Invariants under simultaneous conjugation of SL 2 matrices


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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 mtuples 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 Wellknown 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 (semistable 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 nonnilpotent matrices. Proof. One checks: for all ane parts A, φ restricted to A is a good quotient. Can we do better than this?
24 Semistable and stable points Semistable 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 semistable 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 Semistable 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 semistable are spanned by positive weight vectors (4 Up to change of basis, points that are not stable are spanned by nonnegative weight vectors
26 Semistable 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: nonsemistable points are in span of ( : nilpotent matrices All points are nonstable!
27 Semistable 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 semistable i 1 matrix is nilpotent and the pair is reducible, i.e., can be conjugated to: (( 0 (, 0.
28 Semistable 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 Semistable 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 welldened 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 Semistable and stable points More than 2 matrices Ane cover of (Q... Q nn is always possible Stable, semistable 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 5dimensional 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/semistable 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 Semisimple 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.
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