# Why is the Hochschild homology of k[t] just k[t] in degrees 0 and 1?

Background: the Hochschild homology of an associative algebra is the homology of the complex

$$\ldots \longrightarrow A \otimes A \otimes A \longrightarrow A \otimes A \longrightarrow A$$

where the last two differentials are $$a \otimes b \otimes c \mapsto ab \otimes c-a \otimes bc + ca \otimes b$$ and $a \otimes b \mapsto ab-ba$, and you can guess the rest. More generally, it's "derived coinvariants": take a projective resolution of your algebra, then coinvariants of that.

For $k[t]$, the Hochschild homology is concentrated in degrees 0 and 1, and in both of those degrees it's $k[t]$. I know that I can go look up Loday (\S 3.2.2) and find a calculation, but I'd like a better explanation.

I know that the zero-th Hochschild homology $\operatorname{HH}_0(k[t])$ must just be $k[t]$, because the zero-th Hochschild homology is just coinvariants, and $k[t]$ is commutative.

What I'd like is a "good" explanation for $\operatorname{HH}_1(k[t])$.

Edit: Ben has a simple explanation below. Let me also rephrase the question, hoping for more. Here are a few things: If $A$ is semisimple, then $\operatorname{HH}_{\ast}(A)$ is concentrated in degree 0. Is there something about $k[t]$ that ensures it's concentrated in degrees 0 and 1? Conversely, can I conclude anything about $A$ from the fact that $\operatorname{HH}_{\ast}(A)$ is zero above $\ast=1$?

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You were just trying to bait me with that one, weren't you? – Ben Webster Oct 7 '09 at 19:32
I was rather assuming that you'd answer :-) – Scott Morrison Oct 7 '09 at 20:25

Another way to write the Hochschild homology is as follows:

take A as a bimodule over itself, take a free resolution as a bimodule, and then apply the functor of coinvariants ($M \mapsto M/\langle rm-mr|r\in A\rangle$).

Your definition used the "bar-complex" resolution of the form $\to A \otimes A \otimes A \to A \otimes A$ but k[t] has a much nicer resolution as a bimodule over itself, the Koszul resolution.

This is of the form $k[t] \otimes k[t] \to k[t] \otimes k[t]$ with the map given by $1 \otimes t - t \otimes 1$, so when you apply coinvariants, you get two copies of $k[t]$ with trivial differential.

Actually all Koszul algebras have a nice resolution of the diagonal bimodule, and thus its easier to compute their Hochschild homology, though in general, they don't always have trivial differential after applying coinvariants.

EDIT: For the later question, probably the best answers you'll get are from HKR, though just noting that the global dimension of $k[t] \otimes k[t]$ is 2 gets you halfway there.

EDIT AGAIN: Actually, any Koszul algebra has its Hochschild homology bounded above by its global dimension. This is clear from the existence of the diagonal Koszul resolution.

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If you're looking at Hochschild homology of COMMUTATIVE algebras (as in your example) there's a very simple general geometric explanation, the Hochschild-Kostant-Rosenberg theorem: the Hochschild homology of A is is simply the exterior algebra of differential forms (placed in homological degree). Said more formally: Sym Omega^1[1] -- symmetric algebra on the shifted module of 1-forms (which makes it an exterior algebra). The nice thing about this form of the answer is it doesn't require A to be smooth, if you interpret Omega^1_A correctly - namely as the cotangent complex of A.. this is due to Quillen in the gorgeous short paper "On the (co)homology of commutative rings". Anyway this gives a general explanation of the above examples.

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Could this be explained in detail in your new paper, please? – Ilya Nikokoshev Oct 23 '09 at 20:11
(let's assume char=0 before I get into trouble) – David Ben-Zvi Oct 23 '09 at 20:14
Ilya -- yes, there's a very cute (IMHO) very short proof of this there. Basically Hochschild homology is (by definition - once you understand the definition) functions on the derived loop space of Spec R. But maps into an affine variety always factor through the affinization of your space (Spec of global functions). The affinization of the circle (in char zero) is Spec k[e]/e^2=0, where degree(e)=1. So maps from the affinization are just the tangent bundle of Spec R, with a shift (tangent complex, if R is singular). Hence functions on it are the symmetric algebra of one-forms with a shift. – David Ben-Zvi Oct 23 '09 at 20:18
This is definitely char. 0 - and let me doubly emphasize Quillen's paper, it's a great place to learn "derived" thinking in general. – Tyler Lawson Oct 24 '09 at 2:39
This is discussed in some detail in my new paper with David Nadler, which appeared as arxiv.org/abs/1002.3636 – David Ben-Zvi Feb 23 '10 at 3:56

Maybe it's wrong of me to add this as a separate answer, but it is a separate thought.

If you read the paper Geordie Williamson and I wrote on the subject, you'll see that our theorems identify the Hochschild homology of k[t] with the equivariant cohomology of the circle with a trivial circle action. Thus, the bits in degree 0 and 1 correspond to the cohomology of the circle, and the fact that you get k[t] in each corresponds to the fact that k[t] is the cohomology of the classifying space of the circle as a topological group.

There's actually a generalization of this to all Soergel bimodules (explained in the paper above) which says that the Hochschild homology of any irreducible Soergel bimodule has this form where the cohomology of the circle is replaced by the intersection cohomology of the closure of a Bruhat cell BwB in the corresponding group.

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I think this is the "deeper" explanation. Some directions where one might go from there even deeper, I believe, is in arxiv.org/abs/0706.0322 (which I'll be happy to discuss) – Ilya Nikokoshev Oct 7 '09 at 19:56
Well, depth is a strange concept. I mean, certainly this is for me the more interesting answer, but I'm not sure to what degree it's "really about" Hochschild homology. Anyways, I'm glad I put both. – Ben Webster Oct 7 '09 at 20:07

Is there something about k[t] that ensures it's concentrated in degrees 0 and 1.

Yes: it's a free associative algebra - that's why you only have 0th and 1st homology.

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More generally, it is an hereditary algebra. – Mariano Suárez-Alvarez Nov 17 '09 at 3:41

HH^2(A) is equal to deformations of the multiplication of A modulo trivial deformations (see e.g. http://arxiv.org/pdf/hep-th/9408064v2), so if there were a way to see that k[t] had only trivial deformations without calculating HH^2, then maybe this would answer your revised question. HH^k is similarly related to k-deformations (whose definition I forget; it might require an A-infinity point of view), so maybe this approach would work for higher Hochschild cohomology as well.

I'm satisfied by Ben's first answer, though.

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HH^k controls deformations of higher A-infinity products. – Ben Webster Oct 13 '09 at 15:18