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

As far as I know, Morse theory yields much information on the topology of smooth manifolds; in particular, it can be used to prove Artin's vanishing (that the singular cohomology of smooth complex variety of dimension n vanishes in degrees >n). My question is: are there any ideas how to extend any of the consequences of Morse theory to (the study of the etale homology of) algebraic varieties (over fields of arbitrary characteristic)? In particular, what is the relation between Morse homology and Lefschetz pensils?

share|improve this question
add comment

1 Answer

up vote 17 down vote accepted

A Morse function is a map of a manifold to the real line locally equivalent to: $$f(x_1,\ldots, x_n)=-x_1^2\ldots -x_k^2+ x_{k+1}^2+\ldots+x_n^2$$ for some $k$. In other words, for which the singularities are as simple as possible. While a Lefschetz pencil is a map of a smooth projective variety to the projective line local analytically given by $f=x_1^2+\ldots+x_n^2$. So in this sense, they are very analogous. There are differences, however. Given a Morse function $f:X\to \mathbb{R}$, the collection of the above numbers $k$, called indices, determine the homotopy type.[the number of cells in a complex homotopic to $X$]. For "Artin's" vanishing, it is enough to choose an $f$, where these indices are bounded by dimension. (Details can be found in the first few pages of Milnor's Morse theory.) I'm not aware that there is any complete substitute on the other side, given say a Lefschetz pencil $f:X\to \mathbb{P}^1$. However, the pencil does give a way to calculate the (etale) cohomology of $X$ in terms of the critical points of $f$ and the monodromy, or more formally in terms of the direct images $R^if_*\mathbb{Q}_\ell$. And this is quite powerful.


Perhaps it would be more instructive give a Morse-like pseudo-proof of Artin's theorem. By "pseudo" I mean that there is step which I can't justify without a lot more effort than this is worth. Let $X\subset \mathbb{A}^n$ be an irreducible affine variety of dimension $n$ over an algebraically closed field. By generic projection, we get a nonconstant morphism $f:X\to \mathbb{A}^1$ which will play the role of our Morse function. Suppose that (*) the direct images $R^jf_*\mathbb{Q}_\ell$ were constructible and commuted with base change. Then by induction, $R^jf_*\mathbb{Q}_\ell=0$ for $j>n-1$. From the Leray spectral sequence, we need only prove that $H^i(\mathbb{A}^1,F)=0$ for $i>1$ and any constructible sheaf $F$, but this is easy. Note that if $f$ were proper, (*) would be automatic. In general, one could get around it by generic base change [SGA 41/2, p 236] and devissage.

share|improve this answer
    
In your first paragraph, one should probably qualify the statement that the indices determine the homotopy type of $X$. –  Mariano Suárez-Alvarez Oct 9 '12 at 23:30
    
Yes, right..... –  Donu Arapura Oct 10 '12 at 0:03
    
There are 60 homotopy types of compact manifolds of dimension 16 on which there is a Morse function with 3 critical points, so there is more information needed than the orders of the critical values and their indices. This is what I mean —I should have been more explicit. –  Mariano Suárez-Alvarez Oct 10 '12 at 0:44
    
(The numbers come from Eells+Kuiper's Manifolds which are like projective planes) –  Mariano Suárez-Alvarez Oct 10 '12 at 0:46
    
OK, I hadn't realized this. –  Donu Arapura Oct 10 '12 at 1:02
show 1 more comment

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.