The following definitions are from lecture notes of Némethi. A surface singularity $(X,0)$ is defined by $$(X,0) = (\{ f_1 = \ldots = f_m=0 \}) \subset \mathbb (\mathbb{C}^n,0),$$ where $f_i : (\mathbb{C}^n ,0) \to (\mathbb{C},0)$ are germs of analytic functions with $$r(p) = \mathrm{rank} \left [ \frac{\partial f_i}{\partial z_i} (p) \right ]_{i=1, \ldots, m; j=1, \ldots, N} = N-2$$ for any generic or smooth point $p$ of $X$.

If $r(0) = N-2$, then $(X,0)$ is analytically isomorphic to $\mathbb (\mathbb{C}^2,0)$. The singularity $(X,0)$ is called normal, if any bounded holomorphic function $f: X - \{ 0\} \to \mathbb C$ can be extended to a holomorphic function defined on $X$.

Then there is an algorithm in Appendix 1 for finding the resolution graphs of singularities in $\mathbb{C}^3$ with equation of the form $g(x,y) + z^n$ (i.e. suspensions of curve singularities in $(\mathbb C^2,0)$).

I couldn't understand the steps of that algorithm. My question is that

  1. Is there anyone to describe this algorithm for example Brieskorn sphere $\Sigma(2,3,4)$?

  2. Is it possible to find any Magma or Sage code for this purpose?

  • $\begingroup$ Can you point out what you do not understand in the algorithm? Also, the algorithm only works for suspensions of curve singularities in $(\mathbb{C}^2,0)$. $\endgroup$ Commented Mar 24, 2020 at 19:09
  • $\begingroup$ Nemethi points out that one may use this algorithm to verify minimal resolution graphs of Example 1.20. We may think our singularity as $\{ g + z^4 \} \subset (\mathbb{C}^3,0)$ where $g=x^2 +y^3$. I couldn't follow the Step 1 a - multiplicity part of the algorithm. $\endgroup$
    – user150450
    Commented Mar 25, 2020 at 12:11

1 Answer 1


About question 2: I think that the software Singular has this feature; it's well-documented, and if you look for resolution graph you should find the reference.

About question 1: well, I must admit that that algorithm is not pleasant, and that it took me a while to work out the case of $\Sigma(2,3,4)$.

Anyway, here we go. I can't draw graphs here (but if anyone can tell me that we can embed \xygraphs, next time I can make an effort), so things will have to be less pictorial than I'm comfortable with. I'll use greek letters for vertices to avoid clashing with Némethi's letters. So, we fix $g(x,y) = x^2+y^3$ and $n=4$.

First off, the resolution graph of $g$ has three vertices $\alpha, \beta, \gamma$, and an arrowhead $\delta$. The self-intersections are $e_\alpha = -3$, $e_\beta=-2$, $e_\gamma = -1$. ($\gamma$ is the central vertex, corresponding to the last blow-up, $\alpha$ was the first blow-up, and $\beta$ the second.)

From the formula (*) in the appendix, relating multiplicities with self-intersections, and using that $m_\delta = 1$, we obtain that $(m_\alpha, m_\beta, m_\gamma) = (2,3,6)$. We the compute $(d_\alpha, d_\beta, d_\gamma) = (2,1,3)$.

  • Step 1(a): from the computations above, $\alpha$ is covered by two vertices of multiplicity 1, while $\beta$ and $\gamma$ are covered by one vertex of multiplicity 3 each. Call them $\tilde\alpha_1, \tilde\alpha_2, \tilde\beta, \tilde\gamma$. The genus formula says that they're all rational ($\tilde g = 0$).

  • Step 1(b): the edge $(\alpha,\gamma)$ lifts to strings connecting $\tilde\alpha_i$ to $\tilde\gamma$, each of type $G(1,3,2)$; which means just a single vertex $\varepsilon_i$ with $(e_{\varepsilon_i}, m_{\varepsilon_i}) = (-2,2)$ (and two arrowheads ending at $\tilde\alpha_i$ and $\tilde\gamma$). The edge $(\beta,\gamma)$ lifts to a string of type $G(6,3,4)$, which is a single vertex $\zeta$ with $(e_{\zeta}, m_{\zeta}) = (-2,3)$ (and two arrowheads ending at $\tilde\beta$ and $\tilde\gamma$.

  • Step 1(c): the arrowhead $\delta$ lifts to a string of type $G(6,1,4)$, which is again a single vertex $\eta$ with $(e_{\eta}, m_{\eta}) = (-2,2)$ (and two arrowheads, one ending at $\tilde\gamma$ and the other one free).

At this point, the graph is star-shaped with four legs: its center is at $\tilde\gamma$ (with weight unspecified); two legs with weights $(-2,?)$ (ending at $\tilde\alpha_1, \tilde\alpha_2$), one with weights $(-2,?)$ (ending at $\tilde\beta$), one with weight $-2$ and an arrowhead.

  • Step 2: we now compute the missing self-intersections using (*). We obtain $e_{\tilde\alpha_i} = -2$, $e_{\tilde\beta} = e_{\tilde\gamma} = -1$.

  • Step 3: we drop the arrowhead, and blow down $\tilde\gamma$ and then $\zeta$ (which, after blowing down $\tilde\gamma$, can be blown down). The graph we obtain is $E_6$, as expected.

  • 1
    $\begingroup$ This is a great answer, thank you! $\endgroup$
    – user150450
    Commented Mar 29, 2020 at 10:30

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