Upper bounds for the second largest eigenvalue in terms of degree? I am looking for upper bounds on the second largest eigenvalue, $\lambda_2(G)$ of a given graph $G$, with respect to minimum/maximum degrees of the graph. I looked around for some existing bounds most I found were with respect to the order of $G$.
Can anyone point me to something?
 A: There exists arbitrarily large $d$-regular graphs with a cut edge. A direct inspection shows that the expansion tends to $0$. By the Cheeger inequality, $λ_2$ is arbitrarily close to $d=λ_1$.
There could be no bounds except $λ_2(G)<λ_1(G)$; in other words, $λ_2(G)$ can be arbitrarily close to $λ_1(G)$.
A: If $G$ is a d-regular graph, then the second-largest eigenvalue of $A(G)$ is at least $2\sqrt{d-1} + o(1)$, where the $o(1)$ term goes to 0 as the number of vertices gets large.
If the number of vertices isn't much larger than $d$ then the second-largest eigenvalue can be much smaller. If $G$ is a complete graph on $d$ vertices where each vertex is adjacent to itself, then the second-largest eigenvalue of $A(G)$ is 0.
An upper bound for the second-largest eigenvalue is $d$ itself (if $G$ is $d$-regular but not connected).
ETA: I can add a bit more to my answer. The second-largest eigenvalue f a $d$-regular graph gives some sort of measure as far as the expansion of the graph i.e., how many vertices need to be cut to disconnect a large set of vertices from the rest of the graph. A $d$-regular graph that is disconnected will have $d$ as an eigenvalue with multiplicity at least 2. A d-regular connected graph with a cut vertex that bisects the graph nearly in half, will have a second-largest eigenvalue quite close to $d$ something like $d-O(1/n)$ where $n$ is the number of vertices. A $d$-regular expander where $d$ is $\theta(1)$ will have the second-largest eigenvalue no larger than $1-\epsilon)d$ for some $\epsilon \in \theta(1)$.
