Given two graphs $G=(V\_1,E\_1)$ and $H=(V\_2,E\_2)$, the tensor product of $G$ and $H$ is the graph $G \times H = (V,E)$, where $V=V\_1 \times V\_2$ is the Cartesian product of the $V\_i$ and 
$ (u,v) \ E \ (u',v') \Leftrightarrow u E\_1 u' \wedge v E\_2 v'$.


Is anyone aware of a characterization of which $G,H$ give rise to a planar tensor product $G \times H$?



When is the tensor product of two graphs planar?
Yuichiro Fujiwara's comments seem to answer the question in full so I am making it an answer. I quote below from [Kronecker products and local joins of graphs](http://cms.math.ca/cjm/v29/cjm1977v29.0255-0269.pdf) by M. Farzan and D. A. Waller, *Can. J. Math.* **29** (1977), 255–269.



> 
> By a *1-contraction* of $G$ we mean the removal from $G$ of each vertex of degree 1 (and its incident edge).
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> 
> 5.3 THEOREM. Let $G\_1$ and $G\_2$ be connected graphs with more than four vertices. Then $G\_1\times G\_2$ is planar if and only if either
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> 
> (i) one of the graphs is a path and the other one is 1-contractible to a path or a circuit, or
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> (ii) one of them is a circuit and the other is 1-contractible to a path.
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> 


This leaves open only the case in which at least one of the graphs has fewer than five vertices. Proposition 5.4, which I won't bother reproducing here, covers some of these cases. [Beaudou et al.](https://hal.archives-ouvertes.fr/hal-00387303/document) addresses the case $G=K\_2$. A complete solution appears to be still an open problem.


