SU(2) representations and a large surgery formula

Abstract

A knot $K\subset S^3$ is called $SU(2)$-abundant if it satisfies two conditions: first, for all but finitely many $r\in\mathbb{Q}\backslash\{0\}$, there exists an irreducible representation $π_1(S^3_r(K))\to SU(2)$; second, any slope $r=u/v\neq 0$ for which $S^3_r(K)$ admits no irreducible $SU(2)$ representation must satisfy $Δ_K(ζ^2)= 0$ for some $u$-th root of unity $ζ$. We show that if a nontrivial knot $K\subset S^3$ is not $SU(2)$-abundant then it is a prime knot whose Alexander polynomial $Δ_K(t)$ has coefficients restricted to $\{-1,0,1\}$. This implies, in particular, that all hyperbolic alternating knots are $SU(2)$-abundant. Our proof hinges on a large surgery formula connecting instanton knot homology $KHI(S^3,K)$ and framed instanton homology $I^\sharp(S^3_n(K))$ for integers $n$ satisfying $|n|\ge 2g(K)+1$. Using this technique, we derive several interesting results in instanton Floer homology: for any Berge knot $K$, the spaces $KHI(S^3,K)$ and $\widehat{HFK}(S^3,K)$ have identical dimension; for any dual knot $K_r\subset S^3_r(K)$ of a Berge knot $K$ with $r> 2g(K)-1$, we prove $\dim_\mathbb{C}KHI(S^3_r(K),K_r)=|H_1(S^3_r(K);\mathbb{Z})|$; and for any genus-one alternating knot $K$ and any $r\in\mathbb{Q}\backslash\{0\}$, the spaces $I^\sharp(S^3_r(K))$ and $\widehat{HF}(S_r^3(K))$ have equal dimension.