Framings for 2-surgeries on 4-manifolds

Question

I'm interested in doing $2$-surgeries to $\sharp^k S^1 \times S^3$. That is to the manifold obtained from applying $1$-surgeries to $S^4$.

1. Since $\pi_1(O(3)) = \mathbb{Z}_2$, there are two possible framings up to equivalence. Is it possible to distinguish between them in a natural way (similar to how orientability of the resulting manifold works in some other cases)? Maybe something coming from complex geometry?

2. One more or less natural way I'm thinking about is using Akbulut's convention in the following way. Let $M$ be the manifold obtained by $1$ and $2$ surgeries applied to $S^4$. The handle decomposition of $M$ can be described as follows. There are $k$ $1$-handles. There is a $2$-handle for every $2$-surgery. Then we double this $2$-handlebody to obtain $M$, thus getting twice as many $2$-handles: for every original $2$-handle there will be another handle whose attaching circle (nullhomotopic in $1$-handlebody) is linked with the attaching circle of the original handle. Using Akbulut's dotted circle convention we can assign an integer to every framing of $2$-handles. The handles corresponding to doubling will have framing $0$. If there are no $1$-handles, then even integers will correspond to the same manifold, and odd to the other one (as explained in Gompf and Stipsicz). Does the same hold in the presence of $1$-handles?

3. If we obtain the answer, can we see which framing corresponds to the manifold obtained by taking the boundary of a neighbourhood of a $2$-complex embedded in $\mathbb{R}^5$? Like in the following construction: finite generated group realized as fundamental group of manifolds , Constructing 4-manifolds with fundamental group with a given presentation.

Answer 1

You are right that there are two possible ways to perform a 1-surgery, but I don't think that there is a way to choose between them. Think for instance about the simple case where you do a 1-surgery to $S^3 \times S^1$ along the circle $\{p\} \times S^1$. There are two possible choices, but there is no way to choose between them, since there is a self-diffeomorphism of $S^3 \times S^1$ that sends one to the other. They both yield the same manifold $S^4$.

However, you can still resolve this ambiguity in an elegant and simple way by using cohomology and Stiefel - Whitney classes.

You are constructing a 4-manifold $M$ as the boundary of a 5-manifold $W$ obtained with 0-, 1-, and 2-handles, which is in turn obtained by thickening a 2-complex $X$. The 5-dimensional thickenings $W$ of a 2-complex $X$ are in natural 1-1 correspondence with the elements of $H^2(X, \mathbb Z/_{2\mathbb Z})$ via the second Stiefel - Whitney class (for a proof, see this paper of Hambleton, Kreck, and Teichner).

That is, for every $\alpha \in H^2(X, \mathbb Z/_{2\mathbb Z})$ there is precisely one 5-dimensional thickening $W$ of $X$ such that $w_2(W) = \alpha$. The boundary $M=\partial W$ of course will have $w_2(M) = i^*(w_2(W))$. Note that $i^*\colon H^2(W) \to H^2(M)$ is injective (often not surjective). This is a simple picture to remember.

In particular there is always precisely one thickening $W$ that is spin, that is with $w_2(W)=0$, and also precisely one boundary 4-manifold $M = \partial W$ obtained in this way that is spin. This is the one that you obtain from any embedding of $X$ in $\mathbb R^5$, since in that case $W$ is parallelizable.

When there are no 1-handles, the 2-complex is a bouquet of spheres and hence $H^2(X)$ is a product of one $\mathbb Z/_{2\mathbb Z}$ for each 2-handle. This is the simple case.