Matrix-unit rotation for a k–Mat_n(k) Morita pair

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.