I asked this question some weeks ago at mathoverflow, but I got no reply.
Here, by 3D-grid of sidelength $k$ I mean the graph $G=(V,E)$ with $V= \{1,\ldots,k\}^3$ and $E=\{( (a,b,c) ,(x,y,z) ) \mid |a-x|+|b-y|+|c-z|=1 \}$, i.e., the nodes are placed at 3-dimensional integer coordinates between 1 and $k$, and a node is connected to the at most 6 other nodes that differ in precisely one coordinate by one.
What is the name of this graph? I'll use 3D grid, but perhaps 3D mesh or 3D lattice are what other people are used to.
What is the treewidth or pathwidth of this graph? Is this already published somewhere?
I know already that $tw(G) = (3/4) k^2 + O(k)$, i.e. it is really smaller than $k^2$. To me, this suggests that the standard arguments showing that the $k\times k$ 2D-grid has treewidth and pathwidth $k$ will not easily generalize.
To see this, we consider a path decomposition that "sweeps" the grid using mainly node-sets of the form $S_c= \{(x,y,z)\mid x+y+z = c\}$. Observe $|S_c| \leq (3/4) k^2 + O(k)$, $S_{3/2 k}$ being the largest such set. The sets between $S_c$ and $S_{c+1}$ are created by sweeping with a line and need $O(k)$ additional nodes to be separators. More precisely, use the sets $S_{c,d} = \{(x,y,z)\mid (x+y+z = c \wedge x \leq d ) \vee (x+y+z = c \wedge x \geq d ) \}$ as a path decomposition of $G$.
I also have an idea for a proof that shows $tw(G) = \Omega(k^2)$, but that is not finished yet.