7
$\begingroup$

Let $f : X\rightarrow Y$ be a proper flat morphism (of schemes) with connected fibers over a smooth projective curve $Y$ over $\mathbb{C}$. Let $X_{y_0}$ denote a smooth fiber over $y_0\in Y$.

If $f$ is smooth, then the topological Euler characteristic is multiplicative: $\chi(X) = \chi(Y)\cdot\chi(X_{y_0})$. In general, there is a formula for the topological Euler characteristic which includes some correction terms for the singular fibers: $$\chi(X) = \chi(Y)\chi(X_{y_0}) + \sum_{y\in Y}(\chi(X_y) - \chi(X_{y_0})$$

Does there exist an analogous formula for the holomorphic/algebraic Euler characteristic? I'm in particular interested in the case where $X\rightarrow Y$ is a family of curves over $Y$ (so $\dim X = 2$). To be precise, I'm looking for a formula for $\chi(X,\mathcal{O}_X)$ in terms of geometric invariants of $Y$ and some fibral data. By fibral data I mean geometric properties which are local on the base $Y$ (e.g. geometric invariants of fibers, monodromy around singular fibers acting on the cohomology of a smooth fiber...etc).

As a nonexample, Riemann-Roch relates this Euler characteristic to the intersection number of $X_{y_0}$ with $K - X_{y_0}$ (which can be understood locally on $Y$) plus $\chi(X,\mathcal{O}_X(X_{y_0}))$ (which does not appear to be local on $Y$).

Added in edit: To state a more precise question -- Is the holomorphic Euler characteristic $\chi(X,\mathcal{O}_X)$ determined by $Y$ and data which is local on $Y$? Stated another way, do there exist maps $f : X\rightarrow Y$ and $f' : X'\rightarrow Y$ (satisfying the above conditions) such that

  1. there is a covering in the etale topology $\{U_i\rightarrow Y\}_{i\in I}$ and isomorphisms $\phi_i : X_{U_i}\rightarrow X'_{U_i}$, and
  2. $\chi(X,\mathcal{O}_X)\ne \chi(X',\mathcal{O}_{X'})$.

Here I'm happy to assume that $X,X'$ are smooth over $\mathbb{C}$ (but I still want to allow singular fibers of $f,f'$). If it changes the answer, I'm also interested in the variation where $\{U_i\}$ is instead an open covering in the analytic topology. I am also interested in results where we assume that $X,X'$ are moreover general type. If the euler characteristic is not determined by $Y$ and data local on $Y$, then I'm also interested in any contraints such data might impose on $\chi(X,\mathcal{O}_X)$.

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ I am trying to understand how your "nonexample" is not an example. What kind of formula do you envision when $X$ equals $Y\times Z$ if you are not allowing $\chi(Z,\mathcal{O}_Z)$ as part of your formula? $\endgroup$
    – Jason Starr
    Commented Apr 9, 2022 at 0:31
  • $\begingroup$ Are you willing to assume that $X$ (not $f$) is smooth? $\endgroup$
    – Piotr Achinger
    Commented Apr 9, 2022 at 7:42
  • 1
    $\begingroup$ @JasonStarr I imagine $\chi(X_{y_0},\mathcal O_{X_{y_0}})$ would be allowed as it is 'fibral', but not $\chi(X,\mathcal O_X(X_{y_0}))$ as it is 'global' on $X$. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Apr 9, 2022 at 14:12
  • $\begingroup$ @JasonStarr Remy is right - $\chi(X_{y_0},\mathcal{O}_{X_{y_0}})$ would be fine, but $\chi(X, \mathcal{O}_X(X_{y_0}))$ wouldn't be, unless it can somehow be expressed in terms of data which is local on $Y$. $\endgroup$
    – Will Chen
    Commented Apr 9, 2022 at 15:09
  • $\begingroup$ @PiotrAchinger Sure, I'd be happy for an answer in the case of smooth $X$. $\endgroup$
    – Will Chen
    Commented Apr 10, 2022 at 20:07

1 Answer 1

Reset to default
5
$\begingroup$

I'm not entirely sure what would constitute an answer. But here a few simple observations. Let me focus on what you seem be interested in, namely a projective family of connected curves over smooth projective curve $f:X\to Y$. Let me also assume $X$ smooth. Let $h$ be the genus of $Y$, and $g$ the genus of the general fibre. Then from the Leray spectral sequence, Riemann-Roch and Grothendieck duality $$\chi(\mathcal{O}_X)= (1-g)(1-h) - \deg R^1f_*\mathcal{O}_X=\chi(\mathcal{O}_Y)\chi(\mathcal{O}_{X_{y_0}})+\deg f_*\omega_{X/Y}$$ A theorem of Fujita then gives an inequality $$\chi(\mathcal{O}_X)\ge \chi(\mathcal{O}_Y)\chi(\mathcal{O}_{X_{y_0}})$$ which is something. To say more, one would need to compute $\deg f_*\omega_{X/Y}$. But I think this depends on more than local information at the bad fibres, which is what I think you are asking. The reason I say this, is that this degree can be positive and variable for Kodaira surfaces, which have no bad fibres at all. [Added Comment I guess there are various things called Kodaira surfaces. The ones I have in mind are of general type, and hence algebraic. See page 220 of Compact Complex Surfaces by Barth, Hulek, Peters and Van de Ven for further details.]

Improve this answer
$\endgroup$
3
  • $\begingroup$ Somehow the point is that the Euler characteristic of a rank $n$ vector bundle is not just $n\chi(X,\mathcal O_X)$, but also depends on global data like the degree. By contrast (and somewhat to my surprise), for local systems the situation is much easier: $\chi(X,\mathcal L) = \operatorname{rk}(\mathscr L)\chi(X,\underline{\mathbf Q})$. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Apr 10, 2022 at 23:29
  • $\begingroup$ Thank you for this! It looks like Kodaira surfaces are not algebraic? Do you know of an algebraic example? (I've also edited the question to make it more precise). $\endgroup$
    – Will Chen
    Commented Apr 14, 2022 at 17:54
  • $\begingroup$ @WillChen Yes, I added more explanation. $\endgroup$
    – Donu Arapura
    Commented Apr 14, 2022 at 22:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .