Equations in simple algebras: variations on a theme of A. Borel. Boris Kunyavskĭı Bar-Ilan University RAGE conference Emory University, May 18, 2011

Transcription

1 Equations in simple algebras: variations on a theme of A. Borel Boris Kunyavskĭı Bar-Ilan University RAGE conference Emory University, May 18, 2011

2 This talk is based on a joint work with Tatiana Bandman (Bar-Ilan University), Nikolay Gordeev (St. Petersburg State Pedagogical University), and Eugene Plotkin (Bar-Ilan University) (see preprint MPIM )

3 MAIN THEME

4 MAIN THEME EQUATIONS in GROUPS = EQUATIONS in ASSOCIATIVE ALGEBRAS

5 EQUATIONS in GROUPS EQUATIONS in ASSOCIATIVE ALGEBRAS

6 EQUATIONS in (SEMI)SIMPLE ALGEBRAIC GROUPS EQUATIONS in MATRIX ALGEBRAS

7 EQUATIONS in (SEMI)SIMPLE ALGEBRAIC GROUPS EQUATIONS in (SEMI)SIMPLE LIE ALGEBRAS EQUATIONS in MATRIX ALGEBRAS

8 WHAT DO WE (ALGEBRAISTS) MEAN BY EQUATIONS?

9 WHAT DO WE (ALGEBRAISTS) MEAN BY EQUATIONS? Usually, we think of an expression of the form P(a, b, c,..., x, y, z) = Q(a, b, c,..., x, y, z) where a, b, c,... are constants,

10 WHAT DO WE (ALGEBRAISTS) MEAN BY EQUATIONS? Usually, we think of an expression of the form P(a, b, c,..., x, y, z) = Q(a, b, c,..., x, y, z) where a, b, c,... are constants, i.e., some fixed elements of some algebra A where we live;

11 WHAT DO WE (ALGEBRAISTS) MEAN BY EQUATIONS? Usually, we think of an expression of the form P(a, b, c,..., x, y, z) = Q(a, b, c,..., x, y, z) where a, b, c,... are constants, i.e., some fixed elements of some algebra A where we live; x, y, z,... are unknowns,

12 WHAT DO WE (ALGEBRAISTS) MEAN BY EQUATIONS? Usually, we think of an expression of the form P(a, b, c,..., x, y, z) = Q(a, b, c,..., x, y, z) where a, b, c,... are constants, i.e., some fixed elements of some algebra A where we live; x, y, z,... are unknowns, i.e., we can substitute elements of A instead of these symbols,

13 WHAT DO WE (ALGEBRAISTS) MEAN BY EQUATIONS? Usually, we think of an expression of the form P(a, b, c,..., x, y, z) = Q(a, b, c,..., x, y, z) where a, b, c,... are constants, i.e., some fixed elements of some algebra A where we live; x, y, z,... are unknowns, i.e., we can substitute elements of A instead of these symbols, or, in highbrow language, some fixed set of generators of the free algebra F A in some fixed variety of algebras containing A (say, variety of groups);

14 WHAT DO WE (ALGEBRAISTS) MEAN BY EQUATIONS? Usually, we think of an expression of the form P(a, b, c,..., x, y, z) = Q(a, b, c,..., x, y, z) where a, b, c,... are constants, i.e., some fixed elements of some algebra A where we live; x, y, z,... are unknowns, i.e., we can substitute elements of A instead of these symbols, or, in highbrow language, some fixed set of generators of the free algebra F A in some fixed variety of algebras containing A (say, variety of groups); P, Q are polynomials,

15 WHAT DO WE (ALGEBRAISTS) MEAN BY EQUATIONS? Usually, we think of an expression of the form P(a, b, c,..., x, y, z) = Q(a, b, c,..., x, y, z) where a, b, c,... are constants, i.e., some fixed elements of some algebra A where we live; x, y, z,... are unknowns, i.e., we can substitute elements of A instead of these symbols, or, in highbrow language, some fixed set of generators of the free algebra F A in some fixed variety of algebras containing A (say, variety of groups); P, Q are polynomials, i.e., elements of F A as above,

16 WHAT DO WE (ALGEBRAISTS) MEAN BY EQUATIONS? Usually, we think of an expression of the form P(a, b, c,..., x, y, z) = Q(a, b, c,..., x, y, z) where a, b, c,... are constants, i.e., some fixed elements of some algebra A where we live; x, y, z,... are unknowns, i.e., we can substitute elements of A instead of these symbols, or, in highbrow language, some fixed set of generators of the free algebra F A in some fixed variety of algebras containing A (say, variety of groups); P, Q are polynomials, i.e., elements of F A as above, or, in simple words, expressions containing signs of operations allowed in A and subject to laws of the corresponding variety (e.g., if A is commutative, we identify xy and yx)

17 WHAT DO WE (these authors) MEAN BY EQUATIONS?

18 WHAT DO WE (these authors) MEAN BY EQUATIONS? We are more restrictive: we only consider equations of the form P(x, y, z,... ) = a where a A is fixed and P is a polynomial, as above, but constant entries are NOT allowed in the left-hand side.

19 WHAT DO WE (these authors) MEAN BY EQUATIONS? We are more restrictive: we only consider equations of the form P(x, y, z,... ) = a where a A is fixed and P is a polynomial, as above, but constant entries are NOT allowed in the left-hand side. E.g., if A is a matrix algebra, we consider equations XY YX = C but NOT BX XB = C.

20 MAIN QUESTIONS

21 MAIN QUESTIONS Given a polynomial P, is the equation solvable a) for all a A, P(x, y, z,... ) = a

22 MAIN QUESTIONS Given a polynomial P, is the equation solvable a) for all a A, or, at least, b) for a typical a A? P(x, y, z,... ) = a

23 GROUP CASE

24 GROUP CASE Let w(x 1,..., x d ) be an element of the finitely generated free group F d = F(x 1,..., x d ) (i.e., a word in x i and x 1 i ), and let a group G be given. Is the equation solvable a) for all g G, or, at least, b) for a typical g G? w(x 1,..., x d ) = g

25 BOREL s THEOREM

26 BOREL s THEOREM If K is a field, G is a connected semisimple algebraic K-group, and w 1, then the corresponding word map w : G d G is dominant.

27 This means that the image of the map is Zariski dense

28 This means that the image of the map is Zariski dense (i.e., for a typical right-hand side the word equation is solvable).

29 TOWARDS an INFINITESIMAL ANALOGUE G g

30 TOWARDS an INFINITESIMAL ANALOGUE G g G g g G

31 A STRAIGHTFORWARDLY OPTIMISTIC TRANSFER If K is a field, G is a connected semisimple algebraic K-group, g = Lie(G) and P is a nonzero Lie monomial, then the corresponding map P : g d g is dominant

32 A STRAIGHTFORWARDLY OPTIMISTIC TRANSFER If K is a field, G is a connected semisimple algebraic K-group, g = Lie(G) and P is a nonzero Lie monomial, then the corresponding map P : g d g is dominant IS WRONG!

33 COUNTER-OPTIMISTIC EXAMPLE Let P(X, Y, Z) = [[[[[Z, Y ], Y ], X ], Y ], [[[[Z, Y ], X ], Y ], Y ]], g = sl(2, K) (K is a field of characteristic zero).

34 COUNTER-OPTIMISTIC EXAMPLE Let P(X, Y, Z) = [[[[[Z, Y ], Y ], X ], Y ], [[[[Z, Y ], X ], Y ], Y ]], g = sl(2, K) (K is a field of characteristic zero). Then P(X, Y, Z) 0 on g (Razmyslov).

35 TOWARDS an INFINITESIMAL ANALOGUE G g

36 TOWARDS an INFINITESIMAL ANALOGUE G g A MORE SOBER POINT of VIEW

37 TOWARDS an INFINITESIMAL ANALOGUE G g A MORE SOBER POINT of VIEW G g

38 A REFINED QUESTION (VARIATION # 1) For a given element P(X 1,..., X d ) of the finitely generated free Lie K-algebra L d over a given algebraically closed field K, and a given semisimple Lie algebra g over K, is the map P(X 1,..., X d ): g d g dominant under the condition that P(X 1,..., X d ) is not an identity on g?

39 MAIN THEOREM

40 MAIN THEOREM Theorem 1. Let L(R, K) be a Chevalley algebra. If char(k) = 2, assume that R does not contain irreducible components of type C r, r 1 (here C 1 = A 1, C 2 = B 2 ).

41 MAIN THEOREM Theorem 1. Let L(R, K) be a Chevalley algebra. If char(k) = 2, assume that R does not contain irreducible components of type C r, r 1 (here C 1 = A 1, C 2 = B 2 ). Suppose that a polynomial P(X 1,..., X d ) is not an identity of the Lie algebra sl(2, K).

42 MAIN THEOREM Theorem 1. Let L(R, K) be a Chevalley algebra. If char(k) = 2, assume that R does not contain irreducible components of type C r, r 1 (here C 1 = A 1, C 2 = B 2 ). Suppose that a polynomial P(X 1,..., X d ) is not an identity of the Lie algebra sl(2, K). Then the induced map P : L(R, K) d L(R, K) is dominant.

43 IS THERE a WAY OUT IF P 0 on SL(2)?

44 Yes! IS THERE a WAY OUT IF P 0 on SL(2)?

45 IS THERE a WAY OUT IF P 0 on SL(2)? Yes! Theorem 2. Let L(R, K) be a Chevalley algebra corresponding to an irreducible root system R, and suppose that R C r if char(k) = 2. Suppose that the map P : L(R, K) d L(R, K) is dominant for R = A 2 and B 2. Then P is dominant for every L(R, K), r > 1.

46 IDEA of PROOF

47 IDEA of PROOF General construction (Chevalley, Springer, Steinberg,... )

48 IDEA of PROOF General construction (Chevalley, Springer, Steinberg,... ) Let L = L(R, K) be a Chevalley algebra, H a Cartan subalgebra, W the Weyl group.

49 IDEA of PROOF General construction (Chevalley, Springer, Steinberg,... ) Let L = L(R, K) be a Chevalley algebra, H a Cartan subalgebra, W the Weyl group. There is a G-equivariant dominant morphism π : L Y where Y is an affine variety and the map satisfies the following condition: π = π H : H Y π 1 (π(h)) = Wh where W is the Weyl group, which acts naturally on H.

50 If p = char(k) is a good characteristic for G, this can be made more explicit.

51 If p = char(k) is a good characteristic for G, this can be made more explicit. There is an isomorphism π : L/G H/W, and the quotient H/W is isomorphic to A r.

52 If p = char(k) is a good characteristic for G, this can be made more explicit. There is an isomorphism π : L/G H/W, and the quotient H/W is isomorphic to A r. Model example (g = sl(n)):

53 If p = char(k) is a good characteristic for G, this can be made more explicit. There is an isomorphism π : L/G H/W, and the quotient H/W is isomorphic to A r. Model example (g = sl(n)): To each matrix we associate the collection of coefficients of the characteristic polynomial.

54 PROOF of THEOREM 1

55 PROOF of THEOREM 1 We want to show that P : L d L is dominant.

56 PROOF of THEOREM 1 We want to show that P : L d L is dominant. Denote I = P(L d ).

57 PROOF of THEOREM 1 We want to show that P : L d L is dominant. Denote It turns out that I = P(L d ). P is dominant π(i) = Y.

58 1st case: R = A n.

59 1st case: R = A n. We identify L(A r, K) with sl(r + 1, K), the algebra of (r + 1) (r + 1)-matrices with zero trace.

60 1st case: R = A n. We identify L(A r, K) with sl(r + 1, K), the algebra of (r + 1) (r + 1)-matrices with zero trace. We fix the chain of subalgebras L 1 L 2 L r = sl(r + 1, K) where L i 1 = sl(i, K) is the subalgebra embedded in the i i upper left corner of the matrix algebra L i = sl(i + 1, K).

61 1st case: R = A n. We identify L(A r, K) with sl(r + 1, K), the algebra of (r + 1) (r + 1)-matrices with zero trace. We fix the chain of subalgebras L 1 L 2 L r = sl(r + 1, K) where L i 1 = sl(i, K) is the subalgebra embedded in the i i upper left corner of the matrix algebra L i = sl(i + 1, K). We also fix, for each i, the subalgebra H i L i of diagonal matrices in L i.

62 MAIN IDEA

63 MAIN IDEA

64 MAIN IDEA

65 Thus it is enough to find a semisimple n n matrix with zero trace and without zero eigenvalues which lies in the image of the map P.

66 KEY OBSERVATION (DELIGNE SULLIVAN)

67 KEY OBSERVATION (DELIGNE SULLIVAN) We may assume that the transcendence degree of K is sufficiently large because this does not have any influence on dominance of P.

68 KEY OBSERVATION (DELIGNE SULLIVAN) We may assume that the transcendence degree of K is sufficiently large because this does not have any influence on dominance of P. Then we may also assume that there exist a subfield F K and a division algebra D r+1 M r+1 (K) with centre F such that D r+1 F K = M r+1 (K).

69 KEY OBSERVATION (DELIGNE SULLIVAN) We may assume that the transcendence degree of K is sufficiently large because this does not have any influence on dominance of P. Then we may also assume that there exist a subfield F K and a division algebra D r+1 M r+1 (K) with centre F such that D r+1 F K = M r+1 (K). The algebra D r+1 is dense in M r+1 (K).

70 KEY OBSERVATION (DELIGNE SULLIVAN) We may assume that the transcendence degree of K is sufficiently large because this does not have any influence on dominance of P. Then we may also assume that there exist a subfield F K and a division algebra D r+1 M r+1 (K) with centre F such that D r+1 F K = M r+1 (K). The algebra D r+1 is dense in M r+1 (K). Hence the set [D r+1, D r+1 ] is dense in [M r+1 (K), M r+1 (K)] = sl(r + 1, K).

71 On the other hand, [D r+1, D r+1 ] D r+1.

72 On the other hand, [D r+1, D r+1 ] D r+1. Thus the set S r+1 = D r+1 sl(r + 1, K) is dense in sl(r + 1, K).

73 On the other hand, [D r+1, D r+1 ] D r+1. Thus the set S r+1 = D r+1 sl(r + 1, K) is dense in sl(r + 1, K). Therefore the restriction of P to Sr+1 d is not the zero map.

74 On the other hand, [D r+1, D r+1 ] D r+1. Thus the set S r+1 = D r+1 sl(r + 1, K) is dense in sl(r + 1, K). Therefore the restriction of P to Sr+1 d is not the zero map. Then there exist s 1,..., s d S r+1 such that s = P(s 1,..., s d ) 0.

75 On the other hand, [D r+1, D r+1 ] D r+1. Thus the set S r+1 = D r+1 sl(r + 1, K) is dense in sl(r + 1, K). Therefore the restriction of P to Sr+1 d is not the zero map. Then there exist s 1,..., s d S r+1 such that s = P(s 1,..., s d ) 0. Since s 1,..., s d D r+1, we have s D r+1.

76 On the other hand, [D r+1, D r+1 ] D r+1. Thus the set S r+1 = D r+1 sl(r + 1, K) is dense in sl(r + 1, K). Therefore the restriction of P to Sr+1 d is not the zero map. Then there exist s 1,..., s d S r+1 such that s = P(s 1,..., s d ) 0. Since s 1,..., s d D r+1, we have s D r+1. As there are no nonzero nilpotent elements in division algebras, all elements of D r+1 are semisimple, so we may assume s H r.

77 On the other hand, [D r+1, D r+1 ] D r+1. Thus the set S r+1 = D r+1 sl(r + 1, K) is dense in sl(r + 1, K). Therefore the restriction of P to Sr+1 d is not the zero map. Then there exist s 1,..., s d S r+1 such that s = P(s 1,..., s d ) 0. Since s 1,..., s d D r+1, we have s D r+1. As there are no nonzero nilpotent elements in division algebras, all elements of D r+1 are semisimple, so we may assume s H r. Since s has no zero eigenvalues, we are done.

78 REMAINING CASES R 6= An

79 REMAINING CASES R A n R has a subsystem R which has the same rank as R and decomposes into a disjoint union of irreducible subsystems R = i R i where each R i is a system of type A ri.

80 VARIATION # 2: FROM DOMINANCE TO SURJECTIVITY

81 VARIATION # 2: FROM DOMINANCE TO SURJECTIVITY The map g g g induced by a Lie polynomial may be dominant but not surjective.

82 VARIATION # 2: FROM DOMINANCE TO SURJECTIVITY The map g g g induced by a Lie polynomial may be dominant but not surjective. Example.

83 VARIATION # 2: FROM DOMINANCE TO SURJECTIVITY The map g g g induced by a Lie polynomial may be dominant but not surjective. Example. P = P(X, Y ) = [[[X, Y ], X ], [X, Y ], Y ]]: sl(2, K) sl(2, K) sl(2, K)

84 VARIATION # 2: FROM DOMINANCE TO SURJECTIVITY The map g g g induced by a Lie polynomial may be dominant but not surjective. Example. P = P(X, Y ) = [[[X, Y ], X ], [X, Y ], Y ]]: sl(2, K) sl(2, K) sl(2, K) If {e, f, h} is the standard basis of sl(2): [e, f ] = h, [h, e] = 2e, [h, f ] = 2f, one can show that in Im(P) there are no elements of the form me or mf with m 0.

85 ENGEL POLYNOMIALS

86 ENGEL POLYNOMIALS We call E m (X, Y ) = [[... [ X, Y ], Y ],..., Y ] L }{{} 2 m times an Engel polynomial of degree (m + 1).

87 ENGEL POLYNOMIALS We call E m (X, Y ) = [[... [ X, Y ], Y ],..., Y ] L }{{} 2 m times an Engel polynomial of degree (m + 1). We call m a i E i (X, Y ) L 2, i=1 where a i K, a generalized Engel polynomial.

88 Theorem 3. Let P(X, Y ) L 2 be a generalized Engel polynomial of degree (m + 1), and let P : L(R, K) 2 L(R, K) be the corresponding map of Chevalley algebras.

89 Theorem 3. Let P(X, Y ) L 2 be a generalized Engel polynomial of degree (m + 1), and let P : L(R, K) 2 L(R, K) be the corresponding map of Chevalley algebras. If R does not contain irreducible components of types R = A 1, B r, C r, F 4 if char(k) = 2, R = G 2 if char(k) = 3,

90 Theorem 3. Let P(X, Y ) L 2 be a generalized Engel polynomial of degree (m + 1), and let P : L(R, K) 2 L(R, K) be the corresponding map of Chevalley algebras. If R does not contain irreducible components of types R = A 1, B r, C r, F 4 if char(k) = 2, R = G 2 if char(k) = 3, and K > m R,

91 Theorem 3. Let P(X, Y ) L 2 be a generalized Engel polynomial of degree (m + 1), and let P : L(R, K) 2 L(R, K) be the corresponding map of Chevalley algebras. If R does not contain irreducible components of types R = A 1, B r, C r, F 4 if char(k) = 2, R = G 2 if char(k) = 3, and K > m R, then the image of P contains (L(R, K) \ Z(L(R, K)) {0}.

92 Theorem 3. Let P(X, Y ) L 2 be a generalized Engel polynomial of degree (m + 1), and let P : L(R, K) 2 L(R, K) be the corresponding map of Chevalley algebras. If R does not contain irreducible components of types R = A 1, B r, C r, F 4 if char(k) = 2, R = G 2 if char(k) = 3, and K > m R, then the image of P contains (L(R, K) \ Z(L(R, K)) {0}. Moreover, if P is an Engel polynomial, then the same is true under the assumption K > R +.

93 PROTOTYPES The case P(X, Y ) = [X, Y ] was settled by G. Brown (1963).

94 PROTOTYPES The case P(X, Y ) = [X, Y ] was settled by G. Brown (1963). The case P(X, Y, Z) = [X, Y,..., Y, Z], g = sl(n), was treated by R. Thompson (1966).

95 KEY TOOL: PRESCRIBED GAUSS DECOMPOSITION

96 KEY TOOL: PRESCRIBED GAUSS DECOMPOSITION Suppose we are not in the cases appearing in the list of exceptions.

97 KEY TOOL: PRESCRIBED GAUSS DECOMPOSITION Suppose we are not in the cases appearing in the list of exceptions. Write L = H + U and fix an arbitrary non-central element h H.

98 KEY TOOL: PRESCRIBED GAUSS DECOMPOSITION Suppose we are not in the cases appearing in the list of exceptions. Write L = H + U and fix an arbitrary non-central element h H. Then for every non-central element l L(R, K) there is g G such that g(l) h + U. (Gordeev, 2006)

99 KEY TOOL: PRESCRIBED GAUSS DECOMPOSITION Suppose we are not in the cases appearing in the list of exceptions. Write L = H + U and fix an arbitrary non-central element h H. Then for every non-central element l L(R, K) there is g G such that g(l) h + U. (Gordeev, 2006) (Actually we mostly need a particular case h = 0 used by Brown.)

100 WHAT ABOUT CENTRAL ELEMENTS?

101 WHAT ABOUT CENTRAL ELEMENTS? Theorem 4. Let P m (X, Y ) L 2 be an Engel polynomial of degree m, and let P : L(R, K) 2 L(R, K) be the corresponding map of Chevalley algebras. Then for m big enough the image of P contains no nonzero elements of Z(L(R, K)).

102 MORE VARIATIONS

103 MORE VARIATIONS ON THE GROUP SIDE OF THE BRIDGE:

104 MORE VARIATIONS ON THE GROUP SIDE OF THE BRIDGE: WORD MAPS ON (SEMI)SIMPLE GROUPS

105 MORE VARIATIONS ON THE GROUP SIDE OF THE BRIDGE: WORD MAPS ON (SEMI)SIMPLE GROUPS The commutator map G G G, (x, y) [x, y] is surjective for nice groups, e.g.,

106 MORE VARIATIONS ON THE GROUP SIDE OF THE BRIDGE: WORD MAPS ON (SEMI)SIMPLE GROUPS The commutator map G G G, (x, y) [x, y] is surjective for nice groups, e.g., connected semisimple compact topological groups (M. Gotô, 1949)

107 MORE VARIATIONS ON THE GROUP SIDE OF THE BRIDGE: WORD MAPS ON (SEMI)SIMPLE GROUPS The commutator map G G G, (x, y) [x, y] is surjective for nice groups, e.g., connected semisimple compact topological groups (M. Gotô, 1949), complex Lie groups (Pasiencier Wang, 1962)

108 MORE VARIATIONS ON THE GROUP SIDE OF THE BRIDGE: WORD MAPS ON (SEMI)SIMPLE GROUPS The commutator map G G G, (x, y) [x, y] is surjective for nice groups, e.g., connected semisimple compact topological groups (M. Gotô, 1949), complex Lie groups (Pasiencier Wang, 1962), linear algebraic groups over an algebraically closed field (Ree, 1964)

109 MORE VARIATIONS ON THE GROUP SIDE OF THE BRIDGE: WORD MAPS ON (SEMI)SIMPLE GROUPS The commutator map G G G, (x, y) [x, y] is surjective for nice groups, e.g., connected semisimple compact topological groups (M. Gotô, 1949), complex Lie groups (Pasiencier Wang, 1962), linear algebraic groups over an algebraically closed field (Ree, 1964) some simple groups over reals (Doković, 1986)

110 MORE VARIATIONS ON THE GROUP SIDE OF THE BRIDGE: WORD MAPS ON (SEMI)SIMPLE GROUPS The commutator map G G G, (x, y) [x, y] is surjective for nice groups, e.g., connected semisimple compact topological groups (M. Gotô, 1949), complex Lie groups (Pasiencier Wang, 1962), linear algebraic groups over an algebraically closed field (Ree, 1964) some simple groups over reals (Doković, 1986) and more general fields (R. Thompson, 1961, 1962)

111 MORE VARIATIONS ON THE GROUP SIDE OF THE BRIDGE: WORD MAPS ON (SEMI)SIMPLE GROUPS The commutator map G G G, (x, y) [x, y] is surjective for nice groups, e.g., connected semisimple compact topological groups (M. Gotô, 1949), complex Lie groups (Pasiencier Wang, 1962), linear algebraic groups over an algebraically closed field (Ree, 1964) some simple groups over reals (Doković, 1986) and more general fields (R. Thompson, 1961, 1962) finite (nonabelian) simple groups (Ore s problem) (Liebeck O Brien Shalev Tiep, 2010)

112 MORE GENERAL WORD MAPS ON FINITE SIMPLE GROUPS

113 MORE GENERAL WORD MAPS ON FINITE SIMPLE GROUPS various counterparts of dominance to express the notion of typical word equation (Shalev et al.) surjectivity of Engel maps on PSL(2, q) (Bandman Garion Grunewald, 2010)

114 WHAT ABOUT NOT SO NICE SIMPLE GROUPS?

115 WHAT ABOUT NOT SO NICE SIMPLE GROUPS? The question on the existence of a simple group not every element of which is a commutator remained open for a long time.

116 WHAT ABOUT NOT SO NICE SIMPLE GROUPS? The question on the existence of a simple group not every element of which is a commutator remained open for a long time. First examples of such groups appeared in geometric context (Barge Ghys, 1992) where the groups under consideration were infinitely generated.

117 WHAT ABOUT NOT SO NICE SIMPLE GROUPS? The question on the existence of a simple group not every element of which is a commutator remained open for a long time. First examples of such groups appeared in geometric context (Barge Ghys, 1992) where the groups under consideration were infinitely generated. Later on there were constructed finitely generated groups with the same property (Muranov, 2007).

118 WHAT ABOUT NOT SO NICE SIMPLE GROUPS? The question on the existence of a simple group not every element of which is a commutator remained open for a long time. First examples of such groups appeared in geometric context (Barge Ghys, 1992) where the groups under consideration were infinitely generated. Later on there were constructed finitely generated groups with the same property (Muranov, 2007). These are counter-examples in very strong sense: the so-called commutator width, defined as supremum of the minimal number of commutators needed for a representation of a given element as a product of commutators, may be arbitrarily large or even infinite (Muranov).

119 WHAT ABOUT WORD MAPS ON RANDOM GROUPS?

120 WHAT ABOUT WORD MAPS ON RANDOM GROUPS? If G is a non-elementary word-hyperbolic group and w is any nontrivial word, then one cannot hope to have a typical element of G in the image of the word map induced by w (D. Calegary Maher, 2010).

121 ON THE BRIDGE: OTHER VARIATIONS FOR LIE ALGEBRAS

122 ON THE BRIDGE: OTHER VARIATIONS FOR LIE ALGEBRAS It would be interesting to understand the situation with infinite-dimensional simple Lie algebras (as well as with finite-dimensional algebras of Cartan type in positive characteristics).

123 ON THE BRIDGE: OTHER VARIATIONS FOR LIE ALGEBRAS It would be interesting to understand the situation with infinite-dimensional simple Lie algebras (as well as with finite-dimensional algebras of Cartan type in positive characteristics). The first question is whether every element of such an algebra can be represented as a Lie product of two other elements.

124 One could try to extend some of results to the case where the ground field is replaced with some sufficiently good ring.

125 One could try to extend some of results to the case where the ground field is replaced with some sufficiently good ring. One has to be careful: there are rings R such that not every element of sl(n, R) can be represented in the form [X, Y ] (M. Rosset S. Rosset, 2000).

126 SYSTEMS of EQUATIONS

127 SYSTEMS of EQUATIONS It would be interesting to consider a more general set-up when we have a polynomial map P : L d L s.

128 SYSTEMS of EQUATIONS It would be interesting to consider a more general set-up when we have a polynomial map P : L d L s. Some dominance results were obtained for the multiple commutator map P : L L d L d given by the formula P(X, X 1,..., X d ) = ([X, X 1 ],..., [X, X d ]) (Gordeev Rehmann, 2001).

129 SYSTEMS of EQUATIONS It would be interesting to consider a more general set-up when we have a polynomial map P : L d L s. Some dominance results were obtained for the multiple commutator map P : L L d L d given by the formula P(X, X 1,..., X d ) = ([X, X 1 ],..., [X, X d ]) (Gordeev Rehmann, 2001). Similar questions in group setting were also discussed for a particular case w = (w 1, w 2 ): G 2 G 2 by Breuillard Green Guralnick Tao (2010).

130 EQUATIONS WITH CONSTANTS

131 EQUATIONS WITH CONSTANTS It would be interesting to consider maps P with some fixed X i = A i.

132 EQUATIONS WITH CONSTANTS It would be interesting to consider maps P with some fixed X i = A i. What about dominance? surjectivity?

133 On the RING SIDE of the BRIDGE

134 On the RING SIDE of the BRIDGE VARIATIONS on a THEME of KAPLANSKY

135 On the RING SIDE of the BRIDGE VARIATIONS on a THEME of KAPLANSKY Let P(X 1,..., X d ) K X 1,..., X d be an associative, noncommutative polynomial (i.e., an element of the finitely generated free associative algebra).

136 On the RING SIDE of the BRIDGE VARIATIONS on a THEME of KAPLANSKY Let P(X 1,..., X d ) K X 1,..., X d be an associative, noncommutative polynomial (i.e., an element of the finitely generated free associative algebra). Let P : M n (K) d M n (K) denote the corresponding map.

137 On the RING SIDE of the BRIDGE VARIATIONS on a THEME of KAPLANSKY Let P(X 1,..., X d ) K X 1,..., X d be an associative, noncommutative polynomial (i.e., an element of the finitely generated free associative algebra). Let P : M n (K) d M n (K) denote the corresponding map. Assume it is not identically zero.

138 On the RING SIDE of the BRIDGE VARIATIONS on a THEME of KAPLANSKY Let P(X 1,..., X d ) K X 1,..., X d be an associative, noncommutative polynomial (i.e., an element of the finitely generated free associative algebra). Let P : M n (K) d M n (K) denote the corresponding map. Assume it is not identically zero. 1st new phenomenon: it may be central (i.e., its image may coincide with the set of scalar diagonal matrices).

139 On the RING SIDE of the BRIDGE VARIATIONS on a THEME of KAPLANSKY Let P(X 1,..., X d ) K X 1,..., X d be an associative, noncommutative polynomial (i.e., an element of the finitely generated free associative algebra). Let P : M n (K) d M n (K) denote the corresponding map. Assume it is not identically zero. 1st new phenomenon: it may be central (i.e., its image may coincide with the set of scalar diagonal matrices). The question on the existence of such polynomials is attributed to Kaplansky. It was answered in the affirmative by Formanek and Razmyslov.

140 Assume the map P to be noncentral.

141 Assume the map P to be noncentral. Is it dominant?

142 Assume the map P to be noncentral. Is it dominant? No: the image of P may coincide with sl(n, K). Are there other obstructions to dominance?

143 Assume the map P to be noncentral. Is it dominant? No: the image of P may coincide with sl(n, K). Are there other obstructions to dominance? Suppose that P is noncentral and contains a matrix with nonzero trace.

144 Assume the map P to be noncentral. Is it dominant? No: the image of P may coincide with sl(n, K). Are there other obstructions to dominance? Suppose that P is noncentral and contains a matrix with nonzero trace. Is it dominant?

145 Assume the map P to be noncentral. Is it dominant? No: the image of P may coincide with sl(n, K). Are there other obstructions to dominance? Suppose that P is noncentral and contains a matrix with nonzero trace. Is it dominant? No, even for n = 2! (Kanel-Belov Malev Rowen, 2010)

146 Assume the map P to be noncentral. Is it dominant? No: the image of P may coincide with sl(n, K). Are there other obstructions to dominance? Suppose that P is noncentral and contains a matrix with nonzero trace. Is it dominant? No, even for n = 2! (Kanel-Belov Malev Rowen, 2010) Example. P(X, Y ) = (XY YX ) + (XY YX ) 2.

147 WE GO AHEAD ACROSS the BRIDGE

148 WE GO AHEAD ACROSS the BRIDGE The same inductive argument as in the proof of Theorem 1 shows that if P(X 1,..., X d ) is not identically zero on M 1 (K) d then the map P is dominant for all n.

149 WE GO AHEAD ACROSS the BRIDGE The same inductive argument as in the proof of Theorem 1 shows that if P(X 1,..., X d ) is not identically zero on M 1 (K) d then the map P is dominant for all n. In the situation where P(X 1,..., X d ) is identically zero on K d, one can consider the induction base n = 2 and prove that if the restriction of P to M 2 (K) d is dominant then so is P.

150 WE GO AHEAD ACROSS the BRIDGE The same inductive argument as in the proof of Theorem 1 shows that if P(X 1,..., X d ) is not identically zero on M 1 (K) d then the map P is dominant for all n. In the situation where P(X 1,..., X d ) is identically zero on K d, one can consider the induction base n = 2 and prove that if the restriction of P to M 2 (K) d is dominant then so is P. The assumption made above holds, for instance, for any semi-homogeneous, non-central polynomial having at least one 2 2-matrix with nonzero trace among its values (Kanel-Belov Malev Rowen).

151 WE GO AHEAD ACROSS the BRIDGE The same inductive argument as in the proof of Theorem 1 shows that if P(X 1,..., X d ) is not identically zero on M 1 (K) d then the map P is dominant for all n. In the situation where P(X 1,..., X d ) is identically zero on K d, one can consider the induction base n = 2 and prove that if the restriction of P to M 2 (K) d is dominant then so is P. The assumption made above holds, for instance, for any semi-homogeneous, non-central polynomial having at least one 2 2-matrix with nonzero trace among its values (Kanel-Belov Malev Rowen). If, under the same assumptions on P, Im(P) lies in sl(n, K), then Im(P) = sl(n, K).

152 THANKS FOR YOUR ATTENTION!

153 QUIZ

154 QUIZ Locate Bridge

155 QUIZ Locate Bridge Highway

156 QUIZ Locate Bridge Highway Mountain road

157 QUIZ Locate Bridge Highway Mountain road Approximate solutions are welcome!

158 QUIZ Locate Bridge Highway Mountain road Approximate solutions are welcome! Prize for the 1st correct answer:

159 QUIZ Locate Bridge Highway Mountain road Approximate solutions are welcome! Prize for the 1st correct answer: Bonus drink after tomorrow banquet.

160 QUIZ Locate Bridge Highway Mountain road Approximate solutions are welcome! Prize for the 1st correct answer: Bonus drink after tomorrow banquet. Answers to be disclosed after the open questions session tomorrow evening.

161 QUIZ Locate Bridge Highway Mountain road Approximate solutions are welcome! Prize for the 1st correct answer: Bonus drink after tomorrow banquet. Answers to be disclosed after the open questions session tomorrow evening. THANKS AGAIN!