Detecting a "bad map" in Fintushel-Stern knot surgery

Question

Background

Let $X$ be a simply-connected smooth 4-manifold which contains a smoothly embedded torus $T$ with trivial normal bundle (in other words, $T^2\times D^2\subset X$). Let $K$ be a knot in $S^3$, and let $\nu(K)$ be a tubular neighbourhood of $K$. Note that $\partial(T^2\times D^2)=T^2\times S^1=\partial([S^3-\nu(K)]\times S^1)$.

Fintushel and Stern define the knot surgery of $X$ to be the manifold $$X_K=(X-T^2\times D^2)\cup_{\varphi}([S^3-\nu(K)]\times S^1),$$ where $\varphi$ is "any map that preserves the homology class $[pt\times \partial D^2]$" (or as it is often equivalently described, so that the meridian $pt\times\partial D^2$ of the torus coincides with the longitude of $K$). Because $[S^3-\nu(K)]\times S^1$ has the same homology as a tubular neighborhood of T in X, and because the gluing preserves $[pt \times \partial D^2]$, the homology and intersection form of $X_K$ will agree with that of $X$ (the intersection form being a homeomorphism invariant of smooth simply-connected 4-manifolds). If it is also assumed that $X - T^2\times D^2$ is simply-connected, then $\pi_1(X_K) = 1$; so $X_K$ will be homeomorphic to $X$.

My Question

Is it obvious (or can at least be easily 'detected') what a "bad" choice of map would yield here? That is, you perform the gluing with a map which doesn't preserve that homology class --- can we determine how the result would differ from the "correct" type of map. For example, maybe the resulting space now has a singularity of some kind, or maybe the homeomorphism type is no longer the same as $X$, etc.

On some level I guess this is also asking 'how prolific are the maps that satisfy the homology condition' or 'how restrictive is the property on $\varphi$'?

Answer 1

There is nothing inherently "bad" with other choices. My guess is that Fintushel and Stern chose this identification for three reasons: first, they can give a nice formula for how the Seiberg–Witten invariants of $X$ change under knot surgery; second, they don't need to worry about spin structures; third, all other choices can be reduced to a generalised log tranform (see below) plus a followed by a "standard" knot surgery.

In general, $X$ will stay simply-connected as long as $X\setminus T$ is: this is because $\pi_1(E_K \times S^1)$ is normally generated by $\pi_1$ of its boundary (here $E_K$ is the knot exterior), and Seifert–van Kampen tells you that $X_K$ is always simply-connected. As for the homeomorphism type, there's some care to be taken with spin structures (which spin structures on $T^3$ extend to $X\setminus T$? which ones extend to $E_K \times S^1$? what happens when we glue?) when $X\setminus T$ is spin—these issues do not arise with the choice you're describing, so that's one extra reason for it. When $X\setminus T$ is not spin, then $X$ and $X_K$ are homeomorphic by Freedman: they're both simply-connected by the remark above, neither of them is spin since $X\setminus T$ isn't, and they have the same signature by Novikov additivity.

Finally, choosing any other gluing (and these are in bijection with primitive vectors in $\mathbb{Z}^3$) should factor through doing a generalised log transform (i.e. surgering out a neighbourhood of $T$ and gluing back a $T^2\times D^2$ with some diffeomorphism of the boundary) and then performing the "usual" knot surgery on the 2-torus $T^2 \times \{0\}$ in the newly–glued-in $T^2\times D^2$. This generalised log transform operation is described in some detail in Gompf and Stipsicz's 4-manifolds and Kirby calculus, in Section 8.3. If you understand generalised log surgery well enough (and Fintushel and Stern certainly did), then all other choices of gluing maps can be just viewed as the standard one, in a different (but not so different) 4-manifold.