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Let $U$ be a smooth quasi-projective variety over $\mathbf{C}$. Let $U^{\infty}$ be $U$ but thought of as a smooth manifold.

Q1: Is there a simple proof (so it should avoid Hironaka's desingularization) that shows that $H_{dR}^*(U^{\infty},\mathbf{R})$ is finite dimensional?

Q2: Do we always have some kind of "triangulation of finite type" of $U^{\infty}$ that would explain the finiteness of its De Rham cohomology groups?

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    $\begingroup$ $H_{dR}^*(U^{\infty},\mathbf{R})$ looks like the usual $C^\infty$ de Rham cohomology, which is finite dimensional by the standard de Rham theorem. What you mean is probably the algebraic de Rham cohomology $H_{dR}^*(U,\mathbf{C})$. $\endgroup$
    – abx
    Commented Jan 26, 2014 at 18:01
  • $\begingroup$ Does quasi-projective imply quasi-compact in the Zariski topology? Anyhow, it is not clear to me that you can find a finite open good cover for $U$... $\endgroup$
    – Hugo Chapdelaine
    Commented Jan 26, 2014 at 20:56
  • $\begingroup$ For the Zariski topology, yes. $\endgroup$
    – abx
    Commented Jan 26, 2014 at 20:58
  • $\begingroup$ Ok nice. So then it is enough to show that on a smooth affine variety, the De Rham cohomology is finite dimensional. Do we always have the existence of a FINITE open good cover (in the topological sense) on smooth affine varieties? $\endgroup$
    – Hugo Chapdelaine
    Commented Jan 26, 2014 at 21:20

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I think there are proofs which are much easier. For example, you can try to compute the cohomology using the Morse theory. For that you need existence of Morse functions having finitely many critical points. However, if a Morse function is real algebraic (and you can always find such), it always has finitely many critical points as follows from Whitney's theorem:

Whitney, H., {\em Elementary structure of real algebraic varieties}, Ann. Math., 66 (1957), 545--556.

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  • $\begingroup$ Hi Misha thanks a lot for the reference. Is it possibl using these ideas to show that each smooth affine algebraic variety admits a finite open good cover (in the topological sense) (if such a result is true of course)? $\endgroup$
    – Hugo Chapdelaine
    Commented Jan 31, 2014 at 16:10
  • $\begingroup$ Morse theory gives you a cell decomposition for a manifold; to go from this to something like a triangulation is a purely topological problem. It is quite often non-trivial, however. $\endgroup$
    – Misha Verbitsky
    Commented Feb 1, 2014 at 15:10
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Re Q2,

MR0374131 (51 #10331) Reviewed
Hironaka, Heisuke
Triangulations of algebraic sets. Algebraic geometry (Proc. Sympos. Pure Math., Vol. 29,
Humboldt State Univ., Arcata, Calif., 1974), pp. 165–185. Amer. Math. Soc., Providence, R.I., 1975.
14B99 (32B20 57C15)

In case you are wondering: no, this does not use resolution of singularities. Hironaka's point in writing the article was to demonstrate that resolution is unnecessary for triangulation.

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    $\begingroup$ But the proof -- at least, Grothendieck's proof -- that de Rham and topological cohomology are isomorphic uses resolution of singularities. How to avoid this? $\endgroup$
    – Edgardo
    Commented Jan 26, 2014 at 16:32
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    $\begingroup$ @Edgardo: I agree. That is why I specify that I am only addressing Q2, about existence of triangulations. $\endgroup$
    – Jason Starr
    Commented Jan 26, 2014 at 18:26
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Usually, rather than triangulating algebraic varieties, it is better to find cylindrical decompositions. These are finite cellular decompositions that exist for every real semi-algebraic set and thus, restricting scalars, for every complex variety. See Algorithms in Real Algebraic Geometry for more.

If you want a proof of finiteness of $H^{\ast}_{DR}$ without resolution of singularities and without topological tools, you can see Monsky, Finiteness of de Rham Cohomology. I wouldn't call it simple, though!

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