Let $X \subset \mathbb{P}^3$ be a singular surface. For simplicity take $X$ to be a cone over a smooth conic. And let $C \subset X$ be a curve passing through the singular point, with normalization morphism \begin{equation*} \nu: \tilde{C} \rightarrow C. \end{equation*}

My question is: how can we determine the degree of the ramification divisor of the morphism $\nu$?

And specifically, in the case of a quadric, let's just consider the case where $\tilde{C} \subset \tilde{X}$, where $\pi: \tilde{X} \rightarrow X$ is the minimal resolution of $X$ and $\tilde{C}$ intersects the exceptional $(-2)$-curve transversely. Can we calculate the degree of ramification in this case?

My thought was to look at the exact sequence \begin{equation*} 0 \rightarrow \Omega_X \rightarrow \Omega_X^{\vee \vee} \rightarrow Q \rightarrow 0, \end{equation*} where the cokernel $Q$ is supported at the singular point and then pull this back via $\pi$ \begin{equation*} \pi^* \Omega_X \rightarrow \pi^*(\Omega_X^{\vee \vee}) \rightarrow \pi^* Q \rightarrow 0. \end{equation*}

and then restrict this to $\tilde{C}$. So to make this work, I need to calculate $Q$ and $\pi^* Q$. Is there a good reference for this calculation? Or a better way to understand the ramification more directly?



1 Answer 1


I think there is no ramification in the case you are interested in.

The surface $\tilde X$ is the minimal ruled surface $\mathbb F_2$. The map $\tilde X \to X$ contracts the $-2$-curve $E$ to a point and maps the fibers of the ruling of $\mathbb F_2$ to lines in $\mathbb P^3$. So at every point of $E$ the map has differential of rank 1 and the kernel of the differential is the tangent space to $E$, because $E$ is contracted.

If $E$ intersects $\tilde C$ transversally, by the previous remarks, $\tilde C \to C$ has non zero differential at $x$.

  • $\begingroup$ Thank you for your helpful comment, and I believe your method is superior to calculations with the cotangent bundle. However, I am interested in the case where the $\tilde{C}.E > 1$, so that this is indeed a singular point of $C$. $\endgroup$
    – Jordan
    Nov 21, 2012 at 18:30
  • $\begingroup$ In fact $E\tilde C=1$ is not necessary, it is enough that the intersection of $E$ and $\tilde C$ are transversal. I've edited my answer accordingly. $\endgroup$
    – rita
    Nov 21, 2012 at 19:16
  • $\begingroup$ Thank you, your explanation does make plenty of sense. For some reason, I had thought the situation was more complicated for space curve singularities, but I suppose not in this case. $\endgroup$
    – Jordan
    Nov 21, 2012 at 21:06

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