Do we get a instanton $S^{3}$ if we do $1/n$ surgery on a knot in $S^{3}$?

Question

Consider the following question: Let $K\subset S^{3}$ be a nontrivial knot, and let $Y$ be the manifold obtained by doing $1/n$-surgery ($n\geq1$). Is it possible that the instanton Floer homology of $Y$ vanishes?

The answer seems to be no. Indeed by Gordon-Luecke's theorem, $Y$ is not $S^{3}$, while as far as I know, people don't have any example of an integer homology 3-sphere other than $S^{3}$ having vanishing instanton Floer homology.

My question is: is there a rigorous proof of this statement? (for example, using instanton suture Floer homology?)

Answer 1

Kronheimer and Mrowka first proved that 1-surgery on a knot has non-trivial instanton Floer homology using Witten's conjecture (see their paper http://arxiv.org/abs/math/0311489). They later proved this using sutured instanton Floer homology as you suggested (see Section 7 of http://arxiv.org/abs/0807.4891).

As far as I know, for $n$ at least 2, this is unknown in general. K-M's first proof of Property P uses a long exact sequence in instanton Floer homology relating the instanton Floer homologies of $S^3$, $S^3_1(K)$, and $S^3_0(K)$. Some total speculation would be to try to construct a long exact sequence relating the instanton Floer homologies of $S^3$, $S^3_{1/n}(K)$, and $S^3_0(K)$ with some sort of twisted coefficients (analogous to Theorem 9.14 in http://arxiv.org/abs/math/0105202) and try to repeat their argument.

Answer 2

This follows now from Theorem 1.15 of this paper for $n>1$. To clarify, Theorem 1.15 is stated for framed instanton homology $I^\#(S^3_{1/n}(K))$. If instanton Floer homology $I(S^3_{1/n}(K))=0$, then by Definition 1 of Froyshov, $\hat{I}(S^3_{1/n}(K))=0$ since it is a subquotient (notice I am not using Froyshov’s notation, see section 9.1 for the relation between Scaduto’s and Froyshov’s notation). Then by Theorem 1.3 (which relates $\hat{I}$ and $I^\#$ for homology 3-spheres) $dim\ I^\#(S^3_{1/n}(K))=1$, and hence $S^3_{1/n}(K)$ is an instanton $L$-space (Definition 1.13). Then Theorem 1.15 implies that $1/n \geq 2g-1$, a contradiction if $n >1$.

For $n=1$ (following Steven Sivek’s comment below), then $g=1$ by the above inequality. It is known that an $L$-space knot of genus 1 is the right-handed trefoil knot $T$ (see Theorem 1.15 and Remark 1.7). Then $S^3_1(T)$ is the Poincaré homology 3-sphere (see Proposition 7.12). But the Floer homology of the Poincaré homology 3-sphere is non-zero since (for example), the Euler characteristic is twice the Casson invariant which is $1$, as proved originally by Floer (even though it is an instanton homology $L$-space).

Answer 3

In fact, if $C = \text{rank}(I_*(S^3_0(K)))$, we have $$f(k) = \text{rank}(I_*(S^3_{1/k}(K))) = kC.$$ This holds with any coefficient field.

Floer's exact sequence gives us an exact triangle relating $I_*(S^3_0) \to I_*(S^3_{1/k}) \to I_*(S^3_{1/(k+1)}).$ As a result, we have $f(k+1) \le f(k) + C$, and $f(1) = f(0) = C$.

But there's also the exact triangle of Culler, Daemi, and Xie, displayed as (1.4) here, relating $I_*(S^3_{1/(k-1)}) \to I_*(S^3_{1/k})^2 \to I_*(S^3_{1/(k+1)})$. From this we have $2f(k) \le f(k-1) + f(k+1)$ for $k \ge 2$ (and for $k=1$ we get $f(2) = 2f(1)$).

Now $f(k) = kC$ can be proven by induction.

As a result, it follows that in Floer's triangle the connecting map $I_*(S^3_0) \to I_*(S^3_{1/k})$ is always zero, and in Culler-Daemi-Xie's triangle the connecting map $I_*(S^3_{1/(k+1)}) \to I_*(S^3_{1/(k-1)})$ is zero, as well.

At this point, if you know that the displayed maps in the CDX triangle above preserve the $\Bbb Z/2$-grading, you can argue by another induction using the CDX sequence that $I_*(S^3_{1/k}(K)) \cong I_*(S^3_1(K))^{\oplus k}$ as $\Bbb Z/2$-graded vector spaces.