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Let $X\to Y$ be a $d:1$ ramified covering map from a smooth complex projective variety $X$ to a smooth complex projective variety $Y$.

Question 1. What does the smoothness imply on the branch locus in $X$ or its image in $Y$? Does it imply that its a normal crossing?

Question 2. If not, assuming $Y=\mathbb{P}^n$, can we always deform $X$ to get a normal-crossing branch locus in $\mathbb{P}^n$ (or at least for its preimage in $X$)?

Comment 1. Regarding the nice answer of Tony, you may assume $X$ is simply-connected.

Comment 2. By normal crossing, I mean the reduced branch locus to be normal crossing.

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    $\begingroup$ By definition, the branch locus is the locus of critical values of $f \colon X \to Y$, hence it is in $Y$. The set of critical points (which is in $X$) is called the ramification locus. $\endgroup$
    – Francesco Polizzi
    Commented Nov 27, 2013 at 12:22
  • $\begingroup$ Did you want to add the hypothesis that the $d$-to-$1$ morphism is Galois / normal and has Abelian group of deck transformations? $\endgroup$
    – Jason Starr
    Commented Nov 27, 2013 at 18:25
  • $\begingroup$ Well, I am not sure. The kind of fibrations I am interested in are when $X$ is a smooth Calabi-Yau threefold and $Y$ is $\mathbb{P}^3$. I don't care about the complex structure of $X$, so I am looking for a generic member of its complex moduli space. If there are theorems for those particular type of fibrations, I would appreciate if you recall that. $\endgroup$
    – Mohammad Farajzadeh-Tehrani
    Commented Nov 27, 2013 at 20:25

2 Answers 2

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The branch locus in $Y$ need not be a normal crossings divisor even when $Y$ is a projective space. Suppose $X$ is a smooth complex projective surface with non-abelian fundamental group. By Noether normalization we can find a finite morphsim $f : X \to \mathbb{P}^{2}$. The branch locus $B \subset \mathbb{P}^{2}$ will be a divisor by the Zariski-Nagata purity theorem but this divisor can never be normal crossings. If it was, then the fundamental group $\pi_{1}(\mathbb{P}^{2}-B)$ will be abelian by Fulton's nodal curve theorem. Since $\pi_{1}(X-f^{-1}(B)) \subset \pi_{1}(\mathbb{P}^{2}-B)$ is of finite index, and $\pi_{1}(X-f^{-1}(B))$ surjects onto $\pi_{1}(X)$ e.g. by the Fulton-Lazarsfeld connectedness theorem, it will follow that $\pi_{1}(X)$ is abelian which is a contradiction. In general the only implication of the smoothness will be the purity of branch loci: the branch locus has to be of pure dimension one.

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    $\begingroup$ Thanks. You mentioned a goof set of related theorems. At the very end of your answer, your sentence is incomplete: " it will follows that $\pi_1(X)$ ?". Actually, I am only interested in the case where $X$ is a simply-connected threefold. What can be said in the simply-connected case. $\endgroup$
    – Mohammad Farajzadeh-Tehrani
    Commented Nov 27, 2013 at 3:09
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Even in the simply connected case there is no hope.

Let $X$ be a smooth cubic surface in $\mathbb{P}^3$ and take a general projection $f\colon X \to \mathbb{P}^2$. Then $f$ is a branched triple cover and it is well known that the branch locus $B \subset \mathbb{P}^2$ is a sextic curve with six ordinary cusps lying on a conic.

There is no way to deform this cover (mantaining $X$ smooth) in order to obtain a branch locus which is normal crossing.

Notice that $X$ is a rational surface, hence $\pi_1(X)=\{1\}$.

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  • $\begingroup$ Very nice example. Could please recall one reference for this example. By the way, Fermat cubic surface in $\mathbb{P}^3$, $x_0^3+x_1^3+x_2^3+x_3^3=0$ via the map $[x_0,x_1,x_2,x_3]\to [x_1,x_2,x_3]$ is cyclic cover of $\mathbb{P}^2$ ramified along a smooth cubic curve. It is not obvious that the covering you are mentioning is not deformable to this one. $\endgroup$
    – Mohammad Farajzadeh-Tehrani
    Commented Nov 27, 2013 at 14:14
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    $\begingroup$ This is very classical, Zarisky surely knew this. For a modern treatment and other examples, see Miranda's paper Triple covers in Algebraic Geometry, section 10. In your example the schematic branch locus must be counted with multiplicity two, since there is total ramification. So it is not reduced, hence not normal crossing. It must be a sextic curve in any case. $\endgroup$
    – Francesco Polizzi
    Commented Nov 27, 2013 at 14:19
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    $\begingroup$ In other words, your example is a deformation of the general cover, in which the six-cuspidal sextic degenerates to a smooth cubic counted twice. $\endgroup$
    – Francesco Polizzi
    Commented Nov 27, 2013 at 14:22
  • $\begingroup$ That's fine, what I care about is the underlying reduced divisor. I want the reduced branch locus to be normal crossing, otherwise you are absolutely right. May be I should have said that in my question. $\endgroup$
    – Mohammad Farajzadeh-Tehrani
    Commented Nov 27, 2013 at 14:32

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