14

I'm currently trying to learn about Hochschild homology of differential graded algebras. After reading the definition, the notion of Hochschild homology is somewhat unmotivated and myterious to me. What is the motivation to define Hochschild homology and what are some nice examples?

I'm particularly interested in the Hochschild homology of truncated polynomial algebras $$k[x]/(x^{n+1})$$ where $k$ is a field of characteristic zero and $x$ is of some degree $d$.

Are there any nice references for Hochschild homology?

Jendrik Stelzner
  • 17,081
Dave
  • 1,853
  • 1
    You can try chapter 9 in Weibel's An Introduction to Homological Algebra, he does quite a lot there. Although not that much in terms of motivation. On the other hand, the definition seems pretty natural to me and not that dissimilar from e.g. de Rham cohomology or simplicial methods (and indeed there are connections to both as you can learn in Weibel). – Marek Oct 03 '13 at 13:32
  • 2
    You might take a look at the Hochschild-Kostant-Rosenberg THM: http://ncatlab.org/nlab/show/Hochschild-Kostant-Rosenberg+theorem –  Oct 03 '13 at 13:47
  • Also, Loday's Cyclic Homology is a nice reference. –  Oct 03 '13 at 13:48
  • Writing down the Hochschild homology (HH) of the abstract $k$-algebra $k[x]/(x^{n+1})$ shouldn't be to hard. But I don't know how the HH of a DGA is defined (maybe it equals the HH of an abstract algebra if the DGA is concentrated in a single degree like yours ?). So can you please give the definition of HH of a DGA ? – Ralph Oct 03 '13 at 15:03
  • Thank you all for the references! – Dave Oct 03 '13 at 20:22
  • @Ralph: I found the definition of Hochschild homology of a DGA in chapter 5.9 of the book Algebraic models in Geometry by Y.Félix, J. Oprea and D. Tanré. Yes, I think if the DGA is concentrated in a single degree, then it should be the same as the HH of an algebra. – Dave Oct 03 '13 at 20:28
  • Yes, but I was of course wrong because your DGA isn't concentrated in a single degree as long as $d> 0$. – Ralph Oct 14 '13 at 17:53
  • To add to @user55407, the HKR can apparently be generalized to the derived de-Rham complex. So, given a complete intersection ring $R[x_1,\ldots,x_n]/(f_1,\ldots,f_k)$ you can associate a dga via the Koszul complex $K_\bullet(f_1,\ldots,f_k)$. From here you can compute the cotangent complex $\mathbb{L}$ and the derived de-Rham complex $\wedge^\bullet \mathbb{L}$. The derived de-Rham complex should be (quasi-)isomorhic to the hochschild homology. – 54321user Jun 02 '17 at 19:21
  • Also, it may be overkill, but check out https://arxiv.org/abs/0710.1937; there are useful theorems for computing $HH$ of dgas. – 54321user Jun 02 '17 at 19:35

1 Answers1

7

Set $R = k[x]/(x^{n+1}),\,u=x\otimes 1-1\otimes x,\,v=\sum_{i=0}^n x^i\otimes x^{n-i} \in R^e := R \otimes_k R$.

First, let's recall from Weibel (Ex. 9.1.4) that in the ungraded case a projective resolution of $R$ over $R^e$ is given by the periodic complex $$\cdots \xrightarrow[]{v} R^e \xrightarrow[]{u} R^e \xrightarrow[]{v} R^e \xrightarrow[]{u} R^e \xrightarrow[]{\mu} R \to 0$$

Now suppose $R$ is a DGA with $\deg(x)=d$ and zero differentials. The latter implies that the notions of the Hochschild homology of $R$ as DGA and as graded algebra agree. Hence we can compute the Hochschild homology of $R$ by a projective resolution of $R$ over $R^e$ in the category of graded $R^e$-modules.

For a graded $R^e$-module $M$ let $\Sigma^kM$ be the shifted graded $R^e$-module given by $(\Sigma^kM)_i := M_{i-k}$. Set $e_k := (0,\ldots,1\otimes 1,\ldots 0) \in (\Sigma^kR^e)_k$. Then $\Sigma^kR^e=R^e\cdot e_k$ is a free graded $R^e$-module (in particular it's a projective object in the category of graded $R^e$-modules).

Taking into account $\deg u = d, \,\deg v=nd$, we can adjust the projective resolution from Weibel above and find the following projective resolution of $R$ over $R^e$ (taken in the category of graded $R^e$-modules): $$\cdots \to \Sigma^{(n+1)d}R^e \xrightarrow[]{d_2} \Sigma^dR^e \xrightarrow[]{d_1} R^e \to R \to 0$$ $$\cdots \to \Sigma^{(n+1)di}R^e\xrightarrow[]{d_{2i}}\Sigma^{(n+1)di-nd}R^e \xrightarrow[]{d_{2i-1}}\Sigma^{(n+1)d(i-1)}R^e\to\cdots $$ where $d_{2i}: e_{(n+1)di} \mapsto v\cdot e_{(n+1)di-nd},\,d_{2i-1}: e_{(n+1)di-nd} \mapsto u \cdot e_{(n+1)d(i-1)}$.

Now $HH_\ast(R,M)$ can be computed by tensoring this complex with $M$ (over $R^e$) and taking the homology. Using the relation $M \otimes_{R^e}\Sigma^kR^e=\Sigma^k M$ we obtain, for example, for $M=R$ the complex $$\displaystyle\cdots \to \Sigma^{(n+1)di}R\xrightarrow[]{d_{2i}}\Sigma^{(n+1)di-nd}R \xrightarrow[]{0}\Sigma^{(n+1)d(i-1)}R\to\cdots $$ where $d_{2i}: e_{(n+1)di} \mapsto (n+1)x^n\cdot e_{(n+1)di-nd}$. Hence

If $n+1$ is invertible in $k$ then (as graded $R$-module) $$HH_{2i}(R,R)=\Sigma^{(n+1)di}Rx,\quad HH_{2i-1}(R,R)=\Sigma^{(n+1)di-nd}R/(x^n), \quad H_0(R,R)=R.$$

Ralph
  • 1,880
  • Thank you very much! It looks very interesting, I'll have to think about it some more. – Dave Oct 15 '13 at 15:02
  • 1
    You wrote that "[zero differentials] implies that the notions of the Hochschild homology of $R$ as a DGA and as a graded algebra agree." This seems to disagree with Loday's discussion in section 5.3.4 of Cyclic Homology: he writes "Note that if $\delta = 0$, then $HH$ and $HC$ of $(A,0)$ are not equal a priori to $HH$ and $HC$ of $A$ since the last face $d_n$ is not the same in the two cases. However if $A$ is concentrated in degree 0, then both theories agree." – JHF Mar 29 '16 at 14:55
  • This looks great.Now my question is what does this compute? Is it seeking some holes? topological invariants etc, are there analogs of betti numbers and what not. It seems easy enough to construct but what does it mean? –  Mar 30 '16 at 09:37
  • @JHF I think the following: Given a DGA R, one can either compute its HH as of a DGA, or one can forget the differential and the grading and compute its HH as of an associative algebra. This is what Loday compares in section 5.3.4; they indeed do not agree (because of the signs) even when $d=0$. However, in the post above, the term "HH of a graded algebra" is used, which is, I suppose, by definition exactly HH of R as of a DGA with zero differential. One can even define HH of an $A_\infty$-algebra and, I think, the above is a special case of this definition. – Pavel Aug 01 '18 at 11:25