CYCLIC HOMOLOGY Contents 1. Introduction [01/08/18] 1 2. Derived
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CYCLIC HOMOLOGY BORIS TSYGAN Contents 1. Introduction [01/08/18]1 2. Derived tensor product, traces [01/10/18]5 3. Where does B come from? [01/12/18]7 3.1. Noncommutative forms8 4. January 17, 20179 5. January 19, 2017 11 6. January 22, 2018 13 1. Introduction [01/08/18] Let A be an associative algebra over a ring k. Most of the well-established theory is for k a field of characteristic zero, but we will try to also focus on the case of Z and characteristic p. Later on in the course we will try to get to the case of dg-categories or ring spectra. In particular our viewpoint throughout the course will be to try to understand A as \functions" on some noncommutative space. We will especially focus on trying to understand differential forms (or their analogs). Let us establish some inuition. Let A be a k-algebra for k some (unital) commu- tative algebra. To think of elements of A as functions on some space X you might think of, for example, matrix-valued functions. Then what would differential forms on X be? Well it would be something written as n ! = a0da1da2 ··· dan 2 Ω (X); where the symbols dai will not (super)commute as they do in the classical case. Thus we think of, crudely, our space of n-forms as A ⊗ A¯⊗n where A¯ = A=k · 1. Here we are quotienting by the unit because we expect d(1) = 0. Classically recall we have a de Rham differential d :Ωn ! Ωn+1, which is given d : a0da1 ··· dan 7! 1da0da1 ··· dan: We can do this in our situation as well by taking A ⊗ A¯⊗n −!d A ⊗ A¯⊗n+1 to send a0 ⊗ a1 ⊗ · · · 7! 1 ⊗ a0 ⊗ a1 ⊗ · · · . But this is quite dull|it does not even see the product in A (there is no Leibniz rule). We have that d2 = 0 of course, but there is no cohomology except H0 =∼ k. In other words, in this crude formulation the Poincar´elemma holds for all of our noncommutative spaces X. Let's keep this in mind and try something else. Define now ⊗n Cn(A) = Cn(A; A) = A ⊗ A¯ : Date: Winter 2018. 1 2 BORIS TSYGAN We define differential b : Cn(A) ! Cn−1(A) n−1 X j a0 ⊗ a1 ⊗ · · · ⊗ an 7! (−1) a0 ⊗ · · · ⊗ ajaj+1 ⊗ · · · ⊗ an j=0 n + (−1) ana0 ⊗ a1 ⊗ · · · ⊗ an−1: One checks directly that this is well-defined and that b2 = 0. Let's look at some low degrees explicitly: ··· A ⊗ A¯⊗2 A ⊗ A¯ A The last arrow sends a0 ⊗ a1 7! a0a1 − a1a0 and the next arrow sends a0 ⊗ a1 ⊗ a2 7! a0a1 ⊗ a2 − a0 ⊗ a1a2 + a2a0 ⊗ a1: How do we think about this complex? Well let's say X is a manifold and A is some algebra of (matrix-valued) functions on X. A matrix-valued function can be reduced to a scalar function: a0 7! tr(a0). Moreover given two such functions we can produce a one-form a0 ⊗ a1 7! tr(a0da1). Notice that b(a0 ⊗ a1 ⊗ a2) is sent to zero under this reduction. In general, we construct HKR n Cn(A; A) −−−! Ω (X) 1 a ⊗ · · · ⊗ a 7! tr(a da ··· da ) 0 n n! 0 1 n called the Hochschild-Kostant-Rosenberg map. This is a map of complexes if we equip the de Rham complex with the zero differential (and treat it homologically instead of cohomologically). Of course this is all in the specific case where A is an algebra of matrix-valued functions. Can we think of the Cn(A) as forming a cohomological complex? Our naive differential d from before does not play nicely with b (it does not supercommute). There is however a construction of Rinehart in the '60s B : Cn(A) ! Cn+1(A): n X nj B(a0 ⊗ · · · ⊗ an) = (−1) 1 ⊗ aj ⊗ · · · ⊗ an ⊗ a0 ⊗ · · · ⊗ aj−1: j=0 This is a differential such that B2 = bB + Bb = b2 = 0: Exercise 1. Check that (over characteristic zero) HKR ◦ B = ddR ◦ HKR: For example we have a commutative diagram 1 ⊗ a0 ddR tr(a0) B ddR a0 tr a0 There seems to be a sign and so on. error for n odd. CYCLIC HOMOLOGY 3 We can construct a sort of double complex that starts like C0 C1 C2 C3 C0 C1 C2 C0 C1 C0 We define then per Y CPn(A) = CCn (A) = Cj(A) j≡n mod 2 with differential b + B. We call this the periodic cyclic complex. It is a periodic complex (since b + B squares to zero) of order 2 and its homology the periodic cyclic homology of A. To give you a flavor of the sort of things you get, consider the following result. Theorem 2 (Feigin-Tsygan '85). Let A be a finitely generated commutative unital algebra over C. Let X = specmA ⊂ CN . Then for n 2 Z ∼ Y j HPn(A) = H (X; C); j≡2 mod 2 where on the right we have singular cohomology. In particular periodic cyclic homology is very rigid: Theorem 3 (Goodwillie). If Q ⊂ k and I,! B A where I is a nilpotent ideal, then ∼ HP•(B) −! HP•(A): There is also a complex known as the negative cyclic complex − b+B − − ···! CC1 (A) −−−! CC0 (A) ! CC−1(A) !··· − whose homology is the negative cyclic homology HC• (A) for the index in Z. This complex is perhaps the least used of the ones we have introduced some for. In the notation of Ezra Getzler, let u be a formal variable of degree −2. Then we can write neatly all of the above as: per −1 CC• (A) = C•(A)[[u ; u]] with differential b + uB and − CC• (A) = C•(A)[[u]] with differential b + uB and C•(A) = C•(A) with differential b. Let's turn to some extensions. Let M be an A-bimodule. Then take ⊗n Cn(A; M) = M ⊗ A¯ 4 BORIS TSYGAN with differential X b(a0 ⊗ a1 ⊗ · · · ⊗ an) = ±a0 ⊗ · · · ajaj+1 ⊗ · · · + ∓ana0 ⊗ · · · where a0 2 M. In particular note that b(m ⊗ a1) = ma1 − a1m. In particular, HH0(A; M) = M=ham − ma j a 2 A; m 2 Mi = M ⊗A⊗Aop A: More generally we have the following. Proposition 4. Hochschild homology computes the derived tensor product ∼ A⊗Aop HHn(A; M) = Torn (A; M): This in turn motivates us to look for invariant definitions for the other homologies defined above. CYCLIC HOMOLOGY 5 2. Derived tensor product, traces [01/10/18] Recall from last time, given an associative unital algebra A over k we defined for n ≥ 0 ⊗n Cn(A; A) = Cn(A) = A ⊗ A¯ where A¯ = A=k · 1. We defined two differentials b : Cn(A) ! Cn−1(A) B : Cn(A) ! Cn+1(A) given by n−1 X j b(a0 ⊗ a1 ⊗ · · · ⊗ an) = (−1) a0 ⊗ · · · ⊗ ajaj+1 ⊗ · · · ⊗ an j=0 n + (−1) ana0 ⊗ a1 ⊗ · · · ⊗ an−1 n X nj B(a0 ⊗ · · · ⊗ an) = (−1) 1 ⊗ aj ⊗ · · · ⊗ an ⊗ a0 ⊗ · · · ⊗ aj−1: j=0 Recall that we had b2 = bB + Bb = B2 = 0. Remark 5. Notice that B2 = 0 for a stupid reason, as it places the unit in a position other than the first tensor slot. Of course, we could construct a variant ⊗n+1 where Cn(A) = A , in which case we modify B as X X B(a0 ⊗ · · · ⊗ an) = ±1 ⊗ − ∓ ± − ⊗1 where by the dash we mean what we have in the formula above. The sign in the second sum is the sign of the permutation. ⊗n Recall that we had similar definitions Cn(A; M) = M ⊗ A¯ for when M is an A-bimodule. In this case we saw that ∼ H0(A; M) = M=[A; M] = M ⊗A⊗Aop A: Theorem 6. We have, for when A is a flat k-module (e.g. k is a field), ∼ A⊗Aop Hn(A; M) = Torn (A; M): Proof. We have a bar resolution of A as an A ⊗ Aop-module: 0 0 0 ··· A ⊗ A¯⊗2 ⊗ A b A ⊗ A¯ ⊗ A b A ⊗ A b A ⊗n In particular we let Bn = A ⊗ A¯ ⊗ A for n ≥ 0 where n 0 X j b (a0 ⊗ · · · ⊗ an+1) = (−1) a0 ⊗ · · · ⊗ ajaj+1 ⊗ · · · ⊗ an+1: j=0 02 op One finds that b = 0 and moreover that each Bn is an A ⊗ A -bimodule via multiplication on the left and right. It is clear that b0 is a morphism of bimodules. Notice that this resolution is acyclic|it has no homology. In particular there are work this out maps s : Bn ! Bn+1 given s(a⊗ ··· an+1) = 1 ⊗ a0 ⊗ · · · ⊗ an+1 such that b0s + sb0 = id : 6 BORIS TSYGAN op The flatness condition in the hypothesis gives us that Bn is a flat A⊗A -bimodule. Hence the bar resolution is a flat A ⊗ Aop-bimodule resolution of A. Tensoring now with M on the left, we find that ∼ M ⊗A⊗Aop Bn(A) = Cn(A; M): Explicitly, we map Cn(A; M) ! M ⊗A⊗Aop Bn(A) via m ⊗ (a1 ⊗ · · · ⊗ an) 7! m ⊗ (1 ⊗ a1 ⊗ · · · ⊗ 1): Careful about M being a The differential b0 induces the differential b. right-bimodule? Before proceeding with systematic theory, let's look at how to motivate the differential B. First let's look at the trace property of HH0(A; M). Observe that algebras form a 2-category: for every pair A; B of algebras we have the category C(A; B) of (A; B)-bimodules. We have a composition ⊗B : C(A; B) × C(B; C) !C(A; C) which is reasonably associative (not quite strict, but coherently).