TITLE: Edge Colorings of Directed Graphs which Respect an Involution 
QUESTION [6 upvotes]: Let G be a graph and let C be a set of coloring. Suppose that there is an involution $\phi$ from C to C. We can think about the element of C as the nonzero elements of some Abelian group and $\phi(x)=-x$. (For concreteness we can consider especially the special case  that C corresponds to the cyclic group, so that $\phi$ has no fixed points when |C| is even and one fixed point when |C| is odd.) 
Consider a colorings where for every edge {u,v} in G we color the directed edge from u to v with some color c and the directed edge from v to u with -c. We will also require that two edges with the same tail and two edges with the same head must have different colors. If $\phi$ is the identity this is a usual edge coloring of $G$. 
My question is if such colorings were considered and what is known about them. For example, is there an analog of Vizing theorem?  (Vizing theorem asserts that the edges of every graph with maximum degree d can be colored by d+1 colors.)

REPLY [2 votes]: It seems to me that such colorings are closely related to 2-factorizations of graphs. Indeed every color class represents a subgraph which is a disjoint union of directed paths and cycles. Let's denote by $C(x)$ the set of edges colored $x\in G$. We can see that the underlying graphs of $C(x)$ and $C(-x)$ coincide, but they have opposite orientation. When $2x\neq 0$ then the underlying subgraph must have maximum degree 2. When $2x=0$ then $C(x)$ is a matching.
It's possible to show that a graph with $n$ vertices and maximum degree $\Delta$ can be written as a union of at most $\lfloor \frac{\Delta+1}{2}\rfloor$ subgraphs of maximum degree 2. Each of these subgraphs is a union of paths and cycles, so we can assign two colors to the corresponding directed edges. This shows that there is always a coloring satisfying the properties in the question coming from $\mathbb Z/(\Delta+1)\mathbb Z$. This would be the most straightforward analog of Vizing's theorem. Notice that $\Delta+1$ cannot be improved.