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]**



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> **Question:** Can we say anything about the number of critical points of this Lefschetz pencil?
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> Can we give lower/upper bounds, for example involving the dimension/Betti numbers of $X$ and/or $\widetilde{X}$?
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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.




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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.



Bounding the number of critical points in a Lefschetz pencil
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 $$3c\_1(\mathcal{O}\_{\mathbb{P}^N}(1)|\_X)^2 - 2c\_1(\mathcal{O}\_{\mathbb{P}^N}(1)|\_X)\cdot c\_1(T\_X) + c\_2(T\_X).$$


