1. Affine Grassmannian for G a. Gr Ga = lim A n. Intuition. First some intuition. We always have to rst approximation

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1 PROBLEM SESSION I: THE AFFINE GRASSMANNIAN TONY FENG In this problem we are proving: 1 Affine Grassmannian for G a Gr Ga = lim A n n with A n A n+1 being the inclusion of a hyperplane in the obvious way Intuition First some intuition We always have to rst approximation In this case, this tells us that Gr G (k) = G(k((t)))/G(k[[t]]) Gr Ga (k) = k((t))/k[[t]] = lim t n k[[t]]/k[[t]] n Now t n k[[t]]/k[[t]] is a k-vector space on the coecients of t n,, t 1, which looks like A n (k) It's worth making clear why this is wrong To promote this to a proof, we would have to make the argument with k replaced by an arbitrary k-algebra R Although Gr G = LG/L + G, this does not mean that Gr G (R) = LG(R)/L + G(R) = G(R((t))/G(R[[t]]) Rather, the left side is the sheacation of the right side (eg for the étale topology) Of course, when R is ah algebraically closed eld then no such sheacation is necessary Onto the proof So what does Gr Ga look like anyway? By denition, its functor of point is given by R {(E, β)} where E is a G a -bundle over R[[t]] and β is a trivialization of E over R((t)) What is a G a -bundle anyway? When confused about such questions, we can always nd one answer by embedding into GL n, for instance as the unipotent group ( 1 t t 1) This makes it clear that a G a -bundle is the same as an extension of trivial bundles: 1 O S E O S 1 Date: September 29,

2 2 TONY FENG Now let's unravel what this tells us over S = Spec R[[t]] We're looking at pairs (E, β) where E is a rank 2 locally free modules M over R[[t]], with an extension structure 0 R[[t]] M R[[t]] 0 with a trivialization β over R((t)) Of course all such M are abstractly trivial, but the point is that β rigidies the situation Thanks to β, we may assume that M R[[t]] R((t)) = R((t)) 2, with the splitting being the coordinate functions This means that M itself is described by an R[[t]]-lattice inside R((t)) 2, which can be expressed as a matrix ( ) 1 f 1 This matrix is ambiguous up to change of basis that preserves the extension structure, which amounts to multiplication by a unipotent matrix over R[[t]] This means that f (which is a priori an element of R((t))) is well-dened up R[[t]], ie is determined by its class in R((t))/R[[t]] So we have concluded that indeed Gr Ga (R) = R((t))/R[[t]] = lim A n (R) 2 Affine Grassmannian for G m 21 The loop group First we are going to investigate the loop group for G m We have LG(R) = R((t)) We can canonically decompose the units as follows Let f = a i t i R((t)) Let n be the rst index such that a n is invertible (This obviously exists, since the image of f in R/N (R)((t)) had better be invertible) Multiplying by a 1 n t n, we can assume that this index is 0 and the corresponding coecients is 1 Then write f = n p where n is the polar part (negative powers of t, with nilpotent coecients) and p tr[[t]] By standard algebra, 1 + n is invertible (formally expand 1/(1 + n)) So f = (1 + n)(1 + p 1 + n ) = (1 + n)(1 + p(1 n + n2 )) We need to be careful, since the second factor need not have purely positive exponents of t Let f = (1 + p(1 n + n 2 )) = (n p ) We will argue that f is less bad in some sense than f, so that iterating this process eventually produces a factorization of the desired form Note that for i > 0, the coecient of t i in f is a combination of coecients for strictly less negative exponents of f, or polynomials in positive degree of coecients which are at least as negative Therefore, we can assign a score to Laurent series which measures the degree of nilpotency of the polar part Laurent series for which the negative coecients

3 PROBLEM SESSION I: THE AFFINE GRASSMANNIAN 3 are more nilpotent have lower scores We leave it to the reader to formalize this idea, and verify that the operation f f decreases the score, so that after a nite number of iterations the nilpotent part disappears n 1+p Since is also invertible (again, formally expand) So we have expressed f as a product of f tr[[t]] and f, a polar part with nilpotent coecients This is obviously unique The conclusion is that R((t)) can be uniquely expressed as has nilpotent coecients, 1 + n 1+p with r R and f +, f as above This gives a decomposition f = r t n f + f LG m = Z Gm W Ŵ where W is the subscheme parametrizing possibiliies for f +, and W is the indsubscheme parametrizing f 22 Some geometric properties We have just seen that LG m = Z Ŵ The functor W assigns to R the set of nite tuples (r i ) with each r i nilpotent This is the inductive limit of schemes W n,n parametrizing tuples (r 1,, r n ) with r n i = 0, which is represented by k[x 1,, x n ]/(x n 1, xn n), which is evidently non-reduced However, each W n is formally smooth To see this, it suces to check that W 1,n (R) = {r R: r n = 0} is formally smooth, ie if I is a nilpotent ideal in R then W 1,n (R) W 1,n (R/I) But this is clear: for any r W 1,n (R/I), choose any lift r of r in R 23 The arc group and ane Grassmannian From this argument we deduce in particular that the arc group is embedded as you expect, so L + G m = Gm W, Gr Gm = Z W Note that π 0 Gr Gm = Z, as expected! Note that the map LG m LG a is not an open embedding The nite layer subschemes of Ŵ are closed, cut out by the equations satised by the nilpotent elements of t n k[[t]]/k[[t]] 3 Relative positions 31 Dominant coweights in GL n Let's remind ourselves about the dominant coweights of GL n, and their natural ordering

4 4 TONY FENG Denition 31 A dominant coweight of GL n is a coweight that pairs non-negatively with any positive coroot If µ, µ are coweights of GL n, then we say that µ µ (sometimes denoted µ µ ) if µ µ is a sum of positive coroots Here's the picture: each root cuts out a hyperplane in the Euclidean space X (T ) R The connected components int the complement of these hyperplanes are cones called the Weyl chambers The Weyl group acts transitively on the set of Weyl chambers Let's unwind what this means for GL n We have of course X (T ) = Z n, with The roots are (m 1,, m n ) diag(t m 1,, t mn ) χ ij : diag(t m 1,, t mn ) t m i m j The standard positive roots, corresponding to the choice of upper-triangular Borel, are χ ij for i > j The positive coroots are then the tuples(m 1,, m n ) with m i = 1, m j = 1 for some pair i > j, and all other entries 0 The conclusion is that the order relation among the m i is exactly dened by: (m 1,, m n) > (m 1,, m n ) if you can trade down entries in i for entries in j, with i > j, to get from (m 1,, m n) to (m 1,, m n ) If one wants to write this out in gory detail: m 1 m 1 m 1 + m 2 m 1 + m 2 m m n = m m n 32 Loci of relative position Let X = Spec R, and E 1, E 2 two vector bundles of rank n over R[[t]], with an identication over R((t)) For each x Spec R, we get by restriction two vector bundle over κ(x)[[t]], which have some relative position β x Then we claim that X µ := {x X : β x µ} is closed TONY: [this is trickier than it seems TBC] 33 General groups For general G, for each dominant coweight µ we get a highest weight representation V µ of G This turns a principle G-bundle E into a vector bundle V E µ Then we can translate the notion of relative position for G into one for vector bundles (for each µ): Lemma 32 Gr G, µ (k) is the set of (E, β) such that for each λ, the map extends to a map β µ : V E µ V E µ k[[t]] k((t)) V µ k k((t)) t λ,µ V µ k k[[t]]

5 PROBLEM SESSION I: THE AFFINE GRASSMANNIAN 5 Indeed, by choosing trivializations we can represent β by an element g G(k((t))), which is well-dened up to left and right multiplication by G(k[[t]]) The corresponding element for the vector bundle Vµ E is an element of GL n (k((t))), which is the image of g by the dening map G GL(V µ ) The largest exponent which appears is then λ, µ so that the map does extend as claimed 4 Minuscule Schubert varieties Let us recall the meaning of minuscule weights We say that a dominant cocharacter µ of G is minuscule if µ 0 and for every positive root α, we have µ, α 1 Example 41 For GL n the dominant miniscule coweights are µ 1 =(1, 0,, 0) µ 2 =(1, 1, 0,, 0) µ n =(1, 1,, 1, 0) ν 1 =(0, 1,, 1) ν 2 =(0,, 0, 1,, 1) ν n =(0,, 0, 1) (Depending on your convention) The Schubert cell for µ i parametrizes lattices Λ with k[[t]] n Λ t 1 k[[t]] n with dim k Λ/k[[t]] n = i This is just the (normal!) Grassmannian Gr(i, n) The Schubert cell for ν i parametrizes lattices Λ k[[t]] n of corank 1, ie such that dim k k[[t]] n /Λ = is 1-dimensional over k (These are what we might have called elementary upper modications of type 1) Miniscule coweights are minimal for the ordering on dominant coweights, Gr µ = Gr µ for any such µ Therefore, Gr µ is a homogeneous space under L + G, hence is smooth Quasi-minuscule Schubert varieties A quasi-minuscule weight is a dominant coweight µ such that µ, α 2 for all positive roots α An example is the positive coroot of SL 2 Let's examine this case We're supposed to see that the corresponding (closed) Schubert variety is the projective cone over a quadric curve in P 2 The quasi-minuscule weight of SL 2 is the positive coroot ( ) t µ : t t 1

6 6 TONY FENG Therefore, the Schubert variety parametrizes lattices Λ such that which is commensurate with O, so tk[[t]] 2 Λ t 1 k[[t]] 2 dim k (Λ/tk[[t]] 2 ) = 2 and dim k (t 1 k[[t]] 2 /Λ) = 2 Note that Λ is determined by its image Λ in tk[[t]] 2 The ltration denes a ltration on Λ Consider tk[[t]] 2 k[[t]] t 1 k[[t]] 2 Λ k[[t]] 2 k[[t]] 2 /tk[[t]] 2 Its image is 1-dimensional, so the possibilities for this image are parametrized by a P 1 The image of Λ in t 1 k[[t]] 2 /k[[t]] 2 is also 1-dimensional, and together with the choice of the rst step of the ltration determines Λ However, one line is excluded as a possible image, namely the one corresponding to the rst choice under the isomorphism k[[t]] 2 /tk[[t]] 2 t 1 k[[t]] 2 /k[[t]] 2 via multiplication by t