Hochschild Hyperhomology and Cyclic Tensor Invariance — Examples

The grading puts $R_d=k{x}^{d\mathrm{/}2}$ for even $d\ge 0$ and $R_d=0$ for odd $d$, so $R$ is graded with $1_R$ of internal degree $0$; the regular bimodule is $k$-central, and by [F2] the shift $R\{4\}$ is again a graded $k$-central $R$-bimodule, with $(R\{4\}{)}_d=R_{d-4}$ and $1$ of internal degree $4$. Hence $F$, with $F^0=R$ in cochain degree $0$, $F^1=R\{4\}$ in cochain degree $1$ and all other terms zero, is a bounded cochain complex of graded $k$-central $R$-bimodules whose differential $d_F=0$ is a degree-zero bimodule map.

By [F3] the termwise groups are $H{H}_j(R,F^i)=H_j(C_{\bullet }(R,F^i))$ with $C_j(R,F^i)=F^i{\otimes }_k{R}^{{\otimes }_{k}j}$, and the induced map $H{H}_j(R,d_F^0)$ is induced by the zero chain map, hence is zero; so by [F4] the termwise complex is the two-term complex $H{H}_j(R,F^0)\stackrel{0}{\to }H{H}_j(R,F^1)$ concentrated in cochain degrees $0$ and $1$, with $H^0=H{H}_j(R,F^0)$, $H^1=H{H}_j(R,F^1)$ and all other cohomology zero.

By [F5] the total complex is $T^n(R,F)={⨁}_{i-j=n,j\ge 0}{C}_j(R,F^i)$ with $D=d_F+(-1{)}^{i}b$; since $d_F=0$ the differential acts on $C_j(R,F^i)$ by $(-1{)}^{i}b$ alone, so $T^{\bullet }(R,F)$ is the direct sum of the two column complexes $(C_{\bullet }(R,F^0),b)$ and $(C_{\bullet }(R,F^1),-b)$, which have the same cycles and the same boundaries, and hence the same homology, as $(C_{\bullet }(R,F^0),b)$ and $(C_{\bullet }(R,F^1),b)$. Therefore $H{H}^{\mathrm{h}\mathrm{y}\mathrm{p}\mathrm{e}\mathrm{r},n}(R,F)\cong {⨁}_{i-j=n,i\in \{0,1\},j\ge 0}H{H}_j(R,F^i)$ as graded $k$-modules.

Applying [F1] with $n=1$ gives $H{H}_0(R,R)\cong R^{\left(\genfrac{}{}{0px}{}{1}{0}\right)}\{0\}=R$ and $H{H}_1(R,R)\cong R^{\left(\genfrac{}{}{0px}{}{1}{1}\right)}\{2\}=R\{2\}$, while $H{H}_j(R,R)=0$ for $j>1$.

By [F3] and [F7] the chain module $C_j(R,R\{4\})=(R\{4\}){\otimes }_k{R}^{{\otimes }_{k}j}$ is identified with $C_j(R,R)\{4\}=(R{\otimes }_k{R}^{{\otimes }_{k}j})\{4\}$ by the identity on elementary tensors, a degree-zero $k$-linear isomorphism; the faces of [F3] use only the left action, the right action and the multiplication of $R$, none of which the shift changes, so the identification intertwines the Hochschild boundaries and induces $H{H}_j(R,R\{4\})\cong H{H}_j(R,R)\{4\}$ for every $j$ as graded $k$-modules.

Combining steps 1.4 and 1.5 with $F^0=R$ and $F^1=R\{4\}$: $H{H}_0(R,F^0)\cong R$ and $H{H}_1(R,F^0)\cong R\{2\}$, while $H{H}_0(R,F^1)\cong H{H}_0(R,R)\{4\}\cong R\{4\}$ and $H{H}_1(R,F^1)\cong H{H}_1(R,R)\{4\}\cong R\{2\}\{4\}=R\{6\}$ by [F2]; moreover $H{H}_j(R,F^i)=0$ for $j>1$ and $i=0,1$, so these four groups are the only nonzero termwise groups.

The four nonzero termwise groups sit in bidegree $(i,j)$ with their free $R$-generators in internal degrees $0,2,4,6$: $(0,0)$ has $R$, $(0,1)$ has $R\{2\}$, $(1,0)$ has $R\{4\}$, and $(1,1)$ has $R\{6\}$. Since the total index of [F5] is $n=i-j$, they project to total cochain degrees $0$, $-1$, $1$ and $0$ respectively, placing $R$ and $R\{6\}$ in total degree $0$, $R\{2\}$ in total degree $-1$ and $R\{4\}$ in total degree $1$.

By step 1.3 the hyperhomology is the direct sum of the termwise groups computed in step 3.1, so $H{H}^{\mathrm{h}\mathrm{y}\mathrm{p}\mathrm{e}\mathrm{r},0}(R,F)\cong R\oplus R\{6\}$, $H{H}^{\mathrm{h}\mathrm{y}\mathrm{p}\mathrm{e}\mathrm{r},-1}(R,F)\cong R\{2\}$ and $H{H}^{\mathrm{h}\mathrm{y}\mathrm{p}\mathrm{e}\mathrm{r},1}(R,F)\cong R\{4\}$, with all other total degrees zero; the two free generators in total degree $0$ have internal degrees $0$ and $6$.

By [F6] the filtration yields a spectral sequence with $E_1^{i,-j}=H{H}_j(R,F^i)$ and $E_2^{i,-j}=H^i(H{H}_j(R,F^{\bullet }))$, and step 1.2 identifies the second page as $E_2^{0,0}=R$, $E_2^{0,-1}=R\{2\}$, $E_2^{1,0}=R\{4\}$, $E_2^{1,-1}=R\{6\}$ and zero elsewhere; every differential $d_r$ with $r\ge 2$ has empty target because $E_2$ is supported only in the columns $i=0,1$, so $E_2=E_{\mathrm{\infty }}$. The abutment of [F6] then reads ${\mathrm{gr}}^{0}H{H}^{\mathrm{h}\mathrm{y}\mathrm{p}\mathrm{e}\mathrm{r},0}=R$, ${\mathrm{gr}}^{1}H{H}^{\mathrm{h}\mathrm{y}\mathrm{p}\mathrm{e}\mathrm{r},0}=R\{6\}$, ${\mathrm{gr}}^{0}H{H}^{\mathrm{h}\mathrm{y}\mathrm{p}\mathrm{e}\mathrm{r},-1}=R\{2\}$ and ${\mathrm{gr}}^{1}H{H}^{\mathrm{h}\mathrm{y}\mathrm{p}\mathrm{e}\mathrm{r},1}=R\{4\}$. The degree-zero extension is split because the zero differential makes the total complex the direct sum of the two Hochschild column complexes, as shown in 1.3; their free generators have internal degrees $0$ and $6$.

Collecting: the four nonzero termwise groups are $H{H}_0(R,F^0)\cong R$, $H{H}_1(R,F^0)\cong R\{2\}$, $H{H}_0(R,F^1)\cong R\{4\}$ and $H{H}_1(R,F^1)\cong R\{6\}$, with free-generator degrees $0,2,4,6$ at $(i,j)=(0,0),(0,1),(1,0),(1,1)$, projecting to total cochain degrees $0,-1,1,0$; the zero differential makes the total complex a direct sum of the two Hochschild complexes, giving $H{H}^{\mathrm{h}\mathrm{y}\mathrm{p}\mathrm{e}\mathrm{r},0}\cong R\oplus R\{6\}$, $H{H}^{\mathrm{h}\mathrm{y}\mathrm{p}\mathrm{e}\mathrm{r},-1}\cong R\{2\}$ and $H{H}^{\mathrm{h}\mathrm{y}\mathrm{p}\mathrm{e}\mathrm{r},1}\cong R\{4\}$ with all other total degrees zero, and the $E_2$ page equals the $E_{\mathrm{\infty }}$ page. In contrast to this split situation, for a general bounded $F$ the cochain-index pieces only filter the hyperhomology, as in [F6], where neither degeneration of the spectral sequence nor a splitting of the $(i,j)$ pieces is asserted.

The row space $M$ is finitely generated by its standard row basis. Let $\pi :B\to M$ take the first row and define $\sigma :M\to B$ by $\sigma (v)=e_0^{\mathsf{T}}v$, the matrix whose first row is $v$ and whose other rows are zero. Both maps are right $B$-linear, and $\pi \sigma (v)=v$. Thus $M$ is a direct summand of the free right $B$-module $B$, so it is finite projective. The column space $N$ is free of rank $n$ as a right $k$-module.

The row-column pairing $(x,y)\mapsto xy$ is bilinear and balanced over $B$, since $(xb)y=x(by)$ by associativity. It induces $\mu :M{\otimes }_{B}N\to k$ with $\mu (e_i\otimes e_j^{\mathsf{T}})={\delta }_{ij}$. The balance relations give, for every $i,j$, $e_i\otimes e_j^{\mathsf{T}}=e_0{E}_{0i}\otimes e_j^{\mathsf{T}}=e_0\otimes E_{0i}{e}_j^{\mathsf{T}}={\delta }_{ij}{e}_0\otimes e_0^{\mathsf{T}}$. Here $0\le i,j<n$. The tensors on the left span $M{\otimes }_{B}N$, so this quotient is spanned by $e_0\otimes e_0^{\mathsf{T}}$; its image under $\mu$ is $1$, hence $\mu$ is an isomorphism. For the other tensor product, the elementary tensors $e_j^{\mathsf{T}}\otimes e_i$ form a $k$-basis of $N{\otimes }_{k}M$, and column-row multiplication sends them to the matrix-unit basis $E_{ji}$ of $B$. Thus $N{\otimes }_{k}M\cong B$ by an explicit basis-to-basis isomorphism.

Every commutator in $B$ has trace zero by [F2], so $[B,B]\subseteq \ker (\mathrm{tr})$. Conversely, if $i\mathrm{\ne }j$, then $E_{ij}=[E_{ii},E_{ij}]$ and $E_{ii}-E_{jj}=[E_{ij},E_{ji}]$ by [F1]. The off-diagonal units and the diagonal differences span the trace-zero subspace: for a diagonal matrix $\mathrm{diag}(d_0,\dots ,d_{n-1})$ with ${\sum }_i{d}_i=0$, it is ${\sum }_{i=0}^{n-2}{d}_i(E_{ii}-E_{n-1,n-1})$. Thus $\ker (\mathrm{tr})\subseteq [B,B]$. Since $\mathrm{tr}(E_{00})=1$, trace is onto $k$, so $B\mathrm{/}[B,B]\cong k$ in every characteristic; no division by $n$ is used. For $A=k$, the commutator subspace is zero, giving $H{H}_0(k,k)\cong k$.

The cyclic rotation of [F4], in bar degree zero, sends the class of $e_i\otimes e_j^{\mathsf{T}}$ to the class of $e_j^{\mathsf{T}}\otimes e_i$, which by step 1.2 corresponds to the matrix unit $E_{ji}$. By [F1] and [F2], $\mathrm{tr}(E_{ji})={\delta }_{ij}$, while the scalar corresponding to $e_i\otimes e_j^{\mathsf{T}}$ under the identification of step 1.2 is $e_i{e}_j^{\mathsf{T}}={\delta }_{ij}$. Hence the rotation matches the two identifications: the scalar product of the row and column vectors and the trace of the corresponding matrix unit agree.

By step 1.1 the pair $(M,N)$ satisfies the hypotheses of [F4], so the cyclic rotation gives an isomorphism $H{H}_0(k,M{\otimes }_{B}N)\cong H{H}_0(B,N{\otimes }_{k}M)$, natural in the pair. Under the identifications $M{\otimes }_{B}N\cong k$ and $N{\otimes }_{k}M\cong B$ of step 1.2 and the coinvariant description of [F6], this becomes the isomorphism $k\cong B\mathrm{/}[B,B]\cong k$ whose two composites are computed on classes by step 2.1 and step 1.3: the rotation sends $[e_i\otimes e_j^{\mathsf{T}}]$ to $[E_{ji}]$, and the trace of $E_{ji}$ is ${\delta }_{ij}$, which is exactly the class of the scalar product. For higher Hochschild degrees the same rotation is applied to the terms of the bounded complexes concentrated in cochain degree zero, and the derived-cyclicity isomorphism of [F5] applies with $M$ and $N$ regarded as complexes concentrated in cochain degree zero, in hyperhomology degree $-p$, giving $H{H}_p(k,k)\cong H{H}_p(B,B)$ for every $p\ge 0$ (the coefficient complexes are concentrated in cochain degree zero) and showing that the agreement is not special to degree zero.

The complexes $M$ and $N$ are each concentrated in cochain degree $1$ with zero differential, so each is bounded with terms that are free of rank one over $k$; the only nonzero summand of $\mathrm{Tot}(M{\otimes }_{k}N{)}^r$ is at $r=1+1=2$, where it is $k{\otimes }_{k}k\cong k$ by [F3], and the differential is zero because both input differentials vanish. Similarly $\mathrm{Tot}(N{\otimes }_{k}M{)}^2\cong k$ with zero differential. The terms are finite projective over the opposite algebra $k^{\mathrm{o}\mathrm{p}}=k$, so the hypotheses of [F1] and [F2] hold.

In the notation of [F1] the first block is ${\mathrm{Bar}}_0(k){\otimes }_{k}M$ in bar degree $p=0$ and cochain degree $i=1$, and the second block is ${\mathrm{Bar}}_0(k){\otimes }_{k}N$ in bar degree $q=0$ and cochain degree $l=1$. The rotation formula $(-1{)}^{(i-p)(l-q)}$ of [F1] therefore reads $(-1{)}^{(1-0)(1-0)}=(-1{)}^1=-1$ on the unique summand; the termwise map of [F2] reads $(-1{)}^{il}=(-1{)}^{1\cdot 1}=-1$ on the same summand, so the two formulations of the sign agree. Applying the rotation twice multiplies $(-1{)}^{il}(-1{)}^{li}=(-1{)}^2=1$, so the square of the rotation is the identity here.

The coefficient differentials vanish. The sign is evaluated on the degree-zero Hochschild class, which is a cycle because $b_0=0$. This does not make the chain-level compatibility checks in higher bar degrees vacuous; those are part of [F1], and this example uses only the induced map on $H{H}_0$. The only surviving Hochschild degree is $j=0$: by [F4] (equivalently [F6]) the Hochschild complex of the ground field has $H{H}_0(k,k)=k$ and $H{H}_j(k,k)=0$ for $j\ge 1$, because the alternating boundary is $0$ for odd $j$ and an isomorphism for even $j$. Total degrees are computed by $n=i-j$: with $j=0$ and the tensor factor concentrated in cochain degree $2$, the only nonzero hyperhomology is in total cochain degree $2$.

Collecting: the derived cyclicity isomorphism and the termwise cyclicity isomorphism both carry the class of the unique summand by the factor $-1$, so the rotation is a nontrivial automorphism of the one-dimensional vector space in total degree $2$, not merely a sign that could be removed by choosing a different basis; and applying it twice is the identity. This verifies the sign instance of the cyclic comparison and exhibits the necessity of the Koszul sign in the convention $D=d_{\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{l}\mathrm{e}\mathrm{x}}+(-1{)}^{i}b$; it does not reprove the general comparison theorem.