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# Top Algebraic Geom II MATH 683

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This 6 page Class Notes was uploaded by Henderson Lind II on Tuesday September 8, 2015. The Class Notes belongs to MATH 683 at University of Oregon taught by Victor Ostrik in Fall. Since its upload, it has received 49 views. For similar materials see /class/187185/math-683-university-of-oregon in Mathematics (M) at University of Oregon.

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Date Created: 09/08/15

ACTIONS OF ALGEBRAIC GROUPS7 I l EXAMPLES OF ACTIONS 11 De nitions Let G be an affine algebraic group with unit 6 E G and mul tiplication morphism m G X G A G De nition 11 An action of G on an algebraic variety X is a regular morphism a G X X A X such that ae7 I z for all z E X and the diagram commutes GxGxXLGxX idgtlta a Gxx x De nition 12 Let G acts on variety X An orbit is a subset GI C X where It is easy to see that the orbits form a disjoint partition of the variety X Basic Problem Given an action classify the orbits 12 Examples Base eld k is algebraically closed of characteristic zero Example 0 GLn acts on the vector space V 16 There are two orbits here Example 1 GLn acts on symmetric n X n matrices via A gt gt gAg The orbits are in bijection with symmetric bilinear forms on space of dimension n Exercise 1 How many orbits are in example 1 Example 2 GL7 acts on skew symmetric n X n matrices via A gt gt gAg The orbits are in bijection with skew symmetric bilinear forms on space of dimension n Exercise 2 How many orbits are in examp e 2 Example 3 GLn gtlt GLm acts on Matnxm via A gt gt gAh l The orbits are in bijection with linear maps from a space of dimension n to a space of dimension In Exercise 3 How many orbits are in example 3 Example 4 GL7 acts on Matnxn by conjugation A gt gt gAg l The orbits are in bijection with linear maps from a space V of dimension n to itself Exercise 4 What are the orbits in example 4 Example 5 GLn acts on triples EFH E Matnxn satisfying the equation H7 E 2E7 HF 72F EF H by conjugation EF7 G gt gt gEg 1gFg 1gHg 1 Exercise 5 Classify the orbits in examp e 5 Example 6 GLV acts on V V V HomV V V in obvious way The orbits are in bijection with algebras possibly nonassociative of dimension dimV The subset of associative commutative7 Lie etc algebras is invariant under this action It is quite difficult to classify orbits even if dimV 2 Example 7 GLV acts on V V The orbits are in bijection with bilinear forms on space of dimension dimV The classification of orbits in general is not known to me The classification is known at least in the case when form is non degenerate 2 ACTIONS OF ALGEBRAIC GROUPS I Example 8 OV acts on V Exercise 6 Classify the orbits in example 8 Example 9 SpV acts on V Exercise 7 Classify the orbits in example 9 Example 10 0V acts on 520 The orbits are in bijection with pairs of quadratic forms one of which is nondegenerate Example 11 PGLnH acts on 1 Exercise 8 a Classify orbits of PGLnH on P X P b Classify orbits of PGLn1 on P X P X P Example 12 Let V be a vector space of dimension 2 Then G PGLV acts on lP S V The orbits are in bijection with effective divisors of degree n on Example 13 GLn acts on the ag space Fln points of Fln are ags 0 F0 C F1 C C Fn ls with dimFl There is only one orbit here Bruhat decomposition says that the orbits of GLn on Fln gtlt Fln are in bijection with the symmetric group Sn More generally7 let G be a reductive group with ag space GB Bruhat decomposition says that Giorbits on GB gtlt GB orbits of B on GB are in bijection with the Weyl group W 13 Basic result on orbits Recall that a subset of algebraic variety is locally closed if it is open in its Zariski closure The following fundamental result is a simple consequence of Chevalley theorem Theorem 13 An orbit for an action of algebraic group is locally closed Remark 14 The theorem does not hold if we replace algebraic groups by Lie groups Counterexample the group R acts on two dimensional torus lRQZ2 via za7b azb z Notice that the Zariski closure E of an orbit G1 is invariant under the action of G thus it is a union of GI and orbits of smaller dimension Corollary 15 a the dimension of an orbit equals to the dimension of its Zariski c osure b any action admits a closed orbit Proof a is a general property of locally closed subsets for b take any orbit of minimal possible dimension D We introduce the following order relation on the set of orbits 01 S 02 if and only if 01 is contained in 62 Exercise 9 Find the closure relation for orbits in exercises 18 Any orbit is a homogeneous space for G Hence we have Lemma 16 For any Giorbit O we have dimO S dimG 2 QUIVERS AND PROBLEMS OF LINEAR ALGEBRA 21 Examples of problems of linear algebra lnformally7 a problem of linear algebra is a problem of classi cation of situations involving nite dimensional vector spaces and operators between them we will give a formal de nition later ACTIONS OF ALGEBRAIC GROUPS I 3 Example 21 Classify linear operators f V A W up to isomorphism Here f 39 V A W is isomorphic to V A W if there are isomorphisms v V A V and w W A W such that the diagram commutes f V gt W In this case the situation is completely described by 3 integers dimV7 dimW and dimfV Example 22 Classify vector spaces V together with two subspaces W1 C V and W2 C V up to isomorphism Here triple VW1W2 is isomorphic to triple V W1W2 if there is an isomorphism v V A V such that vW1 W1 and vW2 In this case the situation is completely described by 4 integers dimV7 dimW17 dimW2 and dimW1 W2 Example 23 Classify linear operators f V A V up to isomorphism Here f V A V is isomorphic to f V A V if there is isomorphism v V A V such that the diagram commutes VV In this case the situation is completely described by the theory of Jordan normal form Thus it is still manageable but continuous parameters are required Example 24 Classify vector spaces V together with four subspaces Wi C V7 i 127 37 4 up to isomorphism In this case the situation we have some continuous parameters for example in special case when dimV 2 and dimWZ 1 the problem reduces to classi ca tion of 4 points on the projective line which are classi ed by double ratio Questions a What about three subspaces b ls this problem manageable in the sense of Example 23 Example 25 Classify pairs of linear operators g V A V up to isomorphism This problem is generally considered as completely hopeless as nobody was able to give a reasonable solution to this lnformally7 problem of linear algebra is called nite if the situation is completely described by discrete parameters7 wild if it contains as a subproblem Example 25 and tame if it is not wild or nite we will give more formal de nition later 22 Quivers and associated problems of linear algebra By de nition a quiver is just a nite directed graph where edges joining vertex with itself are allowed Here is a formal de nition De nition 26 A quiver Q is a quadruple 1 E7 8 t where I and E are nite sets and st E A I are given maps 4 ACTIONS OF ALGEBRAIC GROUPS I With any quiver Q IEst we associate the following problem of linear algebra Problem 1 Classify all collections of nite dimensional vector spaces E I and linear operators f8 V5 8 A V48 e E E up to isomorphism Here the collection Vi7 f8 is isomorphic to collection Vl7 if there are isomorphisms vi Vi A Vi such that vtefe f v5e for all e E E De nition 27 A collection Vhae as above is called a representation of quiver Q The vector dimVZi61 E ZIgt0 is called dimension of such representation Example 28 Let Sm be a quiver with one vertex and m loops joining it with itself The corresponding problem of linear algebra is to classify mituples of linear operators from a vector space to itself In particular7 the quiver 51 corresponds to Example 23 and the quiver 52 corresponds to Example 25 Example 29 Let Pm be the quiver with two vertices and m arrows from the rst vertex to the second one The corresponding problem of linear algebra is to classify mituples of linear operators from a vector space to another one Example 210 Let Vm be the quiver with ml vertices 01 7m and m arrows l A 07 2 A 07 m A 0 The corresponding problem of linear algebra is essentially to classify mituples of linear subspaces of a given vector space Here is a reformulation of the problem above in terms of actions of algebraic groups Let us x vector spaces Vi what is the same7 let us the dimension vector 1 dimVi E Z o Consider the group G Gd Hi6 CLOi and the vector space H Hd GlingHomV5e7 V463 The group G acts on the space H in the following way 90 39 feeeE 95efegeleEE Problem 1 is equivalent to the following Problem 2 For any 1 E Z120 classify the orbits of Gd on Hd Remark 211 The action of Gd on Hd is not effective Namely7 the normal subgroup AQm Aidm g acts trivially on Hd Exercise 7 Prove that the group GdAQm acts effectively on Hd 23 Finite7 tame and wild De nition 212 a A quiver is called nite if for any 1 E Z120 the group Gd acts on Hd with nitely many orbits b A quiver is called wild if the corresponding problem of linear algebra contains the problem 52 as a subproblem c A quiver is called tame if it is neither nite nor wild Remark 213 One shows that for an action of linear algebraic group on an af ne variety we have either nitely many orbits7 or family of orbits parametrized by some algebraic variety we say that we have 77continuous parameters77 in the second case Thus in this situation discrete nite ACTIONS OF ALGEBRAIC GROUPS I 5 24 Quadratic form and potentially nite graphs Let Q IEst be a quiver Consider the following matrix indexed by the set 7 e e Else we 7 j e e Else Me 7 239 Notice that the matrix aij is symmetric and depends only on the underlying undi rected graph of the quiver For every i E I we introduce the formal variable 11 and consider the following quadratic form 1 41 i ai r 16 116 Example 214 For quiver Sm the form 41 l 7 m12 for quiver Pm the form 41 1 1 7 7211112 for quiver Vm the form 41 20 11 7 10 11 De nition 215 We say that a quiver or underlying undirected graph is positive semi de nite if the form 41 considered as a form over R is We say that a quiver or underlying undirected graph is negative if it is neither positive de nite nor positive semi de nite Example 216 a Quiver Sm is positive de nite for m 07 positive semi de nite for m 1 negative for m 2 2 b Quiver Pm s positive de nite for m 17 positive semi de nite for m 2 negative for m 2 c Quiver Vm s positive de nite for m S 3 positive semi de nite for m 4 negative for m 2 5 Proposition 217 Assume that Q is a nite quiver Then Q is positive de nite Proof For any 1 E ZIgt0 the group Gd should have an open orbit on Hd in deed Hd U0 0 U0 6 since Hd is irreducible Hd 6 for some 0 Hence dide dimO S dimGd 7 1 see Remark 211 and dimGd 7 dide gt 0 Crucial observation is that dim Gd 7 dide 4d It follows that 4 is positive de nite B Our goal now is to prove converse any positive de nite graph is nite 25 Classi cation of positive semi de nite graphs In what follows a sub graph means full subgraph subset of vertices and all edges joining them All graphs are supposed to be connected Lemma 218 Assume a graph contains a negative subgraph Then it is negative Thus we see that a positive semi de nite subgraphs does not contain vertices with more than one selfloops and no two vertices are joined by more than two edges De nition 219 We say that a graph is balanced if there is vector 1 1iie 6 Rio such that 41 0 The following Lemma is crucial Lemma 220 Assume a graph contains a proper balanced subgraph Then it is negative 6 ACTIONS OF ALGEBRAIC GROUPS I Proof Let J be the set of vertices of balanced subgraph and let 0 g J be another vertex which is joined with at least one vertex from J Let Ii e be a vector as in de nition of balanced graphi Consider the following vector yi Ii ifi 6 J7 yo IO and yi 0 ifi g J U Clearly qy 17 mzg 7 IO edgesjommgolmdiej 11 where m is the number of selfloops in vertex 0 Clearly this expression is negative for suf ciently small positive 10 Notice that 51 and P2 are balanced Thus if a positive semi de nite graph contains a selfloop it should be equal to 51 if it contains a multiple edge it should be equal to Pgi Thus from now on we consider only graphs without selfloops and multiple edgesi Lemma 221 Assume that vector 1 6 Rio has the following property 21139 ijoinedwithizj Then 41 0 and our graph is balanced Exercise Prove this Lemma Now we can come up with examples of balanced graphs An7 Dn7 E57 E77 Egi Exercise Check that all these graphs are positive semi de nite

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