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The converse is pretty obvious. If G is a cycle, then it is isomorphic to it's line graph. How to prove that if L(G) is isomorphic to G, then G is a cycle...?

P.S.- Assume G is connected

idpd15
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2 Answers2

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A vertex of $G$ of degree $d_i$ contributes $\binom{d_i}2$ edges to $L(G)$. Then with $n$ denoting the common number of vertices and edges of $G$ and $L(G)$,

$$ \sum_id_i=2n\;, $$

$$ \sum_i\frac{d_i(d_i-1)}2=n\;, $$

so

$$ \sum_id_i^2=4n $$

and thus

\begin{align} \operatorname{Var}(d)&=E[d^2]-E[d]^2 \\ &=\frac{\sum_id_i^2}n-\left(\frac{\sum_id_i}n\right)^2 \\ &=4-4 \\ &=0\;. \end{align}

Thus all vertices in both graphs have the same degree $2$, which is only the case in a union of cycle graphs.

joriki
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  • Thanks man superb. I mean i knew the relations that you have used. But i didn't think about using the concept of variance to show that the degrees are equal(especially solving a question in graphs!!). How did you come up with this? Do you know of some other questions where this can be used?? – idpd15 Jun 20 '16 at 11:15
  • @idpd15: Concepts from probability are often useful in graph theory; e.g. both examples in the Wikipedia article on the probabilistic method are about graphs. In the present case, I realised that the degree sum is fixed and we can count the edges in $L(G)$ by adding up $\binom{d_i}2$, which is a convex (quadratic) function of $d_i$. If $\sum_id_i$ is fixed, the sum over a convex function of $d_i$ is minimal if all $d_i$ are equal. I could have just written that, but I translated it into the variance speak to make it more concrete. – joriki Jun 20 '16 at 11:22
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Let $G$ be a connected graph on $n$ vertices and $m$ edges and suppose $G$ is isomorphic to its line graph $L(G)$. Then, $m=n$ and so $G$ is a connected graph having $n$ edges. This implies $G$ is of the form $T+e$, where $T$ is a tree and $e$ is an edge not in $T$. If $G=T+e$ is the $n$-cycle graph, then we are done.

So suppose $T+e$ is not a cycle. Then, $T+e$ has a vertex of degree 3 or more. Three of the edges incident to this vertex in $G$ form a cycle of length 3 in $L(G)$. Also, the unique cycle in $G$ obtained by adding edge $e$ to the tree $T$ induces a cycle in $L(G)$. Hence, $L(G)$ has at least two cycles. We showed that $G$ has only one cycle and that $L(G)$ has two or more cycles, which is a contradiction because $G \cong L(G)$. So this case is impossible, and $T+e$ must be a cycle.

Ashwin Ganesan
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