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Let $X$ be a Kahler manifold and $Z\subset X$ be a smooth hypersurface. How to compute the Hodge diamond of the double covering $Y\to X$ ramified over $Z$? (And what I have to know? Would the map $H^*(X)\to H^*(Z)$ be enough?)

P.S. I tried the Gysin sequence, but it looks like there are many loose ends.

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The answer goes back to Esnault and Viehweg: $$H^q(Y,\Omega_Y^p)= H^q(X,\Omega_X^p)\oplus H^q(X,\Omega_X^p(\log Z)\otimes L^{-1})$$ where $L$ is the anti-invariant part of the direct image of $O_Y$ to $X$ under the natural $\mathbb{Z}/2$ action.

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  • $\begingroup$ Donu, that is cool! Could you give a reference for this statement? $\endgroup$
    – Dmitri Panov
    Commented Oct 18, 2012 at 2:29
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    $\begingroup$ @Dmitri: I included an answer that shows a computation of this plus one reference to an intermediate statement. I bet that what Donu is referring to is in Esnault-Viehweg's Inventiones paper on logarithmic de Rham complexes, and also probably in their book on Vanishing Theorems (see reference in my answer), but I have not searched for it, so I don't know for sure. $\endgroup$
    – Sándor Kovács
    Commented Oct 18, 2012 at 7:24
  • $\begingroup$ Sorry, things have been very busy. Anyway, my thanks to Sandor for providing more details. $\endgroup$
    – Donu Arapura
    Commented Oct 18, 2012 at 14:12
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This is more like a comment that in order to get a positive answer you need to specify more data.

Let us consider the simplest situation when $Z$ is empty, so $H^{\ast}(X)\to H^{\ast}(Z)$ is the zero map. It might happen in this situation, that there is more than one (unramified) double cover of $X$ and moreover double covers have different Betti numbers. Of course Hodge diamonds will be different as well.

Example. Consider the elliptic curve $E=\mathbb C/(\mathbb Z+i\mathbb Z)$. Let $X=(E\times E)/\sigma$ be the quotient of $E\times E$ by the following fixed point free involution: $(x,y)\to (-x,y+\frac{1}{2})$. Then $X$ has several unramified double covers. One obvious cover is $E\times E$ that we started with. A different cover of $X$ can be obtained from $E\times E$ by taking a double cover of the first factor and then taking a quotient by the lift of $\sigma$. This cover has same topology as $X$ and hence different $b_1$ from $E\times E$. Indeed, $b_1(X)=2\ne 4=b_1(E\times E)$.

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  • $\begingroup$ You misunderstood the question a bit. I consider a specific covering (the covering) which comes from $H_1(X\setminus Z)\to H_0(Z)$. $\endgroup$
    – Alex Gavrilov
    Commented Oct 18, 2012 at 13:36
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    $\begingroup$ In general THE covering does not exist: you have to specify a line bundle $L$ (cf. Donu's answer) such that $2L$ is linearly equivalent to $Z$. Diffent choices of $L$ may give different values of the invariants. $\endgroup$
    – rita
    Commented Oct 18, 2012 at 15:47
  • $\begingroup$ Alex, indeed, as Rita says THE covering does not exist always. To give an explicit example with $Z\ne 0$ consider an elliptic curve $E$ and let $Z$ be a divisor consisting of two points. There is clearly more than one double cover. The covering space is of course a genus $2$ curve. Now, to see that among these covers one can not single out a specific on just notice that the space of all degree two covers of the elliptic curve with varying pairs of branching points is connected. $\endgroup$
    – Dmitri Panov
    Commented Oct 19, 2012 at 8:46
  • $\begingroup$ Let me explain what I mean. There exists a unique map $H_0(Z)\to \mathbb{Z}\slash 2$ which is nontrivial on all the connected components of $Z$. There exists a (unique) map $H_1(X\setminus Z)\to H_0(Z)$. Thus, we have the map $\pi_1(X\setminus Z)\to \mathbb{Z}\slash 2$. It determines an (unramified) covering of $X\setminus Z$ which can be compactified to get a covering of $X$. (Certainly, there may be many other coverings as well). I may be mistaken about the uniqueness of this construction but your example did not convince me. $\endgroup$
    – Alex Gavrilov
    Commented Oct 19, 2012 at 12:57
  • $\begingroup$ Alex, I claim that there is no canonical map $H_1(X\Z)\to H_0(Z)$ (this is what happen in the example of 2-torus minus two points that I proposed to you). How do you get the canonical map? $\endgroup$
    – Dmitri Panov
    Commented Oct 19, 2012 at 16:42
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This started as a competing answer, but now it is just a computation of what Donu has stated already. It might still be useful for some.

First let's introduce some notation: $\pi:Y\to X$ is the double cover and $Z'=(\pi^*Z)_{\mathrm{red}}$ is the reduced pre-image of $Z$.

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In the situation of the question we have that $$ \pi_*\Omega_Y^p(\log Z')\simeq \Omega_X^p(\log Z) \oplus \big( \Omega_X^p(\log Z) \otimes \mathscr L^{-1} \big) \tag{$\star$} $$ where $\mathscr L$ is (as Donu already said) the anti-invariant part of the direct image of $\mathscr O_Y$ to $X$ under the natural $\mathbb{Z}/2$ action.

Since $\pi$ is finite, all higher direct images vanish and hence we have a similar isomorphism for cohomology: $$ H^q(Y,\Omega_Y^p(\log Z'))\simeq H^q(X, \pi_*\Omega_Y^p(\log Z'))\simeq H^q(X, \Omega_X^p(\log Z) )\oplus H^q(X, \Omega_X^p(\log Z) \otimes \mathscr L^{-1}) $$ by (3.22) of Esnault-Viehweg, Lectures on Vanishing Theorems.

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If one is interested in Hodge numbers of the open manifolds $X\setminus Z$ and $Y\setminus Z'$, then this should be good. Otherwise we need to connect these to the non-logarithmic sheaves. For that probably the best tool is the following short exact sequence:

$$ 0 \to \Omega_X^p \to \Omega_X^p(\log Z)\to \Omega_Z^{p-1} \to 0. $$

(The existence of this short exact sequence is a simple exercise, or can be found in (2.3) of ibid.

There is of course an equivalent one on $Y$ with $Z'$:

$$ 0 \to \Omega_Y^p \to \Omega_Y^p(\log Z')\to \Omega_{Z'}^{p-1} \to 0. $$

Aha!

Until this point I thought that I was going to get a different answer than Donu and that was the main reason I even started writing, but now it seems that I might get from this what Donu stated.

The point is, $\pi$ induces an isomorphism $Z'\to Z$ and hence the right hand side of the two short exact sequences are the same. So if we add $\Omega_X^p(\log Z) \otimes \mathscr L^{-1}$ to the first short exact sequence and push-forward the second short exact sequence, then we get $$ 0 \to \Omega_X^p \oplus \big( \Omega_X^p(\log Z) \otimes \mathscr L^{-1} \big) \to \Omega_X^p(\log Z)\oplus \big( \Omega_X^p(\log Z) \otimes \mathscr L^{-1} \big)\to \Omega_{Z}^{p-1} \to 0. $$ and

$$ 0 \to \pi_*\Omega_Y^p \to \pi_*\Omega_Y^p(\log Z')\to \pi_*\Omega_{Z'}^{p-1} \to 0. $$

Now, since $\pi|_{Z'}:Z'\to Z$ is an isomorphism and by $(\star)$ we get that these two short exact sequences are the same, so we have

$$ \pi_*\Omega_Y^p\simeq \Omega_X^p \oplus \big( \Omega_X^p(\log Z) \otimes \mathscr L^{-1} \big) \tag{$\star$} $$

and we get

$$ H^q(Y,\Omega_Y^p)\simeq H^q(X, \pi_*\Omega_Y^p)\simeq H^q(X, \Omega_X^p )\oplus H^q(X, \Omega_X^p(\log Z) \otimes \mathscr L^{-1}) $$ as stated by Donu.

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  • $\begingroup$ Doesn't this generalize quite easily to degree $d$ coverings? If I'm not mistaken, one still has $\pi^* \Omega_X(\log Z)) \cong \Omega_Y(\log Z')$? $\endgroup$
    – maxo
    Commented Sep 22 at 14:55
  • $\begingroup$ Yes, absolutely. You just need all the powers of $\mathscr L^{-1}$ up to $d-1$. (The question was for double covers) $\endgroup$
    – Sándor Kovács
    Commented Sep 23 at 15:07

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