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.

Product of two quotient spaces - an frequent and (in)famous mistake

The first edition of Bourbaki's General Topology (chapter I, §9, p. 56) contains the following theorem.

Proposition 3. Soient $E$, $F$ deux espaces topologiques, $R$ une relation d'équivalence dans $E$, $S$ une relation d'équivalence dans $F$. L'application canonique de l'espace produit $(E/R) \times (F/S)$ sur l'espace quotient $(E\times F)/(R\times S)$ est un homéomorphisme.

It is followed by a very convincing proof. However, the theorem is wrong. The subsequent editions give an example where the spaces are not homeomorphic, even when one of the equivalence relation is equality.

I finally understood where the mistake is. It is in the very statement! Indeed, there is a canonical map, say $h$, between those two spaces, but it goes the other way round, namely from $(E\times F)/(R\times S)$ to $(E/R)\times (F/S)$. This map is continuous, as it should be. But Bourbaki, assuming that the natural canonical map goes the other way round, pretended that $h^{-1}$ is continuous, and embarked in proving that its reciprocal bijection, $h$, is also continuous, what it is...

There are cases where one would like this theorem to holds, for example when one discusses topologies on the fundamental group. Indeed, the fundamental group of a pointed space $(X,x)$ is a quotient of the space of loops based at $x$ on $X$ for the pointed-homotopy relation, hence can be endowed with the quotient of the topology of compact convergence (roughly, uniform convergence on compact sets). Multiplication of loops is continuous. However, the resulting group law on $\pi_1(X,x)$ need not be.

The mistake appears in the recent litterature, see for example this paper, or that one (which has been even featured as «best AMM paper of the year» in 2000...). MathScinet is not aware of the flaws in those papers... Fortunately, MathOverflow is!

The van Kampen Theorem

Let me recall the statement of this theorem.

Theorem. Let $X$ be a topological space, let $U,V$ be connected open subsets of $X$ such that $W=U\cap V$ is connected and let $x$ be a point of $U\cap V$. Then, the fundamental group $\pi_1(X,x)$ is the amalgamated product $\pi_1(U,x) *_{\pi_1(W,x)} \pi_1(V,x)$, that is, the quotient of the free product of the groups $\pi_1(U,x)$ and $\pi_1(V,x)$ by the normal subgroup generated by the elements of the form $i_U(c)i_V(c)^{-1}$, where $i_U$ and $i_V$ are the natural injections from the groups $\pi_1(U,x)$ and $\pi_1(V,x)$ respectively in their free product.

The classical proof of this result in topology books relies decomposes a loop at $x$ as a product of loops at $x$ which are either contained in $U$, or in $V$.

(In fact, van Kampen proves a theorem which is quite different at first sight.)

It has been long recognized that there is a completely different approach is possible, from which all loops are totally absent. For this proof we make a supplementary assumption, namely that our spaces are « semi-locally simply connected » : Any point $a$ of $X$ has a neighborhood $A$ such that the morphism $\pi_1(A,a)\to \pi_1(X,a)$ is trivial.

When $X$ is a connected slsc space together with a point $x$, the theory of the fundamental group is related to the theory of coverings,under the form of an equivalence of categories between coverings of $X$ and sets with an action of $\pi_1(X,x)$. The equivalence of categories is explicit; it maps a covering $p\colon Y\to X$ to the fiber $p^{-1}(x)$ on which $\pi_1(X,x)$ acts naturally via the path-lifting property of coverings (given $y\in p^{-1}(x)$, any loop $c$ at $x$ lifts uniquely to a path with origin $y$, the endpoint of which is $c\cdot y$).

Given this equivalence, one can prove the van Kampen Theorem very easily in two steps. First of all, one observes that it is equivalent to have a covering of $X$ as to have a covering of $U$ and a covering of $V$ together with an identification of these coverings above $W$. A covering of $U$ corresponds to a set $A$ with an action of $\pi_1(U,x)$; a covering of $V$ corresponds to a set $B$ with an action of $\pi_1(V,x)$; an identification of these coverings above $W$ corresponds to a bijection from $A$ to $B$ which is compatible with the two actions of $\pi_1(W,x)$ acting on $A$ via the morphism $\pi_1(W,x)\to \pi_1(U,x)$ and on $B$ via the morphism $\pi_1(W,x)\to \pi_1(V,x)$. It is harmless to assume that $A=B$ and that the bijection from $A$ to $B$ is the identity. Now, a covering of $X$ corresponds to a set $A$ together with two actions of the groups $\pi_1(U,x)$ and $\pi_1(V,x)$ such that the two actions of $\pi_1(W,x)$ are equal. This is precisely the same as a set $A$ together with an action of the amalgamated product $\pi_1(U,x)*_{\pi_1(W,x)}\pi_1(V,x)$. CQFD.

The same proof applies and allows to describe the fundamental group of an union of spaces in more general contexts. For example, let us use the same method to understand the fundamental group of the circle $\mathbf S_1$. It is clear that a circle is nothing but an interval $ [0,1]$ of which the two endpoints are glued, and a covering of the circle corresponds to a covering $p\colon X\to [0,1]$ of the interval $[0,1]$ together with an identification of the fibers at $0$ and~$1$. Now, a covering of the interval can be written as a product $A\times [0,1]$ (where $A$ is the fiber at $0$, say). Consequently, identifying the fibers at $0$ and $1$ means giving yourself a bijection of $A$ to $A$. In other words, a covering of the circle « is »  a set $A$ together with a permutation of $A$, in other words, a set $A$ with an action of the additive group $\mathbf Z$. Moreover, the obvious loop is the image by the glueing map $[0,1]\to\mathbf S_1$ of the obvious path joining $0$ to $1$ so that this loop is the generator of $\pi_1(\mathbf S_1,p(0))$.

Observe that the latter example is not an instance of the van Kampen Theorem. One could get it via a groupoid-version of van Kampen.

All of this is more or less explained in the following texts:

• Adrien and Régine Douady, Algèbre et théories galoisiennes, Cassini 2005.
• Ronald Brown, Topology and Groupoids, Booksurge Publishing, 2006.
• I remember having read an old Bourbaki Tribu from the 50sby Cartan, Eilenberg and/or Weil explaining this, but I cannot find it anymore on the archive. :-(