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I am now trying to construct a $Z_2\times Z_2$-cover over $\mathbf{P}^n$. From the paper of Pardini, we need line bundles $L_1$, $L_2$, $L_3$ and divisors $D_1$, $D_2$, $D_3$ which satisfies the following 6 relations.

$$2L_i\equiv D_j+D_k\quad \textrm{and} \quad L_i+L_j\equiv L_k+D_k$$

However, the paper introduces 'reduced' data, which only need the following 2 relations.

$$2L_1\equiv D_2+D_3\quad \textrm{and} \quad 2L_2\equiv D_1+D_3$$

For an example, let $n=2$ and take $D_i$ to be the locus $X_i=0$. Then, for fiber coordinates $y_1$, $y_2$, and local coordinates $x_2$, $x_3$ (on $U_1$ where $X_1\neq 0$), if I am right, the equation of the bidouble cover can be written by

$$y_1^2=x_2x_3,\quad y_2^2=x_3.$$

By calculating its Jacobian, the bidouble cover should be singular at $(x_2,x_3,y_1,y_2)=(0,0,0,0)$, which is false. So I am wondering that is it wrong to think of such local equations from the 'reduced' data.

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    $\begingroup$ The equations you write give a ramification of order 4 along $x_3=0$, which is not what you want. The correct equations are $y_2^2=x_2$, $y_3^2=x_3$. $\endgroup$
    – abx
    Commented Jan 10, 2017 at 8:52

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As abx note in the comments, you miswrote the equations. Still the singular point remains.

The point is that the "reduced" data works well algebraically, as indeed you can deduce $L_3$ from the other data. But if you try to "reduce" analogously the equations you are eliminating some variables and that correspond geometrically to a projection which, in this case, is just a birational morphism, whose image has a singular point.

I give you here what are the "best", in my opinion, equations for this bidouble cover, determinantal equations that can be used to describe efficiently most bidouble covers. I apologize in advance for possible mistakes, because I'm in a rush, having exams to do right now. I'll check this post better in few hours.

The bidouble cover you consider can be see as a determinantal locus in ${\mathbb P}^5$, by taking variables $x_1,x_2,x_3,y_1,y_2,y_3$ and considering the locus given by the condition that the symmetric matrix $$ \begin{bmatrix} x_1&y_3&y_2\\ y_3&x_2&y_1\\ y_2&y_1&x_3 \end{bmatrix} $$ be of rank $1$.

You can recognize the three equations $y_i^2=x_ix_j$ giving a tridouble cover, the three involutions being given by $y_i\mapsto -y_i$, that has two irreducible components, and the remaining three equations $x_iy_i=y_jy_k$ determine one of the two components, invariant by three involutions, obtained by changing sign simultaneously to two of the $y_j$.

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  • $\begingroup$ Thanks for pointing out that the double cover $X'$ constructed by 'reduced' data is just birational to the one $X$ from 'standard data'. From another MO article, mathoverflow.net/questions/60998/…, I have seen that the bidouble cover can be constructed by a composition of double coverings, say $Y$. However, when I draw the ramification divisor, it seems that the bidouble cover $X$ constructed by standard data(the 6 relations) cannot be isomorphic to $Y$(but they can be birational with the same divisors $D_i$). Is it right? $\endgroup$
    – user190964
    Commented Jan 10, 2017 at 11:48
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    $\begingroup$ I'm not sure what you mean by $Y$, but I think here you are wrong, this factorization is biregular. A bidouble cover $X \rightarrow Z$ can be factored as a composition of two double covers in three different way, say $X \rightarrow Y_i \rightarrow Z$: if $L_1,L_2,L_3,D_1,D_2,D_3$ are the building data of $X \rightarrow Z$, then $f_i \colon Y_i \rightarrow Z$ is the double cover branched on $D_j+D_k$ (here $\{i,j,k\}=\{1,2,3\}$) whereas the double cover $X \rightarrow Y_i$ is branched on the pull-back of $D_i$. $\endgroup$
    – Roberto Pignatelli
    Commented Jan 10, 2017 at 12:21
  • $\begingroup$ Your $Y$ is the one I want to ask. My doubt is due to asymmetry of $Y$. Along the ramification divisors on $Y$, there are two points which corresponds to $D_2\cap D_3$(following your index), whereas there is only one for $D_1\cap D_2$. I think I made a mistake... I will check it again carefully. I really appreciate of your answer. $\endgroup$
    – user190964
    Commented Jan 10, 2017 at 12:37

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