Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $k$ be an algebraically closed field. Let $X/k$ be a smooth projective variety. For a suitable embedding in $\mathbb{P}^{n}$ we can form a Lefschetz pencil $\widetilde{X} \to D = \mathbb{P}^{1}$.

[Edit]: In response to Jason Starr's comment: I assume that every singular fibre of a Lefschetz pencil has a single ordinary double point (and is otherwise nonsingular). [/Edit]

Question: Can we say anything about the number of critical points of this Lefschetz pencil?

Can we give lower/upper bounds, for example involving the dimension/Betti numbers of $X$ and/or $\widetilde{X}$?

Asking Google gives some results for symplectic manifolds. I could not find anything related to algebraic varieties.

Notation: Let us fix the notation $j \colon U \to D$ for the smooth locus of $f$, and $i \colon S \to D$ the complement of $U$ in $D$. (So $S$ is the subset of $D$ with singular fibres.) Furthermore, $d$ is the dimension of $X/k$, hence also of $\widetilde{X}/k$. It is customary to write $n$ for the dimension of the fibres, so $d = n+1$. Let us write $q$ for the number of critical points, so $q = \#S(k)$. Finally fix a prime number $\ell$, invertible in $k$.

Motivation/baby case: (Please take in mind that I am a beginner with perverse sheaves, so the following might be totally wrong.) If the vanishing cycles are zero (a special case, implying $d$ is even) the number of critical points, $q$, has to be less then $\dim \mathrm{H}^{d}(\widetilde{X}, \mathbb{Q}_{\ell})$. I think this can be proven using the Leray spectral sequence for perverse sheaves (so that we have $\mathrm{E}_{2}$-degeneration). One can prove that ${}^{p}\mathrm{R}^{d}f_{*}\mathbb{Q}_{\ell} = (\mathrm{R}^{d-1}f_{*}\mathrm{Q}_{\ell})[1] \oplus i_{*}\mathbb{Q}_{\ell}(-d/2)$. The critical points then contribute to the dimension of $\mathrm{E}_{2}^{0,d} = \mathrm{H}^{0}(D, {}^{p}\mathrm{R}^{d}f_{*}\mathbb{Q}_{\ell})$. Using the $\mathrm{E}_{2}$-degeneration, we see that $\mathrm{E}_{2}^{0,d}$ is a direct summand of $\mathrm{H}^{d}(\widetilde{X}, \mathbb{Q}_{\ell})$, proving that $q$ is less than the $d$-th Betti number.


Probably this has been investigated before, in particular in the case that the vanishing cycles are not zero. If so, I would be very happy with a reference to the literature.

share|improve this question
1  
What is your definition of "Lefschetz pencil"? In particular, do you assume that every singular fiber has a single ordinary double point (and is otherwise nonsingular)? –  Jason Starr May 9 at 18:35
    
@JasonStarr – Exactly. Are there variations on the definition? I will edit into the post. –  jmc May 9 at 18:46

1 Answer 1

Using the Thom-Porteous formula, and assuming the standard definition of "Lefschetz pencil", the number of singular fibers is precisely $$c_{n+1}(\Omega_{X/k}\otimes_{\mathcal{O}_X}\mathcal{O}_{\mathbb{P}^N}(1)|_X) + c_1(\mathcal{O}_{\mathbb{P}^N}(1)|_X)\cdot c_n(\Omega_{X/k}\otimes_{\mathcal{O}_X}\mathcal{O}_{\mathbb{P}^N}(1)|_X).$$

Edit. Just to spell this out in terms of the usual Chern classes, $c_q(T_X)$, the formula is $$ \sum_{q=0}^{n+1} (n+2-q)(-1)^q c_1(\mathcal{O}_{\mathbb{P}^N}(1)|_X)^{n+1-q}\cdot c_q(T_X). $$ So, for instance, when $n+1$ equals $1$, i.e., $X$ is a curve, the number is $$2c_1(\mathcal{O}(1)_{\mathbb{P}^N}|_X) - c_1(T_X) = 2g(X)-2 + 2\text{deg}_X(\mathcal{O}_{\mathbb{P}^N}(1)|_X).$$ Similarly, when $n+1$ equals $2$, i.e., $X$ is a surface, the number is $$3c_1(\mathcal{O}_{\mathbb{P}^N}(1)|_X)^2 - 2c_1(\mathcal{O}_{\mathbb{P}^N}(1)|_X)\cdot c_1(T_X) + c_2(T_X).$$

share|improve this answer
    
Cool! Thanks for the answer. To clarify: Do you mean $X$ or $\widetilde{X}$? And besides that: Does this also work in characteristic $p > 0$? What I saw so far (a bit of Thom and Porteous, but not say, Fulton's paper) seems to be restricted to characteristic $0$. –  jmc May 9 at 19:02
1  
@jmc: I mean $X$, not $\widetilde{X}$. This does work in characteristic $0$, but only if this is an "honest" Lefschetz pencil. In positive characteristic, in order to obtain an "honest" Lefschetz pencil, sometimes you need to re-embed by the Veronese 2-uple map. If the dimension of $X$ is $1$ and you, further, want that all fibers are irreducible, sometimes you need to re-embed by the Veronese 3-uple map. –  Jason Starr May 9 at 20:21
1  
"This does work in characteristic $0$," --> "This does work in characteristic $p$," –  Jason Starr May 9 at 20:45
    
– Thanks for the additional edit + comment. So this is not immediately linked to Betti numbers. One more question: Do I understand it correctly that the number of singular fibres only depends on $X$ and the embedding into some $\mathbb{P}^{N}$? So after fixing an embedding, it does not depend on the axis of the Lefschetz pencil, right? –  jmc May 10 at 6:44
    
@jmc: "... it does not depend on the axis of the Lefschetz pencil, right?" That is correct: the number of singular fibers in a Lefschetz pencil depends only on the intersection numbers formed from Chern classes of $X$ and powers of the first Chern class of $\mathcal{O}_{\mathbb{P}^N}(1)|_X$. You are also correct that this description of the number is not immediately linked to Betti numbers. However, you can describe the number using Euler characteristics of $\widetilde{X}$ and the fibers. –  Jason Starr May 10 at 18:29

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.