module S of Sk µ µ χ : the character of the Specht module S of Sn−k In R, we have λ µ M τ τ χ · χ = cλ,µ χ τ`n τ for some nonnegative integers cλ,µ ∈ N, called the Littlewood- Richardson numbers. It is well-known that R is isomorphic to the algebra of symmetric τ functions in infinitely many commuting variables; cλ,µ’s can be computed via Schur’s symmetric functions. τ • For which k ∈ N, do we have c(k),(n−k) ∈ {0,1} for every τ ` n? Equivalently, when is the induced trivial character indSn 1 multiplicity free? Sk ×Sn−k Sk ×Sn−k Definition A subgroup H 6 G is called a Gelfand subgroup if the induced G trivial character indH 1H is multiplicity free. Remark G The induced character indH 1H can be viewed as the character of the left G-module of C-valued functions on G/H, C[G/H] = {f : G/H → C}. To really understand an algebra, one needs to understand its structure constants. In particular, we want to know when the Littlewood-Richardson numbers are as simple as 0 or 1. Of course, as a function of λ,µ, and τ, this question is already too general, so, we ask a more specific question: Definition A subgroup H 6 G is called a Gelfand subgroup if the induced G trivial character indH 1H is multiplicity free. Remark G The induced character indH 1H can be viewed as the character of the left G-module of C-valued functions on G/H, C[G/H] = {f : G/H → C}. To really understand an algebra, one needs to understand its structure constants. In particular, we want to know when the Littlewood-Richardson numbers are as simple as 0 or 1. Of course, as a function of λ,µ, and τ, this question is already too general, so, we ask a more specific question: τ • For which k ∈ N, do we have c(k),(n−k) ∈ {0,1} for every τ ` n? Equivalently, when is the induced trivial character indSn 1 multiplicity free? Sk ×Sn−k Sk ×Sn−k To really understand an algebra, one needs to understand its structure constants. In particular, we want to know when the Littlewood-Richardson numbers are as simple as 0 or 1. Of course, as a function of λ,µ, and τ, this question is already too general, so, we ask a more specific question: τ • For which k ∈ N, do we have c(k),(n−k) ∈ {0,1} for every τ ` n? Equivalently, when is the induced trivial character indSn 1 multiplicity free? Sk ×Sn−k Sk ×Sn−k Definition A subgroup H 6 G is called a Gelfand subgroup if the induced G trivial character indH 1H is multiplicity free. Remark G The induced character indH 1H can be viewed as the character of the left G-module of C-valued functions on G/H, C[G/H] = {f : G/H → C}. How many lines are there that intersect four lines in 3-space? L1 L2 L3 L4 Littlewood-Richardson numbers appear in the solutions of some perceivable geometric problems. L1 L2 L3 L4 Littlewood-Richardson numbers appear in the solutions of some perceivable geometric problems. How many lines are there that intersect four lines in 3-space? Littlewood-Richardson numbers appear in the solutions of some perceivable geometric problems. How many lines are there that intersect four lines in 3-space? L1 L2 L3 L4 Sometimes there are no solutions! Sometimes there are no solutions! • • There can be one solution! • • There can be one solution! • There can be one solution! • • There can be one solution! • There can be one solution! • • There can be one solution! • • • • • • There can be two solutions! • • • • There can be two solutions! • • • There can be two solutions! • • • There can be two solutions! • • There can be two solutions! • • • • There can be two solutions! • • • • There can be two solutions! • • • • • There can be infinitely many solutions! There can be infinitely many solutions! • There can be infinitely many solutions! • Anarchism? 3 3 No! If we consider ourselves living inside P instead of C , 3 all of that disorder disappears.. Here, P is the set of all 4 1-dimensional vector subspaces of C . At first, it appears as if there is no uniform answer. 3 3 No! If we consider ourselves living inside P instead of C , 3 all of that disorder disappears.. Here, P is the set of all 4 1-dimensional vector subspaces of C . At first, it appears as if there is no uniform answer. Anarchism? 3 3 If we consider ourselves living inside P instead of C , 3 all of that disorder disappears.. Here, P is the set of all 4 1-dimensional vector subspaces of C . At first, it appears as if there is no uniform answer. Anarchism? No! 3 Here, P is the set of all 4 1-dimensional vector subspaces of C . At first, it appears as if there is no uniform answer. Anarchism? 3 3 No! If we consider ourselves living inside P instead of C , all of that disorder disappears.. At first, it appears as if there is no uniform answer. Anarchism? 3 3 No! If we consider ourselves living inside P instead of C , 3 all of that disorder disappears.. Here, P is the set of all 4 1-dimensional vector subspaces of C . 3 3 P C = {(1,u1,u2,u3): ui ∈ C} (1,0,0,0) x3 x1 (0,1,v2,v3) 2 3 3 P = P \ C x2 4 C \{0} x0 x3 x1 x2 3 P (0,1,v2,v3) 2 3 3 P = P \ C 4 C \{0} x0 3 C = {(1,u1,u2,u3): ui ∈ C} (1,0,0,0) x3 x1 x2 3 P (0,1,v2,v3) 2 3 3 P = P \ C 4 C \{0} x0 3 C = {(1,u1,u2,u3): ui ∈ C} (1,0,0,0) x3 x1 x2 (0,1,v2,v3) 2 3 3 P = P \ C 4 C \{0} x0 3 3 P C = {(1,u1,u2,u3): ui ∈ C} (1,0,0,0) x3 x1 x2 (0,1,v2,v3) 4 C \{0} x0 3 3 P C = {(1,u1,u2,u3): ui ∈ C} (1,0,0,0) x3 x1 2 3 3 P = P \ C x2 4 C \{0} x0 3 3 P C = {(1,u1,u2,u3): ui ∈ C} (1,0,0,0) x3 x1 (0,1,v2,v3) 2 3 3 P = P \ C x2 4 C \{0} x0 3 3 P C = {(1,u1,u2,u3): ui ∈ C} (1,0,0,0) x3 x1 (0,1,v2,v3) 2 3 3 P = P \ C x2 2 2 P C = {(0,1,v2,v3): vi ∈ C} x2 (0,0,d,e) 1 = 2 \ 2 x3 P P C 3 C \{0} x2 x1 x3 (0,0,d,e) 1 2 2 P = P \ C 3 C \{0} 2 2 P C = {(0,1,v2,v3): vi ∈ C} x2 x1 x3 1 2 2 P = P \ C 3 C \{0} 2 2 P C = {(0,1,v2,v3): vi ∈ C} x2 (0,0,d,e) x1 x3 3 C \{0} 2 2 P C = {(0,1,v2,v3): vi ∈ C} x2 (0,0,d,e) x1 1 = 2 \ 2 x3 P P C n n+1 I Identify C with {(1,u1,...,un): ui ∈ C }. The map n n ι : C ,→ P defined by (1,u1,...,un) 7→ `(1,u1,...,un) is an open n n embedding. Then, the lines in C are lines in P . I There is a natural chain of closed imbeddings, 1 2 n {pt} ,→ P ,→ P ,→ ··· ,→ P . Furthermore, this chain of subvarieties give a natural cellular decomposition, n n n−1 n n−1 1 P = C t P = C t C t ··· t C t {pt}. Summary: n n+1 I P is the set of 1-dimensional subspaces of C . n+1 n π : C \{0} −→ P (u0,...,un) 7−→ `(u0,...,un) I There is a natural chain of closed imbeddings, 1 2 n {pt} ,→ P ,→ P ,→ ··· ,→ P . Furthermore, this chain of subvarieties give a natural cellular decomposition, n n n−1 n n−1 1 P = C t P = C t C t ··· t C t {pt}. Summary: n n+1 I P is the set of 1-dimensional subspaces of C .