Hochschild Homology and Diagonal Koszul Resolutions — Examples

For $n\ge 1$, write a pure tensor in $C_n(k,k)$ as $c_0\otimes \cdots \otimes c_n$. The multilinear product map ${\varphi }_n(c_0\otimes \cdots \otimes c_n)=c_0\cdots c_n$ is well-defined by [F7] and is an isomorphism by repeated application of the tensor-unit maps [F8]; its inverse sends $c$ to $c\otimes 1\otimes \cdots \otimes 1$. Indeed, the balancing relations let each scalar factor move to the first slot, so the composite with ${\varphi }_n$ is the identity in either order. For $n=0$, use the identity $C_0(k,k)=k$. Thus identify every chain group with $k$.

Under these identifications every face ${\delta }_i^{(n)}$ preserves the product of the scalar entries: the first and last formulas in [F2] use the regular scalar actions, while each internal formula multiplies two scalars. Hence each face is ${\mathrm{id}}_k$, so by [F3], $b_n={\sum }_{i=0}^n(-1{)}^i{\mathrm{id}}_k$; pairing consecutive terms gives $b_n=0$ for odd $n$ and $b_n={\mathrm{id}}_k$ for even $n\ge 2$. Also $b_0=0$. In particular, $b_1=0$ and $b_2={\mathrm{id}}_k$.

Since $C_0(k,k)=k$ and $b_0=b_1=0$, the degree-zero homology is $k$. Equivalently, [F5] gives the quotient by the span of $ab-ba$, which is zero because $k$ is commutative. If $j>0$ is even, then $b_j={\mathrm{id}}_k$, so the cycle group is zero. If $j$ is odd, then $b_j=0$ and $b_{j+1}={\mathrm{id}}_k$, so every cycle is a boundary. By [F4], these are exactly $H{H}_0(k,k)\cong k$ and $H{H}_j(k,k)=0$ for $j>0$.

Let $K$ be the diagonal Koszul complex for the zero-variable polynomial ring $R=k$. By [F6], $K$ is $k$ in degree zero and zero in positive degrees, with identity augmentation. The maps $p:C_{\bullet }(k,k)\to K$ and $i:K\to C_{\bullet }(k,k)$ are identity in degree zero and zero in positive degrees for $p$, and the degree-zero identity inclusion for $i$. Define $h_n:C_n(k,k)\to C_{n+1}(k,k)$ to be the identity under [F1]'s scalar identifications when $n$ is odd and zero when $n$ is even; put $h_{-1}=0$. For $n=0$, $b_1{h}_0+h_{-1}{b}_0=0=\mathrm{id}-ip$. For $n>0$, the parity formulas in step 2.1 give $b_{n+1}{h}_n+h_{n-1}{b}_n={\mathrm{id}}_{C_n}$. Thus $h$ is a chain homotopy from ${\mathrm{id}}_{C_{\bullet }}$ to $ip$, and $C_{\bullet }(k,k)$ is chain-homotopy equivalent to the empty diagonal Koszul complex. Its homology matches the calculation in step 3.1. This direct comparison uses no general AC-bearing polynomial comparison theorem and no choice.

Specialize the diagonal complex to one variable. [F1, F3, given] It has the two terms $R^e{\theta }_1\cong R^e\{2\}$ in homological degree one and $R^e$ in degree zero, with differential $d({\theta }_1)=u_1=x\otimes 1-1\otimes x$ and augmentation $\mu :R^e\to R$. Thus the displayed augmented complex is $0⟶R^e\{2\}\stackrel{x\otimes 1-1\otimes x}{\to }{R}^e\stackrel{\mu }{\to }R⟶0$. The polynomial Hochschild theorem applies to this diagonal complex under AC [F4].

Tensor with the regular bimodule and compute the only differential. [F2, F3, step 1.1, algebra] The theorem gives the coefficient complex $R\{2\}\stackrel{d}{\to }R$. Writing the degree-one generator as ${\theta }_1$, for $m\in R$ its differential is $d(m{\theta }_1)=xm-mx=0$, since the left and right actions on the regular bimodule agree. This is a degree-zero map because $x$ and ${\theta }_1$ both have internal degree $2$.

Read the homology of the two-term zero-differential complex. [F1, F2, step 2.1, algebra] There is no term above degree one. Since $d=0$, every degree-one element is a cycle and there are no degree-one boundaries, so $H_1=R\{2\}$. In degree zero, there are no degree-zero boundaries and all of $R$ is a cycle, so $H_0=R$. For $j\ge 2$ the chain term is zero, giving $H{H}_j=0$. Applying [F2] gives the displayed Hochschild groups.

Check the endpoints, shift, and exact use of AC. [F1, F2, F3, step 2.1, step 3.1, given] The empty wedge is the degree-zero basis and carries shift $\{0\}$; the top wedge ${\theta }_1$ has internal degree $2$ and gives $R\{2\}$. The outgoing degree-zero differential is zero by convention, the incoming degree-one map is zero by the explicit calculation, and there are no terms in degrees $j\ge 2$. AC is used only to invoke the polynomial Hochschild theorem in step 1.1; the one-variable differential and homology calculation use no choice. The example is a computation, not an equivalence. [F1, F2, F3, step 1.1, step 2.1, step 3.1, algebra] $\mathrm{\square }$

Verify the twisted right action and the bimodule hypotheses. [F2, given] Define $\sigma (f(x))=f(-x)$. Since $\sigma (fg)=\sigma (f)\sigma (g)$, $\sigma (1)=1$, and ${\sigma }^2=\mathrm{id}$, it is a unital ring automorphism of $R$. Put $m\cdot f=m\sigma (f)$. Then $m\cdot 1=m$ and $(m\cdot f)\cdot g=m\sigma (f)\sigma (g)=m\sigma (fg)=m\cdot (fg)$, so this is a unital right action. In particular $m\cdot x=-mx=-xm$. For $r,f\in R$, $r(m\cdot f)=rm\sigma (f)=(rm)\cdot f$, using commutativity of $R$; hence the left and right actions commute. Since $\sigma (q)=q$ for $q\in \mathbb{Q}$, the two scalar actions agree. Thus $M$ is a $\mathbb{Q}$-central $R$-bimodule, as required by [F1].

Write the one-variable coefficient complex and compute its map. [F1, F2, step 1.1, algebra] Under the AC premise [F3], [F1] computes $H{H}_{\bullet }(R,M)$ by the two-term complex $M{\theta }_1\stackrel{d}{\to }M$. Its differential is $d(m{\theta }_1)=xm-m\cdot x=xm-(-xm)=2xm$. There are no terms above degree one.

Compute the kernel and cokernel of multiplication by $2x$. [F1, step 2.1, algebra] If $0\mathrm{\ne }f(x)=a_d{x}^d+\cdots$ with $a_d\mathrm{\ne }0$, then $2xf(x)$ has leading coefficient $2a_d\mathrm{\ne }0$ in $\mathbb{Q}$, so $d$ is injective. Because $2$ is a unit, its image is $2x\mathbb{Q}[x]=x\mathbb{Q}[x]$. Evaluation at zero $\epsilon :\mathbb{Q}[x]\to \mathbb{Q}$ is surjective and has kernel $x\mathbb{Q}[x]$: a polynomial with zero constant coefficient is divisible by $x$. Thus $\mathrm{coker}d\cong \mathbb{Q}$ and $\ker d=0$, giving $H{H}_0(R,M)\cong \mathbb{Q}$ and $H{H}_1(R,M)=0$. The terms above degree one vanish, so $H{H}_j(R,M)=0$ for $j\ge 2$.

Check grading, endpoints, and the exact AC use. [F1, F2, step 2.1, step 3.1, given] If ${\mathrm{deg}}_{\mathrm{i}\mathrm{n}\mathrm{t}}x=2$, then $f(x)\mapsto f(-x)$ preserves degree, and the Koszul generator ${\theta }_1$ has internal degree $2$; hence the degree-one term is the corresponding shift of $M$ and $d$ has internal degree zero. Degree zero is the cokernel computed in step 3.1; degree one is the kernel, with no incoming term from degree two; all higher terms are zero. AC is used only to apply the polynomial Hochschild theorem in step 2.1. The action, leading-term injectivity, and evaluation quotient are explicit and use no choice. The example is a computation, not an equivalence. [F1, F2, step 1.1, step 2.1, step 3.1, algebra] $\mathrm{\square }$

Specialize the diagonal deletion formula to the ordered two-variable wedge. [F1, F3, given] For $m{\theta }_x\wedge {\theta }_y$, deleting the first factor contributes $(xm-mx){\theta }_y$ with positive sign; deleting the second contributes $-(ym-my){\theta }_x$. For degree one, deleting either singleton gives $d_1(a{\theta }_x+b{\theta }_y)=(xa-ax)+(yb-by)$. This gives exactly the two displayed maps with the stated wedge orientation.

Verify that the displayed maps compose to zero. [F1, F2, step 1.1, algebra] Set $u_x(m)=xm-mx$ and $u_y(m)=ym-my$. The bimodule laws in [F2] give $u_x{u}_y(m)=(xy)m-(xm)y-(ym)x+m(xy)$ and $u_y{u}_x(m)=(yx)m-(ym)x-(xm)y+m(yx)$. Because $xy=yx$ and the left and right actions commute, these expressions are equal. Hence $d_1{d}_2(m{\theta }_x\wedge {\theta }_y)=-u_x{u}_y(m)+u_y{u}_x(m)=0$, which checks the mixed-product cancellation.

Compute the regular-coefficient subcase. [F1, F4, step 1.1, step 2.1, algebra] If $M=R$ with its regular bimodule structure, commutativity gives $u_x(m)=u_y(m)=0$ for every $m$. Thus both maps vanish. The terms are $R$ in degree zero, $R{\theta }_x\oplus R{\theta }_y$ in degree one, and $R{\theta }_x\wedge {\theta }_y$ in degree two; there are no higher terms. If $\mathrm{deg}x=\mathrm{deg}y=2$, their internal shifts are respectively $0$, $2$, and $4$. By [F4], this gives $H{H}_0(R,R)=R$, $H{H}_1(R,R)=R\{2{\}}^2$, $H{H}_2(R,R)=R\{4\}$, and $H{H}_j(R,R)=0$ for $j\ge 3$.

Check endpoints, zero input, grading, and AC use. [F1, F2, F3, step 2.1, step 3.1, given] Degree zero is the empty wedge term $M$, and degree two is the single top wedge $M{\theta }_x\wedge {\theta }_y$; the degree-two differential has zero composite with $d_1$ by step 2.1, and there is no degree-three term. If $M=0$, every term and map is zero. For the graded regular-coefficient subcase in step 3.1, use the standard grading with $\mathrm{deg}x=\mathrm{deg}y=2$ and give each wedge generator internal degree $2$; the displayed maps preserve total internal degree. The general coefficient statement does not require a grading on $M$. AC is used only to apply the polynomial Hochschild theorem and its regular-coefficient corollary; the sign and commutator calculations use no choice. This example asserts no biconditional. [F1, F2, F3, F4, F5, step 1.1, step 2.1, step 3.1, given] $\mathrm{\square }$