Polynomial Hochschild homology from the diagonal Koszul complex

Identify HH with bar-tensor homology through Tor. [F1, F5, F6, F9, F10, F14, F15, given] AC implies DC by [F9]--[F10]. By [F1], it suffices to compute the balanced enveloping-algebra Tor group. The right-resolution definition [F14] and bar resolution [F5] compute it using $H_j({\mathrm{Bar}}_{\bullet }(R){\otimes }_{S}M)$, and the natural chain isomorphism [F6] and the definition [F15] identify this homology with $H{H}_j(R,M)$.

The bar and diagonal Koszul complexes resolve the same right $S$-module. [F3, F4, F5, given, algebra] Let $K_{\bullet }=K(u_1,\dots ,u_n;S)$. By [F3], $K_{\bullet }\to R$ is a projective resolution as a left $S$-module. The ring $S=R{\otimes }_{k}R$ is commutative, so the same modules and maps are a projective resolution of $R$ as a right $S$-module. Thus [F5] and $K_{\bullet }$ are projective resolutions of the same right $S$-module.

Tensor the diagonal Koszul resolution with $M$ and compute its differential. [F2, F3, F11, given, algebra] Write $u_i=x_i\otimes 1-1\otimes x_i$. The degree-$j$ Koszul term is free over $S$ on the symbols ${\theta }_I$ with $\mathrm{\mid }I\mathrm{\mid }=j$. Tensoring over $S$ with $M$ identifies each basis copy $S{\otimes }_{S}M$ with $M$, so $K_j{\otimes }_{S}M\cong {⨁}_{\mathrm{\mid }I\mathrm{\mid }=j}M{\theta }_I\cong M{\otimes }_k{\mathrm{\Lambda }}_k^j({\theta }_1,\dots ,{\theta }_n)\mathrm{.}$ By [F2], $u_i$ acts on $m\in M$ as $(x_i\otimes 1)m-(1\otimes x_i^{\mathrm{o}\mathrm{p}})m=x_{i}m-m{x}_i$. Applying the Koszul differential [F11] therefore gives exactly the displayed formula. The operators $m\mapsto x_{i}m-m{x}_i$ commute: expand their composites and use commutativity of the $x_i$ on each side and commutation of the two bimodule actions. Hence terms deleting a fixed pair of wedge factors cancel in opposite orders, so $d^2=0$, also directly confirming that these are chain groups.

Compare the projective resolutions and tensor their homotopies with $M$. [F2, F7, F8, step 1.1, step 1.2] Apply [F7] to obtain comparison maps in both directions over $R$, with composites homotopic to the respective identity maps. Tensoring those maps and homotopies over $S$ with the left $S$-module $M$ preserves the chain-map and homotopy identities. By [F8], the induced homology maps are inverse. Consequently $H_j({\mathrm{Bar}}_{\bullet }(R){\otimes }_{S}M)\cong H_j(K_{\bullet }{\otimes }_{S}M)\mathrm{.}$ The maps are independent of $M$, so this comparison is natural in coefficient bimodule maps.

Identify top homology with the centralizer of $R$ in $M$. [F3, F11, step 1.3, algebra] For $n\ge 1$, the degree-$n$ term has the single basis wedge ${\theta }_1\wedge \cdots \wedge {\theta }_n$ and there is no degree-$(n+1)$ term. Thus $H_n({\mathcal{K}}_{\bullet }(M))=\ker d_n$. Its differential is $d_n(m{\theta }_1\wedge \cdots \wedge {\theta }_n)={\sum }_{i=1}^n(-1{)}^{i-1}(x_{i}m-m{x}_i){\theta }_1\wedge \cdots \wedge \widehat{{\theta }_i}\wedge \cdots \wedge {\theta }_n\mathrm{.}$ The target has the distinct displayed basis wedges, so this is zero exactly when $x_{i}m=m{x}_i$ for each generator $x_i$. Since these generators generate the polynomial algebra, that is equivalent to $rm=mr$ for every $r\in R$: the equality extends from generators to their products by induction and then to polynomial linear combinations by additivity. If $n=0$, the condition is vacuous because $R=k$ and $M$ is $k$-central; the degree-zero complex is $M$ with zero differential, so $H_0=M$.

Construct degree-zero comparison maps by homogeneous lifts. [F2, F3, F4, F9, F11, step 1.2, step 2.1] The comparison in step 2.1 can be chosen to preserve internal degree. Indeed, $R^{{\otimes }_{k}q}$ has its homogeneous monomial basis. The map $R^{{\otimes }_{k}q}{\otimes }_{k}S⟶{\mathrm{Bar}}_q(R),(a_1\otimes \cdots \otimes a_q)\otimes (a\otimes b^{\mathrm{o}\mathrm{p}})⟼b\otimes a_1\otimes \cdots \otimes a_q\otimes a$ is an isomorphism of right $S$-modules. Its inverse sends $b\otimes a_1\otimes \cdots \otimes a_q\otimes a$ to $(a_1\otimes \cdots \otimes a_q)\otimes (a\otimes b^{\mathrm{o}\mathrm{p}})$; it is right $S$-linear because $(a\otimes b^{\mathrm{o}\mathrm{p}})(c\otimes d^{\mathrm{o}\mathrm{p}})=ac\otimes (db{)}^{\mathrm{o}\mathrm{p}}$ and the right bar action sends the outer slots $(b,a)$ to $(db,ac)$. Thus these monomials give a homogeneous free basis for every bar term; for $q=0$ the middle tensor is $k$ and the basis is $\{1\}$. The Koszul terms are free on homogeneous wedge symbols by [F3], and [F11] makes their differentials and augmentations degree-zero maps. Recursively construct a comparison map $f:P_{\bullet }\to Q_{\bullet }$ in either direction. At degree zero, for each homogeneous free generator $g\in P_0$, choose a homogeneous lift in $Q_0$ of its image in $R$ under the augmentation of $P_0$. At degree $q>0$, after $f_{q-1}$ is defined, the element $z=f_{q-1}(d_{P}g)$ is a cycle because $d_P^2=0$ and the already constructed maps commute with the differentials. Exactness of $Q_{\bullet }\to R$ gives a preimage $y\in Q_q$ with $d_{Q}y=z$. Since $z$ is homogeneous and $d_Q$ has internal degree zero, the component of $y$ in the degree of $z$ is also a preimage. Choose such homogeneous lifts using AC [F9], and extend $S$-linearly on the free basis. This constructs degree-zero comparison maps in both directions.

Construct degree-zero homotopies between comparison composites and identities. [F7, F8, F9, F11, F12, F13, step 1.1, step 3.1] The homotopies between the comparison composites and the identity maps can also be chosen degree-zero. For either composite $u$ and identity $v$ on a resolution $P_{\bullet }$, set $s_{-1}=0$. At degree zero, $u_0-v_0$ has zero augmentation because both maps lift $1_R$, so on each homogeneous generator choose a homogeneous preimage under $d_P$. At degree $q>0$, after $s_{q-1}$ is defined, put $r_q=(u_q-v_q)-s_{q-1}{d}_P$. The chain-map identities and the homotopy equation in degree $q-1$ give $d_P{r}_q=0$. Exactness makes $r_q(g)$ a boundary for each homogeneous generator $g$; taking the required-degree component of a preimage gives a degree-zero lift $s_q(g)$. AC chooses the lifts on all homogeneous free generators and $S$-linear extension gives degree-zero homotopies. [F7, F9, step 1.1, step 3.1] After tensoring with graded $M$, [F12]--[F13] give the total internal grading on $K_{\bullet }{\otimes }_{S}M$; the degree-zero comparison maps and homotopies therefore induce an internal-degree-preserving isomorphism on homology.

Combine the comparisons to obtain the natural graded homology isomorphism. [F6, F11, F12, step 1.1, step 2.1, step 4.1, step 1.3, step 2.2, step 3.1] Combining steps 1.1, 1.3, and 2.1 gives the asserted homology isomorphism. For a bimodule map $f:M\to N$, the tensor maps $1\otimes f$ commute with the fixed comparison maps and with the bar-to-Hochschild chain isomorphism [F6], so the isomorphism is natural. In the graded case, each ${\theta }_i$ has degree two; therefore a summand $M{\theta }_I$ has the internal degree of $M$ shifted by $2\mathrm{\mid }I\mathrm{\mid }$, and the isomorphism built in steps 3.1 and 4.1 preserves this degree. Step 2.2 establishes the top-degree centralizer description.

Check zero, one, empty and endpoint cases, and record AC use. [F3, F5, F7, F9, F10, F11, F14, step 2.1, step 3.1, step 4.1, step 1.3, step 2.2] If $M=0$, all terms and homology groups vanish. If $n=0$, then $R=S=k$, the exterior algebra has only its empty wedge in degree zero, and the complex is $M$ in degree zero with zero differential; the diagonal resolution is the identity resolution and the comparison above gives $H{H}_0(k,M)=M$ and $H{H}_j(k,M)=0$ for $j>0$. If $n=1$, the only nonzero differential is $m{\theta }_1\mapsto x_{1}m-m{x}_1$, with positive sign, and the formula yields its kernel and cokernel in degrees one and zero. For every $n$, there are no terms above degree $n$, so $H{H}_j(R,M)=0$ for $j>n$. The empty wedge gives ${\mathcal{K}}_0(M)=M$ and the differential out of degree zero is zero. The only choice principle used is AC: it supplies the bar projectivity through [F5], implies DC for [F7], and selects homogeneous lifts in steps 3.1 and 4.1; the monomial basis and the finite Koszul wedge basis are explicit. [F3, F5, F9, F10, F11, step 2.1, step 1.3] $\mathrm{\square }$