Understanding cobordisms constructed from a Heegaard triple

In the paper by Ozsvath and Szabo, they consider a 4-manifold with boundary which is constructed from a Heegaard triple, i.e., a closed Riemann surface $\Sigma$ with three collections of simple closed curves $\alpha, \beta, \gamma$, each of which generates a half-dimensional sublattice of $H_1(\Sigma)$ (so that each of $\alpha, \beta,\gamma$ determine a handlebody with boundary $\Sigma$ by collapsing the curves). Let's denote this data as $(\Sigma, \alpha,\beta,\gamma)$. Furthermore let's denote the handlebodies obtained from $\alpha,\beta,\gamma$ as $U_\alpha,U_\beta,U_\gamma$. The construction is as follows:

Let $\Delta$ be a (topological) triangle, and let the three sides of $\Delta$ be $e_\alpha,e_\beta,e_\gamma$, resp. Then we can consider a space

$X_{\alpha,\beta,\gamma}=\frac{(\Delta\times \Sigma )\sqcup (e_\alpha\times U_\alpha)\sqcup (e_\beta\times U_\beta)\sqcup (e_\gamma\times U_\gamma) }{e_\alpha\times \Sigma \sim e_\alpha\times \partial U_\alpha, e_\beta\times \Sigma \sim e_\beta\times \partial U_\beta, e_\gamma\times \Sigma \sim e_\gamma\times \partial U_\gamma}$, where each of the gluing $e_i\times \Sigma \sim e_i \times \partial U_i$ is identifying $\partial U_i$ with $\Sigma$ in obvious way.

Then it is obvious that $X_{\alpha,\beta,\gamma}$ is a 4-manifold with boundary $Y_{\alpha,\beta},Y_{\beta,\gamma},Y_{\gamma,\alpha}$ where $Y_{\alpha,\beta}=U_\alpha\cup U_\beta$ is the 3-manifold obtained from the Heegaard diagram $(\Sigma,\alpha,\beta)$, and so on. (In terms of the natural projection $X_{\alpha,\beta,\gamma}\to \Delta$, the boundaries are the preimage of 3 vertices of $\Delta$.)

In the above, I was careless on the orientations, but it is also easy to see that with the natural choice of orientations on the 3-manifold obtained from a Heegaard diagram and the natural choice of the orientation of product manifolds, $\partial X_{\alpha,\beta,\gamma}=-Y_{\alpha,\beta}-Y_{\beta,\gamma}+Y_{\alpha, \gamma}$, i.e., $X_{\alpha,\beta,\gamma}$ can be thought of as a cobordism from $Y_{\alpha,\beta}\sqcup Y_{\beta,\gamma}$ to $Y_{\alpha,\gamma}$.

Then Example 8.1. claims that if we choose $\gamma$ to be an isotopic copy of $\beta$, then the cobordism $X_{\alpha,\beta,\gamma}$ is obtained from the identity cobordism $Y\times [0,1]$ by deleting a regular neighborhood of $U_\beta \times \left\{\frac{1}{2}\right\}$, or equivalently a regular neighborhood of a bouquet of $g$ circles. (where $Y=Y_{\alpha,\beta}$)

Q. How can you visualize this example? I mean, of course $Y_{\beta,\gamma}=\#^g (S^1\times S^2)$ in this case, which bounds a boundary connected sum of $g$ copies of $S^1\times B^3$, but it is not obvious to me that $X_{\alpha,\beta,\gamma}$ becomes $Y\times [0,1]$ after filling out the $Y_{\beta,\gamma}$ boundary by it.

After filling in the boundary with a handlebody you can basically flatten the handlebody by isotopy to collapse $U_\beta$ and $U_\gamma$. If you think of each of $U_\beta$ and $U_\gamma$ as a thickened 3d handlebody then you push them together to get an even thicker 3d handlebody. So your original cobordism was a Y shape, but you push two of the spokes together to get something with only two spokes, and both "boundary sides" are determined by the same two sets of curves (technically one side by $\alpha$ and $\beta$ curves and the other side by $\alpha$ and $\gamma$ curves but the $\beta$ and $\gamma$ curves are isotopic).

• May I ask an elaboration of the collapsing of $U_\beta$ and $U_\gamma$ in your first two sentences? That part is essentially what I don't really make it concrete in my mind. Mar 1 '18 at 2:26
• @cjackal Let $C$ be some curve in $\Sigma$ that bounds a disk in $U_\beta$ (basically, some $\beta$ curve up to slides). $C$ of course also bounds in $U_\gamma$. These two disks come together to form a 2-sphere lying in the boundary of the four dimensional handlebody $Y_{\beta,\gamma}$. Now, every 2-sphere in the boundary of a 4-dimensional handlebody bounds a unique 3-ball in the interior of that handlebody up to isotopy (Laudenbach and Poenaru). This 3-ball can be collapsed, effectively identifying the disk in $U_\beta$ with the disk in $U_\gamma$.
– Carl
Mar 1 '18 at 2:43
• You're basically extending this construction over the whole of $Y_{\beta,\gamma}$, and noting also that the spheres/balls we used are essential. You can picture this by going down a dimension and imagining flattening a standard 3d handlebody to a horizontal cross section. Collapsing a meridional disk identifies the top boundary arc with the bottom boundary arc, since these arcs come together to form the $S^1$ boundary of the disk.
– Carl
Mar 1 '18 at 2:46
• Okay, now I got that the disks (which bound $\beta$-curves or $\gamma$-curves) are isotoped through the 4-d handlebody attached to $Y_{\beta,\gamma}$. Then there's a 3-ball in each $U_\beta$ and $U_\gamma$ which is left after removing the (thickened) disks, which results in two 3-balls in $Y_{\beta,\gamma}=\partial \# (S^1\times D^3)$. Can they also isotope to each other by the same isotopy? Mar 1 '18 at 3:11
• @cjackal Yes. The 4d handlebody is just a 3d handlebody $H$ cross an interval (with the sides $\partial H \times I$ quotiented out) and you're just isotoping from $H \times \{0\}$ to $H \times \{1\}$
– Carl
Mar 1 '18 at 3:45