Is the F4 lattice bipartite? Is the $F_4$ lattice (i.e. 4-dimensional body-centered hypercubic lattice, spanned by the simple roots of $F_4$) bipartite? And if so, what is a good explicit partition of the vertices?
 A: No such labeling can exist because this lattice contains triangles such as $\lbrace(0,0,0,0), \phantom.(2,0,0,0), \phantom.(1,1,1,1)\rbrace$.  This is predictable from the root diagram, which contains the $A_2$ root system.  [Note that to identify this "body-centered hypercubic lattice" with a root lattice we must use the norm $\langle x,x\rangle = \frac12(x_1^2+x_2^2+x_3^2+x_4^2)$ so that the minimal vectors satisfy $\langle r,r \rangle = 2$.]
Indeed the nearest-neighbor graph of this lattice is not even $3$-colorable, because it has $4$-cliques such as the regular tetrahedron
$$
T = \lbrace(0,0,0,0), \phantom.(2,0,0,0), \phantom.(1,1,1,1), \phantom.(1,1,1,-1)\rbrace.
$$
This, too, is predictable from the root diagrams once we know that the $F_4$ lattice is the same as the $D_4$ lattice (the $F_4$ roots are vectors of norm $2$ and $4$ in $D_4$): the root diagram of $D_4$ contains $A_3$.  In general a root lattice has $n$-dimensional simplices spanned by roots if and only if it contains $A_n$.
There does exist a $4$-coloring: the vectors of norm $4$ (the long $F_4$ roots) span a sublattice $L_0$ of index $4$ isometric with $2^{1/2} D_4$, and we can assign each of its cosets a different color.
In fact this $4$-coloring is unique up to color permutation.  To prove this, first observe that each of the vertices of $T$ must have a different color.  The same is true for the tetrahedron obtained from $T$ by replacing $(2,0,0,0)$ by $(0,2,0,0)$, so these two minimal vectors must have the same color.  They differ by a long root of $F_4$, and all the long roots are equivalent under lattice isometries.  Hence any two vectors that differ by a long root have the same color, whence each coset of $L_0$ must have constant color, QED.
