Importance: Medium ✭✭
Author(s): DeVos, Matt
Recomm. for undergrads: no
Prize: bottle of wine (DeVos)
Posted by: mdevos
on: June 22nd, 2007
Conjecture   There exists a fixed constant $ c $ (probably $ c=4 $ suffices) so that every embedded (loopless) graph with edge-width $ \ge c $ has a 5-local-tension.

The edge-width of an embedded graph is the length of the shortest non-contractible cycle.

Definition   Let $ G $ be a directed graph, let $ \Gamma $ be an abelian group, and let $ \phi : E(G) \rightarrow \Gamma $. Define the height of a walk $ W $ to be the sum of $ \phi $ on the forward edges of $ W $ minus the sum of $ \phi $ on the backward edges of $ W $ (edges are counted according to multiplicity). We call $ \phi $ a tension if the height of every closed walk is zero, and if $ G $ is an embedded graph, we call $ \phi $ a local-tension if the height of every closed walk which forms a contractible curve is zero. If in addition, $ \Gamma = {\mathbb Z} $ and $ 0 < \phi(e) < k $ for some $ k \in {\mathbb Z} $, we say that $ \phi $ is a $ k $-tension or a $ k $-local-tension. If we reverse an edge $ e $ and replace $ \phi(e) $ by $ -\phi(e) $, this preserves the properties of tension or local-tension. Accordingly, we say that an undirected graph (embedded graph) $ G $ has a $ k $-tension ($ k $-local-tension) if some and thus every orientation of it admits such a map.
Proposition   A graph has a $ k $-tension if and only if it is $ k $-colorable.
Proof   To see the "if" direction, let $ f : V(G) \rightarrow \{0,\ldots,k-1\} $ be a coloring, orient the edges of $ G $ arbitrarily, and defining $ \phi : E(G) \rightarrow {\mathbb Z} $ by the rule $ \phi(uv) = f(v) - f(u) $. It is straightforward to check that $ \phi $ is a $ k $-tension. For the "only if" direction, let $ \phi : E(G) \rightarrow {\mathbb Z} $ be a $ k $-tension. Now choose a point $ u \in V(G) $ and define the map $ f : V(G) \rightarrow {\mathbb Z}_k $ by the rule that $ f(v) $ is the height of some (and thus every) walk from $ u $ to $ v $ modulo $ k $. Again, it is straightforward to check that this defines a proper $ k $-coloring.

For graphs on orientable surfaces, local-tensions are dual to flows. More precisely, if $ G $ and $ G^* $ are dual graphs embedded in an orientable surface, then $ G $ has a $ k $-local-tension if and only if $ G^* $ has a nowhere-zero $ k $-flow. On non-orientable surfaces, there is a duality between $ k $-local-tensions in $ G $ and nowhere-zero $ k $-flows in a bidirected $ G^* $. Based on this duality we have a couple of conjectures. The first follows from Tutte's 5-flow conjecture, the second from Bouchet's 6-flow conjecture.

Conjecture  (Tutte)   Every loopless graph embedded in an orientable surface has a 5-local-tension.
Conjecture  (Bouchet)   Every loopless graph embedded in any surface has a 6-local-tension.

So although, graphs on surfaces may have high chromatic number, thanks to some partial results toward the above conjectures, we know that they always have small local-tensions. For orientable surfaces, there is a famous Conjecture of Grunbaum which is equivalent to the following.

Conjecture  (Grunbaum)   If $ G $ is a simple loopless graph embedded in an orientable surface with edge-width $ \ge 3 $, then $ G $ has a 4-local-tension.

On non-orientable surfaces, it is known that there are graphs of arbitrarily high edge-width which do not admit 4-local-tensions (see [DGMVZ]). However, it remains open whether sufficiently high edge-width forces the existence of a 5-local-tension. Indeed, as suggested by the conjecture at the start of this page, it may be that edge-width at least 4 is enough. Edge-width 3 does not suffice since the embedding of $ K_6 $ in the projective plane does not admit a 5-local-tension.


*[DGMVZ] M. DeVos, L. Goddyn, B. Mohar, D. Vertigan, and X. Zhu, Coloring-flow duality of embedded graphs. Trans. Amer. Math. Soc. 357 (2005), no. 10 MathSciNet

* indicates original appearance(s) of problem.