User lasse rempe-gillen - MathOverflow

Belyi functions on non-compact surfaces; or: Building Riemann surfaces from equilateral triangles

2013-01-10T12:09:11Z

Some background on (compact) Belyi surfaces

$\newcommand{\Ch}{\hat{\mathbb{C}}}$ A compact Riemann surface $X$ is called a Belyi surface if there exists a branched covering map $f:X\to \Ch$ such that $f$ is branched over at most three points of $\Ch$. Here $\Ch$ denotes the Riemann sphere; we can and will take the three points to be $0$, $1$ and $\infty$.

(Recall that $f$ is a branched covering map if, for every $a\in\Ch$, there is a simply-connected neighborhood $U$ of $a$ such that $f$ maps every component of $f^{-1}(U)$ like $z\mapsto z^d$ for some $d\geq 1$. The function is branched over $a$ if $d>1$ for every such $U$.)

Equivalently, $X$ is a Belyi surface if it can be created by glueing together finitely many equilateral triangles together (defining a complex structure at the vertices in the obvious manner).

Every Belyi function is uniquely determined, up to a conformal change of variable, by its line complex, also known as a dessin d'enfant. This is a finite graph that essentially tells us how to form the surface by glueing together triangles.

In particular, the set of Belyi surfaces is countable. (This also follows from Belyi's famous theorem, which states that Belyi surfaces are exactly those that are definable over a number field.) So, in a nontrivial moduli space of Riemann Surfaces, such as the space of complex tori, most surfaces are not Belyi.

Belyi functions on non-compact surfaces

It seems natural to extend this notion to noncompact surfaces.

Definition. Let $X$ be a non-compact Riemann surface. An analytic function $f:X\to\Ch$ is called a Belyi function if $f$ is a branched covering whose only branched points lie over $0$, $1$ and $\infty$, and if $f$ has no removable singularities at the punctures of $X$.

Question. On which non-compact surfaces do Belyi functions exist?

This question seems extremely natural and came up in discussions among Bishop, Epstein, Eremenko and myself. Again, a Belyi function is uniquely determined by its line complex, which is now an infinite graph. Since the space of these graphs is totally disconnected, one might expect that not every non-compact surface supports a Belyi function. However, we discovered that this initial intuition is wrong:

Theorem. For every non-compact Riemann surface $X$, there is a Belyi function $f:X\to\Ch$. In particular, every non-compact Riemann surface can be built by glueing together equilateral triangles.

Note that, for non-compact surfaces, as pointed out by Misha in the comments, the existence of a Belyi function is formally stronger than being built from triangles, assuming that we allow vertices to be incident to infinitely many triangles. (Such vertices would not correspond to points in the resulting surface, as we have no way of defining a complex structure near these.) To get a Belyi function, we should assume that every vertex is incident to only finitely many triangles, so that each triangle is compactly contained in the resulting surface.

(We can also prove the existence of what one might call "Shabat functions", which have two critical values and omit $\infty$.)

My question:

Have such Belyi functions, and particular the problem of their existence on arbitrary non-compact surfaces, previously appeared in the literature?

(I would also be interested to hear whether our results might be of interest outside of one-dimensional complex function theory and complex dynamics. After all, classical Belyi functions and dessins d'enfants are relevant to many areas of mathematics.)

Answer by Lasse Rempe-Gillen for Asymptotic behavior of entire functions

2013-02-19T13:37:22Z

As mentioned in the comments, the asymptotic behavior of $f$ along the real axis doesn't really tell you anything about the function globally.

For example, your function $f$ can behave in any way you like as $x\to-\infty$.

Indeed, the following is implied by Arakeljan's (or Nersesjan's) approximation theorem: Let $A$ be any finite union of disjoint curves tending to infinity, and let $g:A\to\mathbb{C}$ and $\varepsilon:A\to(0,\infty)$ be continuous.

Then there is an entire function $f$ such that $|f(z)-g(z)|<\varepsilon(z)$ for every $z\in A$.

So, for example, let $A=(-\infty,0] \cup [1,\infty)$, let $g$ be the function $e^{-x}/x$ on $[1,\infty)$ and let $g$ be arbitrary on $(-\infty,0]$. Then you can find an entire function that approximates this function arbitrarily closely.

Answer by Lasse Rempe-Gillen for Applications of discrete-time dynamics

2013-02-11T15:44:05Z

A key technique in studying continuous-time systems is provided by Poincaré Sections. You place a hyperplane in your phase space in a suitable manner; instead of studying the full continuous-time system, you can then study the dynamics of the first-return map to this section.

In fact, the famous Hénon map was introduced to capture the behavior of a Poincaré section of the Lorenz attractor. Letting one of the parameters in the Hénon family degenerate actually leads to the real quadratic family, aka the "logistic family" as mentioned above. This is described very nicely by Lyubich in his 2000 AMS Notices article

http://www.ams.org/notices/200009/index.html

There are probably more direct applications of discrete dynamical systems (look for processes that naturally work in discrete steps). However, in my opinion the above is among the strongest "justifications" for studying discrete-time dynamical systems (if such justification is necessary beyond the beauty, and fundamental nature, of the theory itself).

Answer by Lasse Rempe-Gillen for Functions holomorphic on a region minus a Cantor set

2013-01-31T13:52:39Z

Removability with respect to homeomorphisms is different from the removability with respect to bounded functions mentioned in another answer.

In particular, it is not necessary to have Hausdorff dimension at most 1. Indeed, any quasicircle is removable with respect to homeomorphisms, as mentioned by Hrant.

For much more complicated sets, see Jeremy Kahn's thesis "Holomorphic Removability of Julia sets": He shows that many Julia sets of quadratic polynomials are in fact removable.

http://arxiv.org/abs/math/9812164

(As above, we consider the set in question to be compact.)

In particular, he discusses the notion of "absolute area zero": A set $K$ has absolute area zero if there is no conformal isomorphism from the complement of $K$ to the complement of some set with positive area. Any such set is removable, and any Cantor set that is well-surrounded has absolute area zero.

On the other hand, as has been noted elsewhere, there are many examples of sets that are not holomorphically removable. The simplest example of a Cantor set would be a Cantor set of positive measure. More interesting examples are provided by Chris Bishop, as cited in Misha's answer.

EDIT: You may also wish to look at the paper "Removability theorems for Sobolev functions and quasiconformal maps" by Peter Jones and Stas Smirnov, which contains a number of sufficient conditions for conformal removability: http://www.unige.ch/~smirnov/papers/hr-j.pdf

Graczyk and Smirnov use these criteria to prove removability of a large class of Julia sets.

Answer by Lasse Rempe-Gillen for "Explicit" examples of Irrational numbers very well approximated by rationnal numbers

2013-01-21T16:30:52Z

Your question seems somewhat ill-posed. What type of construction exactly do you allow? I suppose that you might ask for the number to be obtained by basic algebraic operations and perhaps some elementary functions from rational numbers, and perhaps some standard transcendental constants such as $\pi$ and $e$. However, to make it a precise question would require some work, and in some sense whatever you come up with would likely be arbitrary.

I strongly believe that no such number is known, though I do not have a definitive reference stating this, and there are people who know more about this. However, it seems that, where the continued fraction expansions are explicitly known (such as for $e$), the growth of the expansion tends to be at most linear, a far cry from what you would need to violate Bryuno's condition.

The next number you might try is probably $\pi$. However, as far as I understand, it hasn't even been proved that its continued fraction expansion is unbounded! Based on experimental data of the initial part of this expansion, it certainly doesn't seem likely that $\pi$ violates Bryuno condition.

Indeed, given that almost every rational number is Bryuno, it seems rather unlikely that any specific one that we might try will fail the condition, unless there is a specific reason for it.

Moreover, imagine yourself undertaking some work in an unrelated field of mathematics, and you come across a new significant constant $\alpha$. Unbeknownst to you, $\alpha$ is Bryuno, which means it likely looks an awful lot like a rational number $\rho$. Unless you know for some other reason that $\alpha$ must be irrational (which would be rather lucky indeed), how likely are you to be able to tell that $\alpha\neq\rho$?

However, most importantly in my view, it is not clear that you are asking the right question. Suppose that by some miracle, you knew that $x:=\sqrt[5]{e+\pi+\zeta(5)}$ is a non-Bryuno number, how would that help you? (I suppose it would say interesting things about the number from a number-theoretic point of view, but I am assuming that is not really what you are after.)

It seems rather difficult to argue that the number $x$ is much more natural than any other (let's say efficiently) computable real number. And the number in Anthony's comment will certainly be very efficiently computable.

Answer by Lasse Rempe-Gillen for What are good methods for detecting parabolic components and Siegel disk components in the Fatou set of a rational function?

2012-12-04T09:28:22Z

"Detecting" whether a neutral cycle exists or not will be difficult in general.

However, when you do know that there is a parabolic cycle (e.g. because you have constructed your map that way), Braverman shows, as quoted by Adam, that you can compute the Julia set in polynomial time. What should be noted here is that the program used (and the time bound) will depend on the parameter in question.

For irrationally indifferent fixed points, things can get even more tricky. But when the rotation number is nice (e.g. some quadratic irrational such as the 'golden mean'), you will have a critical point on the boundary of the Siegel disk. So you can draw the boundary of the Siegel disk by plotting the orbit of the critical point.

In parameter spaces, rather than trying to 'detect' neutral cycles, you may wish to draw the boundaries of hyperbolic components using Newton's method. That is, take a point in the hyperbolic component that you are interested in (where there is an attracting cycle), and then find a curve along which the modulus of the multiplier tends to one. Then you will have found an indifferent parameter. Now you can similarly change the argument of the multiplier, again using Newton's method, and trace this curve. Some care is required near "cusps".

For an example of such a picture, in the family of exponential maps, see Figure 1 in my article with Dierk Schleicher, "Bifurcations in the space of exponential maps" (http://arxiv.org/abs/math/0311480).

Answer by Lasse Rempe-Gillen for Is this a Julia set (and if so, for which function family is it the Julia set)?

2012-12-04T09:08:18Z

The set in question is the bifurcation locus of the family $f_{\lambda}$. It is hence the set of non-normality of the family $$\bigl(\lambda\mapsto f_{\lambda}^n(\lambda/3)\bigr)_{n\in\mathbb{N}};$$

see Theorem 4.2 in McMullen's book "Complex Dynamics and Renormalization".

As has been pointed out, you cannot expect the set to exactly coincide with the Julia set of a rational function. It should be possible to prove this formally. Indeed, your set has complementary components bounded by analytic curves, or regions bounded by curves analytic except at a single cusp. Such curves cannot bound Fatou components of a rational map, unless the map is a Blaschke product and the Julia set is itself a circle.

(I cannot seem to find a reference for this fact right now. However, the boundary of a Siegel disk may be smooth, but cannot be analytic anywhere; otherwise, the conjugacy to a rotation would extend beyond the boundary. On the other hand, boundaries of attracting basins are even known to have Hausdorff dimension strictly greater than one; see Przytycki, "On hyperbolic Hausdorff dimension of the boundary of a basin of attraction for a holomorphic map and of quasirepellers". Showing that the boundary cannot be analytic is much easier, both for attracting and parabolic basins.)

Answer by Lasse Rempe-Gillen for Transcendentality of all irrationals in the Cantor set

2012-11-28T12:07:04Z

This question was asked by Mahler ("Some suggestions for further research", Bull. Austral. Math. Soc. 29 (1984), no. 1, 101–108).

See Adamczewski, Bugeaud, "On the decimal expansion of algebraic numbers" (2005) for some things that are known. I have confirmed with a colleague that this is still very much a (widely) open problem. It does not appear that this problem would be appropriate for undergraduate research.

Of course this is not to say that learning about it would not be beneficial, or that there could not be related research problems that are more tractable. I suggest that you should seek the guidance of an experienced researcher in the field.

Answer by Lasse Rempe-Gillen for Proving a least prime factor

2012-11-26T09:49:37Z

Your problem is the following:

P1: Given $n$ and a prime $p$ such that $p$ divides $n$, does $n$ have a prime factor less than $p$?

However, the condition that $p$ divides $n$ can be removed; that is, your problem is equivalent to:

P2: Given $n$ and a prime $p$, does $n$ have a prime factor less than $p$?

(To solve P2 given an algorithm that solves P1, just apply P1 to the number $n\cdot p$.)

Problem P2 appears very close to the factoring problem: Given $n$ and any number $k$, does $n$ have a prime factor less than $k$? I don't see any reason why restricting $k$ to be prime should make things any easier.

So it would seem highly unlikely that there is a method that improves on testing whether $p/n$ has a prime factor less than $p$.

Answer by Lasse Rempe-Gillen for 13 months and not even one report. what would you do?

2012-11-24T17:03:35Z

I am sorry to hear about your trouble with your paper. Unfortunately, from experience, it is not that unusual to wait 14 months or longer for a referee's report.

It is certainly fine to email the editor politely, as Karl says, and inquire about the status. You could also ask whether the journal has had any preliminary comments on the article from the referee. (I.e., the referee may be willing to state that they think the result is interesting, and they are likely to recommend acceptance if no issues with the paper come to light. Indeed, many journals now do solicit such comments from referees.)

In my view, the editor should be a bit embarrassed about not responding to your request in September. However, if they are doing their job, they are already hassling the referee and trying to get them to agree to a date by which the report will be provided.

If you do not hear back from the editor in a reasonable timeframe after your e-mail, you might want to e-mail the main journal e-mail address, mentioning that you have tried to contact the editor but have not been successful, and asking whether they know something about the progress of the paper. That way, you will make sure that the journal's secretary knows about the issue, and will follow it up. (However, do make sure to be polite and not to sound critical of the editor, which would be counterproductive.)

Finally, you are absolutely right that it would be tough to get a rejection letter after this time - although it is certainly not unheard of. You would hope that, if the referee was going to reject the paper, they would have done so quickly. However, if the report is negative, there is little that the journal can do.

If the report does recommend acceptance, then I would expect that the journal would be inclined to follow the referee's advice.

Answer by Lasse Rempe-Gillen for Hausdorff dimension of Julia sets of quadratics not in the Mandelbrot set.

2012-11-24T13:00:47Z

The Hausdorff dimension of Julia sets of quadratic polynomials has been well-studied, although some questions still remain.

You specifically asked about parameters $c$ that do not belong to the Mandelbrot set. In this case, the map $q_c(z) = z^2 + c$ has a totally disconnected Julia set. Here is what can be said.

1) The Hausdorff dimension is always strictly greater than zero. (This is true for all non-linear, non-constant rational functions, even for meromorphic functions, as proved by Stallard. See e.g. Corollary 2.11 in my paper "Hyperbolic dimension and radial Julia sets of transcendental functions", Proc. Amer. Math. Soc. 137 (2009), 1411-1420.)

2) As $c$ tends to infinity, the Hausdorff dimension of the Julia set tends to zero. This is because the Julia set can be written as the limit set of a conformal iterated function system with two maps, corresponding to the inverse branches of the maps, and these are strongly contracting if $c$ is large.

3) As Alex mentions, Hausdorff dimension does not vary continuously for parameters on the boundary of the Mandelbrot set. In fact, the following is true:

Theorem. Suppose that $c\in \partial M$. Then there is a sequence $(c_n)$ of parameters outside the Mandelbrot set such that $\dim(J(q_{c_n}))\to 2$.

This follows from Shishikura's famous proof that the boundary of the Mandelbrot set has Hausdorff dimension equal to $2$ ("The Hausdorff dimension of the boundary of the Mandelbrot set and Julia sets", Ann. of Math. 147 (1998), no. 2, 225–267). Indeed, he shows that there is a dense set of parameters on the boundary where the hyperbolic dimension equals two. Any nearby parameter will have a Julia set of Hausdorff dimension close to $2$.

On the other hand, there are many parameters on the boundary of the Mandelbrot set where the Hausdorff dimension is strictly less than $2$. So it is not hard to see that the dimension does not depend continuously in the way that you desire.

If we ask about radial limits (i.e., consider the conformal map that takes the complement of the closed unit disk to the complement of the Mandelbrot set, and approach the boundary of the Mandelbrot set along the image of a straight ray), things become more subtle, and I am not sure what exactly is known. However, from what I can remember, it is known that, even for the simple case where $c(t) = 1/4+t$, $t>0$, the Hausdorff dimension of $J(q_{c(t)})$ does not tend to that of $J(q_{1/4})$ as $t\to 0$. (This is the parabolic implosion that Alex mentions.)

Answer by Lasse Rempe-Gillen for the intersection of a sequence of measurable sets

2012-11-14T15:32:53Z

Here is an explicit example that elaborates on Juan's and Gerald's answers. Let $\Omega=[0,1]$.

Let $\Omega_j$ be the union of $2^{j-1}$ intervals of length $2^{-j}$, with the first interval starting at $0$, and all intervals being spaced distance $2^{-j}$ apart.

That is, $\Omega_1 = (0,1/2)$; $\Omega_2 = (0,1/4) \cup (1/2,3/4)$, etc.

Then this sequence is a counterexample to your statement, as the intersection of $n$ of these sets has measure exactly $2^{-n}$.

Beautiful examples of arc-like continua

2012-11-13T21:02:54Z

A continuum is a nonempty compact, connected metric space.

A continuum $X$ is called arc-like if, for every $\varepsilon>0$, there is a continuous and surjective function $f:X\to [0,1]$ such that $f^{-1}(t)$ has diameter less than $\varepsilon$ for every $t\in[0,1]$.

(Equivalently, $X$ is homeomorphic to an inverse limit of arcs with surjective bonding maps.)

Arc-like continua are also called "snake-like" or "chainable" continua. For more background, see Nadler's excellent textbook 'Continuum Theory: An Introduction'.

Examples include the arc, the $\sin(1/x)$-continuum, the Knaster bucket-handle and, perhaps most famously of all, the pseudo-arc (which is the unique hereditarily indecomposable arc-like continuum).

It is easy to make a nice picture of the bucket-handle (and of the $\sin(1/x)$-continuum). As far as I know, there isn't really any good way to make a sensible picture of the pseudo-arc.

I am writing a paper that involves arc-like continua, and I would be interested to know:

Are there other interesting examples of arc-like continua that lend themselves to making nice and illuminating computer pictures?

(Of course we could combine the above examples to create new arc-like continua, but I wouldn't class this as being 'interesting'. Nadler's book has an example of a hereditarily decomposable arc-like continuum that contains no arc, but it would seem difficult to turn this into a sensible picture.)

Any pointers (or, even better, pictures!) would be appreciated. As the question is open-ended, I'm making it Community Wiki.

(In case you are interested, the main result of my paper states that there is a transcendental entire function $f:\mathbb{C}\to\mathbb{C}$ with the following property. If $X$ is an arc-like continuum with a terminal point $x_0\in X$, then there is a component $C$ of the Julia set $J(f)$ such that $C\cup{\infty}$ is homeomorphic to $X$. In particular, the pseudo-arc can appear as (the one-point compactification of) a component of the Julia set of a transcendental entire function.)

Answer by Lasse Rempe-Gillen for Is the generalized Baire space complete?

2012-11-13T21:33:48Z

I think the notion of completeness doesn't make sense for topological spaces; you need at least a uniform structure.

According to the encyclopedia of mathematics, the product of complete uniform spaces is complete:

http://www.encyclopediaofmath.org/index.php/Complete_uniform_space

(I found this by googling "product of complete uniform spaces").

EDIT. I just checked my copy of Kelley's "General Topology". Chapter 6 deals with uniform spaces. In this chapter, Theorem 10 states that the topology of the product uniformity is the product topology, and Theorem 25 states that the product of uniform spaces is complete if and only if each coordinate space is complete. This provides the reference you asked for.

Answer by Lasse Rempe-Gillen for A question about local connectedness

2012-11-13T22:27:13Z

Any indecomposable continuum has the property you desire.

(A continuum is indecomposable if it cannot be written as a union of two proper subcontinua.)

One way to see this is that any indecomposable continuum has uncountably many composants, all of which are mutually disjoint, and all of which are dense in $x$. (Here the composant of a point $x$ in $X$ is the union of all proper subcontinua of $X$ that contain $x$.)

Here is a more direct proof: Suppose that $C$ is a proper subcontinuum of $X$ that is a neighborhood of $x$. Then every connected component of $V := X\setminus C$ contains a point of $C$ in its boundary. (This is known as the 'boundary bumping theorem'.)

If $\overline{V}$ is connected, then $\overline{V}$ and $C$ are proper subcontinua of $X$ whose union is $X$.

If $V$ is disconnected, decompose $V$ into two relatively closed disjoint subsets $A$ and $B$; then $A\cup C$ and $B\cup C$ are the desired subcontinua.

A simple example of an indecomposable continuum is given by the Knaster bucket-handle, see

http://commons.wikimedia.org/wiki/File%3aThe_Knaster_%22bucket-handle%22_continuum.svg .

The solenoid, mentioned in another answer, is another indecomposable continuum. You can also get such examples from "Lakes of Wada" continua. Of course the double Cantor brush given by Victor is not indecomposable (and in fact hereditarily decomposable).

Answer by Lasse Rempe-Gillen for A delicate measure-theoretic question about Jordan curves and arcs in the plane.

2012-11-13T20:12:25Z

The answer should be yes.

Consider $E$ to be a subset of the Riemann sphere $\hat{\mathbb{C}}$. The complement $U$ of $s$ is simply-connected, so we can map it conformally to the unit disk (taking $\infty$ to $0$, say).

Take the preimage of a circle of radius $r$, close to $1$, under this conformal map. This gives you a Jordan curve, and the Jordan domain $Z_r$ enclosed by this curve will have area tending to the area of $s$. So for a suitable value of $r$, this domain will have the desired property.

(Am I am missing something?)

Answer by Lasse Rempe-Gillen for Boundary regularity of quasiconformal homeomorphisms of the unit disk ?

2012-11-03T11:09:59Z

You may be interested in looking at the article "On boundary correspondence under quasiconformal mappings" by V.Ya. Gutlyanskii and V.I. Ryazanov. In particular, their Corollary 1 concerns what can be said when the Beltrami differential extends uniformly continuous to some part of the boundary.

They also mention the example $$ f(z) = z\cdot (1-\log|z|),$$ which defines a quasiconformal function near zero whose complex dilatation is continuous near zero, but which is not differentiable at zero.

Note that this example is real on the real axis, so it easily yields a counterexample to your question (unless I have misunderstood something!).

Answer by Lasse Rempe-Gillen for A Jordan Arc in the unit disk

2012-11-01T17:05:28Z

I know this is a really old question, but I just happened upon it for some reason. I've always liked these kind of plane topology question, so I think I'll add an answer here.

If you are looking for a simple topological proof, then I suggest using the following theorem of Janiczewski's:

Let z and w be points in the plane, and let $A$ and $B$ be compact, connected sets. If neither $A$ nor $B$ separates $z$ from $w$, and $A\cap B$ is connected, then $A\cup B$ also does not separate $z$ from $w$.

Now, to prove that a Jordan arc $J$ does not separate the plane, take two points $z$ and $w$ in $\mathbb{R}^2\setminus J$, and decompose $J$ into small Jordan arcs, each chosen sufficiently small to ensure that none of these separates $z$ from $w$, and apply Janiczewski's theorem repeatedly.

To see that the union of your arc with the unit circle does not separate two points inside the unit disk that are not on your arc, just apply Janiczewski's theorem again.

Answer by Lasse Rempe-Gillen for A question about indecomposable continua.

2012-11-01T12:01:50Z

As pointed out by Jeff, the notion you define may not really be what you are after, since indecomposable continua are not 'indecomposable' in your sense. However, we can ask:

Is there a nontrivial connected metric space $X$ such that $X$ cannot be written as the union of two proper connected subsets?

The answer, as Jeff suggested, is no.

Indeed, let $X$ be a nontrivial connected metric space. If $X$ does not have any cut-points, then clearly we can write $$X = (x\setminus{x_0}) \cup (X\setminus{x_1})$$ for some $x_0\neq x_1$, and are done.

If $X$ does have a cut-point $x_0$, let $A$ and $B$ be open subsets of $X$ such that $$A\cap B = {x_0}; \quad A\setminus{x_0},B\setminus{x_0}\neq\emptyset \quad\text{and}\quad A\cup B = X.$$

We claim that $A$ and $B$ are connected. Indeed, if $U\ni x_0$ is relatively open and closed in $A$, then $U\cup B$ is open and closed in $X$, so we must have $U=A$ (since $X$ is connected).

Regarding your question on the number of proper connected subsets, we can still ask the following question:

If $X$ is any nontrivial connected metric space, what can be said about the cardinality of the set $S$ of proper connected subsets of $X$?

It seems plausible that the set $S$ has at least the cardinality of the continuum, but I wasn't able to find a reference (and haven't thought very deeply about it). Certainly the set $S$ must be infinite.

The Arnol'd family of circle maps - origins and density of hyperbolicity

2011-11-17T14:33:19Z

$\newcommand{\R}{\mathbb{R}}\newcommand{\Z}{\mathbb{Z}}$ The Arnol'd family or standard family of circle maps is defined by $$F_{\mu_1,\mu_2}:\R/\Z\to\R/\Z;\quad t\mapsto t + \mu_1 + \mu_2\sin(2\pi t); \quad \mu_1\in\R, \mu_2>0.$$

Arnol'd considers this family in the paper Small denominators I. Mapping the circle onto itself. The family is studied in Section 12 (in slightly different parametrization).

My first question is whether this is the first instance where the family has been suggested. On the Wikipedia page for this family, it is claimed that the family was introduced by Kolmogorov, but the editor in question has said that this is likely to have been a mistake. However, it is easy to believe that the family, which is very natural, could have been considered elsewhere independently, so I would be interested if anyone knows more about its origin. (I do not read Russian, but have asked a colleague to translate the relevant section in Arnol'd's paper, which does not seem to indicate a prior origin of the family.)

In a recent paper with van Strien, we were able to prove the density of hyperbolicity, i.e. of maps where both critical points belong to the basins of periodic attractors, in the non-invertible region $\mu_2>1/(2\pi)$. Density of hyperbolicity is essentially the central problem in one-dimensional dynamics (see Smale's 11th problem), and the Arnol'd family is one of the most-studied families of one-dimensional maps after (quadratic) polynomials. However, I am not aware of density of hyperbolicity in this family having been explicitly stated as an open problem in the literature before a recent paper of de Melo, Salomão and Vargas, which studied generalizations of this family with larger numbers of critical points. Again, does anyone know of this problem (or related ones, such as rigidity), being raised in the classical literature?

Many thanks for your help!

Answer by Lasse Rempe-Gillen for Domain of Holomorphy

2011-10-27T10:04:58Z

Further to Ben's answer, it might be useful to picture the situation in $\mathbb{C}$. (Of course in $\mathbb{C}$ every domain is a domain of holomorphy, but we can still exhibit the same phenomenon that causes us to need the more complicated definition.)

The principal branch of the logarithm $f := \operatorname{Log}$ is defined on the slit plane $\Omega :=\mathbb{C}\setminus (-\infty,0]$. The function $f$ does not extend holomorphically, or even continuously, to any point of $(-\infty,0]$.

However, let $x\in (\infty,0)$, set $\delta := -x$, and consider the disk $V := D(x,\delta)$ and the half-disk $$ U := {z\in V: \operatorname{Im} z > 0}.$$

Then the restriction of $f$ to $U$ extends analytically to a holomorphic function $g$ on $V$, where $g(z)=f(z)$ if $z\in U$, $g(z) = \log(-z)+i\pi$ if $z$ belongs to the diameter $(2x,0)$ of $V$, and $g(z)=\operatorname{Log}(z) + 2\pi i$ otherwise.

The point is that the slit plane is not a "natural" maximal domain of definition for the function $\operatorname{Log}$. (If we were looking for such a domain, it would be a spiralling Riemann surface spread over the punctured plane ...) In other words, suppose you have a domain of holomorphy for some analytic function, but you only know the germ of this function at some point. Then you can reconstruct the domain uniquely as the maximal domain to which the germ can be analytically continued.

Of course the interesting thing is that there are domains that are not domains of holomorphy, when $n\geq 2$.

Real-analytic manifolds in real-analytic sets

2010-02-02T14:04:13Z

Let $U\subset \mathbb{R}^n$ be open, and let $f:U\to\mathbb{R}$ be real-analytic. We consider the zero set $Z:=f^{-1}({0})$.

For a paper I am writing, I am looking for the best reference to the following basic fact:

If $Z$ has topological dimension equal to $d$, then $Z$ contains a real-analytic manifold of dimension $d$.

I can get this from Lojasiewicz's theorem or similar results, but that is a slightly unwieldy reference, and something probably needs to be said about how exactly one deduces it. Given that the statement is rather simple, I was wondering if someone knows of a more direct reference to this fact.

And to add a mathematical question: This result is obviously much weaker than Lojasiewicz's theorem. Is there a proof that doesn't require developing the full structure theorem?

Many thanks for any pointers!

Answer by Lasse Rempe-Gillen for Parametrization of the boundary of the Mandelbrot set

2010-12-08T14:16:14Z

I am not quite sure what you are asking. The boundary of the Mandelbrot set certainly is not an analytic curve. In fact, a famous result of Shishikura shows that the boundary of the Mandelbrot set has Hausdorff dimension 2.

Indeed, it is not even known whether the boundary is a curve at all (i.e., locally connected): this is currently probably the most famous conjecture in one-dimensional holomorphic dynamics.

If the Mandelbrot set is locally connected, then there is a natural description of the boundary of the Mandelbrot set (as the boundary values of the Riemann map of the complement of $M$); this is also known to be a natural combinatorial description in many ways. However, as mentioned above, this parametrization is not analytic, or even $C^1$.

Answer by Lasse Rempe-Gillen for Attractive Basins and Loops in Julia Sets

2010-11-29T23:20:18Z

It seems you are basically interested in an introduction to complex dynamical systems. The books by Beardon, Milnor and Steinmetz all give good introductions.

Regarding your specific questions:

a) The point at infinity is a superattracting fixed point, but more importantly its immediate basin of attraction - that is, the component of the basin containing the fixed point itself - is completely invariant (invariant under forward and backwards iteration). This is the case for all polynomials (of degree at least two), and is one of the reasons that studying polynomials is easier than studying general rational maps (where e.g. the Julia set - where the dynamics is chaotic - may in fact be the whole Riemann sphere). The basin of infinity supports foliations into "external rays" and "equipotentials", and this allows one to study the Julia set. This idea was introduced by Douady and Hubbard, and is the basis of the famous "Yoccoz puzzle".

b) There are all kinds of possible orbits in the Julia set. This isn't the place to discuss them. Stable orbits of polynomials may converge to a (super)-attracting fixed point, to a parabolic fixed point (where the multiplier is a root of unity), or belong to a rotation domain (a simply connected domain on which the dynamics is conjugate to a rotation).

c) See above. There may be at most one attracting cycle for a quadratic polynomial (apart from infinity), because there is at most one critical point. (This is a theorem of Fatou.) Conjecturally, systems with an attracting cycle are dense in the Mandelbrot set; this is perhaps the most celebrated conjecture in one-dimensional dynamics at the moment.

Fermat for polynomials, as used in the AKS (Agrawal-Kayal-Saxena) algorithm

2010-07-28T16:31:59Z

The basis for the deterministic polynomial-time algorithm for primality of Agrawal, Kayal and Saxena is (the degree one version of) the following generalization of Fermat's theorem.

Theorem

Suppose that P is a polynomial with integer coefficients, and that p is a prime number. Then $(P(X))^p\equiv P(X^p)\ (\mod p)$.

Surely this result was known previously, but I have not been able to find a reference in the literature on the AKS algorithm (which means that the authors also did not know of a reference). Does anyone here know of one?

Furthermore, there is a converse to the lemma in the AKS paper:

Lemma

If n is a composite number, then $(X+a)^n\not \equiv X^n+a\ (\mod n)$ whenever a is coprime to n.

Again, it is easy to generalize this statement. For example, if P is a polynomial which has at least two nonzero coefficients and such that all nonzero coefficients are coprime to n, then $P(X)^n\not\equiv P(X^n)\ (\mod n)$ for composite n.

On the other hand, clearly some conditions are necessary; for example $(3X+4)^6\equiv 3X^6+4\ (\mod 6)$.

Is there a best possible statement? And, again, is there a reference?

Answer by Lasse Rempe-Gillen for Minimum number of contractions needed to obtain a particular invariant set

2010-07-27T10:03:21Z

An interesting question. Of course there is some ambiguity in the formulation "when can we tell".

Certainly in explicit examples, one may be able to apply ad-hoc methods. For example, things are easier for the Sierpinski gasket and carpet, since these have identifiable features in terms of their complementary regions.

For example, if we wish to write the Sierpinski gasket as a union of smaller copies, it should be fairly easy to see that each complementary region must be mapped to another complementary region. But this means that each contraction must correspond to one of the "smaller triangles" that appear in the usual gasket contraction, and we need at least three of these to make up the whole gasket.

The same type of argument should work for the Sierpinski carpet.

EDIT: Let me provide a few additional arguments to illustrate what I mean in the case of the Sierpinski gasket and carpet.

Lemma

Any equilateral triangle contained in the Sierpinski gasket is the triangle surrounding one of the "children" in the usual construction.

Proof

An exercise for the reader. (Hint: Note that the only way for a straight line in the gasket to begin in one of the standard triangles but not end there is to pass through the two points by which it is connected to the rest of the gasket.)

Corollary

If $A$ is an affine similarity that maps the Sierpinski gasket into itself, then $A$ can be written as a finite composition of the three maps from the "standard" IFS that generates the gasket.

A similar argument works for the carpet. Here it is not enough to consider the outer square, but if we add the first generation squares to it, the same works. To state this, let A be the union of the boundaries of nine squares that are joined together to form a larger square; e.g. $$A=\{(x,y)\in[0,3]^2: x\in\{0,1,2,3\} \text{ or } y\in\{0,1,2,3\}\}$$. Let us call any image of A under an affine similarity a "3-by-3 grid".

Lemma

The outer boundary of any nine-by-nine grid contained in the Sierpinski gasquet is the boundary of one of the squares occuring in the usual construction.

Again, I will leave the proof as an exercise. The claim that the usual system is optimal then follows immediately once more.

Simply-connected domain around a curve

2010-07-26T15:38:49Z

In a current project with a colleague, we have come across the following reasonably classical-sounding geometric question. While not vital to our work, it would be interesting if anyone has seen this type of issue discussed before, and if there are any references.

To keep things simple, I will formulate a special case, which appears to retain all issues present in the general case.

Question

Let $\gamma:[0,\infty)\to\mathbb{C}\setminus\{0\}$ be a continuous and injective curve in the punctured complex number plane, with $\gamma(t)\to\infty$ as $t\to\infty$.

Is there a simply-connected domain $V\subset\mathbb{C}\setminus\{0\}$ with $\gamma\subset V$ such that $\gamma$ tends to the boundary of $V$ horocyclically in $V$?

(The latter condition means the following: There is a conformal isomorphism $\phi$ that maps $V$ to the right half plane in such a way that $\operatorname{Re}(\phi(\gamma(t)))\to+\infty$ as $t\to\infty$.)

Remarks

We may additionally suppose that a function $\delta(t)$ is given, and require that $V\subset\bigcup_{t\geq 0}B(\gamma(t),\delta(t))$, where $B(z,\delta)$ is the open disk of radius $\delta$ around $z$. This makes the statement of the question slightly more complicated, but makes the discussion of examples less cumbersome. (It is also part of the more general setup I mentioned above, which additionally replaces $\mathbb{C}\setminus\{0\}$ by an arbitrary open Riemann surface.)
If the curve $\gamma$ is $C^1$, then it is easy to construct such a domain, by taking $V$ to be a (shrinking) "tubular" neighborhood of $\gamma$. In this case, it is even possible to ensure that the convergence is non-tangential, meaning that the argument of $\phi(\gamma(t))$ stays bounded as $t\to\infty$.
It is not too difficult to see that non-tangential convergence cannot be obtained without some regularity assumption. However, perhaps a suitable uniform Lipschitz condition is sufficient.
However, for horocyclic convergence, one can easily make do with much weaker conditions. For example, it seems enough to ask that the curve is $C^1$ on some sequence of intervals tending to infinity, while the behaviour in between can be as bad as you like.
I tend to think that one can probably construct a counterexample for horocyclic convergence, but I may be wrong and in any case it is likely to involve some thought and effort. So I am hoping that someone can tell me that this or a similar question has been discussed in the literature.

Answer by Lasse Rempe-Gillen for Demystifying complex numbers

2010-07-28T09:34:30Z

I always like to use complex dynamics to illustrate that complex numbers are "real" (i.e., they are not just a useful abstract concept, but in fact something that very much exist, and closing our eyes to them would leave us not only devoid of useful tools, but also of a deeper understanding of phenomena involving real numbers.) Of course I am a complex dynamicist so I am particularly partial to this approach!

Start with the study of the logistic map $x\mapsto \lambda x(1-x)$ as a dynamical system (easy to motivate e.g. as a simple model of population dynamics). Do some experiments that illustrate some of the behaviour in this family (using e.g. web diagrams and the Feigenbaum diagram), such as:

The period-doubling bifurcation
The appearance of periodic points of various periods
The occurrence of "period windows" everywhere in the Feigenbaum diagram.

Then let x and lambda be complex, and investigate the structure both in the dynamical and parameter plane, observing

The occurence of beautiful and very "natural"-looking objects in the form of Julia sets and the (double) Mandelbrot set;
The explanation of period-doubling as the collision of a real fixed point with a complex point of period 2, and the transition points occuring as points of tangency between interior components of the Mandelbrot set;
Period windows corresponding to little copies of the Mandelbrot set.

Finally, mention that density of period windows in the Feigenbaum diagram - a purely real result, established only in the mid-1990s - could never have been achieved without complex methods.

There are two downsides to this approach: * It requires a certain investment of time; even if done on a superficial level (as I sometimes do in popular maths lectures for an interested general audience) it requires the better part of a lecture * It is likely to appeal more to those that are mathematically minded than engineers who could be more impressed by useful tools for calculations such as those mentioned elsewhere on this thread.

However, I personally think there are few demonstrations of the "reality" of the complex numbers that are more striking. In fact, I have sometimes toyed with the idea of writing an introductory text on complex numbers which uses this as a primary motivation.

Answer by Lasse Rempe-Gillen for Is there a way to find regions of depth in the Mandelbrot set other than simply poking around?

2010-07-27T12:24:24Z

It is a little bit difficult to answer the question as posed, because there is a question as to what you mean by "depth".

One of the previous answers mentions Misiurewicz points - parameters where the critical orbit is pre-periodic. Examples are the "branch points" and "tips" in the Mandelbrot set. However, these are not likely to be good candidates for what you are looking for. Indeed, rescalings of the Mandelbrot set near such a parameter will converge, with the same limit as corresponding rescalings in the dynamical plane, by a theorem of Tan Lei. This does not seem like what you are looking for.

If you pick a "Feigenbaum point" (an infinitely renormalizable parameter of bounded type, such as the famous Feigenbaum value which is the limit of the period-2 cascade of bifurcations), then Milnor's hairiness conjecture, proved by Lyubich, states that rescalings of the Mandelbrot set converge to the entire complex plane. So there is certainly a lot of thickness near such a point, although again this may not be what you are looking for. It may also prove computationally intensive to produce accurate pictures near such points, because the usual algorithms will end up doing the maximum number of iterations for almost all points in the picture.

In this case you are left with either non-renormalizable parameters (those that do not belong to any small copies of the Mandelbrot set) or infinitely renormalizable parameters with interesting behaviour.

An example of the formal is given by the so-called "Fibonacci parameter"; see "Parameter scaling for the Fibonacci point" by Wenstrom for information (and a picture) about the parameter space structure near this point. Rodrigo Perez investigated the structure of parameter space near such pieces in much more detail.

To imagine creating a zoom sequence for infinitely renormalizable parameters, imagine the following: Begin with a Misiurewicz parameter c_0. Then pick a little Mandelbrot copy M_1 near c_0, and choose a Misiurewicz point c_1 in this little copy. Pick a little Mandelbrot copy M_2 very close to c_1, and again a Misiurewicz point c_2 in there, and so on. Zooming in on the limit parameter is essentially the same as zooming in to the Misiurewicz parameter c_0, then changing track and zooming into the little Mandelbrot copy M_1, centering on c_1, and so on. Of course instead of using Misiurewicz points, you could at various types use any other types of points that are on the boundary of the Mandelbrot set (boundary points of interior components, Feigenbaum parameters, etc).

Local form of a real-analytic function taking values in a Banach space

2010-04-30T17:00:23Z

Let $B$ be an infinite-dimensional Banach space, and let $M\subset\mathbb{R}^n$ be a neighborhood of the origin in $\mathbb{R}^n$.

Suppose that $I:M\to B$ is a real-analytic function with $I(0)=0$ and such that the derivative of $I$ at $0$ has maximal rank.

Is it true that there exist neighborhoods $U,V\subset B$ of $0$ and a real-analytic diffeomorphism $\phi:U\to V$ such that $\phi\circ I$ is the restriction of a linear map $\mathbb{R}^n\to B$? If so, what is a good reference?

EDIT: I asked this question in the real-analytic setting, but might as well have done so in the complex-analytic case.

Comment by Lasse Rempe-Gillen

2013-02-17T14:35:44Z

I hope you have had or will have some success with getting feedback through your friend.

Comment by Lasse Rempe-Gillen

2013-02-15T08:49:33Z

This looks like homework ...

Comment by Lasse Rempe-Gillen

2013-02-12T16:01:55Z

The majority of mathematicians will never switch to constructive mathematics, at least not until a contradiction is discovered in ZFC. :) Even in the (rather unlikely) event that this happens, it is far more likely that some weaker, still non-constructive, set of axioms is going to be used. The mathematical world we work in is too powerful and convenient to give up without good reason. Of course this doesn't mean that the study of intuitionistic logic isn't interesting from a foundational point of view.

Comment by Lasse Rempe-Gillen

2013-02-11T13:14:19Z

As I mention below, the question becomes quite different when asking about "conformal" removability; i.e. removability for conformal homeomorphisms. Of course holomorphic removability in the sense here implies conformal removability, but the converse is far from true. For example, you can easily have conformally removable sets of Hausdorff dimension 2 (but zero Lebesgue measure).

Comment by Lasse Rempe-Gillen

2013-02-07T12:56:36Z

@Zoltan: Since your complex structure is smooth, it would seem that the identity is automatically a diffeomorphism between your surface S with the original structure, and with the new complex structure? Also, you have not answered my question: what do you mean by the curve being analytic? Let me make it more precise. The unit circle is an analytic curve. Take a diffeomorphism of the sphere that maps the unit circle to itself, but not in a real-analytic manner. Pull back the usual complex structure using this diffeomorphism. Does this satisfy your condition or not?

Comment by Lasse Rempe-Gillen

2013-01-31T13:05:01Z

@Zoltan: You should probably clarify what exactly you mean by complex structures here: just a collection of charts? The structure given by a Beltrami differential? Etc. More crucially: what does "$C$ remains analytic" mean? That the map $\gamma$ is analytic, or that the curve has some analytic parametrization? In the former case, the answer is essentially trivial, as noted by Aleksey: the identity map should restrict to be analytic on the curve $C$. In the latter case, I doubt there's a good general answer you can expect - e.g. any homeomorphism preserving $C$ will give such a structure.

Comment by Lasse Rempe-Gillen

2013-01-21T16:35:07Z

If you know some of the other editors, it seems like it would be perfectly reasonable to contact them. Make sure to be diplomatic, because you don't want to come across as impatient or unreasonable, but you do have a right to expect a response to your communications from the editor, even if it is just a short "we are still waiting to hear from the referee". Good luck!

Comment by Lasse Rempe-Gillen

2013-01-17T16:03:46Z

To not have had contact from the editor seems irregular. It sounds like a tricky situation. I'm not sure what your situation is - if you are a recent PhD student, have you spoken to your supervisor? They might be able to contact someone at the journal on your behalf.

Comment by Lasse Rempe-Gillen

2013-01-14T09:02:06Z

This being a question that doesn't have a 'right' answer, and one that is quite open-ended, should it perhaps be Community Wiki?

Comment by Lasse Rempe-Gillen

2013-01-14T08:54:22Z

Thanks Loic - I wasn't aware of Kevin's work on vector fields.As you say, this isn't quite what I was asking for, but I will take a look at this paper.

Comment by Lasse Rempe-Gillen

2013-01-11T14:11:17Z

@Misha, I think I see what you mean - for noncompact surfaces, being built from triangles is indeed (formally) weaker than having a Belyi function on it. To have a Belyi function, every corner should be adjacent to only finitely many triangles. I will add the question to clarify this. (Our theorem establishes the stronger property for all non-compact surfaces, and hence the weaker property also holds.)

Comment by Lasse Rempe-Gillen

2013-01-11T09:30:39Z

Perhaps I should have clarified that, when we build the Riemann surface from infinitely many triangles, we only include those corner points that are contained in only finitely many triangles. Near these, we define a Riemann surface structure in the obvious way. (Near the others, it isn't at all clear how we would define a Riemann surface structure.)

Comment by Lasse Rempe-Gillen

2013-01-11T09:23:45Z

I am not sure that I understand your comment correctly. Any holomorphic map is a "homeomorphism away from branch points". By "branched cover", I mean precisely the definition given in the second paragraph, which is stronger than both of your definitions. (The exponential map is a covering to its image on the plane, but it is not a Belyi function in my sense.)

Comment by Lasse Rempe-Gillen

2012-12-28T16:19:33Z

Hi Alex, I think Mahdi is right - for the exponential map, $f^{-1}(\mathbb{C})=\mathbb{C}$ ... indeed, $f(Y)\subset Y$ is equivalent to $Y\subset f^{-1}(Y)$.

Comment by Lasse Rempe-Gillen

2012-12-26T17:47:51Z

@Aaron - do you have any further questions on the topic, or might you consider accepting an answer?