Saturday, June 11, 2016

Triviality of vector bundles with connections on simply connected varieties

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 XX be a smooth projective complex algebraic variety. Assume that XX is simply connected. Then every vector bundle with an integrable connection on XX is trivial.

Let indeed (E,)(E,\nabla) be a vector bundle with an integrable connection on XX and let us show that it is trivial, namely, that there exist nn global sections e1,,ene_1,\dots,e_n of EE which are horizontal (ei=0\nabla e_i=0) and form a basis of EE at each point.

Considering the associated analytic picture, we get a vector bundle (Ean,)(E^{\mathrm{an}},\nabla) with an integrable connection on the analytic manifold X(C)X(\mathbf C). Let xX(C)x\in X(\mathbf C). By the theory of linear differential equations, this furnishes a representation ρ\rho of the topological fundamental group π1(X(C),x)\pi_1(X(\mathbf C),x) in the fiber ExE_x of the vector bundle EE at the point xx. Saying that (Ean,)(E^{\mathrm{an}},\nabla) is trivial on X(C)X(\mathbf C) means that this representation ρ\rho is trivial, which seems to be a triviality since XX is simply connected.

However, in this statement, simple connectedness means in the sense of algebraic geometry, namely that XX has no non-trivial finite étale covering. And this is why the theorem can be surprising, for this hypothesis does not imply that π1(X(C),x))\pi_1(X(\mathbf C),x)) 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 X(C)X(\mathbf C) is topologically equivalent to a finite cellular space, so that its fundamental group π1(X(C),x)\pi_1(X(\mathbf C),x)  is finitely presented.

Proposition (Malčev, 1940). — Let GG be a finitely generated subgroup of GL(n,C)\mathrm{GL}(n,\mathbf C). Then GG is residually finite: for every finite subset TT of GG not containing {In}\{\mathrm I_n\}, there exists a finite group KK and a morphism f ⁣:GKf\colon G\to K such that TKer(f)=T\cap \operatorname{Ker}(f)=\varnothing.

Consequently, the image of ρ\rho is residually finite. If it were non-trivial, there would exist a non-trivial finite quotient KK of im(ρ)\operatorname{im}(\rho), hence a non-trivial finite quotient of π1(X(C),x)\pi_1(X(\mathbf C),x), which, as we have seen, is impossible. Consequently, the image of ρ\rho is trivial and (Ean,)(E^{\mathrm{an}},\nabla) is trivial.

In other words, there exists a basis (e1,,en)(e_1,\dots,e_n) of horizontal sections of EanE^{\mathrm{an}}. By Serre's GAGA theorem, e1,,ene_1,\dots,e_n are in fact algebraic, ie, induced by actual global sections of EE on XX. By construction, they are horizontal and form a basis of EE at each point. Q.E.D.

It now remains to explain the proof of the proposition. Let SS be a finite symmetric generating subset of GG containing TT, not containing In\mathrm I_n, and let RR be the subring of C\mathbf C generated by the entries of the elements of SS and their inverses. It is a non-zero finitely generated Z\mathbf Z-algebra; the elements of SS are contained in GL(n,R)\mathrm {GL}(n,R), hence GG is a subgroup of GL(n,R)\mathrm{GL}(n,R). Let m\mathfrak m be a maximal ideal of RR and let kk be its residue field; the point of the story is that this field kk is finite (I'll explain why in a minute.) Then the reduction map RkR\to k induces a morphism of groups GL(n,R)GL(n,k)\mathrm{GL}(n,R)\to \mathrm {GL}(n,k), hence a morphism GGL(n,k)G\to \mathrm{GL}(n,k). By construction, a non-zero entry of an element of SS is invertible in RR hence is mapped to a non-zero element in kk. Consequently, SS is disjoint from the kernel of ff, as was to be shown.

Lemma. — Let RR be a finitely generated Z\mathbf Z-algebra and let m\mathfrak m be a maximal ideal of RR. The residue field R/mR/\mathfrak m is finite.

Proof of the lemma. — This could be summarized by saying that Z\mathbf Z is a Jacobson ring: if AA is a Jacobson ring, then every finitely generated AA-algebra KK which is a field is finite over AA; in particular, KK is a finite extension of a quotient field of AA. In the case A=ZA=\mathbf Z,  the quotient fields of Z\mathbf Z are the finite fields Fp\mathbf F_p, so that KK is a finite extension of a finite field, hence is a finite field. Let us however explain the argument. Let KK be the field R/mR/\mathfrak m; let us replace Z\mathbf Z by its quotient A=Z/PA=\mathbf Z/P, where PP is the kernel of the map ZR/m\mathbf Z\to R/\mathfrak m. There are two cases: either P=(0)P=(0) and A=ZA=\mathbf Z, or P=(p)P=(p), for some prime number pp, and AA is the finite field Fp\mathbf F_p;
we will eventually see that the first case cannot happen.

Now, KK is a field which is a finitely generated algebra over a subalgebra AA; let kk be the fraction field of AA. The field KK is now a finitely generated algera over its subfield kk; by Zariski's form of Hilbert's Nullstellensatz, KK is a finite algebraic extension of kk. Let us choose a finite generating subset SS of KK as a kk-algebra; each element of SS is algebraic over kk; let us consider the product ff of the leading coefficients of their minimal polynomials, chosen to belong to A[T]A[T] and let A=A[1/f]A'=A[1/f]. By construction, the elements of SS are integral over KK, hence KK is integral over AA'. Since KK is a field, we deduce that AA' is a field. To conclude, we split the discussion into the two cases stated above.

If P=(p)P=(p), then A=FpA=\mathbf F_p, hence k=Fpk=\mathbf F_p as well, and KK is a finite extension of Fp\mathbf F_p, hence is a finite field.

Let us assume, by contradiction, that P=(0)P=(0), hence A=ZA=\mathbf Z and k=Qk=\mathbf Q. By what precedes, there exists an element fZf\in\mathbf Z such that Q=Z[1/f]\mathbf Q=\mathbf Z[1/f]. But this cannot be true, because Z[1/f]\mathbf Z[1/f] is not a field. Indeed, any prime number which does not divide ff is not invertible in Z[1/f]\mathbf Z[1/f]. This concludes the proof of the lemma.

Remarks. — 1) The theorem does not hold if XX is not proper. For example, the affine line AC1\mathbf A^1_{\mathbf C} is simply connected, both algebraically and topologically, but the trivial line bundle E=OXeE=\mathscr O_X\cdot e endowed with the connection defined by (e)=e\nabla (e)=e is not trivial. It is analytically trivial though, but its horizontal analytic sections are of the form λexp(z)e\lambda \exp(z) e, for λC\lambda\in\mathbf C, and except for λ=0\lambda=0, 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 XX is a projective smooth simply connected algebraic variety over an algebraically closed field of characteristic pp, then every isocrystal is trivial. It is still open, despite beautiful progress by Hélène Esnault, together with Vikram Mehta and Atsushi Shiho.