If there are only 2 handles and no 1 handle in a Kirby Diagram then the intersection form of the underlying simply-connected 4-manifold coincides with the linking form. But what if there’s at least one 1 handle? Can we still compute its intersection form from the linking form?
Edit: For the purposes I'm looking, I'm concerned with simply-connected 4-manifolds, so we know the 2-handles cancel the 1-handles algebraically (say, linking number 1) and there's no torsion in $H^2$. Can we still determine the intersection form in this case?