Suppose I have a graph $G$, with $E$ edges and $V$ vertices. I now construct its line graph, $L(G)$.
The number of vertices in $L(G)$ has to be $E$ by the definition of a line graph. But what's the maximum number of edges in $L(G)$? I'm interested in the asymptotic worst case, that is, the big-O of the edge count in terms of $E$ and $V$.
My current guess is $O(E^2)$, which is the worst case if every edge is adjacent. But if $G$ is anything bigger than $K_3$ then I know this won't be the case. So it seems like it should be possible to reduce this further.
The best asymptotic bound we can put on the number of edges in the line graph is $O(EV)$ (actually, the product $EV$ by itself is an upper bound).
To get this bound, note that each of the $E$ edges of $L(G)$ has degree less than $2V$, since it shares each of its endpoints with fewer than $V$ edges. So the sum of degrees in $L(G)$ is at most $2EV$, and this is twice the number of edges.
We can achieve $\frac12 E(V-2) = \Omega(EV)$ edges in the line graph for a range of values of $E$ relative to $V$, by taking a complete bipartite graph. The complete bipartite graph $K_{a,b}$ has $a+b$ vertices and $ab$ edges. In the line graph $L(K_{a,b})$, there are $ab$ vertices, and each has degree $a+b-2$ (each edge of $K_{a,b}$ is incident to $a-1$ other edges on the side of size $a$ and $b-1$ other edges on the side of size $b$), so the degree sum is $ab(a+b-2) = E(V-2)$. The number of edges in $L(K_{a,b})$ is half that.
(In particular, the star, which is $K_{1,n-1}$, achieves this bound.)
As a consequence of the construction above, we can find a graph $L(G)$ with $V$ vertices and $E$ edges such that $L(G)$ has at least$\frac14E(V-2) = \Omega(VE)$ edges, for any choice of $E$ and $V$ with $V-1 \le E \le \binom{V}{2}$. Just consider the sequence of graphs $$ K_{1,V-1}, K_{2,V-2}, K_{3,V-3}, \dots, K_{\lfloor V/2 \rfloor, \lceil V/2 \rceil}. $$ Each graph has at least half as many edges as the next, and the last graph has more than $\frac12\binom{V}{2}$ edges. So for any $E$, we can find a graph $G$ in this sequence with $V$ vertices and some number of edges between $\frac12E$ and $E$. That $G$'s line graph will have at least $\frac12 (\frac12E)(V-2)$ edges, and if we add edges arbitrarily to get a graph $G'$ with exactly $E$ edges, its line graph will still have at least $\frac14 E(V-2)$ edges.
In graphs where $E < V-1$, the $O(E^2)$ upper bound is better, and we can achieve it by a star on $E+1$ vertices together with $V-E-1$ isolated vertices.