DEFORMATION THEORY MICHAEL KEMENY

DEFORMATION THEORY MICHAEL KEMENY 1. Lecture 1: Deformations of Algebras We wish to first illustrate the theory with a technically simple case, namely deformations of algebras. We follow On the Deformation of Rings and Algebras by Murray Gerstenhaber (Annals 1964) closely. Let k be any field. Definition 1.1. An associative algebra over k is a vector space V over k together with a bilinear map f : V V V such that f is associative, i.e. f(f(a, b), c) = f(a, f(b, c)). Let A = (V, f) be an associative algebra. We say A is a commutative algebra if f is commutative, i.e. f(a, b) = f(b, a). Let V be a vector space over k and let K = k((t)) be the quotient power series field of R = k[[t]], the power series ring. Consider the K vector space There is a natural inclusion of k-vector spaces. V K := V k K. V V K Definition 1.2. A bilinear map f t : V K V K V K is called a one parameter family of deformations of A if (1) f t has the form f t (a, b) = ab + tf 1 (a, b) + t 2 F 2 (a, b) +... where F i : V V V is a bilinear function and a, b V (by linearity, it is enough to specify f t on V ). (2) f t is associative. Note that (ii) says that we have an algebra A t = (V t, f t ) over K = k((t)). In (i), ab denotes the product f(a, b) on A. Condition (i) is in fact a flatness condition in disguise. Remark 1.3. Note that we did not impose any convergence type conditions on (i), which is a purely formal property. On the other hand, if we are very lucky and say F j = 0 for j n 0 + 1, then for any s k we may set t = s and obtain a k algebra with product f s (a, b) = ab + sf 1 (a, b) + s 2 F 2 (a, b) +... + s n 0 F n0 (a, b). In this case we would have an actual (or algebraic) family of algebras in the intuitive sense. In general, it is hard to decide when (formal) one parameter families of deformations come from algebraic families. Date: October 21, 2015. 1

2 MICHAEL KEMENY Goal. Classify one parameter families of deformations of A = (V, f) up to equivalence, under suitable conditions. Definition 1.4. Fix an associative algebra A = (V, f). Let f t, g t : V K V K be two oneparameter families of deformations of A. Then f t g t (in words f t and g t are equivalent), if there exists a linear automorphism Φ t : V K V K of the form for a V, ψ i for a, b V. Φ t (a) = a + tψ 1 (a) + t 2 ψ 2 (a) +... : V V linear maps, such that we have g t (a, b) = Φ 1 t f t (Φ t (a), Φ t (b)) We say f t is trivial if f t f, where f(a, b) = ab is the product on A. If f t is trivial, then in fact A t = (V K, f t ) is isomorphic to A K := A k K (check!). 1.1. Hochschild cohomology for algebras. One parameter families of deformations of associative algebras are described by Hochschild cohomology. Let k be a field and A an associative k-algebra. Let M be an A-A bimodule. Define We have a chain complex C 0 (A, M) := M C n (A, M) := Hom k (A n, M) for n > 0 C n (A, M) := 0 for n < 0 C n 1 (A, M) d C n (A, M) d C n+1 (A, M) with differentials d : C n (A, M) C n+1 (A, M) defined as for n+1 d = ( 1) i i, i=0 r 0 f(r 1,..., r n ) if i = 0 i f(r 0,..., r n ) := f(r 0,..., r i 1 r i,..., r n ) if 1 i n f(r 0,..., r n 1 )r n if i = n + 1. If n = 0, d : M Hom k (A, M) is defined by Exercise 1. Check d 2 = 0. d(m)(a) := am ma. We will do the above exercise for degree zero: d(dm)(r 0, r 1 ) = r 0 (dm)(r 1 ) dm(r 0 r 1 ) + (dm(r 0 ))r 1 = r 0 (r 1 m mr 1 ) (r 0 r 1 m mr 0 r 1 ) + (r 0 m mr 0 )r 1 = 0 The Hochschild cohomology module H n (A, M) is then the cohomology in the middle of Clearly C n 1 (A, M) d C n (A, M) d C n+1 (A, M). H 0 (A, M) = {m M am = ma a A}.

DEFORMATION THEORY 3 In particular, is the centre of A. H 0 (A, A) = {a A ab = ba b A} = Z(A) Exercise 2. A k-linear function f : A M is called a k-derivation if f(rs) = rf(s) + f(r)s. Let Der k (A, M) denote the k-vector space of derivations, and let PDer k (A, M) be the space of principal derivations, i.e. derivations of the form g m : A M, g m (a) = am ma for some fixed m M. Show H 1 (R, M) = Der k (A, M)/PDer k (A, M). If R is a commutative algebra and M and R module, then PDer k (R, M) = 0 and the above exercise shows H 1 (R, M) = Der k (R, M) Hom R (Ω R/k, M), where Ω R/k is the module of Kähler differentials (see Hartshorne). 1.2. Classifying 1-par. Families of Defs. of Algebras. Let f t be a bilinear map V K V K V K satisfying f t (ab) = ab + tf 1 (a, b) + t 2 F 2 (a, b) +... for a, b V, F i ; V V V bilinear. Then the associativity condition for all a, b, c V holds if and only if Comparing coefficients in t we find f t (f t (a, b), c) = f t (a, f t (b, c)) f t (a, b)c + tf 1 (f t (a, b), c) + t 2 F 2 (f t (a, b), c) +... =af t (b, c) + tf 1 (a, f t (b, c)) + t 2 F 2 (a, f t (b, c)) +... F 1 (a, b)c + F 1 (ab, c) = af 1 (b, c) + F 1 (a, bc). Exercise 3. Check the condition above is equivalent to F 1 Ker(d : C 2 (A, A) C 3 (A, A)) := Z 2 (A, A). Definition 1.5. If f t V K V K V K is a 1-par family of defs of A, then F 1 Z 2 (A, A) is called a first order deformation of A. An arbitrary element G Z 2 (A, A) is called integrable if it is a first order deformation, i.e. if G is the differential of some 1-para family of defs g t : V K V K V K. Lemma 1.6. Let f t : V K V K V K be a 1-par family of defs of A, and assume g t is equivalent to f t. Let F 1 resp. G 1 be the associated first order deformations to f t resp. g t. Then [F 1 ] = [G 1 ] H 2 (A, A). Proof. Write f t (a, b) = ab + tf 1 (a, b) +... g t (a, b) = ab + tg 1 (a, b) +... By assumption, there exists Φ t V K V K, Φ t (a) = a + tψ 1 (a) + t 2 ψ 2 (a) +... such that or in other words g t (a, b) = Φ 1 t (f t (Φ t (a), Φ t (b)) Φ t (ab + tg 1 (a, b) +...) = Φ t (a)φ t (b) + tf 1 (Φ t (a), Φ t (b)) +... Expanding this out and collecting coefficients in t we get ψ 1 (a, b) + G 1 (a, b) = ψ 1 (a)b + aψ 1 (b) + F 1 (a, b).

4 MICHAEL KEMENY Hence we see G 1 (a, b) F 1 (a, b) = aψ 1 (b) ψ 1 (ab) + ψ 1 (a)b = (dψ 1 )(a, b). Exercise 4. Verify the following converse. Suppose f t is a 1-par family of defs with differential F 1 Z 2 (A, A). Assume G 1 Z 2 (A, A) has [F 1 ] = [G 1 ] H 2 (A, A). Show there is some g t f t with differential G t. To be precise, show the equivalence is defined by Φ t (a) = a + tψ 1 (a), where G 1 F 1 = dψ 1. As a result of this exercise and the lemma, we call H 2 (A, A) the group of first order deformations of A. We also have Corollary 1.7. Integrability of an element G Z 2 (A, A) depends only on [G] H 2 (A, A). We now give a criterion for [G] H 2 (A, A) to be integrable, that is, realised as the differential of a 1-par family of defs of A. We first record: Exercise 5. Show f t is associative if and only if (1) : V K V K V K of the form f t (a, b) = ab + tf 1 (a, b) + t 2 F 2 (a, b) +... F i (F j (a, b), c) F i (a, F j (b, c)) = 0 for al k = 1, 2,..., where F 0 (a, b) := f(a, b) = ab is the product on A. Note that we have already used Equation (1) for k = 1, where it gives that F 1 C 2 (A, A) must lie in Z 2 (A, A). If we set k = 2 into Equation (1) we get F 1 (F 1 (a, b), c) F 1 (a, F 1 (b, c)) = af 2 (b, c) F 2 (ab, c) + F 2 (a, bc) F 2 (a, b)c = (df 2 )(a, b, c) For any G C 2 (A, A), define the associator G C 3 (A, A) by G(a, b, c) := G(G(a, b), c) G(a, G(b, c)) C 3 (A, A). It is easily checked that if G Z 2 (A, A) then G Z 3 (A, A). From Equation (1) for k = 2 we thus see: Lemma 1.8. A necessary condition for G Z 2 (A, A) to be integrable is [ G] = 0 H 3 (A, A). We wish to show a converse. Firstly, note that Equation (1) can be rewritten as (2) F i (F j (a, b), c) F i (a, F j (b, c)) = (df k )(a, b, c). In the case k = 1, the left hand side is non-existent and the equation should be interpreted as saying df 1 = 0. A necessary and sufficient condition for F 1 := G Z 2 (A, A) to be integrable is that there exist F 1, F 2,... C 2 (A, A) such that Equation (2) holds for all k (F 0 := f, the product on A, as usual). We will prove: Theorem 1.9 (Gerstenhaber). Assume H 3 (A, A) = 0. integrable. To establish Gerstenhaber s Theorem, it suffices to prove the following. Then every element in H 2 (A, A) is

DEFORMATION THEORY 5 Claim 1.10. Let F 1,..., F n 1 C 2 (A, A) satisfy equation (2) for k = 1, 2,..., n 1. Set G n (a, b, c) := F i (F j (a, b), c) F i (a, F j (b, c)) C 3 (A, A). Then G n Z 3 (A, A). Proof of Theorem 1.9 assuming Claim 1.10. Assume H 3 (A, A) = 0. Assume in addition F 1 Z 2 (A, A), i.e. we are assuming Equation (1) holds for k = 1. Then we have that the associator F 1 lies in Z 3 (A, A) (as we saw above) so we can find F 2 C 2 (A, A) with G 2 = F 1 = df 2. By the claim we then have that G 3 Z 3 (A, A), and so there exists F 3 C 2 (A, A) with G 3 = df 3. Using induction, we then define elements F 2, F 3,... C 2 (A, A) satisfying Equation (2) for all k. 2. Lecture 2 To finish the proof of Gerstenhaber s theorem, it remains to prove Claim 1.10. In order to do this, we first introduce some structures on C (A, A) := n C n (A, A). Let f C m+1 (A, A), g C n+1 (A, A), m, n 0. Then one defines f i g C n+m+1 (A, A) by the formula f i g(a 0,..., a i 1, b 0,..., b n, a i+1,..., a m ) := f(a 0,..., a i 1, g(b 0,..., b n ), a i+1,..., a m ). We define f g C n+m+1 (A, A) by m f g = ( 1) ni f i g i=0 We have f g = { f 0 g + f 1 g +... + f m g if n even, f 0 g f 1 g +... + ( 1) m f m g if n odd Note that if f, g C 2 (A, A) then f g C 3 (A, A) is given as f g(a, b, c) = f(g(a, b), c) f(a, g(b, c)). Thus Equation 2 can be rewritten as (3) F i F j = (df k ). The -product is not associative in general. However, we have: Exercise 6. Assume f, g, h C 2 (A, A). Show (1) (f g) g = f (g g) (2) f (g h + h g) = (f g) h + (f h) g

6 MICHAEL KEMENY We next define a cup-product on C (A, A). For f C m (A, A), g C n (A, A) we define f g C n+m (A, A) by f g(a 1,..., a m, b 1,..., b n ) = f(a 1,..., a m )g(b 1,..., b n ). The cup product makes C (A, A) into an associative algebra over k. It is easy to see that if f C m (A, A), g C n (A, A) then Exercise 7. Assume f, g C 2 (A, A). Show d(f g) = df g + ( 1) m f dg. d(f g) = f dg df g + g f f g. Proof of Claim 1.10. Let F 1,..., F n 1 C 2 (A, A) satisfy equation (2). Set G n (a, b, c) := F i F j C 3 (A, A). We need to prove dg n = 0. We have, by the previous exercise dg n = F i df j df i F j + (F j F i F i F j ) = F i df j df i F j by anti-symmetry of the term F j F i F i F j in (i, j). By hypothesis, we further have F i df j df i F j = F α (F β F γ ) (F α F β ) F γ α+β+γ=k α,β,γ>0 All terms with β = γ vanish by Exercise 6 (1). By collecting terms (α, β, γ) and (α, γ, β) for β γ, it suffices to show F α (F β F γ + F γ F β ) ((F α F β ) F γ + (F α F γ ) F β ) = 0. This follows from Exercise 6 (2). 2.1. Infinitesimal automorphisms. We have seen that we may interpret H 2 (A, A) as the group of first order deformations of A, whereas we may interpret H 3 (A, A) as containing all obstructions to integrating first-order deformations. How about H 1 (A, A)? We will show that this is related to infinitesimal automorphisms of the associative algebra A. A one parameter family of automorphisms of A is an automorphism Φ t of A K := A k K expressible in the form Φ t (A) = a + tψ 1 (a) + t 2 ψ 2 (a) +... for a V, ψ i : V V linear. Since Φ t is an automorphisms of algebras which gives Φ t (ab) = Φ t (a)φ t (b) ab + tψ 1 (ab) + t 2 ψ 2 (ab) +... = (a + tψ 1 (a) + t 2 ψ 2 (a) +...)(b + tψ 1 (b) + t 2 ψ 2 (b) +...). Collecting terms in t k, we see that this is equivalent to (4) ψ i (a)ψ j (b) = ψ k (ab)

DEFORMATION THEORY 7 where we set ψ 0 = id. By moving terms with i = 0 or j = 0 to the RHS, and using the cup-product notation for the LHS we can rewrite Equation (4) as (5) ψ i ψ j = dψ k. For k = 1 we interpret this equation as stating dψ 1 = 0 i.e. ψ 1 Z 1 (A, A). By setting k = 2 into Equation (5) we get ψ 1 ψ 1 = dψ 2. Definition 2.1. An element α Z 1 (A, A) is called integrable if there is a one parameter family of automorphisms Φ t with differential α, i.e. Φ t is of the form Φ t (a) = a + tα(a) + t 2 ψ 2 (a) +... One necessary condition for some ψ Z 1 (A, A) to be integrable is that ψ ψ = dφ for some φ C 1 (A, A). Since we have d(ψ ψ) = dψ ψ ψ dψ = 0 this means we need [ψ ψ] = 0 H 2 (A, A). In fact we have many integrability conditions of this nature. Lemma 2.2. Let ψ 1,..., ψ n 1 C 1 (A, A) satisfy Equation (5) for k = 1, 2,..., n 1. Set H n := ψ i ψ j C 2 (A, A). Then H n Z 2 (A, A). Proof. We have dh n = i+j=n = = 0 α+β+γ=n α,β,γ>0 i+j=n by the hypothesis (and associativity of ). [dψ i ψ j ψ i dψ j ] [(ψ α ψ β ) ψ γ ψ α (ψ β ψ γ )] As in the case of deformations of algebras, the lemma gives necessary (and, if worded correctly, sufficient) conditions for integrability of an element ψ 1 Z 1 (A, A). Indeed, for integrability, we must have [ψ 1 ψ 1 ] = 0 H 2 (A, A), so let ψ 2 be such that By the lemma H 3 := ψ 1 ψ 1 = dψ 2. i+j=3 ψ i ψ j Z 2 (A, A). Assume [H 3 ] = 0. Then we can find ψ 3 such that H 3 = dψ 3 and define H 4 and so on. Continuing in this way, we see that if all elements H i arising in this way satisfy [H i ] = 0 H 2 (A, A), the ψ i is integrable. In particular, we have: Corollary 2.3. Assume H 2 (A, A) = 0. Then each element ψ 1 Z 1 (A, A) is integrable.

8 MICHAEL KEMENY Thus the case H 2 (A, A) = 0 is very special. First of all, it implies that the group of first order deformations of A is trivial. Secondly, it implies that all first order automorphisms (or infinitesimal automorphisms ) can be integrated. We further have the following important result, which states that if H 2 (A, A) = 0 then A is rigid, i.e. it has no-nontrivial deformations up to equivalence. Proposition 2.4. Assume H 2 (A, A) = 0 and let f t = t i F i be a one parameter family of deformations of A, where the F i s are extensions of products on V to V K. Then f t is equivalent to the trivial deformation F 0 := f. Proof. As we saw in an exercise last week, since H 2 (A, A) = 0, there exists and element ψ 1 C 1 (A, A) and a one parameter family f 1,t with zero differential, i.e. f 1,t (a, b) = ab + t 2 F 1,t (a, b) +... such that f 1,t is equivalent to f t via φ 1 := id + tψ 1. Next, the associativity of f 1,t gives the equation F 1,i F 1,j = df 1,k, where F 1,1 := 0, and setting k = 2 we see df 1,2 = 0. Let ψ 2 C 1 (A, A) be such that F 1,2 = dψ 2 and set φ 2 := id + t 2 ψ 2. Then φ 2 defines an equivalence from f 1,t to a one parameter family f 2,t of the form f 2,t (a, b) = ab + t 3 F 2,3 (a, b) +.... By associativity, we have df 2,3 = 0. Thus we may continue in this way, and define elements ψ i C 1 (A, A) for all positive integers i, together with one parameter families f i,t of the form such that f i,t is equivalent to f t via setting φ j := id + t j ψ j. Now define β j is the sum of all terms f i,t (a, b) = ab + t i+1 F i,i+1 (a, b) +... φ i := φ i φ i 1... φ 1, φ := β j t j ψ l1 ψ l2... ψ lk with s l s = j and l 1 > l 2 >... > l k > 0 (we set β 0 = id). Note that φ agrees with φ i for all terms of order less than i. Thus φ defines an equivalence from f t to the trivial deformation. Humboldt-Universität zu Berlin, Institut für Mathematik, Unter den Linden 6 10099 Berlin, Germany E-mail address: michael.kemeny@gmail.com