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Let $f \colon X \longrightarrow Y$ be a finite morphism of degree $n \geq 2$ between smooth compact, complex surfaces.

Let $B \subset Y$ be the branch divisor of $f$ and assume that the corresponding branching order is $2$, namely $$f^*B = 2R + R_0,$$ where $R \subset X$ is the ramification divisor and $R_0$ is the residual curve (sometimes this condition is expressed by saying that $f$ is a generic cover).

Question. If the branch locus $B$ is a smooth divisor, is it true that $RR_0=0$?

I know that the answer is yes for $n=2$ (trivially, because $R_0$ is empty in that case) and for $n=3$ (as a consequence of the general theory of triple covers developed by R. Miranda), but what happens for general $n$?

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It seems to me that the intersection is zero in general, i.e. the answer is YES.

Let us prove that $f^{-1}(B)$ is a smooth curve in $X$. This clearly implies the desired result. The proof of smoothness of $f^{-1}(B)$ relies one the following definition and a lemma.

Definition. Let $C$ be a complex curve in a complex surface $S$ and $x\in C$. Let us call the local fundamental group of $S\setminus C$ at $x$ to be $\pi_1(U_x\setminus (C\cap U_x))$ where $U_x$ is a sufficiently small neighborhood of $x$.

I think that the following lemma is correct.

Lemma. A complex analytic curve $C$ in $X$ is smooth at $x\in C$ iff the local fundamental group of $C$ at $x$ is $\mathbb Z$.

If this lemma holds, then it is easy to see indeed that in your question the total perimage of $B$ is smooth, because for any point $x\in B$ and a sufficiently small neighborhood of $U$ of $x$ the map from $f^{-1}(U\setminus (U\cap C))$ to $U\setminus (U\cap C)$ is etale. I.e. all connected components of $f^{-1}(U\setminus (U\cap C))$ have $\pi=\mathbb Z$.

Intuition for the lemma. I think that the lemma can be obtained as a corollary of Mumford's result stating that a point $x$ on a surface is smooth iff its link is the 3-sphere $S^3$. Here is why.

Indeed, suppose that $\pi_1(U_x\setminus (C\cap U_x))=\mathbb Z$. This means that the intersection of $C$ with the boundary of $U_x$ is an unknot. Indeed, the only knot is $S^3$ whose complement has fundamental group $\mathbb Z$ is the unknot.

Now, let us take a double cover of $U_x\setminus (C\cap U_x)$ and take its analytic completion. This is just a double cover of $U_x$ and its boundary is again $S^3$. So this double cover is smooth. Finally we see that $U_x$ is a smooth quotient of a ball by a smooth involution, hence its branching set (i.e., $C$) is a smooth curve.

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  • $\begingroup$ This is interesting, thanks for the suggestion. Do you have any reference for the Lemma? $\endgroup$
    – Francesco Polizzi
    Oct 20, 2016 at 6:13
  • $\begingroup$ Francesco, I don't know the reference to this lemma, but I think it is correct. I added a sketch proof, please have a look. Basically, I think that the link of a point of a plane curve is an uknot iff the curve is smooth at the point (I think I saw this statement somewhere) $\endgroup$
    – aglearner
    Oct 20, 2016 at 7:58
  • $\begingroup$ To prove the lemma: Since $C$ is a curve in a smooth surface and you have a local statement you can reduce to the case of plane curves. For plane curves you can prove the lemma by noting that the first betti number of the Milnor fiber equals the Milnor number of the singularity and that $H^1$ of the Milnor fiber is isomorphic with the abelianization of the local fundamental group. Moreover the Milnor number is zero if and only if $C$ is smooth at $x$. $\endgroup$
    – Remke Kloosterman
    Oct 20, 2016 at 10:04
  • $\begingroup$ Dear Remke, thanks for your comment. In my definition the local fundamental group at a smooth point of $C$ is $\mathbb Z$ (it is $\pi_1$ of the complement to $C$). Naturally $H^1$ of the Milnor fiber in this case is $0\ne \mathbb Z$. So there is some discrepancy with what you write in the comment. Could you please explain the confusion? $\endgroup$
    – aglearner
    Oct 20, 2016 at 10:51
  • $\begingroup$ You are right, I went to fast. The Milnor fiber is the fiber of $U_x\setminus (C \cap U_x)$ to a punctured disc. Now an argument using the Serre spectral sequence should give you $H^1(U_x \setminus (C\cap U_x))$ has rank one if and only if $H^1$ of the Milnor fiber is trivial. (There are also exact sequences which for fibrations relate the fundamental group of the total space, the fiber and the base. See e.g., page 73 of Dimca's book Singularities and Topology of Hypersurfaces. Hence you may be able to avoid spectral sequences) $\endgroup$
    – Remke Kloosterman
    Oct 20, 2016 at 15:55

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