By the way, as Martin Widmer remarked during his recent talk in Bordeaux, o-minimality is a very clever way of formalizing the notion of “being non-pathological”.

]]>

Theorem 1Let be times differentiable functions on . Then

with some .

Here is the Vandermonde determinant.

The proof is based on the more special Schwarz mean value theorem, also mentioned in that entry. It is more convenient for us to state the Schwarz theorem as follows.

Theorem 2 (Schwarz)Let be an times differentiable function on . Denote by the Lagrange interpolation polynomial for at the points (the polynomial of degree at most~ taking value at for .) Then for some we have .

In this form the Schwarz theorem is immediate: the “auxiliary function” vanishes at , and by the Rolle theorem its -th derivative vanishes somewhere on .

To prove Theorem 1 we consider the interpolation polynomials of at . Let be independent indeterminates. Elementary transformations of determinants show that

Applying to the both sides the differential operator

we obtain

Now the Schwarz theorem implies that with a suitably chosen .

]]>

Recently our master student Martin Djukanovic showed me a wonderful short proof of this statement. We argue by induction in the dimension. There is nothing to prove in dimension . Now let be an -dimensional vector space over some complete field, . We fix a basis of and show that any norm on is equivalent to the sup-norm with respect to this basis:

By induction, this holds in dimension . In particular, in dimension any normed space over a complete field is itself complete, because this is obviously true for the sup-norm.

For any we have with . It remains to show that with some other . Assuming the contrary, we find an infinite sequence of non-zero vectors such that as . Taking a subsequence, we may assume that the sup-norm of each is attained at its -th coordinate, and after re-scaling we may assume that the -th coordinate (and thereby the sup-norm) of each is . Thus, each can be written as , where and .

Since , we have as . Thus, the sequence converges to . Hence it is a Cauchy sequence. But is complete by the induction, which implies that converges in . We obtain , a contradiction.

]]>

A standard combinatorial argument shows that for positive integers and

Is it possible to give a “geometric” interpretation (or any other “conceptual” interpretation) of identity (1)?

**Update**

A satisfactory answer was given by Allison and Mark Lewko, see the comments.

]]>

Let me recall that Liouville proved that the infinite sum is transcendental, while Cantor proved that the set of algebraic numbers is countable, but the set of reals is uncountable.

Dear colleagues, Cantor’s proof is less “constructive” than Liuoville’s only on emotional or aesthetic level. Mathematically, **both are equally constructive**.

Indeed, the only way I can imagine of defining a real number “constructively” is providing an algorithm for computing rational approximations to it of any prescribed precision. More formally, call a real number *constructible* if there exists an algorithm (a Turing machine) which, having a natural number at the input, produces at the output a rational approximation to with precision . I do not know another reasonable (non-equivalent) definition of a constructible real number; if anybody knows, I would be happy if he shares it with me.

Of course, the Liouville number is constructible, the good rational approximations being the partial sums of the infinite series.

Further, call a sequence constructible if there exists an algorithm, which, having natural and at the input, produces a rational approximation to with precision at the output.

Now two simple exercises.

- Show that there exists a constructible sequence containing all algebraic numbers.
- Show that for any constructible sequence, there exists a constructible number not contained in this sequence.

A careful examination of Cantor’s proof reveals that its first part solves the first exercise, and the second part solves the second exercise.

I would never waste time for recalling these trivialities in my precious blog if this misunderstanding were not so widespread even among professional mathematicians.

]]>

Bombieri and Pila remark that a similar (and even stronger) statement holds if is algebraic but does not admit polynomial parametrization.

The obvious disadvantage of this result is dependence of the constant in . It turns out that a slightly weaker inequality holds uniformly for all algebraic curves of given degree. Precisely:

Theorem 1let be segment of an irreducible plane algebraic curve of degree contained in the unit square . Then for

Equivalently, if is a segment of an irreducible algebraic curve of degree contained in the square , then

One cannot do better than the exponent , as the example shows.

Due to the uniformity, this result easily extends to arbitrary dimension by projection and slicing (Pila 1995): let be an irreducible affine variety of dimension and degree , contained in the affine space ; then

The proof of Theorem 1 goes along similar lines as that in the transcendence case, but is more involved, because now we have to keep track of the dependence of all parameters in the function . For a function we define the *-norm* by

where is the sup-norm. With this definition we can make inequality (1) of the previous post totally explicit:

The Leibniz identity

shows that

In particular, if then

In the sequel by a *-curve* we mean a plane algebraic curve of degree less than in and less then in . Arguing as in the previous post, we prove the following: let be a sufficiently smooth function, defined on a compact interval . Let be points on the curve defined by , **not lying** on on a -curve. Assume that the coordinates and are rational numbers with denominator ; then

with . (Here unnumbered constants may vary from equation to equation, while numbered constants , , etc. are individual.)

From now on we assume that . If the set does not lie on a single -curve, then the right-hand side of (1) does not exceed the length of . Splitting into subintervals shorter than this right-hand side, it follows immediately that the set lies on at most

-curves. (We have to write to take into account the case when the set lies on a single -curve.)

Now assume that is a -algebraic function; that is, it satisfies a polynomial equation , where is an irreducible real polynomial of -degree and -degree . In this case the curve , defined by has at most intersections with any -curve. Hence

Our next goal is get rid of the norm of . To do this, Bombieri and Pila show that the norm is “large” only on short intervals. Precisely:

Proposition 2Let be a -algebraic function defined on an interval . Let be a positive integer. Then for every there is at most intervals of length at most each such that outside these intervals.

We postpone the proof of this proposition until one of my subsequent posts, and show now how it allows one to replace by in (2). Precisely:

Proposition 3For as in Proposition 2 and any we have

The proof is by induction in . The definition of under which the induction works will appear in the course of the proof.

Applying Proposition 2 with and , we split our interval into intervals of length (call them “short”) where is “large”, and the remaining part where . This remaining part itself splits into at most intervals (call them “long”). Applying to every “long” interval estimate (2), we see that the curve has at most

points with denominator above these “long” intervals.

Now let be a “short” interval, and let be its length. We may assume that with , so that is an integral multiple of :

If then has only point over . If then the number of points with denominator on over is the same as the number of points with denominator on the curve over the interval . **By induction**, the latter quantity is bounded by

Since , the latter quantity does not exceed

points over “short” intervals.

Thus, the total number of points with denominator on is bounded by the sum of (5) and (6). Now let be a so large that for the quantity (6) does not exceed . Then

Now if then (3) holds trivially for , which gives the base of our induction. And if then the right-hand side of (7) does not exceed that of (3), which gives the induction step. **This proves Proposition 3**.

Now we easily complete the **proof of Theorem 1**. Let be the irreducible equation of our curve, so that . We may assume that . Indeed, one may find a unimodular integral matrix with entries non-exceeding in absolute value such that the polynomial is of both -degree and -degree . The image of the unit square under the linear transformation defined by the inverse of this matrix is contained in the square , and it remains to split the latter into unit squares and translate the variables in each.

Further, a compact segment of an irreducible algebraic curve of -degree and -degree splits into compact segments of the type or , where is a -algebraic function. Applying to each of the latter estimate (4), we obtain the result.

(I thank all participants of the *Groupe de travail “Géométrie diophantienne”* in Bordeaux for their precious comments.)

]]>

(As they remark, the same statement holds if is algebraic but does not admit polynomial parametrization: this is a rather easy consequence of the Mordell-Weil theorem.)

This theorem is instrumental in the recent novel proof, by Pila and Zannier (2008), of the Manin-Muford conjecture on torsion points on subvarieties of Abelian varieties.

The argument of Bombieri and Pila is based on the mean value theorem of H.A.Schwarz. It asserts that, given an interval , a function , and points , there exists a point such that

where is the Vandermond determinant. The case is the Lagrange mean value theorem.

Dörge (1927) used this theorem to bound the number of integral points on finite segments of algebraic curves and deduce from this a very simple proof of Hilbert’s irreducibility theorem. The work of Dörge is reproduced in Lang’s *Fundamentals of Diophantine Geometry* (see Theorem 2.1 in Chapter 9), and I advice to read this piece (it is independent of the rest of the book). The argument of Bombieri and Pila is a very beautiful extension of Dörge’s idea.

The starting point of Bombieri and Pila’s argument is yet another “mean value theorem”. Let be an interval, and . Then

with some . Let us prove this for :

by the Lagrange theorem. (We take and .) The proof of the general case is pretty the same, but with Lagrange replaced by Schwarz.

If the interval is compact, we obtain the following consequence: there exists a constant (depending on the functions ) such that for any

Now fix a positive integer . In the sequel by a *-curve* we mean a plane algebraic curve of degree less than in and in . Then points do **not** lie on a -curve if and only if the determinant

is non-zero. (Here is the “vertical” index, and is the “horizontal” index.) If all the coordinates and are rational numbers with denominator , then is a non-zero rational number with denominator at most . We obtain the **lower** estimate

Now let be a sufficiently smooth function on a compact interval and the plane curve . If all our points lie on then can be expressed like with . Applying inequality (1) with the functions instead of , we obtain the following **upper** estimate:

where depends on and . Thus, if our points

- have rational numbers with denominator as coordinates,
- lie on , and
- do not lie on a -curve,

then

The same is true if we take more than points with these properties, because we can always select among them points not lying on a -curve.

Splitting our interval into subintervals of length less than , we obtain the following statement (“the main lemma” of Bombieri and Pila):

the set lies on -curves.

We are ready to prove the theorem of Bombieri and Pila. Let be a compact segment of a plane analytic transcendental curve. We may assume that is of the type . Also, a compactness argument shows that number of intersections of with any -curve is bounded by a constant, depending on and . It follows that for any

Selecting small enough, we complete the proof.

(I thank the participants of the *Groupe de travail “Géométrie diophantienne”* in Bordeaux for their precious comments.)

]]>