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Let $K$ be a field and $\Gamma$ a quiver (=multidigraph) and $K[\Gamma]$ its quiver algebra (free $K$-module on the set of all paths of length $\geq0$ where multiplication is concatenation if endpoints match and $0$ otherwise).

I'm looking for families of $\Gamma$ for which Hochschild homology $HH_\ast(K[\Gamma])$ hasn't yet been computed.

If $\Gamma$ is the directed $n$-cycle (vertices $1,\ldots,n$ and edges $(1,2),(2,3),\ldots,(n-1,n),(n,1)$), or if $\Gamma$ is the $n$-loop (vertex $1$ and edges $(1,1),\ldots,(1,1)$ with $n$ copies), what is $\dim_KHH_k(K[\Gamma])$? How about if $\Gamma$ is the one-point-wedge of finitely many directed cycles $C_{n_1}\vee\ldots\vee C_{n_k}$?

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up vote 6 down vote accepted

The Hochschild homology of all path algebras $A=k\Gamma$ is known. I will assume the quiver is finite.

They are hereditary algebras, so $HH_p(A)$ is zero as soon as $p>0$ provided $A$ is finite dimensional (which in this case is the same thing as $\Gamma$ being acyclic): this follows from the theorem of Keller that Dag Madsen mentioned the other day to you.

In general, there is a very short and simple projective resolution of $A$ as a bimodule, from which one can actually compute with ease. Let $E\subseteq A$ be the subalgebra spanned by the vertices and $V\subseteq A$ be the subspace spanned by the arrows. Then both $A$ and $V$ are $E$-bimodules in a natural way, so we can consider the complex $$0\to A\otimes_EV\otimes_E A\xrightarrow{d}A\otimes_EA$$ with $d$ the unique $A$-bimodule map such that $d(1\otimes v\otimes 1)=v\otimes 1-1\otimes v$. You should be able to check that this is in fact a projective resolution of $A$ as an $A$-bimodule, and use it to compute $HH_*(A)$.

This is to answer Dag's question in the comment. Let's use the above resolution $P$ to compute $HH_*(A)=\operatorname{Tor}_*^{A^e}(A,A)$, so we have to compute $H(A\otimes_{A^e}P)$. The complex $A\otimes_{A^e}P$ is, up to hopefully obvious identifications, $$0\to A\otimes_E V\otimes_E\to A\otimes_E\tag{$\star$}$$ where the notation $A\otimes_E$ means $A\otimes_{E^e}E$ and the differential is the $k$-linear map such that $d(a\otimes v)=[a,v]$. Now $A$ has the set of paths in $\Gamma$ as basis, $A\otimes_E$ has the set of closed paths in $\Gamma$ as basis, and $A\otimes_EV\otimes_E$ has the set of elementary tensors of the form $u\otimes\alpha$ with $u$ and $\alpha$ a path and an arrow in $\Gamma$, respectively, such that $u\alpha$ is a closed path; the image of the map $d$ is then generated by the elements $ua-au$ for all such elementary tensors.

If $u$ is a closed path in $\Gamma$ of positive length, then for all $n\in\mathbb N$ we have $u^n\in A\otimes_E$. There is a $k$-linear map $\epsilon:A\otimes_E\to k$ which maps every closed path to $1$; this map clearly vanishes on the image of $d$ and then, as $\epsilon(u^n)=1$, we see that $u^n$ is not a boundary. Moreover, the set $\{u^n:n\geq1\}$ is linearly independent modulo boundaries because the image of $d$ is a homogeneous subspace of $A\otimes_E$ for the grading by length. It follows that $HH_0(A)$ is infinite dimensional as soon as there is a cycle of positive length in $\Gamma$.

Let $\ell\geq1$ and let us look at the degree $\ell$ component of the complex ($\star$), again for the length graduation, we have $$0\to A_{\ell-1}\otimes_E V\otimes_E\xrightarrow{d_\ell} A_\ell\otimes_E$$ where $A_i$ denotes the subspace spanned by paths of length $i$. The dimension of $A_\ell\otimes_E$ is the number of closed paths of length $\ell$, and the dimension of $A_\ell\otimes_EV\otimes_E$ is the number of pairs $(v,\alpha)$ with $v$ a path, $\alpha$ and arrow and $v\alpha$ a closed path. These two numbers are the same: there is an bijective map from closed paths of length $\ell$ to pairs $(u,\alpha)$ of this form --- simply split off the arrow. It follows that the kernel of $d_\ell$ is zero iff its cokernel is zero. Since we know that $HH_0(A)$ is infinite dimensional, it follows that so is $HH_1(A)$.

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For the algebras you mention in the question doing this computation is very easy (none of them are finite dimensional, so Keller's theorem does not apply to them, of course) Decribing the actual dimensions is a combinatorial problem (notice that the dimension is generally infinite in the case where $\Gamma$ has cycles) For example, in the $n$-loop case, in which the algebra is the free algebra on $n$ generators, the dimensions (of the graded components of a given degree) of the homology are necklace numbers. – Mariano Suárez-Alvarez Mar 12 '14 at 7:52
Is the following true: If there is an oriented cycle in the quiver, and we don't divide up each degree in graded components, then $HH_{\ast}(k \Gamma)$ is infinite dimensional in degree $0$ and $1$, and zero in all other degrees? – Dag Oskar Madsen Mar 12 '14 at 11:22
Mitchell's classical paper on rings with several objects from 72 shows that free categories have Hochschild cohomoogical dimension 1 and gives that resolution. – Benjamin Steinberg Mar 12 '14 at 12:53
Thank you very much for your help, Mariano. Hmm, I'm running out of algebras. Sorry, just one more question, if I may (or would it be better as a new post?): is HH of monomially related commutative algebras $K[x_1,\ldots,x_n]/(x^{a_1},\ldots,x^{a_k})$ or Stanley-Reisner rings been computed? For example, what is HH of the "simplicial sphere" $K[x_0,\ldots,x_n]/(x_0\cdots x_n)$? – Leon Mar 14 '14 at 4:54
You really should look at the literature to see for what sort of álgebras people compute HH. – Mariano Suárez-Alvarez Mar 14 '14 at 4:58

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