branched cover over slice disc I know some examples in 4 dimensions of rational homology balls (meaning that a manifold such that its rational homology groups are as a ball $B^4$) which are branched covers over a slice disc. Is the opposite also true? Is branched cover over a slice disc always a rational ball? 
 A: I'm not sure, but in the case of a double branched cover $W$ I was able to show the rational homology is of the form
$$H_n(W;\mathbb Q)\cong\begin{cases}0&n\geq4\\
\mathbb Q^a&n=3\\
\mathbb Q^{2a}&n=2\\
\mathbb Q^{a}&n=1\\
\mathbb Q&n=0
\end{cases}$$
for some $a$. $W$ is constructed by gluing two homeomorphic $4$-manifolds $A$ and $B$ together. These are formed by splitting along a $3$-manifold $M$ which cobounds a slice disk $\Delta$ and a Seifert surface for the knot. By a relative form of Alexander duality $H_p(A)\cong H^{3-p}(M,\partial M)$. Now looking at the Mayer Vietoris Sequence for $W=A\cup B$, and using the fact that $H_p(M)=H^{3-p}(M,\partial M)$ one sees that $H_2(W;\mathbb Q)\cong H_1(W;\mathbb Q)\oplus H_3(W;\mathbb Q)$. (Use the fact that the Euler characteristic of the sequence is trivial, since it is exact.) On the other hand, $\partial W$ is a rational homology sphere, so that the long exact sequence of the pair $(W,\partial W)$ will give us useful information. Indeed it shows that $H_p(W;\mathbb Q)\cong H_p (W,\partial W;\mathbb Q)$ for $p\leq 3$, whereupon Lefschetz duality tells us that $H_1(W;\mathbb Q)\cong H_3(W;\mathbb Q)$. So all of the homology of $W$ is determined by its first homology. There may be some way to show the first homology vanishes rationally, which is what happens one dimension down for finite branched covers of knots. On the other hand the torus is a $2$-fold branched cover of $S^2$ so branched covers can create rational homology in general.
Update: One can explicitly build a handle decomposition  of the $3$ manifold $M$ by looking at the slice disk. You can build it using only $0$ and $1$ handles, so it only has homology in degree $1$. This pretty quickly forces $H_3(W;\mathbb Q)=0$, and then by my earlier calculation, all of the homology groups are trivial. So this gives a complete proof for the $2$-fold branched case. The $n$-fold case should work similarly. 
