Wednesday, March 22, 2017

Warning! — Theorems ahead

Don't worry, no danger ahead! — This is just a short post about the German mathematician Ewald Warning and the theorems that bear his name.

It seems that Ewald Warning's name will be forever linked with that of Chevalley, for the Chevalley-Warning theorem is one of the rare modern results that can be taught to undergraduate students; in France, it is especially famous at the Agrégation level. (Warning published a second paper, in 1959, about the axioms of plane geometry.)

Warning's paper, Bemerkung zur vorstehenden Arbeit von Herrn Chevalley (About a previous work of Mr Chevalley), has been published in 1935 in the Publications of the mathematical seminar of Hamburg University (Abhandlungen aus dem Mathematischen Seminar der Universität Hamburg), just after the mentioned paper of Chevalley. Emil Artin had a position in Hamburg at that time, which probably made the seminar very attractive; as a matter of fact, the same 1935 volume features a paper of Weil about Riemann-Roch, one of Burau about braids, one of Élie Cartan about homogeneous spaces, one of Santalo on geometric measure theory, etc.

1. The classic statement of the Chevalley—Warning theorem is the following.

Theorem 1. – Let $p$ be a prime number, let $q$ be a power of $p$ and let $F$ be a field with $q$ elements. Let $f_1,\dots,f_m$ be polynomials in $n$ variables and coefficients in $F$, of degrees $d_1,\dots,d_m$; let $d=d_1+\dots+d_m$. Let $Z=Z(f_1,\dots,f_m)$ be their zero-set in $F^n$. If $d<n$, then $p$ divides $\mathop{\rm Card}(Z)$.

This is really a theorem of Warning, and Chevalley's theorem was the weaker consequence that if $Z\neq\emptyset$, then $Z$ contains at least two points. (In fact, Chevalley only considers the case $q=p$, but his proof extends readily.) The motivation of Chevalley lied in the possibility to apply this remark to the reduced norm of a possibly noncommutative finite field (a polynomial of degree $d$ in $d^2$ variables which vanishes exactly at the origin), thus providing a proof of Wedderburn's theorem.

a) Chevalley's proof begins with a remark. For any polynomial $f\in F[T_1,\ldots,T_n]$, let $f^*$ be the polynomial obtained by replacing iteratively $X_i^q$ by $X_i$ in $f$, until the degree of $f$ in each variable is $<q$. For all $a\in F^n$, one has $f(a)=f^*(a)$; moreover, using the fact that a polynomial in one variable of degree $<q$ has at most $q$ roots, one proves that if $f(a)=0$ for all $a\in F^n$, then $f^*=0$.

Assume now that $Z$ contains exactly one point, say $a\in F^n$, let $f=\prod (1-f_j^{q-1})$, let $g_a=\prod (1-(x_i-a_i)^{q-1})$. Both polynomials take the value $1$ at $x=a$, and $0$ elsewhere; moreover, $g_a$ is reduced. Consequently, $f^*=g$. Then
$$ (q-1)n=\deg(g_a)=\deg(f^*)\leq \deg(f)=(q-1)\sum \deg(f_j)=(q-1)d, $$
contradicting the hypothesis that $d<n$.

b) Warning's proof is genuinely different. He first defines, for any subset $A$ of $F^n$ a reduced polynomial $g_A=\sum_{a\in A} g_a=\sum_{a\in A}\prod (1-(x_i-a_i)^{q-1})$, and observes that $g_A(a)=1$ if $a\in A$, and $g_A(a)=0$ otherwise.
Take $A=Z$, so that $f^*=g_Z$. Using that $\deg(f^*)\leq \deg(f)=(q-1)d$ and the expansion
$ (x-a)^{q-1} = \sum_{i=0}^{q-1} x^i a^{q-1-i}$, Warning derives from the equality $f^*=g_Z$
the relations
\[ \sum_{a\in Z} a_1^{\nu_1}\dots a_n^{\nu_n}=0, \]
for all $(\nu_1,\dots,\nu_n)$ such that $0\leq \nu_i\leq q-1$ and $\sum\nu_i <(q-1)(n-d)$. The particular case $\nu=0$ implies that $p$ divides $\mathop{\rm Card}(Z)$. More generally:

Proposition 2. — For every polynomial $\phi\in F[T_1,\dots,T_n]$ of reduced degree $<(q-1)(n-d)$, one has $\sum_{a\in A} \phi(a) = 0$.

c) The classic proof of that result is even easier. Let us recall it swiftly. First of all, for every integer $\nu$ such that $0\leq \nu <q$, one has $\sum_{a\in F} a^\nu=0$. This can be proved in many ways, for example by using the fact that the multiplicative group of $F$ is cyclic; on the other hand, for every nonzero element $t$ of $F$, the change of variables $a=tb$ leaves this sum both unchanged and multiplied by $t^\nu,$ so that taking $t$ such that $t^\nu\neq 1$, one sees that this sum vanishes. It follows from that that for every polynomial $f\in F[T_1,\dots,T_n]$ whose degree in some variable is $\lt;q-1$, one has $\sum_{a\in F^n} f(a)=0$. This holds in particular if the total degree of $f$ is $\lt; (q-1)n$.
Taking $f$ as above proves theorem 1.

2. On the other hand there is a second Warning theorem, which seems to be absolutely neglected in France. It says the following:

Theorem 3. — Keep the same notation as in theorem 1. If $Z$ is nonempty, then $\mathop{\rm Card}(Z)\geq q^{n-d}$.

To prove this result, Warning starts from the following proposition:

Proposition 4. – Let $L,L'$ be two parallel subspaces of dimension $d$ in $F^n$. Then $\mathop{\rm Card}(Z\cap L)$ and $\mathop{\rm Card}(Z\cap L')$ are congruent modulo $p$.

Let $r=n-d$. Up to a change of coordinates, one may assume that $L=\{x_1=\dots=x_{r}=0\}$ and $L'=\{x_1-1=x_2=\dots=x_{r}=0\}$. Let
\[ \phi = \frac{1-x_1^{q-1}}{1-x_1} (1-x_2^{q-1})\cdots (1-x_{r}^{q-1}).\]
This is a polynomial of total degree is $(q-1)r-1<(q-1)(n-d)$. For $a\in F^n$, one has $\phi(a)=1$ if $a\in L$, $\phi(a)=-1$ if $a\in L'$, and $\phi(a)=0$ otherwise. Proposition 4 thus follows from proposition 2. It is now very easy to prove theorem 3 in the particular case where there exists one subspace $L$ of dimension $d$ such that $\mathop{\rm Card}(Z\cap L)\not\equiv 0\pmod p$. Indeed, by proposition 4, the same congruence will hold for every translate $L'$ of $L$. In particular, $\mathop{\rm Card}(Z\cap L')\neq0$ for every translate $L'$ of $L$, and there are $q^{n-d}$ distinct translates.

To prove the general case, let us choose a subspace $M$ of $F^n$ of dimension $s\leq d$ such that
$\mathop{\rm Card}(Z\cap M)\not\equiv 0\pmod p$, and let us assume that $s$ is maximal.
Assume that $s< d$. Let $t\in\{1,\dots,p-1\}$ be the integer such that $\mathop{\rm Card}(Z\cap M)\equiv t\pmod p$. For every $(s+1)$-dimensional subspace $L$ of $F^n$ that contains $M$, one has $\mathop{\rm Card}(Z\cap L)\equiv 0\pmod p$, by maximaility of $s$, so that $Z\cap (L\setminus M)$ contains at least $p-t$ points. Since these subspaces $L$ are in 1-1 correspondence with the lines of the quotient space $F^n/M$, their number is equal to $(q^{n-s}-1)/(q-1)$. Consequently,
\[ \mathop{\rm Card}(Z) = \mathop{\rm Card}(Z\cap M) + \sum_L \mathop{\rm Card}(Z\cap (L\setminus M))
\geq t + (p-t) \frac{q^{n-s}-1}{q-1} \geq q^{n-s-1}\geq q^{n-d}, \]
as was to be shown.

3. Classic theorems seem to an everlasting source of food for thought.

a) In 1999, Alon observed that Chevalley's theorem follows from the Combinatorial Nullstellensatz he had just proved. On the other hand, this approach allowed Brink (2011) to prove a similar result in general fields $F$, but restricting the roots to belong to a product set $A_1\times\dots\times A_n$, where $A_1,\dots,A_n$ are finite subsets of $F$ of cardinality $q$. See that paper of Clarke, Forrow and Schmitt for further developments, in particular a version of Warning's second theorem.

b) In the case of hypersurfaces (with the notation of theorem 1, $m=1$), Ax proved in 1964 that the cardinality of $Z$ is divisible not only by $p$, but by $q$. This led to renewed interest in the following years, especially in the works of Katz, Esnault, Berthelot, and the well has not dried up yet.

c) In 2011, Heath-Brown published a paper where he uses Ax's result to strengthen the congruence modulo $p$ of proposition 4 to a congruence modulo $q$.

d) By a Weil restriction argument, a 1995 paper of Moreno-Moreno partially deduces the Chevalley-Warning theorem over a field of cardinality $q$ from its particular case over the prime field. I write partially because they obtain a divisibility by an expression of the form $p^{\lceil f \alpha\rceil}$, while one expects $q^{\lceil \alpha}=p^{f\lceil\alpha\rceil}$. However, the same argument allows them to obtain a stronger bound which does not involve not the degrees of the polynomials, but the $p$-weights of these degrees, that is the sum of their digits in their base $p$ expansions. Again, they obtain a divisibility by an expression of the form $p^{\lceil f\beta\rceil}$, and it is a natural question to wonder whether the divisibility by $p^{f\lceil\beta\rceil}$ can be proved.

Sunday, February 5, 2017

Counting points and counting curves on varieties — Tribute to Daniel Perrin

$\require{enclose}\def\VarC{\mathrm{Var}_{\mathbf C}}\def\KVarC{K_0\VarC}$
Daniel Perrin is a French algebraic geometer who turned 70 last year. He his also well known in France for his wonderful teaching habilities. He was one of the cornerstones of the former École normale supérieure de jeunes filles, before it merged in 1985 with the rue d'Ulm school. From this time remains a Cours d'algèbre which is a must for all the students (and their teachers) who prepare the agrégation, the highest recruitment process for French high schools. He actually taught me Galois theory (at École normale supérieure in 1990/1991) and Algebraic Geometry (the year after, at Orsay). His teaching restlessly stresses  the importance of examples. He has also been deeply involved in training future primary school teachers, as well as in devising the mathematical curriculum of high school students: he was responsible of the report on geometry. It has been a great honor for me to be invited to lecture during the celebration of his achievements that took place at Orsay on November, 23, 2016.

Diophantine equations are a source of numerous arithmetic problems. One of them has been put forward by Manin in the 80s and consists in studying the behavior of the number of solutions of such equations of given size, when the bound grows to infinity. A geometric analogue of this question considers the space of all curves with given degree which are drawn on a fixed complex projective, and is interested in their behavior when the degree tends to infinity. This was the topic of my lecture and is the subject of this post.

Let us first begin with an old problem, apparently studied by Dirichlet around 1840, and given a rigorous solution by Chebyshev and Cesáro around 1880: the probability that two integers be coprime is equal to $6/\pi^2$. Of course, there is no probability on the integers that has the properties one would expect, such as being invariant by translation, and the classical formalization of this problem states that the numbers of pairs $(a,b)$ of integers such that $1\leq a,b\leq n$ and $\gcd(a,b)=1$ grows as $n^2 \cdot 6/\pi^2$ when $n\to+\infty$,

This can be proved relatively easily, for example as follows. Without the coprimality condition, there are $n^2$ such integers. Now one needs to remove those pairs both of which entries are multiples of $2$, and there are $\lfloor n/2\rfloor^2$ of those, those where $a,b$ are both multiples of $3$ ($\lfloor n/3\rfloor^2$), and then comes $5$, because we have already removed those even pairs, etc. for all prime numbers. But in this process, we have removed twice the pairs of integers both of which entries are multiples of $2\cdot 3=6$, so we have to add them back, and then remove the pairs of integers both of which are multiples of $2\cdot 3\cdot 5$, etc. This leads to the following formula for
the cardinality $C(n)$ we are interested in:

 C(n) = n^2 - \lfloor\frac n2\rfloor^2 - \lfloor \frac n3\rfloor^2-\lfloor \frac n5\rfloor^2 - \dots
+ \lfloor \frac n{2\cdot 3}\rfloor^2+\lfloor\frac n{2\cdot 5}\rfloor^2+\dots
- \lfloor \frac n{2\cdot 3\cdot 5} \rfloor^2 - \dots $.

Approximating $\lfloor n/a\rfloor$ by $n/a$, this becomes

C(n) \approx  n^2 - \left(\frac n2\right)-^2 - \left (\frac n3\right)^2-\left( \frac n5\rfloor\right)^2 - \dots
+ \left (\frac n{2\cdot 3}\right)^2+\left(\frac n{2\cdot 5}\right)^2+\dots
- \left (\frac n{2\cdot 3\cdot 5} \right)^2 - \dots $

which we recognize as

C(n)\approx n^2 \left(1-\frac1{2^2}\right) \left(1-\frac1{3^2}\right)\left(1-\frac1{5^2}\right) \dots

where $\zeta(2)$ is the value at $s=2$ of Riemann's zeta function $\zeta(s)$. Now, Euler had revealed the truly arithmetic nature of $\pi$ by proving in 1734 that $\zeta(2)=\pi^2/6$. The approximations we made in this calculation can be justified, and this furnishes a proof of the above claim.

We can put this question about integers in a broader perspective if we recall that the ring $\mathbf Z$ is a principal ideal domain (PID) and study the analogue of our problem in other PIDs, in particular for $\mathbf F[T]$, where $\mathbf F$ is a finite field; set $q=\operatorname{Card}(\mathbf F)$. The above proof can be adapted easily (with simplifications, in fact) and shows that number of pairs $(A,B)$ of monic polynomials of degrees $\leq n$ such that $\gcd(A,B)=1$ grows as $q^n(1-1/q)$ when $n\to+\infty$. The analogy becomes stronger if one observes that $1/(1-1/q)$ is the value at $s=2$ of $1/(1-q^{1-s})$, the Hasse-Weil zeta function of the affine line over $\mathbf F$.

What can we say about our initial question if we replace the ring $\mathbf Z$ with the PID $\mathbf C[T]$? Of course, there's no point in counting the set of pairs $(A,B)$ of coprime monic polynomials of degree $\leq n$ in $\mathbf C[T]$, because this set is infinite. Can we, however, describe this set? For simplicity, we will consider here the set $V_n$ of pairs of coprime monic polynomials of degree precisely $n$. If we identify a monic polynomial of degree $n$ with the sequence of its coefficients, we then view $V_n$ as a subset of $\mathbf C^{n}\times\mathbf C^n$. We first observe that $V_n$ is an Zariski open subset of $\mathbf C^{2n}$: its complement $W_n$ is defined by the vanishing of a polynomial in $2n$ variables — the resultant of $A$ and $B$.

When $n=0$, we have $V_0=\mathbf C^0=\{\mathrm{pt}\}$.

Let's look at $n=1$: the polynomials $A=T+a$ and $B=T+b$ are coprime if and only if $a\neq b$;
consequently, $V_1$ is the complement of the diagonal in $\mathbf C^2$.

For $n=2$, this becomes more complicated: the resultant of the polynomials $T^2+aT+b$ and $T^2+cT+d$ is equal to $a^2d-abc-adc+b^2-2bd+bc^2+d^2$; however, it looks hard to guess some relevant properties of $V_n$ (or of its complement) just by staring at this equation. In any case, we can say that $V_2$ is the complement in $\mathbf C^4$ of the union of two sets, corresponding of the degree of the gcd of $(A,B)$. When $\gcd(A,B)=2$, one has $A=B$; this gives the diagonal, a subset of $\mathbf C^4$ isomorphic to $\mathbf C^2$; the set of pairs of polynomials $(A,B)$ whose gcd has degree $1$ is essentially $\mathbf C\times V_1$: multiply a pair $(A_1,B_1)$ of coprime polynomials of degree $1$ by an arbitrary polynomial of the form $(T-d)$.
\begin{align}V_2&=\mathbf C^4 - \left( \mathbf C^2 \cup \mathbf C\times V_1\right)\\
&= \mathbf C^4 - \left( \enclose{updiagonalstrike}{\mathbf C^2}\cup \left(\mathbf C\times (\mathbf C^2-\enclose{updiagonalstrike}{\mathbf C})\right)\right)\\
&=\mathbf C^4-\mathbf C^3
if we cancel the two $\mathbf C^2$ that appear. Except that this makes no sense!

However, there is a way to make this computation both meaningful and rigorous, and it consists in working in the Grothendieck ring $\KVarC$ of complex algebraic varieties. Its additive group is generated by isomorphism classes of algebraic varieties, with relations of the form $[X]=[U]+[Z]$ for every Zariski closed subset $Z$ of an algebraic variety $X$, with complement $U=X-Z$. This group has a natural ring structure for which $[X][Y]=[X\times Y]$. Its unit element is the class of the point, $[\mathbf A^0]$ if one wishes. An important element of this ring $\KVarC$ is the class $\mathbf L=[\mathbf A^1]$ of the affine line. The natural map $e\colon \VarC\to \KVarC$ given by $e(X)=[X]$ is the universal Euler characteristic: it is the universal map from $\VarC$ to a ring such that $e(X)=e(X-Z)+e(Z)$ and $e(X\times Y)=e(X)e(Y)$, where $X,Y$ are complex varieties and $Z$ is a Zariski closed subset of $X$.

In particular, it generalizes the classical Euler characteristic, the alternate sum of the dimensions of the cohomology groups (with compact support, if one wishes) of a variety. A subtler invariant of $\KVarC$ is given by mixed Hodge theory: there exists a unique ring morphism $\chi_{\mathrm H}\KVarC\to\mathbf Z[u,v]$ such that for every complex variety $X$, $\chi_{\mathrm H}([X])$ is the Hodge-Deligne polynomial of $X$. In particular, if $X$ is projective and smooth, $\chi_{\mathrm H}([X])=\sup_{p,q} \dim h^q(X,\Omega^p_X) u^pv^q$. If one replaces the field of complex numbers with a finite field $\mathbf F$, one may actually count the numbers of $\mathbf F$-points of $X$, and this furnishes yet another generalized Euler characteristic.

The preceding calculation shows that $e(V_0)=1$, $e(V_1)=\mathbf L^2-\mathbf L$ and $e(V_2)=\mathbf L^4-\mathbf L^3$; more generally, one proves by induction that $e(V_n)=\mathbf L^{2n}-\mathbf L^{2n-1}$ for every integer $n\geq 0$.

Equivalently, one has $e(W_n)=\mathbf L^{2n-1}$ for all $n$. I have to admit that I see no obvious reason for the class of $W_n$ to be equal to that of an affine space. However, as Ofer Gabber and Jean-Louis Colliot-Thélène pointed out to me during the talk, this resultant is the difference of two homogeneous polynomials $p-q$ of degrees $d=2$ and $d+1=3$; consequently, the locus it defines is a rational variety — given $a,b,c$, there is generically a unique $t$ such that $p-q$ vanishes at $(at,bt,ct,t)$.

These three results have a common interpretation if one brings in the projective line $\mathbf P_1$. Indeed, pairs $(a,b)$ of coprime integers (up to $\pm1$) correspond to rational points on $\mathbf P_1$, and if $\mathbf F$ is a field, then pairs $(A,B)$ of coprime polynomials in $\mathbf F[T]$ correspond (up to $\mathbf F^\times$) to elements of $\mathbf P_1(\mathbf F(T))$.
In both examples, the numerical datum $\max(|a|,|b|)$ or $\max(\deg(A),\deg(B))$ is called the height of the corresponding point.

In the case of the ring $\mathbf Z$, or in the case of the ring $\mathbf F[T]$ where $\mathbf F$ is a finite field, one has an obvious but fundamental finiteness theorem: there are only finitely many points of $\mathbf P_1$ with bounded height. In the latter case, $\mathbf C[T]$, this naïve finiteness does not hold. Nevertheless, if one sees $\mathbf P_1(\mathbf C(T))$ as an infinite dimensional variety — one needs infinitely many complex numbers to describe a rational function, then the points of bounded height constitute what is called a bounded family, a “finite dimensional” constructible set.

The last two examples have a common geometric interpretation. Namely, $\mathbf F(T)$ is the field of functions of a projective smooth algebraic curve $C$ over $\mathbf F$; in fact, $C$ is the projective line again, but we may better ignore this coincidence. Then a point $x\in\mathbf P_1(\mathbf F(T))$
corresponds to a morphism $\varepsilon_x\colon C\to\mathbf P_1$, and the formula $H(x)=\deg(\epsilon_x^*\mathscr O(1))$ relates the height $H(x)$ of $x$ to the degree of the morphism $\varepsilon_x$.

Since the notion of height generalizes from $\mathbf P_1$ to projective spaces $\mathbf P_n$ of higher dimension (and from $\mathbf Q$ to general number fields), this suggests a general question. Let $V\subset\mathbf P_n$ be a projective variety over a base field $k$ hat can one say about the set of points $x\in V(k)$ such that $H(x)\leq B$, when the bound $B$ grows to $\infty$?
The base field $k$ can be either a number field, or the field of functions $\mathbf F(C)$ of a curve $C$ over a finite field $\mathbf F$, or the field of functions $\mathbf C(C)$ of a curve over the complex numbers. In the last two cases, the variety can even be taken to be constant, deduced from a variety $V_0$ over $\mathbf F$ or $\mathbf C$.

  1. When $k$ is a number field, this set is a finite set; how does its cardinality grows? This is a question that Batyrev and Manin have put forward at the end of the 80s, and which has attracted a lot of attention since.
  2. When $k=\mathbf F(C)$ is a function field over a finite field, this set is again a finite set; how does its cardinality grows? This question has been proposed by Emmanuel Peyre by analogy with the question of Batyrev and Manin.
  3. When $k=\mathbf C(C)$ is a function field over $\mathbf C$, this set identifies with a closed subscheme of the Grothendieck-Hilbert scheme of $V$; what can one say about its geometry, in particular about its class in $\KVarC$? Again, this question has been proposed by Emmanuel Peyre around 2000.

In a forthcoming post, I shall recall some results on these questions, especially the first one, and in particular explain an approach based on the Fourier summation formula. I will then explain a theorem proved with François Loeser where we make use of Hrushovski–Kazhdan's motivic Fourier summation formula in motivic integration to prove an instance of the third question.

Monday, January 9, 2017

“May you and all your students flourish.”

Francis Su was the former president of the Mathematical Association of America. He just gave a beautiful address at the AMS-MAA Joint Meeting, entitled “Mathematics for Human Flourishing”.

Basically, when asked about the goal of mathematics, the answer is often related to its contribution to the progress of mankind through the advancement of science. Francis Su explicits what the deepest goal of mathematics may be: contribute to the flourishing not only of mankind as a whole, but of each of us as human beings. Starting from Aristotle's view that a well-lived life goes through the exercise of “virtue” — excellence of character leading to the excellence of conduct. He then quotes five basic desires which mathematics help fulfill while cultivating such virtues: play, beauty, truth, justice and love.

Francis Su's address is full of personal stories, encounters, and quotes, and I invite all of you either to watch the video on the Facebook page of the MAA, or to read its transcript on Francis Su's blog.

On the beginning of this New Year, I would just like to conclude this short message by repeating his
final wish: ”May you and all your students flourish!”

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.

Thursday, May 5, 2016

Bourbaki and Felix Klein

A colleague just sent me Xerox copies of a few pages of a 1899 biography of the général Bourbaki. Its author, François Bournand, was the private secretary of Édouard Drumont, an antisemitic writer and journalist. The book would probably not be worth much being mentioned here without its dedication:

À l'abbé Félix Klein
de l'Institut catholique
Hommage respectueux de son dévoué en N.-S.
François Bournand
Professeur d'histoire de l'art à l'École professionnelle catholique

Abbé is abbot, in this context, a catholic priest without a parish; the French initials N.-S. mean Notre Seigneur, Our Lord. It appears that this Félix Klein (note the accent on the e) also has a Wikipedia page.

Friday, April 29, 2016

Roth's theorems

A few days ago,  The Scotsman published a paper about Klaus Roth's legacy, explaining how he donated his fortune (1 million pounds) to various charities. This paper was reported by some friends on Facebook. Yuri Bilu added the mention that he knew two important theorems of Roth, and since one of them did not immediately reached my mind, I decided to write this post.

The first theorem was a 1935 conjecture of Erdős and Turán concerning arithmetic progression of length 3 that Roth proved in 1952. That is, one is given a set $A$ of positive integers and one seeks for triples $(a,b,c)$ of distinct elements of $A$ such that $a+c=2b$; Roth proved that infinitely many such triples exist as soon as the upper density of $A$ is positive, that is:
\[ \limsup_{x\to+\infty} \frac{\mathop{\rm Card}(A\cap [0;x])}x >0. \]
In 1975, Endre Szemerédi proved that such sets of integers contain (finite) arithmetic progressions of arbitrarily large length. Other proofs have been given by Hillel Furstenberg (using ergodic theory) and Tim Gowers (by Fourier/combinatorical methods); Roth had used Hardy-Littlewood's circle method.

In 1976, Erdős strengthened his initial conjecture with Turán and predicted that arithmetic progressions of arbitrarily large length exist in $A$ as soon as
\[ \sum_{a\in A} \frac 1a =+\infty.\]
Such a result is still a conjecture, even for arithmetic progressions of length $3$, but a remarkable particular case has been proved by Ben Green and Terry Tao in 2004, when $A$ is the set of all prime numbers.

Outstanding as these results are (Tao has been given the Fields medal in 2006 and Szemerédi the Abel prize in 2012), the second theorem of Roth was proved in 1955 and was certainly the main reason for awarding him the Fields medal in 1958. Indeed, Roth gave a definitive answer to a long standing question in diophantine approximation that originated from the works of Joseph Liouville (1844). Given a real number $\alpha$, one is interested to rational fractions $p/q$ that are close to $\alpha$, and to the quality of the approximation, namely the exponent $n$ such that $\left| \alpha- \frac pq \right|\leq 1/q^n$. Precisely, the approximation exponent $\kappa(\alpha)$ is the largest lower bound of all real numbers $n$ such that the previous inequality has infinitely many solutions in fractions $p/q$, and Roth's theorem asserts that one has $\kappa(\alpha)=2$ when $\alpha$ is an irrational algebraic number.

One part of this result goes back to Dirichlet, showing that for any irrational number $\alpha$, there exist many good approximations with exponent  $2$. This can be proved using the theory of continued fractions and is also a classical application of Dirichlet's box principle. Take a positive integer $Q$ and consider the $Q+1$ numbers $q\alpha- \lfloor q\alpha\rfloor$ in $[0,1]$, for $0\leq q\leq Q$; two of them must be less that $1/Q$ apart; this furnishes integers $p',p'',q',q''$, with $0\leq q'<q''\leq Q$ such that $\left| (q''\alpha-p'')-(q'\alpha-p')\right|\leq 1/Q$; then set $p=p''-p'$ and $q=q''-q'$; one has $\left| q\alpha -p \right|\leq 1/Q$, hence $\left|\alpha-\frac pq\right|\leq 1/Qq\leq 1/q^2$.

To prove an inequality in the other direction, Liouville's argument was that if $\alpha$ is an irrational root of a nonzero polynomial $P\in\mathbf Z[T]$, then $\kappa(\alpha)\leq\deg(P)$. The proof is now standard: given an approximation $p/q$ of $\alpha$, observe that $q^d P(p/q)$ is a non-zero integer (if, say, $P$ is irreducible), so that $\left| q^d P(p/q)\right|\geq 1$. On the other hand, $P(p/q)\approx (p/q-\alpha) P'(\alpha)$, hence an inequality $\left|\alpha-\frac pq\right|\gg q^{-d}$.

This result has been generalized, first by Axel Thue en 1909 (who proved an inequality $\kappa(\alpha)\leq \frac12 d+1$), then by Carl Ludwig Siegel and Freeman Dyson in 1947 (showing $\kappa(\alpha)\leq 2\sqrt d$ and $\kappa(\alpha)\leq\sqrt{2d}$). While Liouville's result was based in the minimal polynomial of $\alpha$, these generalisations required to involve polynomials in two variables, and the non-vanishing of a quantity such that $q^dP(p/q)$ above was definitely less trivial. Roth's proof made use of polynomials of arbitrarily large degree, and his remarkable achievement was a proof of the required non-vanishing result.

Roth's proof was “elementary”, making use only of polynomials and wronskians. There are today more geometric proofs, such as the one by Hélène Esnault and Eckart Viehweg (1984) or Michael Nakamaye's subsequent proof (1995) which is based on Faltings's product theorem.

What is still missing, however, is the proof of an effective version of Roth's theorem, that would give, given any real number $n>\kappa(\alpha)$, an actual integer $Q$ such that every rational fraction $p/q$ in lowest terms such that $\left|\alpha-\frac pq\right|\leq 1/q^n$ satisfies $q\leq Q$. It seems that this defect lies at the very heart of almost all of the current approaches in diophantine approximations... 

Wednesday, April 13, 2016

Weierstrass's approximation theorem

I had to mentor an Agrégation leçon entitled Examples of dense subsets. For my own edification (and that of the masses), I want to try to record here as many proofs as of the Weierstrass density theorem as I can : Every complex-valued continuous function on the closed interval $[-1;1]$ can be uniformly approximated by polynomials. I'll also include as a bonus the trigonometric variant: Every complex-valued continuous and $2\pi$-periodic function on $\mathbf R$ can be uniformly approximated by trigonometric polynomials.

1. Using the Stone theorem.

This 1937—1948 theorem is probably the final conceptual brick to the edifice of which Weierstrass laid the first stone in 1885. It asserts that a subalgebra of continuous functions on a compact totally regular (e.g., metric) space is dense for the uniform norm if and only if it separates points. In all presentations that I know of, its proof requires to establish that the absolute value function can be uniformly approximated by polynomials on $[-1;1]$:
  • Stone truncates the power series expansion of the function \[ x\mapsto \sqrt{1-(1-x^2)}=\sum_{n=0}^\infty \binom{1/2}n (x^2-1)^n, \] bounding by hand the error term.
  • Bourbaki (Topologie générale, X, p. 36, lemme 2) follows a more elementary approach and begins by proving  that the function $x\mapsto \sqrt x$ can be uniformly approximated by polynomials on $[0;1]$. (The absolute value function is recovered since $\mathopen|x\mathclose|\sqrt{x^2}$.) To this aim, he introduces the sequence of polynomials given by $p_0=0$ and $p_{n+1}(x)=p_n(x)+\frac12\left(x-p_n(x)^2\right)$ and proves by induction the inequalities \[ 0\leq \sqrt x-p_n(x) \leq \frac{2\sqrt x}{2+n\sqrt x} \leq \frac 2n\] for $x\in[0;1]$ and $n\geq 0$. This implies the desired result.
The algebra of polynomials separates points on the compact set $[-1;1]$, hence is dense. To treat the case of trigonometric polynomials, consider Laurent polynomials on the unit circle.

2. Convolution.

Consider an approximation $(\rho_n)$ of the Dirac distribution, i.e., a sequence of continuous, nonnegative and compactly supported functions on $\mathbf R$ such that $\int\rho_n=1$ and such that for every $\delta>0$, $\int_{\mathopen| x\mathclose|>\delta} \rho_n(x)\,dx\to 0$. Given a continuous function $f$ on $\mathbf R$, form the convolutions defined by $f*\rho_n(x)=\int_{\mathbf R} \rho_n(t) f(x-t)\, dt$. It is classical that $f*\rho_n$ converges uniformly on every compact to $f$.

Now, given a continuous function $f$ on $[-1;1]$, one can extend it to a continuous function with compact support on $\mathbf R$ (defining $f$ to be affine linear on $[-2;-1]$ and on $[1;2]$, and to be zero outside of $[-2;2]$. We want to choose $\rho_n$ so that $f*\rho_n$ is a polynomial on $[-1;1]$. The basic idea is just to choose a parameter $a>0$, and to take $\rho_n(x)= c_n (1-(x/a)^2)^n$ for $\mathopen|x\mathclose|\leq a$ and $\rho_n(x)=0$ otherwise, with $c_n$ adjusted so that $\int\rho_n=1$. Let us write $f*\rho_n(x)=\int_{-2}^2 \rho_n(x-t) f(t)\, dt$; if $x\in[-1;1]$ and $t\in[-2:2]$, then $x-t\in [-3;3]$ so we just need to be sure that $\rho_n$ is a polynomial on that interval, which we get by taking, say, $a=3$. This shows that the restriction of $f*\rho_n$ to $[-1;1]$ is a polynomial function, and we're done.

This approach is more or less that of D. Jackson (“A Proof of Weierstrass's Theorem,” Amer. Math. Monthly, 1934). The difference is that he considers continuous functions on a closed interval contained in $\mathopen]0;1\mathclose[$ which he extends linearly to $[0;1]$ so that they vanish at $0$ and $1$; he considers the same convolution, taking the parameter $a=1$.

Weierstrass's own proof (“Über die analytische Darstellbarkeit sogenannter willkurlicher Functionen einer reellen Veranderlichen Sitzungsberichteder,” Königlich Preussischen Akademie der Wissenschaften zu Berlin, 1885) was slightly more sophisticated: he first showed approximation by convolution with the Gaussian kernel  defined by $ \rho_n(t) =\sqrt{ n} e^{- \pi n t^2}$, and then expanded the kernel as a power series, a suitable truncation of which furnishes the desired polynomials.

As shown by Jacskon, the same approach works easily (in a sense, more easily) for $2\pi$-periodic functions, considering the kernel defined by $\rho_n(x)=c_n(1+\cos(x))^n$, where $c_n$ is chosen so that \int_{-\pi}^\pi \rho_n=1$.

3. Bernstein polynomials.

Take a continuous function $f$ on $[0;1]$ and, for $n\geq 0$, set \[ B_nf(x) = \sum_{k=0}^n f(k/n) \binom nk t^k (1-t)^{n-k}.\] It is classical that $B_nf$ converges uniformly to $f$ on $[0;1]$.

There are two classical proofs of Bernstein's theorem. One is probabilistic and consists in observing that $B_nf(x)$ is the expected value of $f(S_n)$, where $S_n$ is the sum of $n$ i.i.d. Bernoulli random variables with parameter $x\in[0;1]$. Another (generalized as the Korovkin theorem, On convergence of linear positive operators in the space of continuous functions, Dokl. Akad. Nauk SSSR (N.S.), vol. 90,‎ ) consists in showing (i) that for $f=1,x,x^2$, $B_nf$ converges uniformly to $f$ (an explicit calculation), (ii) that if $f\geq 0$, then $B_nf\geq 0$ as well, (iii) for every $x\in[0;1]$, squeezing $f$ inbetween two quadratic polynomials $f^+$ and $f_-$ such that $f^+(x)-f^-(x)$ is as small as desired.

A trigonometric variant would be given by Fejér's theorem that the Cesàro averages of a Fourier series of a continuous, $2\pi$-periodic function converge uniformly to that function. In turn, Fejér's theorem can be proved in both ways, either by convolution (the Fejér kernel is nonnegative), or by a Korovkine-type argument (replacing $1,x,x^2$ on $[0;1]$ by $1,z,z^2,z^{-1},z^{-2}$ on the unit circle).

4. Using approximation by step functions.

This proof originates with a paper of H. Kuhn, “Ein elementarer Beweis des Weierstrasschen Approximationsatzes,” Arch. Math. 15 (1964), p. 316–317.

Let us show that for every $\delta\in\mathopen]0,1\mathclose[$ and every $\varepsilon>0$, there exists a polynomial $p$ satisfying the following properties:
  • $0\leq p(x)\leq \varepsilon$ for $-1\leq x\leq-\delta$;
  • $0\leq p(x)\leq 1$ for $-\delta\leq x\leq \delta$;
  • $1-\varepsilon\leq p(x)\leq 1$ for $\delta\leq x\leq 1$.
In other words, these polynomials approximate the (discontinuous) function $f$ on $[-1;1]$ defined by $f(x)=0$ for $x< 0$, $f(x)=1$ for $x> 0$ and $f(0)=1/2$.

A possible formula is $p(x)=(1- ((1-x)/2))^n)^{2^n}$, where $n$ is a large enough integer. First of all, one has $0\leq (1-x)/2\leq 1$ for every $x\in[-1;1]$, so that $0\leq p(x)\leq 1$. Let $x\in[-1;-\delta]$; then one has $(1-x)/2\geq (1+\delta)/2$, hence $p(x)\leq (1-((1+\delta)/2)^n)^{2^n}$, which can be made arbitrarily small when $n\to\infty$. Let finally $x\in[\delta;1]$; then $(1-x)/2\geq (1-\delta)/2$, hence $p(x)\geq (1-((1-\delta)/2)^n)^{2^n}\geq 1- (1-\delta)^n$, which can be made arbitrarily close to $1$ when $n\to\infty$.

By translation and dilations, the discontinuity can be placed at any element of $[0;1]$. Let now $f$ be an arbitrary step function and let us write it as a linear combination $f=\sum a_i f_i$, where $f_i$ is a $\{0,1\}$-valued step function. For every $i$, let $p_i$ be a polynomial that approximates $f_i$ as given above. The linear combination $\sum a_i p_i$ approximates $f$ with maximal error $\sup(\mathopen|a_i\mathclose|)$.

Using uniform continuity of continuous functions on $[-1;1]$, every continuous function can be uniformly approximated by a step function. This concludes the proof.

5. Using approximation by piecewise linear functions.

As in the proof of Stone's theorem, one uses the fact that the function $x\mapsto \mathopen|x\mathclose|$ is uniformly approximated by a sequence of polynomial on $[-1;1]$. Consequently,  so are the functions $x\mapsto \max(0,x)=(x+\mathopen|x\mathclose|)/2 $ and $x\mapsto\min(0,x)=(x-\mathopen|x\mathclose|)/2$. By translation and dilation, every continuous piecewise linear function on $[-1;1]$ with only one break point is uniformly approximated by polynomials. By linear combination, every continuous piecewise linear affine function is uniformly approximated by polynomials.
By uniform continuity, every continuous function can be uniformly approximated by continuous piecewise linear affine functions. Weierstrass's theorem follows.

6. Moments.

A linear subspace $A$ of a Banach space is dense if and only if every continuous linear form which vanishes on $A$ is identically $0$. In the present case, the dual of $C^0([-1;1],\mathbf C)$ is the space of complex measures on $[-1;1]$ (Riesz theorem, if one wish, or the definition of a measure). So let $\mu$ be a complex measure on $[-1;1]$ such that $\int_{-1}^1 t^n \,d\mu(t)=0$ for every integer $n\geq 0$; let us show that $\mu=0$. This is the classical problem of showing that a complex measure on $[-1;1]$ is determined by its moments. In fact, the classical proof of this fact runs the other way round, and there must exist ways to reverse the arguments.

One such solution is given in Rudin's Real and complex analysis, where it is more convenient to consider functions on the interval $[0;1]$. So, let $F(z)=\int_0^1 t^z \,d\mu(t)$. The function $F$ is holomorphic and bounded on the half-plane $\Re(z)> 0$ and vanishes at the positive integers. At this point, Rudin makes a conform transformation to the unit disk (setting $w=(z-1)/(z+1)$) and gets a  bounded function on the unit disk with zeroes at $(n-1)/(n+1)=1-2/(n+1)$, for $n\in\mathbf N$, and this contradicts the fact that the series $\sum 1/(n+1)$ diverges.

In Rudin, this method is used to prove the more general Müntz–Szász theorem according to which the family $(t^{\lambda_n})$ generates a dense subset of $C([0;1])$ if and only if $\sum 1/\lambda_n=+\infty$.

Here is another solution I learnt in a paper by L. Carleson (“Mergelyan's theorem on uniform polynomial approximation”, Math. Scand., 1964).

For every complex number $a$ such that $\mathopen|a\mathclose|>1$, one can write $1/(t-a)$ as a converging power series. By summation, this quickly gives that
\[ F(a) = \int_{-1}^1 \frac{1}{t-a}\, d\mu(t) \equiv 0. \]
Observe that this formula defines a holomorphic function on $\mathbf C\setminus[-1;1]$; by analytic continuous, one thus has $F(a)=0$ for every $a\not\in[-1;1]$.
Take a $C^2$-function $g$ with compact support on the complex plane. For every $t\in\mathbf C$, one has the following formula
\[ \iint \bar\partial g(z) \frac{1}{t-z} \, dx\,dy = g(t), \]
which implies, by integration and Fubini, that
\[ \int_{-1}^1 g(t)\,d\mu(t) = \iint \int \bar\partial g(z) \frac1{t-z}\,d\mu(t)\,dx\,dy = \iint \bar\partial g(z) F(z)\,dx\, dy= 0. \]
On the other hand, every $C^2$ function on $[-1;1]$ can be extended to such a function $g$, so that the measure $\mu$ vanishes on every $C^2$ function on $[-1;1]$. Approximating a continuous function by a $C^2$ function (first take a piecewise linear approximation, and round the corners), we get that $\mu$ vanishes on every continuous function, as was to be proved.

7. Chebyshev/Markov systems.

This proof is due to P. Borwein and taken from the book Polynomials and polynomial inequalities, by P. Borwein and T. Erdélyi (Graduate Texts in Maths, vol. 161, 1995). Let us say that a sequence $(f_n)$ of continuous functions on an interval $I$ is a Markov system (resp. a weak Markov system) if for every integer $n$, every linear combination of $(f_0,\dots,f_n)$ has at most $n$ zeroes (resp. $n$ sign changes) in $I$.

Given a Markov system $(f_n)$, one defines a sequence $(T_n)$, where $T_n-f_n$ is the element of $\langle f_0,\dots,f_{n-1}\rangle$ which is the closest to $f_n$. The function $T_n$ has $n$ zeroes on the interval $I$; let $M_n$ be the maximum distance between two consecutive zeroes.

Borwein's theorem  (Theorem 4.1.1 in the mentioned book) then asserts that if the sequence $(f_n)$ is a Markov system consisting of $C^1$ functions, then its linear span is dense in $C(I)$ if and only if $M_n\to 0$.

The sequence of monomials $(x^n)$ on $I=[-1;1]$ is of course a Markov system.  In this case, the polynomial $T_n$ is the $n$th Chebyshev polynomial, given by $T_n(2\cos(x))=2\cos(nx)$, and its roots are given by $2\cos((\pi+2k\pi)/2n)$, for $k=0,\dots,n-1$, and $M_n\leq \pi/n$. This gives yet another proof of Weierstrass's approximation theorem.