### Forbidden subgraphs and equiangular lines

Technion – Israel Institute of Technology
June 5, 2018
Joint work with Саша Полянский (Alexandr Polyanskii)

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

Forbidden subgraphs

Equiangular lines

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

### 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 $C$
$(\lambda I - A) + \frac{\epsilon\lambda}{1-\epsilon}J \succeq 0$, $\mathrm{rank}(\lambda I - A) \lesssim n$

Claim $\lambda_1(G) \le \lambda$

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

### Further results

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

$E_{1/7}(n) \lesssim \big(\frac{4}{3}+\frac{1}{36}\big)n$

$E_{1/(1+2\lambda)}(n) \lesssim 1.49n$, for $\lambda \neq 1, \sqrt{2}, 2$

Technion – Israel Institute of Technology
jiangzilin@technion.ac.il