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 ω = a0da1da2 ··· dan ∈ 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 ∈ 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 ∈ M. In particular note that b(m ⊗ a1) = ma1 − a1m. In particular,

HH0(A, M) = M/ham − ma | a ∈ A, m ∈ 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). Now a trace on a 1-category C (say the morphisms form sets or k-modules) is, for a fixed k-module V , a map trA : C(A, A) → V f f satisfying the cyclic property trA(gf) = trB(fg) for A −→ B −→ A. A trace functor on a 2-category is, for a fixed category V, a functor

trA : C(A, A) → V subject to the condition that C(A, B) × C(B,A) C(B,A) × C(A, B)

◦ ◦ C(A, A) C(B,B)

trA trB V there is a natural isorphism from the left to right branch. In particular for f ∈ ∼ ObC(A, B) and g ∈ ObC(B,A) we have a natural isomorphism trA(gf) −→ trB(fg). This notion is due to D. Kaledin. Applying this notion to the example of algebras (where V is the category of k-modules), we claim

trA(M) = M/[M,A] = HH0(A, M). ∼ We need to check that we have an isomorphism trA(M ⊗B N) −→ trB(N ⊗A M) where M is an (A, B)-bimodule and N is a (B,A)-bimodule. The idea is that M 7→ C•(A, M) also has a trace property. In other words we have C•(A, M ⊗B ∼ N) −→ C•(B,N ⊗A M). We’re out of time for today, but next time we will explain how to deduce the formula for B from this. We will also indicate an approach from the point of view of differential forms due to Ginzberg and Schedler (part I). Then we return to cyclic homology by giving examples and discussing cyclic objects. CYCLIC HOMOLOGY 7

3. Where does B come from? [01/12/18] Last time we had an (A, B)-bimodule M and a (B,A)-bimodule N. We discussed a trace property

HH0(A, M⊗BN) = (M⊗BN)⊗A⊗Aop A ' (N⊗AM)⊗B⊗Bop B = HH0(B,N⊗AM). Now one might ask about the derived/higher trace property? Do we have something similar on the level of chains? Let’s restrict to the case where we have morphisms f : A → B and g : B → A of algebras.

f A B g

We consider M = B as a (A, B)-bimodule via a·b·b1 := f(a)bb1. This is also known as the graph bimodule f B = Γf . Similarly we take N = A to be a (B,A)-bimodule. ∼ Then f B ⊗B gA = gf A as (A, A)-bimodules. In light of the trace property in the case above we might ask whether there is a ∼ quasiisomorphism C∗(A, gf A) −→ C∗(B, fgB). There is a map sending a0 ⊗ · · · ⊗ an 7→ fa0 ⊗ fa1 ⊗ · · · ⊗ fan, call it f∗. Similarly one defines g∗ the other way:

f∗

C∗(A, gf A) C∗(B, fgB)

g∗

We claim that (gf)∗ − id is nullhomotopic, in line with our hopes for the trace property. For each h : A → A we have maps id, h∗ : C∗(A, hA) → C∗(A, hA). We want a homotopy Bh : C∗(A, hA) → C∗+1(A, hA) such that id −h∗ = [bh,Bh]. It is easy to see that Bh(a0) = 1 ⊗ a0 works as a definition in degree zero. Similarly one finds Bh(a0 ⊗ a1) = 1 ⊗ a0 ⊗ a1 − 1 ⊗ ha1 ⊗ a0. In general,

n X nj Bh(a0 ⊗ · · · ⊗ an) = (−1) 1 ⊗ haj ⊗ · · · ⊗ han ⊗ a0 ⊗ · · · ⊗ aj−1, j=0

2 and Bh = 0. Is it expected that this homotopy should square to zero? Well we have that Don’t understand this part 2 2 [Bh, bh] = 0 whence we have a map Bh : C∗ → C∗+2 of complexes. In general you expect that the powers be only homotopic to zero, but it happens to be serendip- itiously the case that it squares to zero on the nose. If this were not the case we would have gotten something of the form

(1) 2 (2) B = B + uB + u B + ··· on C∗(A, hA)[[u]]. So this is one explanation for where the differential B comes from (taking h = id).

Remark 7. What about the category of algebras? (Derived) morphisms of bimod- ules should play a role too, e.g. we should have a Hochschild cohomology

∗ ∗ HH (A, A) = ExtA⊗Aop (A, A). 8 BORIS TSYGAN

3.1. Noncommutative forms. Let’s look at another motivation for these homol- ogy theories coming from work on noncommutative forms via Ginzburg-Schedler. ¯⊗n nc,n Recall that we were naively thinking of nc forms as A ⊗ A =: ΩA/k where we nc,∗ write a0da1 ··· dan := a0 ⊗· · ·⊗an. More formally, we define ΩA/k a graded algebra over k. Its generators (as an algebra) are in degree 0, elements of A, and in degree 1, symbols da linear in a ∈ A. We have relations d1 = 0 and d(ab) = da · b + a · db. ¯⊗n nc,n We have a k-module map A ⊗ A → ΩA/k sending a0 ⊗ a1 ⊗ · · · 7→ a0da1 ··· , ⊗n which is an isomorphism. Indeed, just define the multiplication on ⊕nA ⊗ A¯ to be given by reduction to standard form, for example:

(a0da1)(b0db1) = a0d(a1b0)db1 − a0a1db0db1. One checks that this yields a well-defined associative multiplication. But once you know that it is an algebra, by the universal property, we have a map back, whence an isomorphism. Next we note that there is a graded derivation d :Ω∗ → Ω∗+1 of degree one sending |ω1| d(ω1ω2) = dω1 · ω2 + (−1) ω1dω2 satisfying d(a) = da and d(da) = 0. One checks that this is well-defined (sending relations to relations). We would like to consider de Rham cohomology, but unfortunately, there is none. ∗ ∗ In fact, the map Ωk/k → ΩA/k is a quasiisomorphism. This is no surprise since the forms know nothing about even the multiplication of A. Suppose now that we have a homomorphism h : A → A of unital algebras. we ∗ ∗ obtain a map h∗ :ΩA/k → ΩA/k sending a 7→ ha, da 7→ d(ha). We claim that id −h∗ is nullhomotopic. One way to prove this is to choose A = k ⊕ A˜ (notice that A˜ −→∼ A¯ from which the homotopy follows easily. Let us instead try to find the homotopy via a formula. In particular, we want ∗ ∗−1 ιh :ΩA/k → ΩA/k such that id −h∗ = [d, ιh]. Considering what happens to a ∈ A one finds that ιh(da) = a − h(a). More generally we find a0da1 7→ a0a1 − h(a1)a0, which happens to be bh. Let’s justify the name ιh. Given a “vector field” D ∈ Der A we can differentiate Pn LD : a0da1 ··· dan 7→ k=0 a0da1 ··· d(Dak) ··· dan. This is the only derivation of degree 0 sending a 7→ Da and da 7→ d(Da) One can think of this as an infinitesimal ∗ ∗−1 version of h∗. Next one can define ιD :Ω → Ω a derivation of degree −1 sending a 7→ 0 and da 7→ Da (check that relations are sent to relations). We have

|ω1| ιD(ω1ω2) = ιDω1 · ω2 + (−1) ω1 · ιDω2. We in fact have a Cartan homotopy formula

LD = [d, ιD]. Now we define explicitly n X j−1 ιD(a0da1 ··· dan) = (−1) a0da1 ··· daj−1D(aj)daj+1 ··· dan. j=1 Notice that this is not in standard form. We’re out of time for today, but next time we will see how to define ιh based on something like this. After that we will begin discussing cyclic objects. CYCLIC HOMOLOGY 9

4. January 17, 2017 Recall that given f : A → A a map of k-algebras we constructed twisted Hochschild chains C∗(A, f A) with differential b = bf . We had two endomorphisms of this chain complex id and f∗ such that

id −f∗ = [bf ,Bf ] 2 where the homotopy Bf was constructed explicitly. Somewhat miraculously, Bf = 0. In particular if f = id we obtain a codifferential B on C∗(A, A) such that bB + Bb = 0. We think of B as the noncommutative analog of the de Rham differential. ∗,nc Another approach to noncommutative differential forms was the notion of ΩA , the notion of universal nc differential forms, which was generated by a and da. Of course, this was a very simple-minded construction, as is evident by the fact that there is no cohomology except for the class generated by the unit 1 ∈ Ω0. Nevertheless we can get something out of this idea. In particular, last time we took a homomorphism f : A → A, which induces an endomorphism f∗ of the differential forms. Again we should have some homotopy ? such that id −f∗ = [d, ?]. For any derivation D ∈ Der A we can define n X LD(a0da1 ··· dan) = Da0da1 ··· dan + a0da1 ··· dDaj ··· dan, j=1 as one expects a derivation to act. Moreover we have some sort of interior multi- plication, n X j−1 ιD(a0da1 ··· dan) = (−1) a0da1 ··· daj−1Dajdaj+1 ··· dan. j=1 It turns out that we obtain Cartan’s usual formulas of calculus:

[d, LD] = 0 [LD, ιE] = ι[D,E][d, ιD] = LD. Consider the map ∆ : A → A ⊗ A given by ∆(a) = 1 ⊗ a − a ⊗ 1. One checks that this satisfies ∆(ab) = ∆a · b + a · ∆b. Formally let us try to write: How to make sense of this formula? n X j−1 ι∆(a0da1 ··· dan) = (−1) a0da1 ··· daj−1(1 ⊗ aj − aj ⊗ 1)daj+1 ··· dan. j=1

n j k This gives us a map ΩA → ⊕j+k=n−1ΩA ⊗ ΩA. We have a diagram

n ι∆ L j k ΩA i+j=n−1 ΩA ⊗ ΩA

ιφ n−1 ΩA

|ω1||ω2| where the vertical arrow is given ω1 ⊗ ω2 7→ ω1ω2 − (−1) f∗ω2ω1. One finds that 2 [d, ιf ] = ±(id −f∗) ιf = 0. 10 BORIS TSYGAN

Specializing at f = id we obtain ι satisfying

2 2 [d, ι] = d = ιf = 0. More explicitly, we have

n−1 X j−1 ι(a0da1 ··· dan) = ±(−1) [aj, daj+1 ··· dan · a0 · da1 ··· daj−1], j=1 where the commutator is taken in Ω∗. This is precisely the Ginzburg-Schedler differential. We can now try to form (double) complexes of the same format as the cyclic complex, etc. In particular, we defined the negative cyclic complex as

(C∗(A, A)[[u]], b + uB), and here we take ∗,nc (ΩA [[u]], ι + ud).

Of course when A is commutative we could take 0 + iddR. In this commutative case, when Q ⊂ k, we have a map from the negative cyclic complex to these usual 1 K ahler differentials given a0 ⊗ · · · an 7→ n! a0da1 ··· dan. This map commutes with the differential. What about a map to the noncommutative complex? Here we take

n X 1 a ⊗ · · · ⊗ a ± da ··· da · a · da ··· da 0 n (n + 1)! j n 0 1 j−1 j=0 where the sign is the sign of the permutation. In the commutative case we call this map HKR and so the noncommutative case we denote by HKRnc. Suppose Q ⊂ k and A is a regular Noetherian commutative algebra. Then HKR is a quasi-isomorphism with respect to both b and b + uB. If we in addition invert the parameter u recall that on the left we obtain the periodic cyclic complex. One finds that ∼ Y j HPn(A) −→ HdR(A) j≡n mod 2 under the same assumptions. In the case when we are not in characteristic 0, we replace de Rham with crystalline cohomology. It is a theorem of Ginzburg and Schedler that HKRnc is a quasi-isomorphism for all A. k ∼ Consider the case A = k[x1, ··· , xn], for any k. We claim that ΩA/k −→ ⊗m Hk(A, A). Recall that we had a free resolution of A given by Bm(A) = A⊗A¯ ⊗A. However there is a subresolution, a much smaller complex, Km(A) ⊂ Bm(A): ( ) X sgn σ A ⊗ (−1) xiσ1 ⊗ · · · ⊗ xiσm ⊗ A

σ∈Sm

op for all i1 < ··· im. We claim that this is still a resolution of A as A ⊗ A -modules m and that Km ⊗A⊗Aop A ' ΩA/k. Maybe in characteristic two we have to be a bit more careful. CYCLIC HOMOLOGY 11

5. January 19, 2017 Recall that last time for A = k[V ] = Sym V with V a free k-module, we had

K`(A) ,→ B`(A) consisting of A ⊗ Λ`V ⊗ A,→ A ⊗ A¯⊗` ⊗ A, sending X a ⊗ v1 ∧ · · · ∧ v` ⊗ b 7→ ±a ⊗ vσ1 ⊗ · · · ⊗ vσ` ⊗ b. The differential on this subcomplex sends

` X j−1 j+` a⊗v1 ∧· · ·∧v` ⊗b 7→ (−1) avj ⊗(· · ·∧vˆj ∧· · · )⊗b−(−1) (· · ·∧vˆj ∧· · ·⊗vjb. j=1

This Koszul differential is just coming from the differential on B`(A). We claim that this inclusion is a quasi-isomorphism. Indeed, we check that

· · · → K1 → K0 → A → 0 is exact. Consider the case where A = k[x]. Then A ⊗ A = k[x, y] and our complex becomes 0 → A ⊗ V ⊗ A → A ⊗ A → A → 0 which is exact because the first map sends v 7→ x − y (this is the Koszul complex ` `−1 of the Now we consider A ⊗A⊗Aop K`(A). The differential A ⊗ Λ V → A ⊗ Λ V is given

` X j−1 a ⊗ v1 ∧ · · · ∧ v` 7→ − (−1) (vja − avj) ⊗ (· · · ∧ vˆj ∧ · · · ). j=1 whence the induced differential is zero, since A = Sym V . From the fact that K is a resolution of A we find that we have a homotopy equivalence A ⊗ K`(A) and A ⊗ B`(A). As an example consider A = khx, yi/(xy − yx + 1), the Weyl algebra in one generator. We have V = kx ⊕ ky ⊂ A¯. In this case we have K` for ` = 0, 1, 2 (since V is rank 2). This is a subcomplex since [y, x] = 1 which is zero in A¯. It is ` also a subresolution of A. Here A ⊗A⊗Aop K`(A) = A ⊗ Λ V . Here the differential is nonzero, since we do not have commutativity. How to compute the homology? Well linearly we can take k[x, y] =∼ A sending xnym 7→ xn · ym. The product can be written explicitly ∞ m m X ∂y f · ∂x g (f ∗ g)(x, y) = . m! m=0 This is a baby example of Poincar´e-Birkhoff-Witt. One notices, then, that our complex is simply the (polynomial) de Rham complex of the plane. More generally, 2n−∗ 2n H∗(An) ' HdR (Ak ).

In characteristic zero in particular, we find that H2n(An) ' k and H6=2n(An) ' 0. The class is represented by 1 ⊗ Alt(x1 ⊗ · · · ⊗ xn ⊗ y1 ⊗ · · · ⊗ yn). What about the algebra A = khV i = T (V ), the tensor algebra of V . In this case we have 0 → T (V ) ⊗ V ⊗ T (V ) → T (V ) ⊗ T (V ) → T (V ) 12 BORIS TSYGAN which is a resolution (left as an exercise). After tensoring on the left with A we obtain 0 → A ⊗ V → A ⊗ 0 sending a ⊗ v 7→ [v, a] = va − av. Then H1(A, A) is the kernel and H0(A, A) is the cokernel A/[A, A]. As an exercise compute B : H0 → H1. Recall that for any A we have that H0(A, A) ' A/[A, A]. If A is noncommutative ∗,nc there is no natural product. On the other hand there is a product on ΩA that respects d (but not ι, which is at most probably a BV∞ operator). Suppose we have two algebras A and B. Then we have a shuffle product

C∗(A, A) ⊗ C∗(B,B) → C∗(A ⊗ B,A ⊗ B) P `(σ) given by sending (a0 ⊗ · · · ⊗ am) ⊗ (b0 ⊗ · · · ⊗ bm) to a0b0 ⊗ σ(−1) σ(a1 ⊗ ··· am ⊗ b1 ⊗ · · · ⊗ bn) where the sum ranges over all σ in (m, n)-shuffles. Here ai := ai ⊗ 1 and bj := 1 ⊗ bj in A ⊗ B. Recall that a shuffle permutes everything preserving the order of the first m and the last n. In the case where A = B and both are commutative one obtains a further map to C∗(A, A). We claim that this a morphism of complexes and in particular a quasi-isomorphism when k is a field (all one really needs is vanishing of some tor group). A corollary is that given any commutative A, one finds that HH∗(A) is a k- ∗ algebra under the shuffle product. One might ask for a map ΩA/k → HH∗(A) of k-algebras. One sends a 7→ a ∈ HH0, da 7→ Ba = 1⊗ ∈ HH1, etc. and one has to check the relation B(ab) − aB(b) − B(a)b ' 0. HKR gives us an isomorphism in general. In the characteristic zero case there is actually a map on the level of complexes. CYCLIC HOMOLOGY 13

6. January 22, 2018 Recall the example of A = T (V ) from last time. We wrote down a two-term chain complex computing the Hochschild homology 0 → T (V ) ⊗ V → T (V ) → 0. What about the differential B going the other way? We remark first that T (V )⊗V is isomorphic to the cokernel of the map b : C2(A, A) → C1(A, A), which is in turn 1,nc 0,nc 1,nc isomorphic to ΩA /[Ω , Ω ]. In detail, this map is b : A ⊗ A¯⊗2 → A ⊗ A¯ sending

a0 ⊗ a1 ⊗ a2 7→ a0a1 ⊗ a2 + a2a0 ⊗ a1 − a0 ⊗ a1a2.

The cokernel is thus the quotient of symbols of the form a0da1 by expressions of the form a0d(a1a2) − a0da1 · a2 − a0a1da2. Hence, with a bit of work, we see that this is isomorphic to Ω1/[Ω0, Ω1]. In our case when A = T (V ), the map of A-bimodules sends a ⊗ v ⊗ b 7→ advb. This map turns out to be an isomorphism. More generally, ∞ •,nc M ΩTV = T (V ) ⊗ V ⊗ T (V ) ⊗ · · · ⊗ V ⊗ T (V ) n=0 holds in fact as algebras, where on the right we use concatenation. To see this notice that the right hand side is thus an associative graded algebra with a map into it from A. Moreover we have a degree +1 differential d given n X d(v1 ··· vn) = v1 ··· vj−1 ⊗ vj ⊗ vj+1 ··· vn j=1 where the non-tensor multiplications are elements of TV . We have a map from the right to the left sending a 7→ a and v1 7→ dv1. Now we have

B : C0/bC1 → C1/bC2 by

v1v2 ··· vn 7→ 1 ⊗ v1v2 ··· vn ∼ 1d(v1 ··· vn). On the right we expand the differential and obtain a sum over all cyclic permutations X X vj+1 ··· vnv1 ··· vj−1 · dvj ∼ (vj+1 ⊗ · · · ⊗ vj−1) ⊗ vj.

Notice that on V ⊗n or any Z/n-module (denoting the order n element as λ) we have two operators V ⊗n −−−→1−λ V ⊗n −→N V ⊗n where N = 1 + λ + ··· + λn−1. Notice that Nλ = λN = 0. Now recalling that H0(A, A) ' ⊕n≥0 coker(1 − λ) and H1(A, A) ' ⊕n≥1 ker(1 − λ), we obtain a map B = ⊕n≥1N. Let’s demonstrate how this works for the negative cyclic complex. Recall that we have diagonals of a double complex that we map downwards between. Let us factor out the subcomplex · · · → C3 → C2 → bC2 in each column. We then actually just get a complex

· · · → C1/bC2 → C0 → C1/bC2 → C0 → · · · which we can rewrite it as · · · → Ω1/[Ω0, Ω1] → Ω0 → Ω1/[Ω0/Ω1] → · · · 14 BORIS TSYGAN where we consider Ω0 in degree zero (and homologically grade). We think of this as the (short) de Rham complex of a noncommutative algebra (also known as the X-complex or Cuntz-Quillen complex). The differential in degree zero is d = B and the in degree 1 given a0da1 7→ a0a1 − a1a0. When A = TV then b = 1 − λ and per ∼ per d = N, and one finds that HC∗ −→ HC∗ (TV ) and similarly for HC∗ for ∗ > 0, since HC0(TV ) = TV/[TV,TV ]. It turns out that the negative cyclic complex projects down to this complex, and we claim that it yields an isomorphism on homology when A = TV . This is because (1) it is an isomorphism on vertical (or Hochschild) homology and (2) because of some comparison of spectral sequences argument, or more concretely by examination of the cone of the projection morphism. Next time we will discuss the case of differential graded algebras as well as an example of THH.