These transparencies are available athttp://math.cornell.edu/ allenk/ Modern developments in Schubert calculus Allen Knutson (Cornell) Winston-Salem, North Carolina AMS conference, September Abstract Schubert calculus begins with the study of incidence conditions on k- planes. I ll recall its utility in eigenvalue inequalities, and describe many natural generalizations, most of which are unsolved. I ll run down the progress that has been made in the st century, and then give combinatorial rules for many of these problems, in terms of counting puzzles made from various puzzle pieces. K
These transparencies are available athttp://math.cornell.edu/ allenk/ Morse theory on the space of projections. LetGr k (C n ) = {k-planes inc n } = Hermitian matrices unitarily equivalent to... be the Grassmannian, bearing the functions ρ H : Gr k (C n ) R, π Tr(πH). Question. What does the Morse decomposition using ρ H look like, forh = diag(n,n,n,...,)? (Here s the standard picture of the Morse decomposition of a torus, where the function used is the height function. The four critical points from top to bottom have a point, R, R, and R gradient-flowing down into them, respectively.)
These transparencies are available athttp://math.cornell.edu/ allenk/ The Bruhatcells X λ,andthe Schubertvarieties X λ. Question. What does the Morse decomposition ofgr k (C n ) usingρ H look like, forh = diag(n,n,n,...,)? Answer. There are ( n k) critical points, the diagonal matricesdiag(λ) whereλhas k s and n k s, and for eachλthe stratum X λ (H) flowing down into λ is a cell: X λ(h) = C #{s before s inλ} Which stratum is a projectionπin? IfV = image(π), take V (V C,V C,V C,...,V C n ) = C k(n k) l(λ) jumps ( [n] So H doesn t matter, just its associated flag of partial sums of eigenspaces does. k ). Corollary[Hersch-Zwahlen,96]. Ifπ X λ (H e ) = X λ (H e), wheree e... e n areh e s eigenvalues, then ρ H (π) e λ +...+e λk, with equality iffimage(π) is a sum of those eigenlines.
These transparencies are available athttp://math.cornell.edu/ allenk/ 3 Inequalities on spectra of sums of Hermitian matrices. Question 3. For whichλ,µ,ν ( ) [n] k is Xλ (H e ) X µ (H f ) X ν (H g ) for allh e,h f,h g? Ifπis in that intersection, and H g = H e +H f, then = Tr(π(H e +H f +H g )) = Tr(πH e )+Tr(πH f )+Tr(πH g ) = ρ He (π)+ρ Hf (π)+ρ Hg (π) e λ +...+e λk +f µ +...+f µk +g ν +...+g νk with equality iffh e,h f,h g all commute withπ, so, can be simultaneously block diagonalized [Totaro 994, Helmke-Rosenthal 995, Klyachko 998]. Moreover, if the intersection is positive-dimensional one can tighten up λ, µ, ν (move s back, s forward) to get a stronger inequality. The inequalities constructed this way give the only conditions on g [Klyachko 998]; one only needs the ones for which the intersection is a single point [Belkale 999]; and one does indeed need all of those [Knutson-Tao-Woodward 4]. So when is the intersection a nonempty set of points (where each point corresponds to ak-plane)? We expect points whenl(λ)+l(µ)+l(ν) = k(n k).
These transparencies are available athttp://math.cornell.edu/ allenk/ 4 Puzzles. Consider the following three puzzle pieces, with edges labeled by s and s. They may be rotated but not reflected (unless and are exchanged). Theorem [K-Tao-Woodward 4]. If l(λ)+l(µ)+l(λ) = k(n k), andx λ (H e ) X µ (H f ) X ν (H g ) has dimension(as expected), then its cardinality (counted with multiplicities) is the number of puzzles with λ, µ, ν clockwise around the outside. Otherwise there are no such puzzles. λ ν µ Fun fact: and appear inside!
These transparencies are available athttp://math.cornell.edu/ allenk/ 5 The cohomology ring of the Grassmannian. Question 3. What doesx λ (H) X µ (H ) look like? Example. Letλ = µ =, sox λ (H) is the space ofcp s touching the projective line L made from the top two eigenlines ofh. Likewise definel fromh. IfL = L, then X λ (H) = X µ (H ) is 3-dimensional. IfL L =, then X λ (H) X µ (H ) = L L, so-dimensional. IfL L = p, then X (H) X (H ) = X (H ) X (H ), union alongx (H ). Theorem. If the codimensions add, the homology class of the result is well-defined. The Poincaré dual classes [X λ ] give a basis of cohomology, and puzzles compute the structure constants in that basis: [X λ ][X µ ] = (#puzzles withνon bottom)[xν ]. λ µ λ µ ν ν ν left to right this time! Even before actually computing these structure constants, one can prove using algebraic geometry that they must be positive. (This uses the homogeneity of the Grassmannian it s not true for the blowup ofcp, for example.)
These transparencies are available athttp://math.cornell.edu/ allenk/ 6 Schubertcalculusand 5 generalizations. We ve turned the question into computing the cohomology ring of the Grassmannian, in the Schubert basis. The Schubert varieties {X λ } give bases of other cohomology theories, and for other homogeneous spaces, suggesting many generalizations: K: K-theory, where[x ] = [X ]+[X ] [X ]. T: Torus-equivariant cohomology, where[x ] = (y y )[X ]. This has coefficients inz[y,...,y n ]. Q: Quantum cohomology, where[x ] = q[x ], with coefficients in Z[q]. F: Larger flag manifoldsgl n (C)/P, isomorphic to spaces of Hermitian matrices with more different eigenvalues. G: (Co)minuscule flag manifolds for other groups, like the Lagrangian Grassmannian. For each and every combination, we can ask to prove Abstract positivity or a Computational rule. (As the K-example shows, even defining what positivity to expect can be subtle.) Note that non-manifestly-positive computational rules are known for every one of these problems, so checking conjectures in small examples is easy.
These transparencies are available athttp://math.cornell.edu/ allenk/ 7 Recent progress on Abstract and Computational positivity in Schubert calculus. K-theory C [Buch ] QG C reduces tofgc [Chaput-Manivel- Perrin 8] KTQG C reduces toktfgc [Buch-Mihalcea ] H (Gr k (C n )) C T-equivariant A see TFG C [K-Tao 3] KFG A [Brion ] [Littlewood-Richardson 934, Thomas 974, Schützenberger 977] Quantum A [Mihalcea 6] C reduces to FC [Buch-Kresch- Tamvakis 3] Flag A see FG C [Coşkun?] FG A [Kleiman 973] TFG A [Graham ] KTFG A [Anderson-Griffeth-Miller ] Groups A see FG C [Thomas-Yong 9] (extending [Pragacz 99, Worley 984] The [BKT] result says that quantum cohomology of Grassmannians can be positively computed inside ordinary cohomology of-step flag manifolds. Work by [Belkale-Kumar 6] and [Ressayre ] shows that for eigenvalue inequalities, what is really relevant is a subproblem of FG called the Belkale- Kumarproduct, in which many coefficients of the actual product are set to.
Puzzle rules for many of these. We already gave the three puzzle pieces used to compute H (Gr k (C n )). For K-theory, one new piece is needed. It increases l(ν) (l(λ)+l(µ)) by, and contributes a factor of. It may not be rotated. K For H T, a backwards vertical rhombus called the equivariant piece is needed. It decreases l(ν) (l(λ)+l(µ)) by, and contributes a factor of y i y j that depends on its location. y y + y y = 3 y y 3 Equivariant K-theory is unsolved, but I ll be talking about a closely related puzzle-solvable problem tomorrow.
These transparencies are available athttp://math.cornell.edu/ allenk/ 9 Puzzlerules for many of these,][. The Belkale-Kumar product on a d-step flag manifold needs puzzle pieces with edge labels,..., d, that otherwise look like the three used in H (Gr k (C n )). Interestingly, one can correspond such a puzzle to a tuple of ( ) d+ ordinary puzzles. [K-Purbhoo] There is no (living) conjecture for puzzles to compute H of flags in general, but there is one for -step flag manifolds, which I gave in 999 and expect to be proven soon. In addition to the 3 + ( 3 ) pieces above, there are two extensible kinds shown at right. This rule was checked by Buch, Kresch, and Tamvakis up to n = 6, which is especially impressive in that the nonpositive rules are only calculable by computer up to aboutn = 9 in practice. Recall that -step flag manifold Schubert calculus includes quantum Schubert calculus of Grassmannians, itself equivalent to tensor products of U q (gl n )- representations at q a root of unity, or fusion products of affinegl n reps.