Deformations of a noncommutative surface of dimension 4

Transcription

1 Deformations of a noncommutative surface of dimension 4 Sue Sierra University of Edinburgh Homological Methods in Algebra and Geometry, AIMS Ghana 2016

2 In this talk, I will describe the work of my student Christopher Campbell on algebras that behave like polynomials in 4 variables. Throughout, k is an algebraically closed field of characteristic 0. Let C = k[x 1,..., x n ]. Why is C a nice ring?

3 Here are some ways that C is nice: 1 C is connected graded: C = p 0 C p with C p C q C p+q for all p, q, and with C 0 = k. 2 C is generated by C 1 : C p = C p 1 for all p 1. 3 C has polynomial growth: more specifically dim C p = ( ) p+n 1 p for all p 0. 4 C has global dimension n < : every C-module M has a projective resolution of length n: an exact sequence 0 P n P n 1 P 1 P 0 C 0 with the P i projective. 5 C is a domain: ab = 0 a = 0 or b = 0 6 C is noetherian: every ideal is finitely generated.

4 Problem For fixed n, find all the noncommutative rings R satisfying (1)-(6). (Note: these criteria are slightly weaker than Artin-Schelter regular) n = 1: we have only R = k[x]. n = 2: R is either a quantum plane or the Jordan plane: k q [x, y] = k x, y /xy qyx for some q k k J [x, y] = k x, y /xy yx x 2 (Here (6) becomes: R is left and right noetherian: all left or right ideals are finitely generated.)

5 For n = 3 the answer is known: Theorem (Artin-Tate-Van den Bergh, Stephenson-Zhang) If R is an algebra that satisfies (1) (4),(6) then R is also a domain (5) and moreover R is Artin-Schelter regular. Such R are classified. For n = 4:??????, although there are many examples. Example Let α be a graded automorphism of k[x 1, x 2, x 3 ], so α GL(3, k). Let R = k[x 1, x 2, x 3 ][x 4 ; α] Notation: if f k[x 1, x 2, x 3 ] then x 4 f = f α x 4. This satisfies (1) (6).

6 Example (almost!) Let σ Aut k(u, v) be defined by Let D = k(u, v)[t; σ]. u σ = uv, v σ = v. Let A be the subalgebra of D generated by x 1 = t, x 2 = ut, x 3 = vt, x 4 = uvt. Proposition (Smith-Zhang) A satisfies (1) (5). In particular, dim A p = ( ) p+3 p. However, (6) fails: A is not noetherian.

7 Note: A has transcendence degree 3 given by u, v, t, but behaves like polynomials in four variables. We say A is a surface because in projective geometry, homogeneous coordinate rings of surfaces have transcendence degree 3. For example k[x, y, z] k(x/z, y/z)[z] is the graded ring associated to P 2.

8 The commutative ring k[x 1, x 2, x 3, x 4 ] has 4 generators and 6 relations: x i x j = x j x i for all 1 i < j 4. If A is like" k[x 1, x 2, x 3, x 4 ] it should also have 6 relations. r 1 : x 1 x 3 = tvt = v σ t 2 = vt 2 = x 3 x 1. r 2 : x 2 x 4 = utuvt = u(uv) σ t 2 = u 2 v 2 t 2 = x 4 x 2. r 3 : x 4 x 1 = uvt 2 = utvt = x 2 x 3. r 4 : x 1 x 2 = uvt 2 = x 2 x 3. r 5 : x 3 x 2 = uv 2 t 2 = x 1 x 4. r 6 : x 4 x 3 = uv 2 t 2 = x 1 x 4. Fact: A = k x 1, x 2, x 3, x 4 /(r 1, r 2, r 3, r 4, r 5.r 6 ).

9 Our goal is now to find other algebras like A. Method one: guess. I will try to deform A slightly, keeping (1) (5). Maybe I should deform σ. Let τ Aut k(u, v) so that u τ k(u), v τ k(v). (Equivalently, let τ Aut (P 1 P 1 ), where means the connected component of the identity.) Let (Note A(I) = A.) A(τ) = k t, ut, vt, uvt k(u, v)[t; στ]. Theorem (Rogalski-S.) There exists τ so that A(τ) satisfies (1) (6). Question Is this all of the algebras like A?

10 Method two: Hochschild cohomology. Imagine there is a variety X parameterising algebras like A. We can find a neighbourhood of X containing A by looking for curves in X that pass through A. That is, we want to find a family of algebras A(s), depending on a parameter s, so that A(0) = A and so that the A(s) satisfy (1) (6) if possible, or at least (1) (5).

11 We are looking for deformations of A. That is, we put a new multiplication s on A[s] so that a s b = ab + µ 1 (a, b)s + µ 2 (a, b)s 2 + (Here ab means multiplication in A, so A(0) = A.) Each µ p : A k A A. We want A(s) to be graded so require µ p (A i A j ) A i+j. Fact: s is associative if and only if for all n we have: ( n ) : µ p (µ q (a, b), c) µ p (a, µ q (b, c)) = 0. p+q=n

12 Definition A map µ 1 : A k A A satisfying ( 1 ) or aµ 1 (b, c) µ 1 (ab, c) + µ 1 (a, bc) µ 1 (a, b)c = 0 a, b, c A is a Hochschild 2-cocycle of A Such a µ 1 defines an associative multiplication s on A As: (a + bs) s (a + b s) = ab + (µ 1 (a, b) + ba + ab )s. The cocycle is a coboundary if this new ring is isomorphic to A[s]/s 2. The 2nd Hochschild cohomology of A is HH 2 (A) = { cocycles }/{ coboundaries }.

13 HH 2 (A) is the space of infinitesimal deformations of A, and can be thought of as the tangent space to A of our variety X of algebras. In fact because we want graded deformations we look just for the degree 2 piece HH 2 2 (A).

14 Theorem (Campbell) The graded infinitesimal deformations of A form an 8-dimensional vector space, depending on parameters a,..., h. The new multiplication on A As satisfies the relations: r 1 : x 3x 1 (1 + a)x 1 x 3 + bx cx 2 1. r 2 : x 4x 2 (1 + d)x 2 x 4 + ex fx 2 2. r 3 : x 4x 1 x 2 x 3 + bx 1 x 4. r 4 : x 1x 2 + (a 1)x 2 x 3 + cx 2 x 1 + gx 2 1 hx 2 2. r 5 : x 3 x 2 + (a 1 + d)x 1 x 4 + ex 3 x 4 + fx 2 x 3. r 6 : x 4x 3 x 1 x 4 + bx 3 x 4 gx 1 x 3 + hx 2 x 4.

15 Recall that we defined elements of HH 2 (A) in terms of the multiplication a s b = ab + µ 1 (a, b)s. How does this compare to the relations above? For example, we had r 4 : x 1 x 2 x 2 x 3 = 0. r 4 : x 1x 2 + (a 1)x 2 x 3 + cx 2 x 1 + gx 2 1 hx 2 2 = 0 And µ 1 : A k A A satisfies: µ 1 (x 1 x 2 x 2 x 3 ) = ax 2 x 3 + cx 2 x 1 + gx 2 1 hx 2 2.

16 What about ( 2 ), giving algebras associative modulo s 3 : µ 1 (µ 1 (a, b), c) µ 1 (a, µ 1 (b, c)) = aµ 2 (b, c) µ 2 (ab, c) + µ 2 (a, bc) µ 2 (a, b)c. Proposition (Campbell) The set of µ 1 HH2 2(A) for which there exists µ 2 satisfying ( 2 ) is: V g = {(a, b, c, d, e, f, g, h) (a, b, c) = (d, e, f )} V q = {(a, b, c, d, e, f, g, h) g = h = 0, rank V u = {(a, b, c, d, e, f, g, h) rank ( a b c d e f ) 1} ( ) a b c 1, a + b + c = 0} d e f

17 Why are these results hard? Formally, define the cobar complex to be: 0 Hom k (A, A) b 1 Hom k (A k A, A) b 2 Hom k (A k A k A, A) b 3... For example, b 2 (f )(a, b, c) = af (b, c) f (ab, c) + f (a, bc) f (a, b)c, so 2-cocycles are those f in ker b 2. In fact HH (A) is the cohomology of the cobar complex. Not helpful for calculating (even if we restrict to HH 2 2 )!

18 To calculate HH 2 (A) need to know how to deduce HH 2 from simply knowing the value of b 2 on the relations of A. The cobar complex comes from the bar resolution of A as an (A A)-bimodule. It turns out that A is a Koszul algebra and so has a much nicer Koszul resolution. However, definitions are given using the bar resolution. Thus to be able to do calculations one must have maps: (bar resolution) (Koszul resolution) The arrow is automatic; finding a splitting that one can calculate takes some work.

19 The cobar complex also clarifies the role of HH 3 as the home of obstructions. Recall that we want to extend (or integrate ) an infinitesimal deformation µ 1 to order 2. We can reframe ( 2 ) as: µ 1 (µ 1 (a, b), c) µ 1 (a, µ 1 (b, c)) = b 2 (µ 2 ), or that [µ 1 (µ 1 (a, b), c) µ 1 (a, µ 1 (b, c))] is 0 in HH 3 (A). If HH 3 (A) were trivial, then any µ 1 integrates to µ 2 ( unobstructed ). Unfortunately, this isn t true, but with some work can still use knowing how the maps behave on the relations to calculate which µ 1 integrate.

20 V g = {(a, b, c, d, e, f, g, h) (a, b, c) = (d, e, f )} Since HH 2 (A) = T A (X) and {A(τ) τ Aut (P 1 P 1 )} X, we expect a map Lie(Aut (P 1 P 1 )) = sl 2 sl 2 HH 2 2 (A). This map can be computed and takes (( ) ( )) a b d e, ( 2d, f, e, 2d, f, e, b, c) V c a f d g. Thus the g in V g stands for geometric: V g corresponds to {A(τ)}. So in some sense we understand V g, although we still do not have a complete description of which A(τ) are noetherian or even satisfy (1) (5).

21 Theorem (Campbell) Let a, c, d, f k, with a, d 1 and af dc = 0. Then the algebra: A(a, c, d, f ) := k x 1, x 2, x 3, x 4 /... r 1 := x 3x 1 (1 + a)x 1 x 3 cx 2 1 r 2 := x 4x 2 (1 + d)x 2 x 4 fx 2 2 r 3 := x 4x 1 (1 + d)x 2 x 3 fx 2 x 1 r 4 := x 1 x 2 x 2 x 3 r 5 := x 3 x 2 (1 + a)x 1 x 4 cx 2 x 3 r 6 := x 4x 3 (1 + a)x 1 x 4 cx 2 x 3 is a connected graded domain, of global dimension 4, with dim A(a, c, d, f ) p = ( ) p+3 p. However, it is never noetherian. In other words, it satisfies (1) (5), but not (6).

22 Fix a, c, d, f and consider the algebras as s varies. A(as, cs, ds, fs) They sweep out a curve in X, and its tangent direction at A is (a, 0, c, d, 0, f, 0, 0) V q = {(a, b, c, d, e, f, g, h) g = h = 0, rank ( ) a b c 1}. d e f

23 Question 1 Do the other directions in V q integrate to give a family of algebras in X? 2 What about V u? (Here u stands for unknown.) Note that for many algebras, all infinitesmal deformations that integrate to order 2 integrate completely. 3 Which of the rings we obtain are noetherian?