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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.

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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 '14 at 18:35
@JasonStarr – Exactly. Are there variations on the definition? I will edit into the post. – jmc May 9 '14 at 18:46

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 $$3c_1(\mathcal{O}_{\mathbb{P}^N}(1)|_X)^2 - 2c_1(\mathcal{O}_{\mathbb{P}^N}(1)|_X)\cdot c_1(T_X) + c_2(T_X).$$

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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 '14 at 19:02
@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 '14 at 20:21
"This does work in characteristic $0$," --> "This does work in characteristic $p$," – Jason Starr May 9 '14 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 '14 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 '14 at 18:29

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