  A knot $K\subset S^3$ is called $SU(2)$-abundant if for all but finitely many
$r\in\mathbb{Q}\backslash\{0\}$, there is an irreducible representation
$\pi_1(S^3_r(K))\to SU(2)$, and the slope $r=u/v\neq 0$ with no irreducible
$SU(2)$ representation must satisfy $\Delta_K(\zeta^2)= 0$ for some $u$-th root
of unity $\zeta$. We prove that a nontrivial knot $K\subset S^3$ 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 $KHI(S^3,K)$ and the
framed instanton homology $I^\sharp(S^3_n(K))$ 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
$KHI(S^3,K)$ and $\widehat{HFK}(S^3,K)$ have the same dimension. Second, for
any dual knot $K_r\subset S^3_r(K)$ of a Berge knot $K$ with $r> 2g(K)-1$, we
show $\dim_\mathbb{C}KHI(S^3_r(K),K_r)=|H_1(S^3_r(K);\mathbb{Z})|$. Third, 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 the same
dimension.

Geometric Topology