User robert kucharczyk - MathOverflow

Answer by Robert Kucharczyk for What are some Applications of Teichmüller Theory?

2012-11-27T17:47:13Z

From the perspective of algebraic geometry, Teichmüller theory is an analytic approach to moduli spaces of curves. To keep things simple, let $g\geqq 2$ be an integer and consider the moduli space $\mathscr{M}_g$ of smooth and complete algebraic curves of genus $g$. This is a fairly complicated object: a Deligne-Mumford stack over the integers, nothing that is easy to describe in any way. The associated complex analytic "space" $\mathscr{M}_g(\mathbf{C})$ is still something with a weird structure: a complex orbifold which is not a manifold. But its universal covering space, which can be identified with the Teichmüller space $\mathscr{T}_g$, is a true complex manifold. It has several nice properties as compared with $\mathscr{M}_g$:

Teichmüller space is biholomorphic to an open domain in $\mathbf{C}^{3g-3}$. This is Bers' embedding theorem.
Teichmüller space is diffeomorphic (forgetting the complex structure) to $\mathbf{R}^{6g-6}$, and there is a very intuitive system of coordinates, called Fenchel-Nielsen coordinates, realising such a diffeomorphism. On the other hand, even when you forget the stack structure, $\mathscr{M}_g$ is a variety of general type for $g\geqq 23$, which means that you can only embed it in projective space $\mathbf{P}^d$ where $d$ is "much" larger than the dimension of $\mathscr{M}_g$, and you need "many" equations to cut out its image. So there is no "economical" algebraic coordinate system on $\mathscr{M}_g$ in general.
Complex geodesics in $\mathscr{T}_g$ for a natural metric, the Teichmüller metric, give families of algebraic curves which have a nice and intuitive geometric description, called Teichmüller disks. In quite a few cases they descend to algebraically defined curves in $\mathscr{M}_g$ which are consequently called Teichmüller curves. One can say much more about their geometric and number-theoretical properties than for general curves in $\mathscr{M}_g$. They form an active area of research in these years.

Another application of Teichmüller theory to moduli spaces of curves is that it gives rise to an isomorphism between the mapping class group $\Gamma_g$ of a closed oriented surface of genus $g$ and the (orbifold) fundamental group of the moduli space $\mathscr{M}_g(\mathbf{C})$, so it provides a link between the topology of moduli spaces and the topology of surfaces.

What are some examples of ingenious, unexpected constructions?

2012-03-11T01:24:10Z

Many famous problems in mathematics can be phrased as the quest for a specific construction. Often such constructions were sought after for centuries or even millennia and later proved impossible by taking a new, "higher" perspective. The most obvious example would be the three geometric problems of antiquity: squaring of circles, duplication of cubes and trisection of angles by ruler and compass alone. Closely related to these three is the construction of regular $n$-gons for general $n$. Later we have the solution of an arbitrary algebraic equation by means of radicals or the expression of the circumference of an ellipse by means of elementary functions. In the 20th century we have Hilbert's tenth problem: find an algorithm to determine whether a given diophantine equation has solutions.

All these constructions turned out to be impossible, but the futile search produced some new and great mathematics: Galois theory, group theory, transcendental numbers, elliptic curves ...

But I am looking for examples which are in some sense the opposite of the above: where somebody turned up with an ingenious construction in a problem where it had been generally believed that no such construction should exist. Ideally, this construction should have made new interesting questions and methods turn up, but I am also interested in isolated results that may just be counted as funny coincidences. To make my point clearer, let me present my own two favourite examples.

(1) Belyi's Theorem: If $X$ is a smooth projective algebraic curve defined over a number field, there exists a rational function on $X$ whose only singular values are $0$, $1$ and $\infty$. --- According to his Esquisse d'un programme, Grothendieck had thought about this problem shortly before but found the statement so bold that he even felt awkward for asking Deligne about it. To put the theorem into context, the converse statement (that every curve which admits a rational function with only these three singular values can be defined over a number field) had been known before by abstract nonsense and is quite straightforward to deduce from deep results in Grothendieck-style algebraic geometry. Belyi's proof, however, was completely elementary, constructive, and tricky. Also it is more important than it might seem at first sight since it opens up a very strict, and equally unexpected, connection between the topology of surfaces and number theory.

(2) Julia Robinson's theorem about the definability of integers: Suppose you want to single out $\mathbb{Z}$ as a subset of $\mathbb{Q}$, using as little structure as possible. The result in question is at least to me absolutely striking. I do not know if it was so unexpected to the experts at that time, but the construction is in any case really ingenious. It says that there exists a first-order formula $\varphi$ in the language of rings (i.e. only talking about elements, not subsets, and using only logical symbols and multiplication and addition, and the symbols $0$ and $1$) such that for a rational number $r$, $\varphi (r)$ is true if and only if $r$ is an integer. Robinson's original formula is $$\varphi (r)\equiv\forall y\forall z(\psi (0,y,z)\wedge\forall x(\psi (x,y,z)\longrightarrow \psi (x+1,y,z))\longrightarrow\psi (r,y,z))$$ with $$\psi (x,y,z) \equiv \exists a\exists b\exists c(2 + x^2yz = a^2 + yb^2-zc^2).$$ Since this is not my area of research I do not attempt to estimate the historical importance of this discovery, but it seems to me that it is of great weight in the intersection of number theory and logic.

So I hope these two examples make it clear what I am after, and I am looking forward to reading your examples.

Answer by Robert Kucharczyk for Fiction books about mathematicians?

2012-07-11T21:41:05Z

One might argue whether the main character D-503 of Yevgeny Zamyatin's We counts as a mathematician. It is a classical dystopian novel, similar in spirit to Brave New World or Nineteen Eighty-Four (but older and arguably better). D-503 is rather an engineer than a mathematician in our sense, but in the novel's setting, due to the almost complete mechanization of human life, this is as close as one can get to being a mathematician. In any case, mathematical concepts play a decent rôle, in particular - as strange as this may sound - the purported challenge to imagination posed by imaginary numbers.

Answer by Robert Kucharczyk for What are some examples of ingenious, unexpected constructions?

2012-03-16T13:28:50Z

Another example that just came to my mind is the "set of all sets that do not contain themselves" in Russell's paradox - it was very simple, as unexpected as anything and turned the ideas about sets completely topsy-turvy.

Answer by Robert Kucharczyk for Geodesics for a Cone Metric

2012-03-10T22:28:04Z

The answer to the question as stated is "no": take for $S$ the plane $\mathbb{C}$ with the metric induced by the differential $z\mathrm{d}z$; then $S^{\ast }=\mathbb{C}\smallsetminus {0}$ and there is simply no geodesic from $1$ and $\mathrm{i}$. Or, if you want closed loops and free homotopy, there is no geodesic in the free homotopy class of the loop around $0$. Intuitively, it is clear what happens: if you draw a curve and try to pull it straight, you are forced to go through the cone point at the origin.

If, however, you consider the surface $S$ itself and extend the definition of a geodesic so as to work on Euclidean metrics with cone points as well, the answer is "yes" if $S$ is complete as a metric space. The usual definition of geodesics in this context is one which works for all metric spaces: locally isometric maps from an interval. The statement you asked for is then a consequence of the Arzelà-Ascoli theorem.

These things are treated in detail in the textbook Quadratic differentials by Kurt Strebel.

Answer by Robert Kucharczyk for Where is number theory used in the rest of mathematics?

2012-03-09T21:40:40Z

Assume you are a (differential) geometer and you want to construct locally symmetric spaces of higher rank. Such a space must have a (globally) symmetric space $X$ as its universal covering space, and this can be written as $X=G/K$ where $G$ is the identity component of the isometry group of $X$ and $K$ is the stabiliser of some point in $X$. To get a locally symmetric space of finite volume, you then have to find a lattice $\Gamma\subset G$, i.e. a discrete subgroup such that $\Gamma\backslash G$ has finite volume with respect to the (right-invariant) Haar measure on $G$. Then if $\Gamma$ is torsion-free you get $\Gamma\backslash X$ as a locally symmetric space.

Now how does one construct such lattices? One method is by arithmetic groups, and it is a matter of taste whether you want to consider them as objects of number theory. Suffice it to say that their study requires a lot of techniques from other areas in number theory in a broad sense. It is quite technical to define an arithmetic group, but it is easy to give some examples that already give you some flavour: $\mathrm{SL}_n(\mathbb{Z})$ as a lattice in $\mathrm{SL}_n(\mathbb{R})$ , similarly $\mathrm{Sp}_{q}(\mathbb{Z})$ in $\mathrm{Sp}_q(\mathbb{R})$ or more elaborate constructions where you start from an algebraic number field and an algebra over that field and take some subgroup of the automorphism group of that algebra. Note that in the two cases I presented to you the lattices are not torsion-free, but you can find finite index subgroups which are torsion-free and therefore give you locally symmetric spaces.

As I said there is a fairly technical definition of an arithmetic lattice in a Lie group, and the first guess of everybody hearing of this for the first time is that this should be something exceptional - why should a "generic" lattice be constructible by number-theoretic methods? And indeed, the example of $\operatorname{SL}_2(\mathbb{R})$ supports that guess. The associated symmetric space $\operatorname{SL}_2(\mathbb{R})/\operatorname{SO}(2)$ is the hyperbolic plane $\mathbb{H}^2$. There are uncountably many lattices in $\operatorname{SL}_2(\mathbb{R})$ (with the associated locally symmetric spaces being nothing other than Riemann surfaces), but only countably many of them are arithmetic.

But in higher rank Lie groups, there is the following truly remarkable theorem known as Margulis arithmeticity:

Let $G$ be a connected semisimple Lie group with trivial centre and no compact factors, and assume that the real rank of $G$ is at least two. Then every irreducible lattice $\Gamma\subset G$ is arithmetic.

Answer by Robert Kucharczyk for Where is number theory used in the rest of mathematics?

2012-03-09T22:05:58Z

Here is another example from dynamical systems. As has already been explained in some other posts, the fine structure of dynamical systems often depends on subtle number-theoretic properties of some involved constants.

Assume you have a domain $G\subseteq\mathbb{C}$ and a holomorphic map $f\colon G\to G$ with a fixed point $z\in G$, and you want to study how successive iterates of $f$ around $z$ behave. For sake of simplicity assume that $z=0$. Now locally around $0$ we can approximate $f$ by a linear function $\zeta\mapsto a\zeta$ where $a=f'(0)$, so a simple heuristic says that if we want to understand high powers of $f$, we should understand high powers of $a$. It is then quite clear that the only interesting case is $|a|=1$, so let us assume $a=\mathrm{exp}2\pi it$. Then there are some relations between the growth of the entries in the continued fraction expansion of $t$ and the behaviour of $f$ around $z=0$ under iteration.

You can read more about this in the book "Complex Dynamics in one Variable" by John Milnor. Keywords are Siegel disks and Brjuno numbers.

How would Hilbert and Weber think about the Langlands programme?

2012-02-28T19:15:53Z

Explanations to a general mathematical audience about the Langlands programme often advertise it as "non-abelian class field theory". They usually begin as follows: a modern style formulation of classical class field theory is to say that for a global field $K$, the Artin map defines an isomorphism from the group of connected components of the idele class group to the Galois group $\operatorname{Gal}(K^{ab}|K)$. Pushing this even further, we see that we have a canonical identification of characters of the idele class group with characters of the absolute Galois group $\operatorname{Gal}(\bar{K}|K)$.

Then people usually go on to say that this should extend to a correspondence between a certain class of $n$-dimensional Galois representations and a certain class of representations of $\operatorname{GL}_n(\mathbb{A}_K)$ (where $\mathbb{A}_K$ denotes the adeles of $K$), and very soon they have disappeared into (to me) far off realms.

While it should be clear from my description that I have no clue whatsoever concerning the Langlands programme, I know a little bit about global class field theory in its traditional formulation. That is, I understand it as a means to describe and classify abelian extensions of $K$ with prescribed ramifications, with the Artin map giving an isomorphism from a ray ideal class group of $K$ (say) to the Galois group of the corresponding ray class field over $K$.

So, my question is:

Do there exist results in the global Langlands programme which give us back some down-to-earth, may be ideal-theoretic, insights about number field extensions? And the same question for yet open questions in the global Langlands programme: would their answers give us some sort of "classical" information?

Do the algebraic integers form a free abelian group?

2011-12-22T19:19:00Z

It is a well-known fact, proved in every introductory textbook on algebraic number theory, that if $K$ is an algebraic number field, i.e. a finite extension of $\mathbb{Q}$, then its ring $\mathcal{O}_K$ of integers is a free abelian group.

Does this statement still hold for arbitrary algebraic extensions of $\mathbb{Q}$? In particular, is the underlying abelian group of the ring $\mathcal{O}_{\overline{\mathbb{Q}}}$ of all algebraic integers free abelian?

Should this be true, I am also interested whether anything is known about the dependence of this statement on the axiom of choice, and similar logical questions.

Why tropical geometry?

2011-12-16T14:49:30Z

Tropical geometry can be described as "algebraic geometry" over the semifield $\mathbb{T}$ of tropical numbers. As a set, $\mathbb{T}=\mathbb{R}\cup { -\infty}$; this is endowed with addition being given by the (usual) maximum of real numbers and multiplication by the (usual) sum of real numbers. Recently there has been a lot of research on this kind of geometry. The obvious question is now: why are people interested in this?

There are three possible motivations I am aware of:

(1) "Polynomial" equations over $\mathbb{T}$ can be interpreted as linear equalities and inequalities over the classical reals. So people working in linear optimization and similar areas can use tropical geometry as an alternative, more algebraic approach that might lead to new methods and insights.

(2) In quantization jargon, one can interpret $\mathbb{T}$ as a "classical limit" of semifields which are all isomorphic to $\mathbb{R}^{\ge 0}$ with the usual addition and multiplication. More precisely, for any $q>0$ set $\mathbb{T}_q=\mathbb{R}\cup{ -\infty }$ and consider the bijection $\log_q:\mathbb{R}^{\ge 0}\to\mathbb{T}_q$ (setting $\log_q(0)=-\infty$). Pushing forward the usual semifield structure on $\mathbb{R}^{\ge 0}$ along $\log_q$, we get a semifield structure on $\mathbb{T}_q$. Then it is easy to check that the semifield $\mathbb{T}$ is, in the obvious sense, the limit of $\mathbb{T}_q$ as $q\to 0$. So if one likes to think in these terms (I do not, but that shall not bother me for now), then real algebraic geometry appears (with a grain of salt) as the quantum version of tropical geometry, which in turn gives tropical geometry an important rôle.

(3) This is only a post hoc justification. Some problems of classical algebraic geometry, mainly in enumerative geometry, have been solved by methods of tropical geometry. The usual strategy is as follows: we have a question in algebraic geometry, to which the answer is supposed to be an integer (say). We then set up an analogous question in tropical geometry, prove that the answers to the two questions agree, and then work on the tropical question, which is usually much simpler to answer.

Besides these three arguments, do you have any other motivations for studying tropical geometry?

Answer by Robert Kucharczyk for Why should the anabelian geometry conjectures be true?

2011-11-12T11:49:44Z

I can only offer a "strengthening" of your friends' explanation. Let me first remark that I am not an expert in this field and I am sure that there are some grave mistakes in my argument. However, it is much too long for a comment, so I post it as an answer.

Let us first consider the simpler case of (co)homology instead of fundamental groups. When talking about étale (say, $\ell$-adic) cohomology together with its Galois action, the transcendental analogue is generally taken to be not just the singular cohomology groups, but these groups together with their Hodge structure.

Similarly, consider a hyperbolic curve $X$ over a number field $K$. For simplicity, assume we are given a $K$-rational base point $x\in X(K)$. The fundamental group one considers is either the group $\pi_1^{et}(X,x)$ as an abstract profinite group, or the group $\pi_1^{et}(X\otimes\overline{K},x)$ together with its action of $\operatorname{Gal}(\overline{K}|K)$. The former can be reconstructed from the latter. The weakest version of Grothendieck's anabelian conjecture for curves says (roughly) that we can reconstruct $X$ from $\pi_1^{et}(X,x)$.

Let me explain why we can reconstruct $X$ from $\pi_1^{et}(X\otimes\overline{K},x)$ with its Galois action. The abelianization of this group with Galois action is just the product over all $\ell$ of the $\ell$-adic Tate modules $T_{\ell }(\operatorname{Jac}X)$. These are the $\ell$-adic analogues of the Hodge structures on first homomology, which bear the same information as the Jacobian itself. Thus it is not surprising (although very difficult!) that we can reconstruct $\operatorname{Jac}X$ from these data, and the Jacobian determines the isomorphism class of the curve by Torelli's theorem. [Edit: As Torsten Ekedahl has pointed out in the comments, it is not true that you can recover an abelian variety from its Tate module.]

Now there are certainly some points where the above argument does not work as simply as presented, but the morals is that the analogue of the arithmetic fundamental group over $\mathbb{C}$ should be the topological fundamental group with a "Hodge structure on groups". I do not know if this has ever been worked out, but there is a good understanding of the "Hodge structure on the nilpotent completion of the fundamental group", introduced by Hain and Zucker.

Subtlety in the definition of the Kobayashi metric

2011-11-10T21:40:36Z

When defining the Kobayashi metric on a connected complex analytic space $X$, one makes the following auxiliary definition:

A holomorphic chain from $x\in X$ to $y\in X$ is a finite sequence of holomorphic maps $f_1,\ldots ,f_n\colon\Delta\to X$ (where $\Delta$ is the unit disk in $\mathbb{C}$) together with points $z_1,\ldots ,z_n,w_1,\ldots ,w_n\in\Delta$ such that $f_1(z_1)=x$, $f_i(z_i)=f_{i+1}(w_{i+1})$ for $1\le i< n$ and $f(w_n)=y$.

The length of a holomorphic chain is, with this notation, is $\sum_{i=1}^nd(z_i,w_i)$ (Poincaré metric on $\Delta$).

Finally the Kobayashi pseudo-distance on $X$ is obtained by setting $d(x,y)=$ infimum of lengths of all holomorphic chains from $x$ to $y$.

This "pseudo-distance" is obviously symmetric, satisfies $d(x,x)=0$ and the triangle inequality. The space is called Kobayashi hyperbolic if $d$ is in addition non-degenerate, i.e. if $d$ is a metric.

Now one could as well begin with the following much simpler construction:

Consider the function $\delta :X\times X\to [0,\infty ]$ with $\delta (x,y)=\inf d(z,w)$, the infimum running over all triples $(f,z,w)$ with $f:\Delta\to X$ holomorphic, $z,w\in\Delta$ and $f(z)=x$, $f(w)=y$.

This is still symmetric and satisfies $\delta (x,x)=0$, but now it is unclear whether

a) $\delta (x,y)$ is finite, i.e. the set of triples $(f,z,w)$ is non-empty;

b) $\delta$ satisfies the triangle inequality.

Clearly, a) and b) together are equivalent to $d=\delta$, and $d$ can be obtained from $\delta$ by an easy construction. Finally, my questions:

Under which circumstances is $d=\delta$?

Is there a simple example where $d\neq\delta$?

What is the logical relation between a) and b)?

Automorphisms of local fields

2011-11-12T13:31:33Z

It is an amusing coincidence (at least it appears to be a coincidence to me) that any completion of the field $\mathbb{Q}$ has trivial automorphism group as an abstract field, i.e. when ignoring the topology.

For $\mathbb{R}$, this is obtained as follows: the condition $x\ge 0$ is equivalent to $\exists y:y^2=x$, hence the ordering on $\mathbb{R}$ is preserved by any field automorphism. Since such an automorphism is the identity on $\mathbb{Q}$, it must be the identity anywhere.

For $\mathbb{Q}_p$, one can look at this very short paper. The trick is as follows: $x\in\mathbb{Q}_p^{\times }$ is a unit in $\mathbb{Z}_p$ if and only if it has an $m$-th root in $\mathbb{Q}_p$ for all $m$ prime to $p(p-1)$. Hence $\mathbb{Z}_p^{\times }$ is set-wise preserved by any field automorphism of $\mathbb{Q}_p$, and it is easy to deduce that such an automorphism must be the identity.

Now if we go on to finite extensions, something dramatic happens in the archimedean case: since $\mathbb{C}$ is algebraically closed, its automorphism group is huge.

So, two questions:

(a) Is there any conceptual explanation why $\operatorname{Aut}_{fields }(K)$ is trivial for any completion $K$ of $\mathbb{Q}$?

(b) What happens for finite extensions of $\mathbb{Q}_p$?

Answer by Robert Kucharczyk for Chevalley's Theorem on Constructible Sets

2011-11-11T21:21:46Z

As Jim Humphreys has pointed out in the comments, you have to either work over an algebraically closed field, or use scheme language. It is clear that over an algebraically closed field your example does not make any problems. So let us look at your example from the point of view of schemes.

What you are considering is the morphism $f:\mathbb{A}^1_K\to\mathbb{A}^1_K$ of $K$-schemes which is obtained as follows: recall that $\mathbb{A}^1_K=\operatorname{Spec} K[x]$, hence to give a morphism as above is the same as to give a homomorphism of $K$-algebras, and our homomorphism is given by $x\mapsto x^2$. Then, the morphism $f$ is not only dominant but surjective as a morphism of schemes.

How does one see this? $\mathbb{A}_K^1$ is the set of prime ideals in $K[x]$. There are two types of these: the prime ideal $(0)$, which defines the generic point, and the maximal ideals. Every maximal ideal is generated by a unique monic irreducible polynomial $f$. Hence if $K$ is not algebraically closed, there are strictly more points in $\mathbb{A}_K^1$ than the generic point and those corresponding to elements of $K$. By this classification of prime ideals it is now an easy exercise to deduce surjectivity of $f$.

However, as you rightly remarked, for $K$ imperfect of characteristic two, the induced map on $K$-rational points is not surjective and the image is "weird" in the sense that it is not of the form $S(K)$ for a subscheme $S\subseteq\mathbb{A}^1_K$. But your argument is wrong. The result from Humphreys that you quote again has to be understood as a statement about schemes, and only in the case of an algebraically closed base field can it be interpreted in the "naive" way as a statement about $K$-rational points.

One correct argument is this: if a subset $S\subseteq\mathbb{A}^1_K$ is constructible (i.e., a subscheme), then it is either finite or the complement of a finite subset. This is because the closed proper subsets of $\mathbb{A}_K^1$ are precisely the finite sets of closed points. But if $K$ is imperfect of characteristic two, than the subset $f(K)\subset K$ is infinite and has infinite complement.

By the way, there is a much easier example: take $f:\mathbb{A}^1_K\to\mathbb{A}^1_K$ as above, but with $K=\mathbb{R}$. Then the image of the induced map $\mathbb{R}\to\mathbb{R}$ is the set of nonnegative reals, clearly not "constructible", by the same reason.

Finally, here's the correct version of Chevalley's theorem:

Theorem (EGA IV, 1.8.4.) Let $f:X\to Y$ be a finitely presented morphism of schemes (any morphism between varieties over a field is of this type). Then the image of any constructible subset of $X$ under $f$ is a constructible subset of $Y$.

If $K$ is an algebraically closed field and $X$ and $Y$ are varieties over $K$ and $f$ is a morphism of $K$-varieties, then $X(K)$ can be identified with the set of closed (!) points on $X$, and you obtain the "naive" version.

analytic vs. algebraic Gauss-Manin connection

2011-02-12T21:17:15Z

There are the following two notions of "Gauss-Manin connection":

The complex-analytic one: let $f:X\to S$ be a smooth family of complex manifolds. Then we obtain a local system $R^nf_{\ast}\mathbb{C}$ of complex vector spaces on $S$, defining a holomorphic vector bundle $\mathcal{V}=R^nf_{\ast}\mathbb{C}\otimes\mathcal{O}_S$ on $S$ with an integrable connection $\nabla :\mathcal{V}\to\mathcal{V}\otimes\Omega_S^1$. Now the vector bundle $\mathcal{V}$ can be identified with the relative de Rham cohomology $\mathcal{H}_{dR}^n(X/S)$ of the family, so we get a connection on the latter.
The algebraic one: let $f:X\to S$ be a smooth morphism of smooth schemes over a field $k$. Now Katz and Oda in "On the differentiation of De Rham cohomology classes with respect to parameters" (J. Math. Kyoto Univ. 8 (1968), pp. 199-213) construct an integrable connection on $\mathcal{H}_{dR}^n(X/S)$ as some boundary map in a certain spectral sequence.

It is implicit in the literature that these two constructions are compatible, i.e. for a smooth family of smooth varieties over the complex numbers, the connection described in 1. is just the analytification of the one in 2. This sounds pretty reasonable as well. But thinking a bit about it, I was unable to come up with an argument, so could perhaps someone give me a hint where to find this or how to do it?

Comment by Robert Kucharczyk

2012-11-28T16:22:18Z

In fact the "naive" fundamental group of $\mathscr{M}_g(\mathbb{C})$ is trivial.

Comment by Robert Kucharczyk

2012-11-28T16:18:55Z

@ Qiaochu: yes.

Comment by Robert Kucharczyk

2012-03-17T15:13:06Z

To be more precise, the expression $a^b$ for arbitrary complex numbers $a,b$ is not well-defined, but the expression $e^b$ is.

Comment by Robert Kucharczyk

2012-03-17T15:11:52Z

That is not the problem, $e^z=1+z+{1\over 2}z^2 +{1\over 6}z^3+\ldots$ is perfectly well-defined.

Comment by Robert Kucharczyk

2012-03-12T07:27:48Z

It would be nice if you explained this in a little more detail.

Comment by Robert Kucharczyk

2012-03-12T07:25:37Z

No, they are great!

Comment by Robert Kucharczyk

2012-03-12T07:24:29Z

Oh, I see ... :-)

Comment by Robert Kucharczyk

2012-03-11T17:47:49Z

Are you sure about the formulation? Usually "multiplicative" means that $\varrho (ab)=\varrho (a)\varrho (b)$ for coprime $a$ and $b$. For such functions, the values $\varrho (p)$ and $\varrho (p^k)$ are in general of course totally unrelated (you can arbitrarily prescribe $\varrho (b)$ for every prime power $b$).

Comment by Robert Kucharczyk

2012-03-11T17:41:02Z

Non-Euclidean geometry certainly belongs here, but I am not sure about complex numbers (was not their discovery rather a struggle with unconscious metaphysical assumptions?). I do not know about the other things you mentioned, but it would be great if you took the time to turn that into an answer, with a few explaining sentences.

Comment by Robert Kucharczyk

2012-03-11T17:38:14Z

That is really nice, in particular 1. I was totally unaware of 1 and 2.

Comment by Robert Kucharczyk

2012-03-11T17:37:04Z

That is a great example: it is ingenious and totally non-straightforward.

Comment by Robert Kucharczyk

2012-03-11T17:35:25Z

While this is certainly very interesting mathematics, would that not rather fit in the first group of examples I mentioned just as a contrast?

Comment by Robert Kucharczyk

2012-03-10T18:42:00Z

What do you mean by writing the two names at the end? Is this a question originally posed by them?

Comment by Robert Kucharczyk

2012-03-10T17:55:47Z

@Asaf: I do not agree. The easiest constructions are certainly those from tilings of the hyperbolic plane by hyperbolic polygons, like triangle groups.

Comment by Robert Kucharczyk

2012-03-10T13:17:06Z

In a similar vein, there is a joke: Why did Bourbaki stop writing textbooks? Because they discovered that Serge Lang is one person.