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 $C\_{n\_1}\vee\ldots\vee C\_{n\_k}$?



Hochschild homology of quiver algebras
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 $HH\_1(A)$.


