0
$\begingroup$

I am reading the section 4.2 in Kollar-Mori, where they provide the explicit equations for Du Val Singularities. In the course of the proof, they reduce to studying the equation $x^2+f(x,y)=0$ in a neighborhood of the origin in $\mathbb{C}^3$, and where $f$ is a power series of order 3. To argue about the form of the leading term $f_3$, they need to consider a top form around the germ of singularity $0 \in S$. They say they can pick \begin{equation} \frac{dy \wedge dz}{x}. \end{equation} At first I could not quite see why. Then, I argued as follows. Taking implicit differentiation we get \begin{equation} 2xdx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz=0. \end{equation} Since we already know we are dealing with an $S_2$ surface, it is enough to show the form is defined in a punctured neighborhood of $0 \in S$. Then, I guessed the implicit function theorem should tell me that where $x$ vanishes $(x,y)$ (or $(x,z)$) are local parameters, and that $\frac{\partial f}{\partial z}$ (or $\frac{\partial f}{\partial y}$) does not vanish, and the above form can be written as \begin{equation} -\frac{dx \wedge dy}{\frac{\partial f}{\partial z}} \quad \left(\mathrm{or}\;-\frac{dx \wedge dz}{\frac{\partial f}{\partial y}}\right). \end{equation} This argument looks a bit cumbersome to me. I was thinking I could try to use adjunction. Adjunction tells me that the canonical sheaf of $0 \in S$ is generated by \begin{equation} \left.\frac{dx \wedge dy \wedge dz}{x^2+f(y,z)}\right|_S, \end{equation} but I don't know how to make it look like a two form.

Question

Is there a better way to write down explicitly a local top form (using adjucntion, or some different strategy as well)? Also, is there a more general approach, not relying on the particular shape of the equation I have in the above example?

$\endgroup$
5
  • 1
    $\begingroup$ Hi, welcome! To make the question more readable, may I suggest that in the end you will put a question in bold and ask a self contained question? The rest of the details will then be distinct from the question. $\endgroup$
    – Amir Sagiv
    Commented Aug 19, 2016 at 17:21
  • $\begingroup$ @AmirSagiv Thank you for the advice, I will edit it to make it more direct! $\endgroup$
    – Stefano
    Commented Aug 19, 2016 at 17:24
  • 1
    $\begingroup$ Wikipedia: Poincaré residue. $\endgroup$
    – abx
    Commented Aug 20, 2016 at 15:41
  • $\begingroup$ @abx Thank you! Do you know where I could read more about it? $\endgroup$
    – Stefano
    Commented Aug 20, 2016 at 15:43
  • 1
    $\begingroup$ For instance Griffiths and Harris, p. 147. It is also in Voisin's book but I don't have the precise reference at hand. $\endgroup$
    – abx
    Commented Aug 20, 2016 at 15:47

0

You must log in to answer this question.