Cyclic cohomology: traces, differentials and symmetry

Written by GPT-6.1 Sol (OpenAI), September 2026, at Ultra. Not yet reviewed. Public domain (CC0).

A trace ignores where a circular product starts: . A higher trace must also remember differentials. Its arguments acquire a sign when their starting point moves. Cyclic cohomology combines these two requirements into a complex. The cyclic category explains why the resulting theory has a derived-functor interpretation and a degree-two periodicity operator.

We start with traces and a concrete differential, then connect differential forms to cyclic cocycles. After that we describe the category that records circular order. This order of ideas makes the algebra visible before the categorical machinery.

We assume associative algebras, tensor products and elementary chain complexes. The basic cochain constructions work over a commutative ring . When averaging over a finite cyclic group is used, we explicitly require to contain . Algebraic duals mean all -linear maps; no continuity is implicit. The examples of smooth functions use complex coefficients and elementary integration. Basic references are [Khalkhali 2007] and [Connes 1983].

1. What a higher trace must satisfy

For an associative -algebra , put

$C^n(A)=\operatorname{Hom}_k(A^{\otimes_k(n+1)},k),\qquad n\geq0.$

The algebra need not be unital in this section. Define

$\begin{aligned} (b\varphi)(a_0,\ldots,a_{n+1}) ={}&\sum_{i=0}^{n}(-1)^i \varphi(a_0,\ldots,a_i a_{i+1},\ldots,a_{n+1})\\ &+(-1)^{n+1}\varphi(a_{n+1}a_0,a_1,\ldots,a_n),\\ (\lambda_n\varphi)(a_0,\ldots,a_n) ={}&(-1)^n\varphi(a_n,a_0,\ldots,a_{n-1}). \end{aligned} \tag{1.1}$

Let be the sum in the first line, without the final term. A cochain is cyclic when . Write for these cochains.

Proposition 1.1. The operators satisfy

$b^2=0,\qquad (b')^2=0,\qquad (1-\lambda_{n+1})b=b'(1-\lambda_n). \tag{1.2}$

Consequently is a cochain complex.

Proof. It is useful to work on tensors first. For , let multiply positions , and let

$d_n(a_0,\ldots,a_n)=(a_na_0,a_1,\ldots,a_{n-1}).$

For ,

$d_i d_j=d_{j-1}d_i. \tag{1.3}$

For disjoint multiplications this just compares the positions after one position has disappeared. For adjacent multiplications it is associativity. If , the same check takes place across the end of the list. For example, the case compares with ; the case compares with .

In , each term , , has sign . Its partner has sign . They cancel. The same pairing with only the nonwrapping faces proves the assertion for , after dualizing.

For the last identity, set and . Direct substitution gives

$d_0t_n=d_n,\qquad d_it_n=t_{n-1}d_{i-1}\quad(1\leq i\leq n).$

Let and . The displayed relations give

$b_nT_n=(-1)^n d_n+T_{n-1}b'_n.$

Subtracting from yields . Dualization is exactly the third identity in (1.2). If , it follows that .

Definition 1.2. The cohomology of is denoted . For a -algebra this is the cyclic cohomology computed by cyclic cochains. Over other coefficient rings the derived cyclic theory will require the full cyclic resolution; the cyclic-invariant complex alone is not its definition.

In degree zero,

$(b\tau)(a_0,a_1)=\tau(a_0a_1)-\tau(a_1a_0).$

Thus is exactly the space of traces. There are no incoming coboundaries in degree zero. This remains true over any coefficient ring.

Example 1.3. On , every trace is a scalar multiple of the matrix trace. Let denote the matrix units. For ,

$E_{ij}=[E_{ii},E_{ij}],\qquad E_{ii}-E_{jj}=[E_{ij},E_{ji}].$

A trace vanishes on the first expression and has equal values on all diagonal matrix units. Conversely, the matrix trace satisfies the trace identity because is commutative. This proves the assertion without a hypothesis on the characteristic of .

2. Differential forms produce cyclic cocycles

A differential graded algebra has an associative graded product and a degree-one map satisfying

$d^2=0,\qquad d(\omega\eta)=d\omega\,\eta+(-1)^{\deg\omega}\omega\,d\eta.$

We do not require the product to be graded commutative. A closed graded trace of degree is a linear functional such that

$\int d\zeta=0\quad(\zeta\in\Omega^{n-1}),\qquad \int\omega\eta=(-1)^{pq}\int\eta\omega \quad(\deg\omega=p,\deg\eta=q,p+q=n). \tag{2.1}$

When , closedness is vacuous. Let be an algebra homomorphism.

Theorem 2.1. The cochain

$\varphi(a_0,\ldots,a_n) =\int \rho(a_0)d\rho(a_1)\cdots d\rho(a_n) \tag{2.2}$

is cyclic and satisfies .

Proof. Write for in this proof. Closedness applied to gives

$0=\int da_n\,a_0\,da_1\cdots da_{n-1} +\int a_n\,da_0\cdots da_{n-1}.$

The graded-trace rule moves past the remaining degree expression, with sign . Therefore

$\int a_n\,da_0\cdots da_{n-1} =(-1)^n\int a_0\,da_1\cdots da_n.$

This is cyclicity. For , cyclicity is automatic and the cocycle assertion is the trace identity.

For , expand each in . The term from cancels the second part from . For , the first part from cancels the second part from . What remains is

$(-1)^n\int a_0\,da_1\cdots da_n\,a_{n+1} +(-1)^{n+1}\int a_{n+1}a_0\,da_1\cdots da_n.$

These cancel by the graded-trace identity with degrees and . No reordering of the individual factors has been used.

Example 2.2. Let . Define a derivation and let extract the coefficient of . Then

$\varphi(a_0,a_1)=\tau(a_0\delta(a_1)),\qquad \varphi(z^p,z^q)=q\,\mathbf1_{p+q=0}. \tag{2.3}$

Since , applying to shows that is antisymmetric. Applying the derivation rule to proves . This cocycle is not a cyclic coboundary: on a commutative algebra is zero, whereas .

The same construction on smooth functions on the circle uses and the normalized integral. Its value on is again . The algebraic example needs no assertion about the cohomology of all smooth functions.

Example 2.3. On the two-dimensional torus, use angles modulo , the orientation , and

$\psi(f_0,f_1,f_2)=\frac1{(2\pi i)^2}\int f_0\,df_1\wedge df_2.$

Stokes' theorem and graded commutativity verify (2.1), so Theorem 2.1 applies. For , ,

$\psi((uv)^{-1},u,v)=1,\qquad \psi((uv)^{-1},v,u)=-1.$

Indeed , and . The two minus signs give the first value. This calculation fixes both the orientation and the normalization.

3. Averaging is a coefficient-dependent step

The signed rotation satisfies , since is even. Put

$N_n=1+\lambda_n+\cdots+\lambda_n^n.$

Lemma 3.1. Let be a -module and let on . If is invertible in , the sequence

$0\longrightarrow\ker(1-T)\longrightarrow V \xrightarrow{1-T}V\xrightarrow{N}V \xrightarrow{1-T}V\xrightarrow{N}\cdots, \qquad N=\sum_{j=0}^{r-1}T^j, \tag{3.1}$

is exact.

Proof. Let . Then , , and is the identity on the invariant submodule. Thus . Also , so . To prove the reverse inclusion, put

$Q=-\frac1r\sum_{j=0}^{r-1}jT^j.$

Multiplying out gives . If , then and . The equalities repeat at every position of (3.1).

For all cochain degrees simultaneously, this argument applies when contains . It does not justify deleting the finite-group resolution over , or in positive characteristic.

Example 3.2. Let , let , and take . Then and . Every differential in the unaugmented part of (3.1) is zero. Each positive cohomology group is . This is a concrete obstruction to the averaging argument, not an assertion that every algebra in characteristic has this cohomology.

4. The category of circular order

The category has objects , . A morphism is an equivalence class of nondecreasing functions satisfying

$f(i+n+1)=f(i)+m+1. \tag{4.1}$

Two functions represent the same morphism if they differ by an integral multiple of . Composition is composition of functions. Changing a representative in the inner function changes the composition by a period in the outer one, so composition is well-defined.

The simplicial category embeds in : extend a nondecreasing map periodically using (4.1). The extension remains nondecreasing across the boundary because . Let be the rotation represented by .

Theorem 4.1. Every morphism has a unique factorization

$f=s\,r_n^a,\qquad s\in\operatorname{Hom}_\Delta([n],[m]), \quad a\in\mathbb Z/(n+1). \tag{4.2}$

The automorphism group of is the cyclic group generated by .

Proof. Fix a lift . Since (4.1) has a positive output period, tends to as tends to . Let be the first integer such that . Then

$0\leq f(p)\leq f(p+n)=f(p-1)+m+1\leq m.$

Thus , , is a simplicial map. Its periodic extension is , giving . Replacing by replaces by ; it changes neither nor modulo .

Conversely, a periodically extended simplicial map has its first nonnegative value at : and . The first nonnegative position of a lift of therefore recovers modulo , and then recovers . This proves uniqueness.

If a class is invertible, the composite of lifts for it and its inverse is a translation by a period. Hence its lift is injective and surjective on . A nondecreasing bijection of is . Formula (4.1) then forces ; translations modulo give exactly the claimed group.

Corollary 4.2. There are

$\bigl|\operatorname{Hom}_\Lambda([n],[m])\bigr| =(n+1)\binom{n+m+1}{n+1}$

morphisms. In particular, has elements. The object is not terminal.

Proof. A nondecreasing list of length with entries in is a multiset of that length from symbols. Counting multiplicities gives the binomial coefficient. Theorem 4.1 contributes the rotations.

Theorem 4.3. The category is isomorphic to its opposite. On lifts an explicit contravariant operation is

$D f(j)=\min\{i\in\mathbb Z:f(i)\geq j\}. \tag{4.3}$

It satisfies and

$D^2 f(i)=f(i-1)+1. \tag{4.4}$

Thus its square is naturally isomorphic to the identity, rather than equal to the identity on lifts.

Proof. The minimum in (4.3) exists by the limits used in Theorem 4.1. The function is nondecreasing, and

$Df(j+m+1)=Df(j)+n+1.$

Changing by an output period changes by an input period. It therefore defines a morphism .

The defining equivalence is

$Df(j)\leq i\quad\Longleftrightarrow\quad j\leq f(i). \tag{4.5}$

Applying it twice shows that exactly when , and hence exactly when . This proves the composition rule. The identity lift has .

For (4.4), the assertion is equivalent to . Its least integer solution is . Accordingly

$D^2 f=r_m f r_n^{-1}.$

Conjugation by the objectwise rotations is an automorphism naturally isomorphic to the identity. Since is bijective on every morphism set, is bijective on every morphism set as well. This proves the isomorphism with the opposite category.

5. An algebra as a functor

For this section is unital. We use covariant functors from the circular-order category just defined. Self-duality permits an equivalent contravariant convention, but one must also transform the maps when changing conventions.

More generally, a cyclic object in a category , in this convention, is a covariant functor . Its morphisms are natural transformations. When is abelian, this functor category is abelian as well: objectwise kernels and cokernels inherit their maps from the universal properties, and objectwise exactness verifies the abelian axioms. This explains where the abelian category needed for Ext comes from.

Let

$A^\natural([n])=A^{\otimes_k(n+1)}.$

For a lift , and each residue modulo , consider the integer fiber . It is either empty or a finite consecutive list. Multiply the factors , read modulo , in that list's integer order to obtain . Use for an empty fiber. Set

$A^\natural(f)(a_0\otimes\cdots\otimes a_n) =y_0\otimes\cdots\otimes y_m. \tag{5.1}$

Every input residue occurs in exactly one such list. A list has at most terms, because increasing the input by increases the output by . Changing the lift by an output period shifts a list by an input period, so (5.1) is independent of the lift.

Proposition 5.1. Formula (5.1) defines a covariant functor , natural in unital algebra homomorphisms.

Proof. The fiber of a composite over an integer is the ordered concatenation of the fibers of over the consecutive integers in . Multiplying in two stages or in one stage agrees by associativity. Empty intermediate fibers contribute the unit. Thus . The identity has single-element fibers and acts as the identity. A unital homomorphism preserves every product and every empty product, proving naturality.

For , all tensor powers canonically identify with , and every morphism acts as the identity. Denote this constant functor by .

Theorem 5.2. Natural transformations are in bijection with traces on . The transformation associated to is

$\tau_n(a_0\otimes\cdots\otimes a_n)=\tau(a_0\cdots a_n). \tag{5.2}$

Proof. The map represented by the periodic extension of the constant simplicial map multiplies in that order. Naturality forces (5.2). For , the other morphism to multiplies . Comparing them forces .

Conversely, grouping a circular list into consecutive blocks preserves its product up to a cyclic change of starting point. A trace is invariant under such a change: move the initial block to the end using . Formula (5.2) therefore commutes with every map (5.1). This proves both existence and uniqueness.

The same construction works for a unital ring regarded as a -algebra. For a nonunital algebra, one first adjoins a unit and uses the relative theory of the augmentation to . A statement that uses empty fibers cannot silently omit this step.

Proposition 5.3. For a covariant functor , a natural transformation to is determined by a linear functional that agrees after the two maps .

Proof. Necessity follows from naturality. For sufficiency, the maps correspond to the possible circular cuts. Two neighboring cuts can be obtained by first mapping , making the single vertex between the cuts one block and the other vertices the second block, and then applying the two maps . The assumed equality therefore makes independent of the chosen . Use this functional as the degree- component. For any , the composite of a chosen map with is a map . Independence of the cut proves naturality. The chosen map also forces this component in any natural transformation, proving uniqueness.

6. A projective resolution that keeps the integral information

Let be a commutative ring. The category of covariant functors is abelian: kernels, images and cokernels are taken at each object. Put

$P_q([n])=R[\operatorname{Hom}_\Lambda([q],[n])]. \tag{6.1}$

Square brackets here mean the free -module on a set. A morphism acts by postcomposition. Evaluation at the identity gives

$\operatorname{Hom}_\Lambda(P_q,E)\simeq E([q]). \tag{6.2}$

Indeed the image of a basis vector must be , where is the image of the identity. Since evaluation preserves surjections, each is projective. Sums of these functors give enough projectives: for each , use the corresponding map .

Let be the simplicial injection omitting . Precomposition defines

$F_i:P_q\to P_{q-1},\qquad F_i(e_f)=e_{f\delta_i}.$

On let be times precomposition by . Set

$D_q=1-T_q,\quad N_q=\sum_{j=0}^qT_q^j,\quad b_q=\sum_{i=0}^q(-1)^iF_i,\quad b'_q=\sum_{i=0}^{q-1}(-1)^iF_i. \tag{6.3}$

The formulas involving faces apply in positive degree. Put .

Lemma 6.1. These operators satisfy

$bD=Db',\qquad b'N=Nb,\qquad DN=ND=0. \tag{6.4}$

For each object, the augmented complex alternating and is exact over , with degree-zero quotient .

Proof. Precomposing by a rotation and then omitting a position gives the same face-rotation relations used in Proposition 1.1. Thus . Also , which gives .

For the norm identity, write for unsigned rotation on any cyclic chain module. Iterating the face-rotation relations gives

$d_i t_q^j= \begin{cases} t_{q-1}^j d_{i-j},&i\geq j,\\ t_{q-1}^{j-1}d_{q+1+i-j},&i<j. \end{cases} \tag{6.5}$

The same formulas hold for the operators . In the expansion of , fix , the power of . The first case of (6.5), with , contributes faces indexed . The second case, with , contributes . Their signs are respectively

$(-1)^{k+(q+1)a}=(-1)^{k+(q-1)a},\qquad (-1)^{k+(q+1)a}=(-1)^{k+(q-1)a}.$

Together they give every face in exactly once. Summing over proves .

For exactness, Theorem 4.1 says that precomposition by the rotations acts freely on the basis of . On an orbit choose the signed basis , . The sign is consistent on returning to , since is even, and .

In this basis consists of vectors with equal coefficients, and is the image of . The kernel of consists of vectors whose coefficients sum to zero, and is generated over by the differences . These are the image of . This proof divides by no integer. It works on each orbit and hence on their direct sum.

Form a first-quadrant double complex with for all . A horizontal boundary lowers : it is for odd , and for positive even . A vertical boundary lowers : it is for even , and for odd . Its total degree is .

Theorem 6.2. The total complex, augmented to the constant functor , is a projective resolution of .

Proof. The squares of both boundaries are zero. The mixed terms cancel by (6.4): at odd they are ; at positive even they are . Thus the sum of the boundaries squares to zero. Each total term is a finite direct sum of projective functors.

Calculate first in the horizontal direction at a fixed object . Lemma 6.1 leaves only the column , with terms . The unique factorization in Theorem 4.1 identifies a basis for this quotient with the simplicial maps . The relation is . Since is again simplicial, the boundary on these basis elements is the ordinary alternating face boundary.

This is the unnormalized chain complex of the ordered simplex with vertices . Here is an explicit augmented contraction. To a nondecreasing list attach the vertex zero at the start:

$h_q(a_0,\ldots,a_q)=(0,a_0,\ldots,a_q),\qquad h_{-1}(1)=(0).$

The first face of returns the original list. Every other face cancels the corresponding term of . Thus on the augmented complex. Its homology is in degree zero and zero above.

For each total degree there are finitely many positions. Filtering by vertical degree and using the horizontal homology therefore computes total homology by a finite filtration in that degree; no infinite-product or convergence assertion is needed. The result at every object is in degree zero. The augmentation sends every basis vector of to , and commutes with postcomposition. Its degree-zero homology is consequently the constant functor. Exactness is objectwise, so this proves the theorem.

This resolution preserves the finite-group information over . It does not replace its horizontal rows by invariant cochains using an unjustified average.

7. Why cyclic cohomology is Ext

For functors , means the cohomology of for a projective resolution of , or equivalently of for an injective resolution of . These are derived functors in the abelian category of functors, not in the category of algebras.

For the next theorem, is a field and is a unital -algebra. Compose the projective resolution of Theorem 6.2 with . It becomes a resolution of the constant contravariant functor. Under the bijection , its degree- terms identify with right representables

$Q_q([n])\simeq k[\operatorname{Hom}_\Lambda([n],[q])].$

Dualizing at each object gives an injective resolution of the constant covariant functor. To see injectivity directly, let . The pairing that evaluates on the identity yields

$\operatorname{Hom}_\Lambda(E,Q_q^\vee) \simeq\operatorname{Hom}_k(E([q]),k). \tag{7.1}$

The right-hand functor is exact because is a field. This proves injectivity without a finite-dimensional hypothesis on . Dualizing the objectwise exact resolution is exact for the same reason. There is no use of an isomorphism between a vector space and its bidual.

Theorem 7.1. If is a field of characteristic zero, then, naturally in unital -algebras,

$H_\lambda^j(A)\simeq \operatorname{Ext}^j_\Lambda(A^\natural,k^\natural). \tag{7.2}$

Proof. Apply to the injective resolution just constructed. By (7.1), the resulting term at is . The integer formula (4.3) gives the maps explicitly. For the injection , has its unique two-element fiber at when , and at in circular order when . Thus the vertical maps are exactly in even columns and in odd columns. Since , the horizontal maps are and , alternately.

Lemma 3.1 now computes horizontal cohomology: it vanishes for positive , and at it is . The remaining vertical differential is . The same finite-filtration argument as in Theorem 6.2 identifies total cohomology with . All constructions commute with unital algebra homomorphisms, proving naturality.

For a unital ring , the expression still defines a derived theory. The preceding injective-resolution proof uses exact linear duality over a field, and its final comparison uses characteristic zero. Neither step can be asserted for an arbitrary ring by leaving the formulas unchanged.

The derived viewpoint also gives bivariant groups and their Yoneda composition. It supplies an algebraic bivariant theory; it does not assert an identification with Kasparov's analytic -groups.

8. The universal degree-two class

Theorem 8.1. For every commutative ring , the Yoneda algebra of the constant functor is

$\operatorname{Ext}^*_\Lambda(R^\natural,R^\natural) \simeq R[\sigma],\qquad |\sigma|=2. \tag{8.1}$

Multiplication by is induced by shifting two columns in the cyclic resolution.

Proof. Apply to Theorem 6.2. Each position is a copy of . Every unsigned face and rotation acts as the identity. In an even column, the vertical differential from degree to is zero for even , and the identity for odd . In an odd column it is minus the identity for even , and zero for odd . Odd columns are acyclic. An even column has cohomology in degree zero only.

The vertical-first filtration leaves one copy of at for each even and no other positions. No higher differential can connect two remaining positions. Consequently Ext is in every even degree and zero in every odd degree. The cochain equal to at and zero elsewhere represents its degree- generator: its vertical differential and its horizontal differential are both zero.

To determine the product, let be the degree-minus-two chain map on the total projective resolution that sends identically to when , and to zero otherwise. Parity is preserved. At , the horizontal boundary lands in a discarded column, exactly as the target has no outgoing horizontal boundary from column zero. Hence commutes with the total boundary. Composing with the augmentation represents the degree-two class .

The chain-map description of the Yoneda product composes lifts of cocycles to maps between projective resolutions. To recall the direction of the maps, let be a resolution and . A cocycle kills , so it factors as , with . Push out the left end of the exact sequence along . Its new left middle term is ; this gives the associated -extension of by .

For a second cocycle , where is projective, lift successively to maps . At each step the preceding chain identity puts the required image in the kernel of the next augmentation or differential, and projectivity lifts it through the corresponding surjection. The cochain represents the splice: the lifts supply comparison maps between the truncated resolutions, and pushing out their left ends gives the two extensions joined at . Applying the same lifting argument to the difference of two lifts produces a chain homotopy; composition with the cocycle then changes the result by a coboundary. Degree-zero products are ordinary composition. Thus composition of lifts computes the Yoneda product with the stated order.

Here is already such a lift. Its -fold iterate represents the cochain at . Thus is the generator in every even degree, proving the ring assertion. For coefficients , precomposing cochains with shifts them two columns and is Yoneda multiplication by the same class.

In particular, the integral statement is . It is stronger than merely listing the abelian groups in each degree: the proof identifies their multiplication.

9. Hochschild cohomology and the exact couple

For a covariant functor , let be the total cochain complex obtained by applying to Theorem 6.2. The vertical maps in its even columns are , and in its odd columns are . Horizontal maps are from even columns and from odd columns. Here ; for the dual resolution associated to an algebra the corresponding rotation is the of (1.1).

The simplicial representables , with their alternating face boundary, resolve the constant -functor: their value at is the ordered simplex contracted in Theorem 6.2. Thus

$H^j(E([0])\xrightarrow b E([1])\xrightarrow b\cdots) \simeq\operatorname{Ext}^j_\Delta(R^\natural,E|_\Delta). \tag{9.1}$

Lemma 9.1. The -complex of a covariant functor is contractible. If is the simplicial surjection repeating the last vertex, a contraction is

$h_q=(-1)^{q-1}E(\sigma_{q-1})\quad(q\geq1),\qquad h_0=0. \tag{9.2}$

Proof. In the projective complex, let times precomposition by . The face-degeneracy relations give

$F_i s_q=-s_{q-1}F_i\quad(0\leq i<q),\qquad F_qs_q=(-1)^q\mathrm{id}.$

Consequently the terms of indexed cancel, and its last term is the identity. Dualizing by gives . The case says .

Theorem 9.2. For every covariant -functor over , there is a natural exact sequence

$\cdots\longrightarrow \operatorname{Ext}^{j-2}_\Lambda(R^\natural,E) \xrightarrow S\operatorname{Ext}^{j}_\Lambda(R^\natural,E) \xrightarrow I\operatorname{Ext}^{j}_\Delta(R^\natural,E|_\Delta) \xrightarrow B\operatorname{Ext}^{j-1}_\Lambda(R^\natural,E) \xrightarrow S\cdots. \tag{9.3}$

The map is multiplication by the class of Theorem 8.1; is restriction. This sequence is the long exact sequence of the associated exact couple.

Proof. Shifting two columns embeds into , where . Let be the quotient consisting of columns zero and one. There is a short exact sequence of cochain complexes

$0\longrightarrow K[-2]\xrightarrow S K\longrightarrow L\longrightarrow0. \tag{9.4}$

Projecting onto its column-zero complex is a cochain map. Its kernel is the column-one -complex, placed one degree higher, with the contraction of Lemma 9.1. Hence identifies with (9.1). The inclusion of simplicial representables into column zero of the cyclic resolution lifts the identity of the constant functor. The resulting map on Hom complexes is projection onto column zero. Thus the map just obtained is the derived restriction map.

The long exact sequence of (9.4) is (9.3), and Theorem 8.1 identifies its with Yoneda multiplication. For completeness, the connecting map is concrete. If satisfies , put . The relation and Lemma 9.1 give . Thus , in columns zero and one, is a cocycle of . Its differential in the full complex has only a column-two component, . Undoing the shift gives

$Bx=[N_{j-1}h_jD_jx]\quad(j\geq1). \tag{9.5}$

It is a cocycle because and . For , is zero. Changing a representative or a lift changes this expression by a coboundary, as follows directly by taking the differential of the changed lift in (9.4). This also proves naturality.

For a unital algebra over a characteristic-zero field, the algebraic cochain bicomplex in Theorem 7.1 has the same two-column argument. Its column-zero cohomology is Hochschild cohomology with coefficients in . Explicitly, identify a multilinear form with the map

$(a_1,\ldots,a_j)\longmapsto \bigl(a_0\longmapsto\varphi(a_0,a_1,\ldots,a_j)\bigr).$

The dual bimodule has action . The first Hochschild coefficient term becomes multiplication , and the last becomes ; the middle terms are the products in (1.1). This checks the identification of differentials, not just of vector spaces. Therefore

$\cdots\to H_\lambda^{j-2}(A)\xrightarrow S H_\lambda^j(A) \xrightarrow I HH^j(A,A^*)\xrightarrow B H_\lambda^{j-1}(A) \xrightarrow S H_\lambda^{j+1}(A)\to\cdots. \tag{9.6}$

This is Connes' cyclic–Hochschild exact sequence. The sign in (9.2) is part of our convention for its connecting map.

10. Why the classifying space is the classifying space of the circle

The nerve of a category has a -simplex for each composable string of arrows. Its geometric realization is its classifying space. We use three facts from algebraic topology: is with cohomology , ; a degree-two integral cohomology class on a CW complex is represented by a map to ; a homology isomorphism between simply connected CW complexes is a homotopy equivalence. References are [Hatcher 2002, Example 4.50, Theorems 3.19 and 4.57, and Corollary 4.33]. These are the topological background for the following argument.

Lemma 10.1. For a small category ,

$\operatorname{Ext}^*_{\mathcal C}(\mathbb Z^\natural,\mathbb Z^\natural) \simeq H^*(B\mathcal C;\mathbb Z) \tag{10.1}$

as rings. Its categorical homology with constant coefficients is .

Proof. A bar resolution of the constant covariant functor has, in degree , one representable for each string . The alternating boundary deletes an end vertex or composes the arrows adjacent to an interior vertex. At an object , its basis is a string followed by an arrow . This is the nerve chain complex of the category of objects over . Its terminal object is . Appending that object, with sign , gives a contraction of the augmented complex: the new last face returns the original string, and the other faces cancel with the contraction of the boundary. The resolution is therefore exact and projective.

Applying Hom into the constant functor makes a degree- cochain a function on strings of arrows. Its differential is the nerve coboundary. Taking the tensor product over the category with the constant contravariant functor instead gives the nerve chain complex. These may be unnormalized nerve complexes; retaining degenerate simplices computes the same simplicial homology.

The Alexander–Whitney diagonal splits a string at each vertex. On cochains it yields

$(\alpha\smile\beta)(x_0\to\cdots\to x_{p+q}) =\alpha(x_0\to\cdots\to x_p) \beta(x_p\to\cdots\to x_{p+q}).$

Evaluating a degree- cocycle on the last arrows gives a lift on the bar resolution, retaining the initial string and composing its representable arrow through the remaining arrows. The boundary terms within that initial string commute with the lift; the other terms cancel by the cocycle equation. Composing with a degree- cocycle gives the displayed cup formula. Hence the projective-resolution description of the Yoneda product agrees with the nerve cup product. Realizing the nerve gives its CW simplicial chain and cochain theory. This proves (10.1), including the product.

Theorem 10.2. There is a homotopy equivalence

$B\Lambda\simeq\mathbb{CP}^\infty=BS^1. \tag{10.2}$

Under it, the class of Theorem 8.1 can be taken to be the degree-two generator.

Proof. All objects have arrows to and from , so the nerve is connected. It is also simply connected. To verify this, choose arrows and . The endomorphism set of consists of the identity, so . In the fundamental groupoid of the nerve, the two-simplex of this composite identifies with . For any arrow , the loop represented by is likewise the identity, because its composite is an endomorphism of . Every edge loop can be expressed in terms of these loops after transporting its vertices to by the . Thus every loop is trivial.

Lemma 10.1 and Theorem 8.1 give . We also need its integral homology, rather than inferring it merely from this cohomology assertion. Tensor the projective resolution of Theorem 6.2 over with the constant contravariant functor. A representable tensors to , so every unsigned face and rotation acts as the identity. In each even column the vertical homology is in degree zero; an odd column is acyclic. The vertical-first filtration gives

$H_{2j}(B\Lambda;\mathbb Z)=\mathbb Z,\qquad H_{2j+1}(B\Lambda;\mathbb Z)=0.$

All these groups are torsion-free. The integral universal coefficient sequence therefore identifies cohomology with the dual of homology in each degree.

Choose representing . Then , and hence . These are generators in every even degree by Theorem 8.1. Both spaces have the homology groups just described, so duality shows that is an isomorphism in every degree. Both are simply connected CW complexes. The homological Whitehead theorem now makes a homotopy equivalence.

The tensor calculation also proves

$\operatorname{Tor}^{\Lambda}_{2j}(\mathbb Z^\natural,\mathbb Z^\natural) =\mathbb Z,\qquad \operatorname{Tor}^{\Lambda}_{2j+1}(\mathbb Z^\natural,\mathbb Z^\natural)=0,$

where Tor uses a constant right functor and a constant left functor. Specifying their variance is necessary for a tensor product over a category.

11. Exercises with solutions

Exercise 11.1 (first steps). Let be any associative algebra. Write out for a one-cochain. Explain why an arbitrary antisymmetric bilinear form on need not be a cyclic cocycle.

Solution.

$(b\varphi)(a_0,a_1,a_2) =\varphi(a_0a_1,a_2)-\varphi(a_0,a_1a_2)+\varphi(a_2a_0,a_1).$

Antisymmetry is precisely cyclicity in degree one, but it imposes no derivation rule on products. For a counterexample use . On the monomial basis set , , and set all other values to zero. Then .

Exercise 11.2 (a trace with a derivation). Let be a trace and a derivation with . Prove directly that is a cyclic one-cocycle, without assuming is commutative.

Solution. Since ,

$\tau(a_0\delta(a_1))=-\tau(\delta(a_0)a_1) =-\tau(a_1\delta(a_0)).$

Expanding leaves

$-\tau(a_0\delta(a_1)a_2)+\tau(a_2a_0\delta(a_1)),$

which is zero by the trace identity.

Exercise 11.3 (a limit of averaging). For , and , compute the positive cohomology of the complex alternating and , starting with .

Solution. Its maps are . The degree-one group is . The degree-two group is . Hence the positive odd groups vanish and every positive even group is . Dividing by in Lemma 3.1 would incorrectly erase these groups.

Exercise 11.4 (constant vertex maps). List the morphisms and compute their action on . Explain why the underlying map on the two vertices does not determine a cyclic morphism.

Solution. The two lifts can be chosen as and . The fiber over zero is respectively and . Their actions are and . Both send both vertex residues to the unique target vertex. Their lifts are not equivalent by an output-period translation, so they are distinct morphisms. This distinction matters when is noncommutative.

Exercise 11.5 (self-duality by calculation). For the first lift in Exercise 11.4 compute and . Check the formula for the square of the duality.

Solution. The inequality has least integer solution , so . Consequently . This equals . The difference between the two floor/ceiling lifts also shows why calling this particular duality strictly involutive would lose the rotation in (4.4).

Exercise 11.6 (two-dimensional normalization). For the cocycle of Example 2.3 and integers , evaluate

$\psi(e^{-i((r+u)x+(s+v)y)},e^{i(rx+sy)},e^{i(ux+vy)}).$

Solution. The product of the exponentials is one. The wedge of the two differentials is . Dividing its integral by gives . The determinant changes sign on interchanging the last two inputs, in agreement with the wedge product.

References

[Khalkhali 2007] Masoud Khalkhali, Lectures on Noncommutative Geometry, arXiv:math/0702140, version 2 (2007). Open lecture notes.

[Connes 1983] Alain Connes, Cohomologie cyclique et foncteurs , Comptes Rendus de l'Académie des Sciences, Série I, 296 (1983), 953–958. Author's online text.

[Hatcher 2002] Allen Hatcher, Algebraic Topology, Cambridge University Press, 2002. Author's open text.