Math-1365 · mathematics · Graph theory

Math-1365

A homomorphism problem for flows

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Statement

Definitions and mathematical statement

Let $M$ and $M'$ be abelian groups written additively, and let $B\subseteq M\setminus\{0\}$ and $B'\subseteq M'\setminus\{0\}$ satisfy $B=-B$ and $B'=-B'$. Their Cayley graphs have vertex sets $M$ and $M'$, respectively, with distinct $x,y$ adjacent when $y-x$ belongs to the indicated set. A graph homomorphism $h:\operatorname{Cay}(M,B)\to\operatorname{Cay}(M',B')$ is an arbitrary vertex map preserving adjacency; it need not be additive or injective.

For a finite graph $G$ with an arbitrary orientation, a $B$-flow is a map $f:E(G)\to B$ such that at every vertex the sum of values on outgoing edges equals the sum on incoming edges, with sums evaluated in $M$. Define a $B'$-flow similarly using $M'$. Symmetry of the allowed sets makes existence independent of the chosen orientation: reversing an edge changes its value to its negative.

The conjecture asks whether existence of a homomorphism between the two Cayley graphs implies the following implication for every finite $G$: \[ G\text{ has a }B\text{-flow} \ \Longrightarrow\ G\text{ has a }B'\text{-flow}. \] The prospective new flow need not be the edgewise image under $h$, since $h$ is not assumed to preserve group addition. The zero-free convention gives the intended nowhere-zero flow problem and keeps the Cayley graphs loopless; otherwise a zero in an allowed set would require a separate loop convention.

Short English statement

Does a homomorphism between two abelian Cayley graphs guarantee that every flow using the first allowed value set can be replaced by one using the second?

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