# Upper bound on Betti numbers of an intersection of hypersurfaces (or quadrics)

I have a problem I have been stuck with since several weeks now, and yet I believe it should be easy to specialists.

Let $k$ be an algebraically closed field, $m$ and $n$ two integers. Let $H_1,\dots,H_m$ be $m$ (smooth if that helps) hypersurfaces of degrees $d_1, \dots,d_m$ in the projective space $\mathbb P^n$ over $k$. Let $X$ be the intersection $H_1 \cap \dots \cap H_n$, and for $\ell$ any fixed auxiliary prime different from the characteristic of $k$, let $b_i$ be the Betti numbers of $X$ (rank of $H^i_\text{ét}(X,\mathbb Z_\ell)$). The question is:

Is there a constant $c(m,d_1,\dots,d_n)$ depending on the number of hypersurfaces $m$ and their degrees $d_i$ but not on the dimension $n$ of the ambient projective space such that all the Betti numbers $b_i$ are less than $c(m,d_1,\dots,d_n)$?

I am actually interested in the case where $b_1=\dots=b_m=2$, that is $X$ is an intersection of $m$ quadrics. For example, when $m=1$ and $d_1=2$, $X$ is a smooth quadrics, the cohomology of $X$ is well-known (see e.g. Miles Reid's thesis) and the Betti numbers are always $0$, $1$ or $2$, independently of the dimension $n-1$ of $X$. Still when $m=1$ but with $d_1$ arbitrary, I know that by Lefschetz's theorem $b_i=0$ or $1$ except for $i=n-1$, and I presume there is a formula for $b_{n-1}$ in terms of the degree $d_1$, but independent of the dimension $n$ -- Yet i couldn't find this formula in my memory nor on the web.

Though I am primarily interested in a general answer (in the case of quadrics), please feel free to assume anything that can help you ($X$ smooth, complete intersection, $m=1$, $m=2$ etc.) or replace the étale cohomology by your preferred one.

PS: after some hesitation, I have added the number theory tag because when the $H_i$ are defined over a finite field $\mathbb F_q$ (and $k$ is the algebraic closure of $\mathbb F_q$) the question is closely related (by Grothendieck's Lefschetz fixed points formula) to bounds on the number of points on $X(\mathbb F)$. For example, a positive answer would imply that for $n$ large enough with respects to $m$ and the $d_i$, $X(\mathbb F_q)$ is not-empty.

EDIT: Quochuan's reply has shown that the answer to the question is "no" for an hyper surface of degree $d>2$. He also pointed out that it was easy to compute the cohomology for a complete intersection. Following his suggestion, and this related question, one sees that the answer is also no for the complete intersection of two or more quadrics. In this case, if I'm not mistaken, the cohomology in middle degree grows polynomially in $n$. Anyway, the bottom line is that the case of quadrics is unique in his genre.

• The statement in your last sentence is already known; see the paper of Katz "On a theorem of Ax". – naf May 14 '15 at 18:34

No. When $m = 1$ and $d \ge 3$ the Betti numbers grow exponentially in $n$; see, for example, this blog post for the computation for smooth hypersurfaces (which is over $\mathbb{C}$, but then we can appeal to the Weil conjectures). As you say, you only need the middle Betti number, or equivalently the Euler characteristic, and you can compute this by computing the Chern classes. This is very classical and I think it even appears in Weil's paper introducing the Weil conjectures, maybe for the special case of Fermat hypersurfaces. The computation for smooth complete intersections is very similar, and beyond that I don't know.
• @Joël: thanks! For the intersection of two or more quadrics you can see, for example, homepages.warwick.ac.uk/~masda/3folds/qu.pdf; it looks like the answer is polynomial in $n$ as Dima says. – Qiaochu Yuan May 14 '15 at 23:58
The case of real quadrics was studied by A.Barvinok: On the Betti numbers of semialgebraic sets defined by few quadratic inequalities, Math.Z. 225(1997). He showed that the sum of Betti numbers of the algebraic set defined by $k$ quadrics in $\mathbb{R}^n$ is bounded by something like $n^{O(k)}$ - well, with some mild extra technical conditions. One can use this to get a similar answer for $\mathbb{C}$ instead of $\mathbb{R}$, see (7.2) in [loc.cit.].