TITLE: When is the tensor product of two graphs planar?
QUESTION [7 upvotes]: 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 $G \times H$?

REPLY [6 votes]: 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 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).
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
(i) one of the graphs is a path and the other one is 1-contractible to a path or a circuit, or
(ii) one of them is a circuit and the other is 1-contractible to a path.

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. addresses the case $G=K_2$.  A complete solution appears to be still an open problem.