On the multiplicity of the second largest eigenvalue of adjacency matrix

Question

Let $G$ be a connected, finite, simple, undirected and vertex-transitive graph with $N$ vertices. Denote by $\mu_0> \mu_1\geq \dots \geq\mu_{N-1}$ all eigenvalues of the adjacency matrix associated with $G$.

My question is:

Does there exist $G$ such that the multiplicity of $\mu_1$ is strictly greater than $1$?

Answer 1

Yes, there are many examples of vertex-transitive graphs whose multiplicity of $\mu_1$ is strictly greater than $1$.

• Complete graph $K_n$ for $n \ge 3$ has spectrum $(n-1)^1,(-1)^{n-1}$.
• Complete bipartite graphs $K_{n,n}$ for $n \ge 2$ has spectrum $\pm n, 0^{2n-2}$.
• Cycles
  • $C_3 = K_3$ has spectrum $2, (-1)^2$
  • $C_4 = K_{2,2}$ has spectrum $2, 0^2, -2$.
  • $C_5$ has spectrum $2, (\frac{-1+\sqrt{5}}{2})^2, (\frac{-1-\sqrt{5}}{2})^2$.
  • $C_6$ has spectrum $2, 1^2, (-1)^2, -2$.
  • For other $n > 6$, $C_n$ has spectrum $$2, (2\cos\frac{2\pi}{n})^2, (2\cos\frac{4\pi}{n})^2, \ldots$$ All multiplicities are $2$, expect that of $\mu_0 = 2$ or $\mu_{n-1} = -2$ when $n$ is odd.
• The $n$-cube graph $Q_n$ ($1$-skeleton of the hypercube $[0,1]^n$) for $n \ge 2$ has spectrum $$n^1, (n-2)^{\binom{n}{1}}, (n-4)^{\binom{n}{2}},\ldots$$
• Platonic graphs - the multiplicity of $\mu_1$ for all five platonic graphs is $3$.
  • The tetrahedral graph is isomorphic to $K_4$, it has spectrum $3^1 (-1)^3$.
  • The cubical graph is isomorphic to $Q_3$, it has spectrum $3^1,1^3,(-1)^3,(-3)^1$.
  • The octahedral graph has spectrum $4^1, 0^3, (-2)^2$.
  • The icosahedral graph has spectrum $5^1,(\sqrt{5})^3, (-1)^5, (-\sqrt{5})^3$.
  • The dodecahedral graph has spectrum $3^1,(\sqrt{5})^3, 1^5, 0^4, (-2)^4, (-\sqrt{5})^3$.