I had 2 kind of dumb questions about the definition of Hochschild homology in terms of the Tor functor:

1 - Let $R$ be a $k$-algebra and $M$ an $R$-bimodule, let $H_*(R,M)$ be the Hochschild homology of $R$ with coefficients in $M$, this homology can also be defined in the following way: the "bar resolution" of $R$ (which is itself an $R$-bimodule) as an $R$-bimodule is:

$B(R,R) = \cdots \xrightarrow{b'} R \otimes R \otimes R \otimes R \xrightarrow{b'} R \otimes R \otimes R \xrightarrow{b'} R \otimes R$, $\space \space \space \space$(1)

here we write $\otimes$ for $\otimes_k$, $B(R,R)_n = R^{\otimes n+2}$ and $B(R,R)_0 = R \otimes R$,

$b'_n(a_o \otimes \cdots \otimes a_{n+1}) = \Sigma_{i=0}^n (-1)^i a_0 \otimes \cdots \otimes a_ia_{i + 1} \otimes \cdots \otimes a_{n+1}$,

and $b' \circ b' = 0$.

Let $M$ be an $R$-bimodule (hence a right $R^e$-module), if we apply the functor $(M \otimes_{R^e}-)$ to the complex (1) we're supposed to get back the Hochschild homology:

$\cdots \xrightarrow{b} M \otimes R \otimes R \otimes R \xrightarrow{b} M \otimes R \otimes R \xrightarrow{b} M \otimes R \xrightarrow{b} M$,

which proves that $H_n(R,M) = Tor_n^{R^e}(M,R)$.

The thing here is when I apply the functor $(M \otimes_{R^e}-)$ to the complex (1) I get back:

$\cdots \xrightarrow{b} M \otimes R \otimes R \otimes R \otimes R \xrightarrow{b} M \otimes R \otimes R \otimes R \xrightarrow{b} M \otimes R \otimes R$,

not the complex (1)

What am I doing wrong?

2 - Another thing, "the bar resolution" of a left $R$-module $M$ (as defined on page 283 of Weibel's "Homological algebra") is

$B(R,M) = \cdots \rightarrow R \otimes R \otimes R \otimes M \rightarrow R \otimes R \otimes M \rightarrow R \otimes M \rightarrow M$,

where $B(R,M)_n = R^{\otimes n+1} \otimes M$ and $B(R,M)_0 = R \otimes M$,

I was wondering, if $M$ is an $R$-bimodule

- Is this resolution also a resolution of $M$ as a left $R^e$-module?

- And is $R \otimes M$ projective as a left $R^e$-module?



2 Answers 2


I will try to answer to the first part of your question by introducing the Hochschild homology, showing how the bar resolution can be used to compute a $\operatorname{Tor}$ of interest and relate it to the Hochschild homology.

Let $R$ be a $k$-algebra and $M$ be an $R-R$ bimodule. With $k$ we denote a field and we write $\otimes$ for $\otimes_k$. The enveloping algebra of $R$ is given by $R^e:=R\otimes R^{op}$, denoting by $R^{op}$ the opposite algebra.

The Hochschild homology $HH(R,M)$ with coefficients in $M$ is the homology of the complex

$$\mathcal C(M,R):~~ 0 \leftarrow M\leftarrow M\otimes R\leftarrow M\otimes R\otimes R\leftarrow \cdots$$ with differential $d=\sum_i (-1)^i \partial_i$ defined in terms of well known face maps $\partial_i$. If we introduce the unnormalized bar resolution $\beta(R,R)$ of $R$, we can write the isormorphism $$\beta(R,R)_n:=R\otimes R^{\otimes n} \otimes R \simeq (R\otimes R^{op}) \otimes R^{\otimes n}=R^{e} \otimes R^{\otimes n}$$

of $R-R$-bimodules for each $n\geq 0$; by $R^e$-flatness of such resolution and by definition of $\operatorname{Tor} $ we arrive at $$\operatorname{Tor}^{R^{e}}(M,R)=H(M\otimes_{R^{e}}\beta(R,R)). $$

The isomorphism $\rho:M\otimes_{R^{e}}\beta_n(R,R)\rightarrow M\otimes R^n$, with $\rho(m,r_0,r_1,\dots,r_n,r_{n+1}):=(r_{n+1}mr_0,r_1,\dots,r_n)$ for all $n\geq 0$ (note that we used the fact that $M$ is an $R-R$-bimodule) gives the identification $$M\otimes_{R^{e}}\beta(R,R)= \mathcal C(M,R), $$

as wished.

  • $\begingroup$ Did it help with notation and stuff? $\endgroup$
    – Avitus
    Aug 17, 2014 at 14:00

Regarding your second question: if $M$ is a bimodule, the complex $B(R,M)$ is indeed a complex of $R$-bimodules (the objects are bimodules and the maps are bimodule maps), and it is exact, but in general it is not projective.

Indeed, $R\otimes M$ is not usually projective.

Notice, in fact, that there are isomorphisms $\hom_{R{-}R}(R\otimes M,N)\cong\hom_{{-}R}(M,N)$ natural in $N$, so that the bimodule $R\otimes M$ is projective iff $M_R$ is projective (for this, we need that $k$ be a field or something along those lines)


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .