$\lambda^* = \sqrt{2 + \sqrt{5}} = $ 2.058171027...

Forbidden subgraphs

Equiangular lines

Spectral radius

Largest eigenvalue $\lambda_1(G)$ of adjacency matrix

$\lambda_1\big($$\big) = 1$, $\lambda_1\big($$\big) = \sqrt{2}$, $\lambda_1\big($$\big) = 2$

$\lambda_1$ is monotone
subgraphs have smaller spectral radius

$F(\lambda) := \{$ graphs with spectral radius $\le \lambda\,\}$

closed under taking subgraphs

Hereditary property

Graph properties closed under taking subgraphs

induced-hereditary, minor-hereditary

Robertson–Seymour theorem
Minor-hereditary property can be defined by
a finite set of forbidden minors

Forbidden subgraphs characterization

$F(\lambda) = \{$ graphs with spectral radius $\le \lambda\,\}$

Question. Can $F(\lambda)$ be defined by
a finite set of forbidden subgraphs?

Answer (J.–Polyanskii). Yes, if and only if $\lambda < \lambda^*$
and $\lambda \not\in \{\alpha_2, \alpha_3, \dots\}$, where $\alpha_m = \beta_m^{1/2}+\beta_m^{-1/2}$ and
$\beta_m$ is the largest root of $x^{m+1} = 1 + x + \ldots + x^{m-1}$

When $\lambda \ge \lambda^*$...

Lemma (Shearer). $\exists G$ s.t. $\lambda_1(G) \in (\lambda, \lambda + \epsilon)$.

Assume $G_1, \dots, G_n$ is a finite f.s.c. for $F(\lambda)$

Choose $G$ s.t. $\lambda < \lambda_1(G) < \lambda_1(G_i)$ for $i \in [n]$

$G_i$ is not a subgraph of $G$

Contradiction

When $\lambda < 2$...

$\lambda_1\big($$\big) = 2$, $\lambda_1\big($$\big) > \lambda$

If $G$ contains neithernor, then

$\implies G$ has $\le 3^n$ vertices

Further forbid graphs outside $F(\lambda)$ with $\le 3^n$ vertices

When $2 \le \lambda < \lambda^*$...

Work harder

Forbidden subgraphs

Equiangular lines

Equiangular lines

Lines through origin in $\mathbb{R}^n$,
angle between every pair is the same

Observation. Equiangular lines $\Leftrightarrow \{\pm \alpha\}$-code

Maximum number of equiangular lines

$E_\alpha(n) := $ maximum cardinality of $\{\pm \alpha\}$-code in $\mathbb{R}^n$

$E_{1/(1+2\lambda)}(n) \approx \frac{k}{k-1}n$

$\lambda_1\big($$\big) = 1$	$E_{1/3}(n) \approx 2n$
$\lambda_1\big($$\big) = \sqrt{2}$	$E_{1/(1+2\sqrt{2})}(n) \approx \frac{3}{2}n$
$\lambda_1\big($$\big) = 2$	$E_{1/5}(n) \approx \frac{3}{2}n$

Connection to forbidden subgraphs

Lemma (Balla et al.).
For fixed $\epsilon > 0$, $E_{1/(1+2\lambda)}(n) \lesssim E_L(n)$,
where $L = \big\{-\frac{1-\epsilon}{\lambda}+\epsilon, \epsilon\big\}$.

Suppose $C$ is $L$-code in $\mathbb{R}^n$, define $G$ on $C$:
$v_i \sim v_j \Leftrightarrow \langle v_i, v_j \rangle < 0$.

$A := $ adjacency matrix of $G$, $M := $ Gram matrix of $C$

$(\lambda I - A) + \frac{\epsilon\lambda}{1-\epsilon}J$ $ = \frac{\lambda}{1-\epsilon}M$ $\succeq 0$

Maximize $\left|C\right|$ s.t. $\mathrm{rank}(M) \le n$, or $\mathrm{rank}(\lambda I - A) \lesssim n$.

$A := $ adjacency matrix of $G$ with vertex set $C$
$(\lambda I - A) + \frac{\epsilon\lambda}{1-\epsilon}J \succeq 0$, $\mathrm{rank}(\lambda I - A) \lesssim n$

Special case: $G$ is connected

If $\lambda_1(G) = \lambda$, then $|C| \ge k$, and Perron–Frobenius says $\mathrm{rank}(\lambda I - A) \ge |C| - 1 \ge \frac{k-1}{k}|C|$;
Otherwise $\mathrm{rank}(\lambda I - A) = |C|$.

General case: work on connected components

$A := $ adjacency matrix of $G$ with vertex set $C$
$(\lambda I - A) + \frac{\epsilon\lambda}{1-\epsilon}J \succeq 0$

Claim $\lambda_1(G) \le \lambda$ $\Leftrightarrow G \in F(\lambda)$

Suppose $\{G_1, \dots, G_n\}$ is a f.s.c. for $F(\lambda)$,
assume that some $G_i \subseteq G$

$(\lambda I - A_i) + \frac{\epsilon\lambda}{1-\epsilon} J \succeq 0$
$\implies \lambda - \lambda_1(G_i) + \frac{\epsilon\lambda}{1-\epsilon}v(G_i) \ge 0$

Punchline
Choose $\epsilon$ so small that above is $< 0$ for all $G_i$

Conjecture beyond $\lambda^*$

$E_{1/(1+2\lambda)}(n) \approx \frac{k}{k-1}n$, for all $\lambda$

In particular, $E_{1/7}(n) \approx \frac{4}{3}n$