Milnor lattice and Du Val singularity I am reading this paper: https://arxiv.org/abs/0810.2687 by A. J. de Jong and Robert Friedman. In the proof of Theorem 4.10, a singularity of the following type shows up $$y^2=x^3+z^{6d-1}.$$  When $d=1$,  this is exactly the type of $E_8$ in Du Val singularities. And the Milnor lattice and Dynkin diagram can be obtained by blowing up. In general, looks like the corresponding Milnor lattice is isomorphic to $(2d-2)U\oplus d(-E_8)$.
I am wondering how to verify this in this case. In particular, can this be deduced from the case when $d=1$? Thanks!
 A: Nikulin, in $\S$2 of his classical paper about lattices explains an algorithm how to compute the Milnor lattice of a function germ with an isolated singularity. Specifically, in Theorem 2.2.2 he says it's sufficient to determine the signature of the from and the discriminant lattice; so to verify the claim it's sufficient to compute these.
For quasi-homogeneous isolated singularities in three variables like $y^2 - x^3 - z^{6d-1}$, in Theorem 2.3.1 he explains how to compute the discriminant form in terms of the star-shaped dual graph of the resolution of singularities. For the signature Nikulin refers to this paper by Steenbrink.
I don't understand this well-enough yet to be able to compute these, may be someone else could help...
A: For singularities of the form $\{z^d = f(x,y)\}$ (which, I believe, are called suspension singularities), the Milnor fibre $M$ has a nice topological description as the $d$-fold cover of $B^4$ branched over the Milnor fibre $F \subset B^4$ of the curve singularity $\{f(x,y) = 0\}$ at $(0,0)$.
Now there are several ways of computing the signature from here:

*

*Lefschetz fibrations: there's a Lefschetz fibration on $B^4$ with page $F$ and a monodromy $\mu = \delta_1 \circ \dots \circ \delta_m$ factorisation as a product of (right-handed) Dehn twists. Now $M$ has a Lefschetz fibration with fibre (diffeomorphic to) $F$ and monodromy (conjugated to) $\mu^d$, factored as $(\delta_1\dots\delta_m)^d$. This factorisation gives (with a bit of work) an explicit basis of the second homology and its intersection form.


*handle diagrams: $F$ is obtained by pushing in a surface in $S^3$ whose boundary is a link. Akbulut and Kirby give you a recipe to build a handle decomposition of the $d$-fold cover of $B^4$ cyclically branched over such objects, which in turn gives a basis for $H_2$ and the intersection form in that basis.


*Levine–Tristram signatures: as explained in Kauffman's book On knots, the Levine–Tristram signatures of a knot $K \subset S^3$ compute the signature of cyclic cover of $B^4$ branched over a surface whose boundary is $K$. (More precisely, you need to look at signatures at the $d$-th roots of unity to know about the $d$-fold cover. In the case at hand, we also get to compute the rank of $H_2$ (which is $(d-1)b_1(F)$). If we know that the link of the singularity is a homology sphere (as in the case at hand), usually the classification of unimodular indefinite forms is enough to conclude.
To stay in the algebro-geometric context, Durfee's paper The signature of smoothings of surface singularities seems relevant.
In the case of your singularities, the first and third approach are quite easy to implement. For the monodromy approach, it is convenient to look at the singularity as a $6k-1$-fold suspension (i.e. $d = 6k-1$, $f(x,y) = x^2+y^3$); here $F$ has genus 1 and the factorisation is the product of two Dehn twist along two curves generating $H_1(F)$. For the signature approach, either of the three choices ($d = 2, d = 3, d = 6k-1$) is feasible; I'd say that the easiest is looking at $d=6k-1$, and then you have to compute signatures of the torus knots $T(2,3)$, which is quite easy. (For instance, this might be done explicitly in Lickorish's An introduction to knot theory.)
