# Tensor product of quivers

As a special case of a general construction I have constructed "accidentally" a tensor product of quivers aka directed multigraphs (aka directed graphs for category theorists). Probably this is already known, therefore I would like to know good references for this construction. I would also like to see pictures or a pictorial description of this construction.

Let $\Gamma = (V,E,s,t)$ and $\Gamma' = (V',E',s',t')$ be quivers. Define a new quiver $\Gamma \otimes \Gamma' = (V^{\otimes},E^{\otimes},s^{\otimes},t^{\otimes})$ as follows: The maps $s,t$ correspond to a map $E \times 2 \to V$, which gives a map $E \times 2 \times E' \to V \times E'$. Similarily, $s',t'$ induce $E \times 2 \times E' \to E \times V'$. The set of vertices is the pushout $V^{\otimes}=(V \times E') \cup_{(E \times 2 \times E')} (E \times V')$. The set of edges is $E^{\otimes} =E \times E'$. The obvious map $E^{\otimes} \times 2 \to V^{\otimes}$ yields source and and target maps $s^{\otimes},t^{\otimes} : E^{\otimes} \rightrightarrows V^{\otimes}$.

Thus, an edge in $\Gamma \otimes \Gamma'$ is a pair $(e,e')$ of edges $e: u \to v, e' : u' \to v'$ in $E$ and $E'$, with source $(u,e') \equiv (e,u')$ and target $(v,e') \equiv (e,v')$.

In fact, this endows the category of quivers with a symmetric monoidal structure, compatible with colimits in each variable. The unit is the quiver $\bullet \rightarrow \bullet$.

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I know it might be tricky to do but is there any chance of a picture..? The only other quiver product I've come across in my travels is the natural quiver version of the square or Cartesian product of graphs - see en.wikipedia.org/wiki/Cartesian_product_of_graphs for example. I don't think what you've got is that - I think the units are different (for the Cartesian product, I think the unit should be the null graph/quiver on one vertex.) –  Jan Grabowski Jan 10 '13 at 15:42
I don't know how to answer this from the perspective of the theory of quivers, but categorically we can see quivers as $\mathrm{Set}$-valued functors on the category with two objects and parallel morphisms between them. Then your tensor product seems to be the Day convolution product (see ncatlab.org/nlab/show/Day+convolution) for a particular monoidal structure on this indexing category. –  Karol Szumiło Jan 10 '13 at 17:56
Another tensor product that has been somewhat studied is the pointwise tensor product, see Ryan Kinser (arXiv:0711.1135) and Martin Herschend ("Tensor products on quiver representations", Journal of Pure and Applied Algebra 212 (2008) 452-469. This is different from yours, though. In Martin's paper he also discusses some other notions of tensor product (but I didn't read carefully). Also, I don't understand exactly what you've written: for e in E and e' in E', what is s(e\otimes e')? –  Hugh Thomas Jan 10 '13 at 22:04
@Jan: Yes, they are different (and therefore I mentioned the unit). Unfortunately, I don't know a picture. @Karol: I have also guessed that, but I don't think that that category with two objects is monoidal. @Hugh: These papers deal with representions of quivers, and are unrelated to my question (there one just uses the tensor product of vector spaces etc. pointwise). Concerning your question, the pushout comes equipped with a map from $E \times 2 \times E'$, i.e. two maps on $E \times E'$. These are source and target maps in the tensor product. I have added a more explicit description above. –  Martin Brandenburg Jan 10 '13 at 22:58
@Martin: I think I can write down some monoidal structure, but I have trouble coming up with a symmetric one. By the way, if $\bullet$ denotes a quiver with one vertex and no edges, do I read it correctly that $\bullet \otimes \bullet = \varnothing$? If so then your monoidal structure is not a Day convolution in the standard sense since then product of representables would be representable. However, it is a theorem of Day that every closed monoidal structure on a category of (co)presheaves is a Day convolution with respect to some promonoidal structure on the indexing category... –  Karol Szumiło Jan 11 '13 at 8:23