Invariants under simultaneous conjugation of SL 2 matrices

Transcription

1 Invariants under simultaneous conjugation of SL 2 matrices Master's colloquium, 4 November 2009

2 Outline 1 The problem 2 Classical Invariant Theory 3 Geometric Invariant Theory 4 Representation Theory 5 Conclusion

3 The Problem The problem Consider the space of m-tuples of matrices with determinant 1 (SL 2 matrices. Forget the basis: for any Q SL 2, (M 1,..., M m (QM 1 Q 1,..., QM m Q 1. We want to know the orbit space: {(M 1,..., M m SL m 2 }/. This space arises in monodromy of order 2 ordinary dierential equations with m + 1 singularities

4 Approaches Approach 1: (Classical Invariant Theory Well-known fact: trace, determinant of matrix are invariant under conjugation, so they tell orbits apart Invariant theory: classify these functions Notation: C[SL m 2 ] are polynomial functions on SL 2... SL 2 Notation: C[SL m 2 ]SL 2 are polynomial functions invariant under conjugation We calculated the structure of C[SL m 2 ]SL 2 using some classical results

5 Approaches Approach 2: Geometric Invariant Theory Assign geometric meaning to X = {(M 1,..., M m SL m 2 }/ in projective space Want: Y that separates the orbits as much as possible: every map ψ that is constant on orbits runs through Y : X φ Y ψ θ Z, (the universal mapping property This can be done for (semi-stable points

6 Approaches Approach 3: Representation Theory More abstractly, the action of SL 2 on the set of matrices can be seen as a representation of SL 2 Idea: forget the specic space SL 2 acts on Question: how does GIT work on representations? Question: does this correspond to our earlier results?

7 Approaches Summary of approaches Want to study: {(M 1,..., M m SL m 2 }/. Approach 1: CIT: Look at invariant function C[SL m 2 ]SL 2 Approach 2: GIT: Construct a quotient map with the universal mapping property Approach 3: RT: How does GIT work on representations?

8 Two matrices A theorem (I Let M 0 be the vector space of traceless matrices. Theorem C[M 0 M 0 ] SL 2 = C[TrM 2 1, TrM2 2, TrM 1M 2 ]. Sketch of proof Let f be an invariant function. Restrict it to the pairs {(( ( } t a 1, a, t 0, t c a so we have a function f C[a, c, t]. Almost any matrix pair can be conjugated to this form.

9 Two matrices A theorem (II Sketch of proof, cont'd Note that (( t t ( a 1, c a But then f C[c, a 2, t 2, at]. Now (( t t ( a 1, c a. t 2 = 1 2 TrM2 1, at = 1 2 TrM 1M 2, etcetera so we have found the invariant function that we restricted, which is a polynomial in TrM 1 2, TrM2 2, TrM 1M 2.

10 Two matrices Traceless, general matrices (I Note that for M SL 2 : ( a M i b i M 1 c i d i = M But then for functions: ( 12 a i 2 1 d i b i 1 c i 2 d i 1 2 a i M 1 + ( 12 a i d i a i d i C[SL 2 SL 2 ] SL 2 = (C[M1, M 2 ] SL 2 C[t 1, t 2 ]/(detm i t2 i = 1. So: we translate our results on traceless matrices

11 Two matrices Traceless, general matrices (II C[SL 2 SL 2 ] SL 2 = (C[M1, M 2 ] SL 2 C[t 1, t 2 ]/(detm i t2 i = 1 = C[TrM 2 1,TrM 2 2,TrM 1 M 2, t 1, t 2 ]/(... Correspondence: ( ( ( 12 a i b i a c i d i SL 2 i 1 2 d i b i Say (X 1, X 2 SL 2 2, then c i 1 2 d i 1 2 a i, 1 2 (a i + d i t i = 1 2 (a i + d i 1 2 TrX i TrM i M j = ( 1 2 a i 1 2 d i ( 1 2 a j 1 2 d j +... TrX i X j 1 2 TrX i TrX j

12 Two matrices Traceless, general matrices (III We are translating C[TrM 2 1,TrM2 2,TrM 1M 2, t 1, t 2 ]: t i = 1 2 (a i + d i TrX i TrM i M j = ( 1 2 a i 1 2 d i ( 1 2 a j 1 2 d j +... TrX i X j 1 2 TrX i TrX j. But for (X 1, X 2 SL 2 2 : TrX 2 i = a 2 i + d 2 i + 2b i c i = a 2 i + d 2 i + 2a i d i 2a i d i + 2 b i c i = (a i + d i 2 2 = (TrX i 2 2. Corollary C[SL 2 SL 2 ] SL 2 = C[TrX 1, TrX 2, TrX 1 X 2 ].

13 Three matrices & general case The general case In [Drensky2000], the complete structure of C[M k 0 ] is calculated. Proof uses invariant theory of SO 3 known in 1947 We wrote it down more clearly, and calculated C[SL m 2 ]SL 2 for any m in the same way as previously

14 Three matrices & general case The invariants for 3 SL 2 matrices Theorem Let a = Trx 1, b = Trx 2, c = Trx 3, d = Trx 1 x 2, e = Trx 1 x 3, f = Trx 2 x 3, g = Trx 1 x 2 x 3 Trx 1 x 3 x 2. Then C[SL 3 2 ]SL 2 = C[a, b, c, d,e, f, g]/(rel, where rel = g 2 + 4a 2 + 4b 2 + 4c 2 + 4d 2 + 4e 2 + 4f 2 +2a 2 bcf + 2abc 2 d + 2ab 2 ce 4ace 4abd 4bcf + 4def b 2 e 2 a 2 f 2 c 2 d 2 2bcde 2acdf 2abef a 2 b 2 c Or: C[SL 3 2 ]SL 2 = C[a, b, c, d,e, f ] + C[a, b, c, d,e, f ] g

15 Embedding into projective space Projective space: an introduction Consider the space R { }. Denote a point by (a : b, where (a : b = (λ a : λ b x R corresponds to (x : 1 corresponds to (1 : 0 (0 : 0 is excluded Geometry on projective spaces is much more elegant: in the projective plane every two lines intersect and other nice properties. Generalize to: P n = {(x 1 :... : x n+1 C n+1 }\(0 :... : 0 /(x1 :...:x n+1 =(λ x 1 :...:λ x n+1.

16 Embedding into projective space Embedding SL 2 in a projective space Obvious choice: add determinant as extra coordinate: let Q = {(a : b : c : d : P 4 ad bc = 2 }, and dene embedding ( a b c d φ (a : b : c : d : 1 = (( a b c d : 1. The conjugation can be extended to the whole of Q: conjugating an element by M SL 2 gives: ( M ( a b c d M 1 :.

17 Embedding into projective space Embedding SL m 2 in a projective space We can send tuples of matrices to tuples of Q's, e.g.: (( ( (( ( a 1 b 1 a c 1 d 1, 2 b 2 a c 2 d 2 1 b 1 a c 1 d 1 : 1, 2 b 2 c 2 d 2 : 2. Question: how is this a projective space? Answer: by considering this as an embedding into P 24 : Q Q P (( ( 24 a 1 b 1 a c 1 d 1 : 1, 2 b 2 c 2 d 2 : 2 (a 1 a 2 : a 1 b 2 : a 1 c 2 : a 1 d 2 : a 1 2 :... : 1 a 2 : 1 b 2 : 1 c 2 : 1 d 2 : 1 2.

18 A good quotient A good quotient Denitions A categorical quotient is (Y,φ that separates the orbits as much as possible: every map ψ that is constant on orbits runs through Y : Q Q φ Y ψ θ Z, Some additional properties: good quotient. If the quotient is good and all orbits go to dierent points, then it is called a geometric quotient.

19 A good quotient Constructing the quotient For ane spaces, the map induced by the algebra of invariants is a good quotient. For example: X = SL 2 SL 2, Y = C 3 : Idea: write φ(m 1, M 2 = (TrM 1,TrM 2,TrM 1 M 2. Q Q = U α : α A a cover by open, ane, dense, SL 2 -stable subsets. For example, {(( ( } a 1 b 1 a c 1 d 1 : 1, 2 b 2 c 2 d 2 : = SL 2 2. If we glue together these quotients, we get a new quotient.

20 A good quotient The ane subset SL 2 SL 2 SL 2 SL 2 is the ane subset where via: (( ( (( ( a 1 b 1 a c 1 d 1, 2 b 2 a c 2 d 2 1 b 1 a c 1 d 1 : 1, 2 b 2 c 2 d 2 : 1. Write invariants of SL 2 SL 2 as functions on Q Q: TrM 1 2TrM 1 1 2, TrM 2 TrM ; TrM 1 M 2 TrM 1M So a good quotient is Y = {(a : b : c : d P 3 d 0}, φ : ((M 1 : 1,(M 2 : 2 ( 2 M 1 : TrM 2 1 : TrM 1 M 2 : 1 2.

21 A good quotient Another subset Subset of Q Q where (a 1 + d corresponds to M 1 SL 2 /(a 1 a 2 1 b 1 c 1 = z 2 1, a 2 d 2 b 2 c 2 = 1 via (( a 1 b 1 c 1 1 a 1,z 1, ( a 2 b 2 c 2 d 2 (( a 1 b 1 c 1 1 a 1 : z 1, ( a 2 b 2 c 2 d 2 : 1. In the classical way: C[M 1 SL 2 ] = C[z 1,TrM 2,TrM 1 M 2 ]. So a good quotient is Y = {(a : b : c : d P 3 d 0}, φ : ((M 1 : 1,(M 2 : 2 (TrM 1 TrM 2 : TrM 1 M 2 : 1 2 : TrM 1 2

22 A good quotient Glueing a quotient Setting 1 2 0, TrM 1 2 0, TrM 2 1 0, TrM 1 M 2 0, we get quotients mapping ((M 1 : 1,(M 2 : 2 to: (TrM 1 2 : TrM 2 1 : TrM 1 M 2 : 1 2 ; (TrM 1 TrM 2 : TrM 1 M 2 : 1 2 : TrM 1 2 ; (TrM 1 TrM 2 : TrM 1 M 2 : 1 2 : TrM 2 1 ; (TrM 2 1 : TrM 1 2 : TrM 1 M 2 : TrM 1 TrM 2. Combination Y = {(a : b : c : d : e ab = cd }\{(0 :... : 0 : 1}; φ maps ((M 1 : 1,(M 2 : 2 to (TrM 1 2 : TrM 2 1 : TrM 1 TrM 2 : 1 2 : TrM 1 M 2.

23 A good quotient Glueing a quotient (II LetY = {(a : b : c : d : e ab = cd }\{(0 :... : 0 : 1}; Is (Y,φ a quotient, where φ maps ((M 1 : 1,(M 2 : 2 to Theorem (TrM 1 2 : TrM 2 1 : TrM 1 TrM 2 : 1 2 : TrM 1 M 2? (Y,φ is a good quotient for the set (Q Q nn of non-nilpotent matrices. Proof. One checks: for all ane parts A, φ restricted to A is a good quotient. Can we do better than this?

24 Semi-stable and stable points Semi-stable and stable points Mumford's Geometric Invariant Theory: for what points does a good quotient exist? Denitions Let V be projective. Then x V is semi-stable if there exists a homogeneous SL 2 -invariant polynomial with strictly positive degree which does not vanish at x; x is stable if the orbit of x in the underlying ane space is closed with maximal dimension. Then: Theorem There exists a good quotient φ : X ss Y and φ restricted to the stable points is a geometric quotient.

25 Semi-stable and stable points Mumford's criterion Theorem (Mumford's criterion for SL 2 (1 Find a basis for V so that {( λ λ 1 } λ C acts on basis elements of V as λ v i = c i λ w i vi (2 A basis vector v i is said to have weight w i (3 Up to change of basis, points that are not semi-stable are spanned by positive weight vectors (4 Up to change of basis, points that are not stable are spanned by non-negative weight vectors

26 Semi-stable and stable points Mumford's criterion for one matrix Calculate action: Weights: (( a b c d ( λ 0 0 λ 1 acts as follows: : (( a λ 2 b λ 2 c d : weight( = 2; weight( = 2; weight( ,( , = 0 So: non-semi-stable points are in span of ( : nilpotent matrices All points are non-stable!

27 Semi-stable and stable points Mumford's criterion for two matrices (I The action of diag(λ,λ 1 gives as result: (( ( : 1, a 1 λ 2 b 1 a 2 λ 2 b 2 : 2. Recall: basis vectors (in P 24 were a 1 a 2, a 1 b 2,..., 1 2. Theorem A pair ((M 1 : 1,(M 2, 2 is not semi-stable i 1 matrix is nilpotent and the pair is reducible, i.e., can be conjugated to: (( 0 (, 0.

28 Semi-stable and stable points Mumford's criterion for two matrices (II The action of diag(λ,λ 1 gives as result: (( ( : 1, a 1 λ 2 b 1 a 2 λ 2 b 2 : 2. Recall: basis vectors (in P 24 were a 1 a 2, a 1 b 2,..., 1 2. Theorem A pair ((M 1 : 1,(M 2, 2 is not stable i 1 matrix is nilpotent or the pair is reducible, i.e., can be conjugated to: (( 0 (, 0.

29 Semi-stable and stable points Our quotient for Q Q Recall: We have a good quotient for (Q Q nn : (Y,φ where Y = {(a : b : c : d : e ab = cd}\{(0 : 0 : 0 : 0 : 1}, φ maps ((M 1 : 1,(M 2 : 2 to (TrM 1 2 : 1 TrM 2 : TrM 1 TrM 2 : 1 2 : TrM 1 M 2. Note that (Q Q ss (Q Q nn (Q Q s φ is a well-dened map (Q Q ss {(a : b : c : d : e ab = cd}. Is it a good quotient? Theorem φ is a geometric quotient from (Q Q s to {(x 1 : x 2 : x 3 : x 4 : x 5 x 1 x 2 = x 3 x 4, x x x 2 5 x 3 x 5 4x 2 4 0}.

30 Semi-stable and stable points More than 2 matrices Ane cover of (Q... Q nn is always possible Stable, semi-stable points can always be classied Image of (Q... Q nn : harder to determine Proving it is a geometic quotient for (Q... Q s?

31 The representation theory of SL2 Representation theory Let a group G act linearly on a vector space V : g v V. Determine possible actions up to isomorphism: forget the space V For good G, one can uniquely write V i irreducible V = V 1 V 2... V n, Classify irreducible representations Apply Mumford's criterion

32 The representation theory of SL2 Representation theory of SL 2 Let [k] = {f C[x 1, x 2 ] f is homogeneous of degree k} of dimension k + 1. For example, [2] = span{x 2 1, x 1x 2, x 2 2 }. SL 2 acts on this as follows: ( a b c d f (x1, x 2 = f (a x 1 + c x 2, b x 1 + d x 2, so ( a b c d x 2 1 = a 2 x ac x 1 x 2 + c 2 x 2 2 ; ( a b c d x1 x 2 = ab x (ad + bc x 1 x 2 + cd x 2 2 ; ( a b c d x 2 2 = b 2 x bd x 1 x 2 + d 2 x 2 2. Then [k] is the unique irreducible representation of dimension k + 1

33 The representation theory of SL2 Mumford's criterion for [k] ( λ 0 Look at weights of action of 0 λ 1 : ( λ 0 0 λ 1 f (x 1, x 2 = f (λ x 1,λ 1 x 2. So the weight of a basis vector x1 k x 2 l is k l. So: if we know the structure of Q Q as a representation, the problem is really simple!

34 Representation theory of Q... Q Representation theory of Q Fact M 0 = [2] as representations. Q was (( embedded in a 5-dimensional vector space: a b V =, c d Choose basis {( ( ( ( 0, ,, }, Theorem V = [2] + 2 [0] as representations.

35 Representation theory of Q... Q Representation theory of Q Q Theorem Let V = ([2] + 2 [0] ([2] + 2 [0], then V = [4] + 5 [2] + 5 [0]. Fact [2] [2] = [4] + [2] + [0] as SL 2 -representations. Proof. ([2] + 2 [0] 2 = [2] [2] + 2 [0] [2] + 2 [2] [0] + 4 [0] [0] = ([4] + [2] + [0] + 2 [2] + 2 [2] + 4 [0] = [4] + 5 [2] + 5 [0].

36 Representation theory of Q... Q Mumford's criterion revisited Unstable points: V = [4] + 5 [2] + 5 [0]. x 4 1 ( x 3 1 x 2 + y 2 1 ( x1 3 x 2 y1 2 ( y2 2, y3 2 ( ( ( y 2 4, y 2 5 ( ( ( 1 0 (, ( 12 0, ( Corresponds to our earlier results!

37 Representation theory of Q... Q Generalization to m > 2 Finding stable/semi-stable points is very easy Harder to keep track of the representation isomorphism Computer algebra packages might help?

38 Representation theory the executive summary Recall: irreducible representations of SL 2 are [k] of dimension k + 1 Weight of basis vector x k 1 x l 2 = k l Let Q Q be embedded in V = A 5 A 5, we constructed an explicit isomorphism V = [4] + 5 [2] + 5 [0]. Verify that Mumford's criterion gives the same result. To generalize, use computer algebra packages?

39 Conclusion & Evaluation In the ane case, C[SL m 2 ]SL 2 was determined This gives rise to a good quotient on the projective (Q... Q ss (Q... Q nn (Q... Q s Semi-simple and stable points can also be interpreted with representation theory. A good quotient for (Q... Q ss? A geometric quotient for (Q... Q s? Embedding with one coordinate added?

40 Appendix Literature W. Drensky. Dening Relations for the Algebra of Invariants of 2 2 Matrices. Algebras and Representation Theory, A. Extra. The invariants of 2 2 matrices, their algebraic relations and the corresponding moduli problem. PhD Thesis, Katholieke Universiteit Nijmegen, See for (the draft of my thesis.