Framings for 2-surgeries on 4-manifolds 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 $S^4$.


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*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?

*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?

*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.
 A: 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.
