A function f:A→B between two ordered sets A and B is called *unimodular* iff it either "goes up, then down", or "goes down, then up".

Getting technical, f is unimodular iff there exists a cut A=I∪J (a disjoint union of two intervals, such that ∀x∈I&y∈J.x<y) for which f is monotone on each of I and J.

Getting even less technical, f is unimodular iff it looks like a smiley or a frowney.

- Any monotone function.
- f(x)=x
^{2} and -f(x).
- cos(x) on the interval [-π,π]

There are many equivalent formulations of this condition. I have chosen the *simplest* one; however, proving its equivalence to other -- unfortunately more common -- formulations is very hard. The property is definitely esoteric, and possibly not for the faint of heart.

Let G=(V,E) be a locally finite (i.e. having finite degree at all vertices) graph which is transitive, and let Γ=Aut(G) be the group of automorphisms of G (i.e. for every x,y∈V, there is some automorphism γ∈Γ of G for which γ(x)=y). For every x∈V, denote as the stabilizer of x

Stab(x) = {γ∈Γ: γ(x)=x}

the set of all automorphisms of G

fixing x. This is a

subgroup of Γ. Because G is locally finite, for every y∈G the set

(Stab(x))(y)={γ(y): γ∈Stab(x)}

is finite.

For instance, if G is the two dimensional square grid, then Γ is finitely generated by the elements

- r - move one unit to the right: r(x,y)=(x+1,y);
- u - move one unit up: r(x,y)=(x,y+1);
- c - rotate clockwise around (0,0): c(x,y)=(y,-x);
- f - flip everything around x=0: f(x,y)=(-x,y).

For any v=(a,b)∈V, Stab(v) is in fact finite, and contains just the identity, the 3 rotations around v, the 2 flips about x=a and about y=b, and a rotation of each. It is D_{4}, the dihedral group with 8 elements.
As Stab(v) is finite, clearly any u∈V can only be mapped to finitely many (in fact exactly 8, unless u=v) images by elements of Stab(v).

FINALLY, after all this preparation, we say that G is unimodular if for every x,y∈V,

|(Stab(x))(y)| = |(Stab(y))(x)|.

Each element does onto other elements as other elements do onto it.

Unimodularity of graphs, once grasped, appears to hold for all graphs. Here are two graphs (nodes to be added when I get a round tuit) which are *not* unimodular: