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

Let indeed $(E,\nabla)$ be a vector bundle with an integrable connection on $X$ and let us show that it is trivial, namely, that there exist $n$ global sections $e_1,\dots,e_n$ of $E$ which are horizontal ($\nabla e_i=0$) and form a basis of $E$ at each point.

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

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

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

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

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

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

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

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

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

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

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

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