Computations

Let $A=k$, the ground ring. Then
$$\begin{align*} \HH_0(k) &= k,\\ \HH_n(k) &= 0,\quad n\geq 1. \end{align*}$$
The periodicity exact sequence () implies that
$$\begin{align*} \HC_{2n}(k) &= k,\\ \HC_{2n+1}(k) &= 0, \end{align*}$$
so also
$$\begin{align*} \rH^{\la}_{2n}(k) &= k,\\ \rH^{\la}_{2n+1}(k) &= 0. \end{align*}$$

Let $A=T(V)$ be the tensor algebra over $V$, that is

$$T(V)=\bigoplus_{n=0}^{\infty} V^{\ox n},\quad (v_1, \hdots, ... ...n+1}, \hdots, v_{n+m})= (v_1, \hdots, v_{n+m})\in V^{\ox(n+m)}$$

Then
$$\begin{align*} \HH_0(T(V)) &= \bigoplus_{m\geq 0} V^{\ox m}/(1-\tau)=\bigoplus_{... ...= \bigoplus_{m\geq 0} (V^{\ox m})^{\bZ/m\bZ}, \\ \HH_1(T(V)) &= 0, \end{align*}$$
where $\tau$ is the cyclic operator without sign.

$$\HC_n(T(V)) = \HC_n(k)\oplus\underbrace{\bigoplus_{m> 0} \rH... ...^{\ox m})}_{\mbox{This is zero in the characteristic 0 case.}}$$

Consider now matrix algebras $M_n(A)$ for a unital associative algebra $A$ over a field $k$. There are isomorphisms
$$\begin{align*} \HH_*(M_r(A)) &\isom \HH_*(A), \\ \HC_*(M_r(A)) &\isom \HC_*(A). \end{align*}$$
The map $A\to M_r(A)$ is given by

$$a\mapsto \twobytwo{a}{0}{0}{0}.$$

In the opposite way $\Tr\:M_r(A)\to A$ we have the trace map

$$\al=[\al_{ij}]\mapsto \sum_{i}\al_{ii}.$$

There is also a trace map $\Tr\:M_r(A)^{\ox(n+1)}\to A^{\ox(n+1)}$

$$\Tr(\al^0, \hdots, \al^n):= \sum_{(i_0, \hdots, i_n)}\al^0_{i_0i_1}\ox\al^1_{i_1i_2}\ox\hdots\ox\al^n_{i_ni_0}$$

We claim that this map commutes with the faces and with the cyclic operator.

Let $k$ be a field and $A$ a commutative $k$-algebra. Define the space of 1-forms on $A$, denoted by $\Om^1_{A/k}=\Om^1_A$, as an $A$-module generated by elements $da$ for every $a\in A$ satisfying following relations
$$\begin{align*} d(\la a+\mu b) &= \la da + \mu db \mbox{ (linearity),}\\ d(ab) &= adb + bda \mbox{ (Leibniz rule).} \end{align*}$$
Define the space of $n$-forms as an $n$-th exterior power of $\Om^1_A$

$$\Om_A^n:=\La^n_A\Om^1_A.$$

Elements of $\Om_A^n$ can be written as $a_0da_1\hdots da_n$, $a_i\in A$, $i=0, \hdots, n$, with the relation

$$dada'=-da'da.$$

Define a differential of an $n$-form as

$$d(a_0da_1\hdots da_n):=1da_0da_1\hdots da_n.$$

$$d\: \Om^{n}_A \to \Om^{n+1}_A,\quad d\circ d=0.$$

Now $\Om_A^{\bullet}$ is a cochain complex and its homology is called deRham cohomology of an algebra $A$

$$\drH(A):=\rH_n(\Om_A^{\bullet}, d).$$

If $A$ is commutative, $M$ an $A$-module, then

$$\rH_1(A; M)=M\ox_A\Om_A^1.$$

There is a map

$$\pi\:C_n(A)=A^{\ox (n+1)} \to \Om_A^n$$

$$(a_0, \hdots, a_n)\mapsto a_0 da_1\hdots da_n$$ (1)

There is a map also in the opposite way

$$\Om_A^n\xrightarrow{\eps_n} \HH_n(A)$$

$$\eps_n(a_0da_1\hdots da_n):=\sum_{\sg\in S_n} \sign(\sg)(a_0, a_{\sg(1)}, \hdots, a_{\sg(n)}).$$ (2)

Passing to Hochschild homology it gives a well defined map $\Om_A^n\to \HH_n(A)$. In charecteristic 0 case the composition of the maps in () and () gives an isomorphism

$$\Om_A^n \to \HH_n(A) \to \Om_A^n.$$

Now we can form a map of bicomplexes

$$\xymatrix{ \vdots \ar[d] & \vdots \ar[d] &\vdots \ar[d] & \v... ...\Om^0 \ar[l]_d \\ \Om^1 \ar[d]_0 & \Om^0 \ar[l]_d \\ \Om^0 }$$

As a corollary we have that for a formally smooth algebra $A$ over characteristic 0 field $k$

$$\HC_n(A)\isom \Om^n_A/ d\Om^{n-1}_A\oplus \drH^{n-2}(A) \opl... ...n-4}(A) \oplus \hdots\oplus \drH^{0}(A)\text{ or }\drH^{1}(A).$$