Cobordism and Kirby calculus It may be a simple question but I wonder to ask:
Is it possible to draw a homology cobordism between $3$-manifolds by using the techniques of Kirby calculus?
At least, for instance, Brieskorn spheres?
 A: There are many examples of the sort, in effect. As far as I know, Akbulut and Kirby (Mazur manifolds, Michigan Math. J. 26 (1979)) proved that $\Sigma(2,5,7)$, $\Sigma(3,4,5)$, and $\Sigma(2,3,13)$ bound contractible 4-manifolds; their work was then extended by Casson and Harer (Some homology lens spaces which bound rational homology balls, Pacific Math. J. 96 (1981)). Stern, Fintushel-Stern, and Fickle have more examples.
I'm sure that the Akbulut-Kirby (and some of the Casson-Harer) examples were done by Kirby calculus.
Additionally, there are also examples of Brieskorn spheres bounding rational homology 4-balls (but not integral ones, because they have non-zero Rokhlin invariant). The first example was $\Sigma(2,3,7)$, and the rational ball was produced by Fintushel and Stern (A $\mu$-invariant one homology 3-sphere that bounds an orientable rational ball, in Four-manifold theory (Durham, NH, 1982) (1984)) by explicit handle moves. This has been extended further; the latest news I have are from a paper of Akbulut and Larson (Brieskorn spheres bounding rational balls, Proc. Amer. Math. Soc. 146 (2018)), where they provide two infinite families of examples: $\Sigma(2,4n+1,12n+5)$ and $\Sigma(3,3n+1,12n+5)$, as well as $\Sigma(2,3,19)$.
Akbulut's book 4-manifolds (Oxford University Press) contains a wealth of examples along these lines (not many more with Brieskorn spheres, I should think).
Finally, I am not aware of (but would be interested in seeing) explicit, non-trivial examples of (rational or integral) homology cobordisms between Brieskorn spheres.
A: As Golla pointed out that since every smooth $4$-manifold has a handle decomposition, you can draw a Kirby diagram. See the following pretty nice picture from Akbulut's lecture notes (now it is a published book titled $4$-manifolds).

Golla listed that lots of Brieskorn spheres which are known to be bound integral or rational homology balls, i.e., they are all integral or rational homology cobordant to $S^3$.
Following the technique of Akbulut and Larson, I also recently found new Brieskorn spheres bounding rational homology balls: $\Sigma (2,4n+3,12n+7)$ and $\Sigma(3,3n+2,12n+7)$, see the preprint.
It is interesting to note that the following Brieskorn spheres bound rational homology balls but not integral homology balls (these $4$-manifolds must contain $3$-handle(s)):


*

*$\Sigma(2,3,7)$, $\Sigma(2,3,19)$,

*$\Sigma(2,4n+1,12n+5)$ and $\Sigma(3,3n+1,12n+5)$ for odd $n$,

*$\Sigma(2,4n+3,12n+7)$ and $\Sigma(3,3n+2,12n+7)$ for even $n$.
On the other hand, we know that every closed oriented $3$-manifold is cobordant to $S^3$ due to the celebrated theorem of Lickorish and Wallace. 
In the following picture, you can see the explicit cobordism from $\Sigma(2,3,13)$ to $\Sigma(2,3,7)$ which is constituted by adding the red $(-1)$-framed $2$-handle. (The knot pictures are from KnotInfo). Here, blow down the red one to get the right-hand side. (Of course, they are not integral homology cobordant.)

Note that the Brieskorn spheres $\Sigma(2,3,6n+1)$ are obtained by $(+1)$-surgery on the twist knots $(2n+2)_1$, see for example Saveliev's book pg. 49-50. Here, $6_1$ is the stevedore knot (the left knot in the figure) and $4_1$ is the figure-eight knot (the right one).
