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SU(2) representations and a large surgery formula

Published 23 Jul 2021 in math.GT | (2107.11005v1)

Abstract: A knot $K\subset S3$ is called $SU(2)$-abundant if for all but finitely many $r\in\mathbb{Q}\backslash{0}$, there is an irreducible representation $\pi_1(S3_r(K))\to SU(2)$, and the slope $r=u/v\neq 0$ with no irreducible $SU(2)$ representation must satisfy $\Delta_K(\zeta2)= 0$ for some $u$-th root of unity $\zeta$. We prove that a nontrivial knot $K\subset S3$ is $SU(2)$-abundant unless it is a prime knot and the coefficients of its Alexander polynomial $\Delta_K(t)$ lie in ${-1,0,1}$. In particular, any hyperbolic alternating knot is $SU(2)$-abundant. The proof is based on a large surgery formula that relates instanton knot homology $KHI(S3,K)$ and the framed instanton homology $I\sharp(S3_n(K))$ for any integer $n$ satisfies $|n|\ge 2g(K)+1$. By the same technique, we can calculate many examples of instanton Floer homology. First, for any Berge knot $K$, the spaces $KHI(S3,K)$ and $\widehat{HFK}(S3,K)$ have the same dimension. Second, for any dual knot $K_r\subset S3_r(K)$ of a Berge knot $K$ with $r> 2g(K)-1$, we show $\dim_\mathbb{C}KHI(S3_r(K),K_r)=|H_1(S3_r(K);\mathbb{Z})|$. Third, for any genus-one alternating knot $K$ and any $r\in\mathbb{Q}\backslash{0}$, the spaces $I\sharp(S3_r(K))$ and $\widehat{HF}(S_r3(K))$ have the same dimension.

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