I would like to discuss today a beautiful theorem of Grothendieck concerning differential equations. It was mentioned by Yves André in a wonderful talk at IHÉS in March 2016 and Hélène Esnault kindly explained its proof to me during a nice walk in the Bavarian Alps last April... The statement is as follows:
Theorem (Grothendieck, 1970). — Let be a smooth projective complex algebraic variety. Assume that is simply connected. Then every vector bundle with an integrable connection on is trivial.
Let indeed be a vector bundle with an integrable connection on and let us show that it is trivial, namely, that there exist global sections of which are horizontal () and form a basis of at each point.
Considering the associated analytic picture, we get a vector bundle with an integrable connection on the analytic manifold . Let . By the theory of linear differential equations, this furnishes a representation of the topological fundamental group in the fiber of the vector bundle at the point . Saying that is trivial on means that this representation is trivial, which seems to be a triviality since is simply connected.
However, in this statement, simple connectedness means in the sense of algebraic geometry, namely that has no non-trivial finite étale covering. And this is why the theorem can be surprising, for this hypothesis does not imply that is trivial, only that is has no non-trivial finite quotient. This is Grothendieck's version of Riemann's existence theorem, proved in SGA 1.
However, it is known that is topologically equivalent to a finite cellular space, so that its fundamental group is finitely presented.
Proposition (Malčev, 1940). — Let be a finitely generated subgroup of . Then is residually finite: for every finite subset of not containing , there exists a finite group and a morphism such that .
Consequently, the image of is residually finite. If it were non-trivial, there would exist a non-trivial finite quotient of , hence a non-trivial finite quotient of , which, as we have seen, is impossible. Consequently, the image of is trivial and is trivial.
In other words, there exists a basis of horizontal sections of . By Serre's GAGA theorem, are in fact algebraic, ie, induced by actual global sections of on . By construction, they are horizontal and form a basis of at each point. Q.E.D.
It now remains to explain the proof of the proposition. Let be a finite symmetric generating subset of containing , not containing , and let be the subring of generated by the entries of the elements of and their inverses. It is a non-zero finitely generated -algebra; the elements of are contained in , hence is a subgroup of . Let be a maximal ideal of and let be its residue field; the point of the story is that this field is finite (I'll explain why in a minute.) Then the reduction map induces a morphism of groups , hence a morphism . By construction, a non-zero entry of an element of is invertible in hence is mapped to a non-zero element in . Consequently, is disjoint from the kernel of , as was to be shown.
Lemma. — Let be a finitely generated -algebra and let be a maximal ideal of . The residue field is finite.
Proof of the lemma. — This could be summarized by saying that is a Jacobson ring: if is a Jacobson ring, then every finitely generated -algebra which is a field is finite over ; in particular, is a finite extension of a quotient field of . In the case , the quotient fields of are the finite fields , so that is a finite extension of a finite field, hence is a finite field. Let us however explain the argument. Let be the field ; let us replace by its quotient , where is the kernel of the map . There are two cases: either and , or , for some prime number , and is the finite field ;
we will eventually see that the first case cannot happen.
Now, is a field which is a finitely generated algebra over a subalgebra ; let be the fraction field of . The field is now a finitely generated algera over its subfield ; by Zariski's form of Hilbert's Nullstellensatz, is a finite algebraic extension of . Let us choose a finite generating subset of as a -algebra; each element of is algebraic over ; let us consider the product of the leading coefficients of their minimal polynomials, chosen to belong to and let . By construction, the elements of are integral over , hence is integral over . Since is a field, we deduce that is a field. To conclude, we split the discussion into the two cases stated above.
If , then , hence as well, and is a finite extension of , hence is a finite field.
Let us assume, by contradiction, that , hence and . By what precedes, there exists an element such that . But this cannot be true, because is not a field. Indeed, any prime number which does not divide is not invertible in . This concludes the proof of the lemma.
Remarks. — 1) The theorem does not hold if is not proper. For example, the affine line is simply connected, both algebraically and topologically, but the trivial line bundle endowed with the connection defined by is not trivial. It is analytically trivial though, but its horizontal analytic sections are of the form , for , and except for , none of them are algebraic.
However, the theorem holds if one assumes moreover that the connection has regular singularities at infinity.
2) The group theoretical property that we used is that on a complex algebraic variety, the monodromy group of a vector bundle with connection is residually finite. It is not always true that the topological fundamental group of a complex algebraic variety is residually finite. Examples have been given by Domingo Toledo in “Projective varieties with non-residually finite fundamental group”, Publications mathématiques de l’I.H.É.S., 77 (1993), p. 103–119.
3) The analogous result in positive characteristic is a conjecture by Johan De Jong formulated in 2010: If is a projective smooth simply connected algebraic variety over an algebraically closed field of characteristic , then every isocrystal is trivial. It is still open, despite beautiful progress by Hélène Esnault, together with Vikram Mehta and Atsushi Shiho.
Saturday, June 11, 2016
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