# Integral surgery on $S^2 \times S^1$

It is a well-known fact that $$S^2 \times S^1$$ can be obtained by $$0$$-surgery on unknot.

What about the $$(-1)$$-surgery on $$S^2 \times S^1$$? It seems the resulting manifold, say $$W$$, bounds contractible manifold.

But I cannot prove it yet or refutes my argument. Any help will be appreciated.

## 1 Answer

This is true, with some points to clarify. First, you are presumably talking about surgery along a knot that generates the first homology (and hence fundamental group) of $$S^1\times S^2$$. Then the result of adding the corresponding 2-handle to $$S^1\times B^3$$ is contractible. The construction (called a `Mazur manifold') goes back to Mazur's paper, A Note on Some Contractible 4-Manifolds, Annals 1961.

The other point is that framing as an integer is not a priori defined for a knot that represents a homology class of infinite order. But fortunately the statement is true for any framing (as in choice of trivialization of the normal bundle). I'd suggest some basic reading about 4-dimensional handle calculus, as in the book of Gompf-Stipsicz.

• Thanks for the reference. To clarify the notion, let me ask you one more question. If we replace $S^2 \times S^1$ by $S^3$ and perform $−1$-surgery on $S^3$ (following your description), may we obtain a manifold bounding rational homology ball? – M. Alessandro Ferrari Apr 20 at 10:03
• Not necessarily; it depends on the knot along which you do that surgery. I actually think these questions are better suited to Mathstackexchange than to this forum. – Danny Ruberman Apr 20 at 13:36