Max degree of vertex related to line graph of regular bipartite graph

Question

I am trying to figure out a relationship between the max degree of the line graph and its chromatic number

What I know for sure is that for any graph G, $\Delta(G) \leq \chi(L(G))$. My confusion is the following: When G is k-regular, $\Delta(G) = k$, and I need to show this is equal to $\chi(L(G))$. I thought about proving it with some inequality, but the only things I know to be true are that $\chi(L(G)) \leq \Delta(L(G)) = 2k - 2$ 2k-2 comes from the fact that the degree of the vertex in L(G) of edge e = (u, v) is $deg(u) + deg(v) - 2$.

What is $\chi(L(G))$? Apparently, it's not the chromatic number. — Ben Grossmann, Apr 20 '17 at 13:24
Thanks for clarifiying. I thought $L(G)$ was the Laplacian, hence the confusion. — Ben Grossmann, Apr 20 '17 at 14:30

Perry Elliott-Iverson · Accepted Answer · 2017-04-20T20:16:56.363

0

Note that $\chi(L(G)) = \chi'(G)$, the chromatic index (or edge chromatic number) of $G$. Now since $G$ is bipartite, the chromatic index of $G$ is equal to $\Delta(G)$. See this previous question:

Edge-coloring of bipartite graphs

But since $G$ is $k$-regular, each edge in $G$ is adjacent with $2k-2$ other edges, and $\Delta(L(G)) = 2\Delta(G)-2$. So we have:

$$\chi(L(G)) = \chi'(G) = \Delta(G) = \frac{\Delta(L(G))}{2}+1$$

edited Apr 20 '17 at 20:16

answered Apr 20 '17 at 14:11

Perry Elliott-Iverson

4,562

I realized I had an error. Each edge is adjacent with 2k-2 other edges, not 2k-1. – Perry Elliott-Iverson Apr 20 '17 at 20:17

Max degree of vertex related to line graph of regular bipartite graph

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