User uncooltoby - MathOverflow

Functions holomorphic on a region minus a Cantor set

2012-05-16T16:51:11Z

Let $X$ and $Y$ be simply connected open regions of $\mathbb{C}$, and let $Z \subset X$ be a Cantor set. Assume we have a homeomorphism $f$ from $X$ to $Y$, which is holomorphic on $X \setminus Z$. Is $f$ necessarily holomorphic on $X$?

Hausdorff dimension of Julia sets of quadratics not in the Mandelbrot set.

2012-11-24T04:26:02Z

What are the bounds on the possible values of the Hausdorff dimension of the Julia sets of quadratics not in the Mandelbrot set? In particular, assume we have a quadratic $q_c: z \mapsto z^2 + c$ on the boundary of the Mandelbrot set $M$ and a curve $\gamma$ parameterized by the closed unit interval and with $\gamma(t)$ in $\mathbb{C} \setminus M$ for $t< 1$ and $\gamma(1) = c$. If the Hausdorff dimension of $q_c$ is $h$, can one assume that the Hausdorff dimension of the Julia set of $q_t: z \mapsto z^2 + \gamma(t)$ tends to $h$ as $t$ tends to $1$?

How to determine bounds on the extremal length around annuli?

2012-08-03T20:53:18Z

I wish to determine bounds for the sum of moduli of a family of topological annuli in the complex plane. Towards that end I would like to ask a question about the closely related concept of extremal length.

Let a quadrilateral $Q$ be a simply connected region of the complex plane, with Jordan curve boundary, and with two disjoint Jordan arcs $\gamma$ and $\gamma'$ making up part of the boundary. The two curves are each parameterised injectively by the closed unit interval, with the image $\gamma(t)$ (or $\gamma'(t)$) moving anticlockwise as $t$ increases. It is well known that for every quadrilateral there exists a unique $k \in \mathbb{R^+}$ and unique conformal homeomorphism $\phi$ mapping $Q$ onto the interior of a geometrical rectangle with vertices $0$, $1$, $1+ki$, and $ki$, and with $\gamma$ and $\gamma'$ mapped to the top and bottom edges respectively. In such a case the extremal length of curves joining $\gamma$ and $\gamma'$ in $Q$ is $k$. If the quadrilateral $Q$ satisfies the further conditions that $Re(\phi(\gamma(t))) = 1-t$ and $Re(\phi(\gamma'(t))) = t$ we say that $Q$ is a rectangle.

We construct a family of annuli by gluing together a collection of rectangles and generic quadrilaterals. For each $n$ let $(R^n_1, R^n_3, \ldots, R^n_{2m-1})$ be an m-tuple of rectangles, with extremal lengths $k_1n, k_3n, \ldots, k_{2m-1}n$ respectively, where each $k_i$ is a positive scalar fixed for all $n$. Likewise, let $S_2, S_4, \ldots, S_{2m}$ be a sequence of generic quadrilaterals. For each $n$, we form an annulus by taking the first element $R^n_1$ in our sequence of rectangles and identifying the edge $\gamma'_1$ of $R^n_1$ with the edge $\gamma_2$ of $S_2$, identifying $\gamma'_1(t)$ to $\gamma_2(t-1)$, for each $t$. We then identify the edge $\gamma'_2$ of $S_2$ with the edge $\gamma_3$ of $R_3$, again identifying $\gamma'_2(t)$ to $\gamma_3(t-1)$, and so on, finally identifying the edge $\gamma'_{2m}$ of $S_{2m}$ with $\gamma_1$ of $R^n_1$.

Intuitively, it seems to me, as $n$ increases one should expect the sections of the annuli made up of rectangles to predominate over those sections made up of generic quadrilaterals. See the figure below, where the red shaded regions represent images of the rectangles. Thus for $n$ large one would expect the extremal length around the annuli $A_n$ to closely approximate the sum $(k_1 + k_3 + \ldots + k_{2m-1})n$. Could someone confirm that this is indeed the case, and if it is the case suggest a reference or some suggestions as to how one might go about proving it?

Functions holomorphic on a region minus a Cantor set - pt.2: Iterated function systems.

2012-07-25T21:17:50Z

This post is a follow up to my previous question enquiring whether it is always possible to extend a homeomorphism conformal on a region $R$ minus a Cantor set to the whole of $R$. From the answers I received it is clear that one can't say in general whether this is possible for some function $f$ without knowing further properties of either $f$ or the Cantor set. Thus I'd like to expand upon the particular cases I'm interested in, namely iterated function systems with Cantor set limit set, and ask whether the question has been investigated in these cases. I'm mainly interested in the cases where one has only two generators, so will describe these in more detail below, but am also interested in the cases where one has an arbitrary finite or infinite number of generators.

Let $R$ be a simply connected open region of $\mathbb{C}$ (if need be we can specify that the boundary $\partial R$ of $R$ is piecewise smooth). Let $R_0, R_1 \subset R$ and let $f_0$ and $f_1$ be homeomorphisms mapping $R$ to $R_0$ and $R_1$ respectively (again, if need be we can specify that the boundaries $\partial R_0$ and $\partial R_1$ are piecewise smooth). Letting $\overline{R}_0$ and $\overline{R}_1$ be the closures of $R_0$ and $R_1$ respectively we stipulate that $\partial R_0$ and $\partial R_1$ are disjoint, and thus $R \setminus (\overline{R}_0 \cup \overline{R}_1)$ is simply connected. Furthermore, we make the assumtion that each of $\partial R_0$ and $\partial R_1$ intersects $\partial R$ at at most one point. This gives us three possibilities, examples of which are shown below.

Let $f_0$ and $f_1$ be homeomorphisms, mapping $R$ onto $R_0$ and $R_1$ respectively. If $f_0$ and $f_1$ are holomorphic I think I'm correct in saying that the limit set $\Lambda$ (a.k.a. the accumulation set) will always be a Cantor set.

Now, assume that we only know that $f_0$ and $f_1$ are holomorphic on $R \setminus \Lambda$, but that we can show that $\Lambda$ is a Cantor set (we can always show this when the boundaries of $R$, $R_0$, and $R_1$ are all pairwise disjoint, and perhaps it is also true for the other two types?). If need be we can also assume that any point of the pairwise intersection of $\partial R$ and $\partial R_0$ or $\partial R_1$ is a fixed point of either $f_0$ or $f_1$ respectively. Can we conclude that they are holomorphic on the whole of $R$?

Adding segments to an annulus - a question regarding the conformal modulus.

2012-07-10T19:45:26Z

Let $A \subset \mathbb{C}$ be an topological annulus, i.e. a region of $\mathbb{C}$ bounded by two disjoint Jordan curves.

Let $B \subset \mathbb{C}$ be a quadrilateral, i.e. a topological disc with four distinct marked points $(z_1,z_2,z_3,z_4)$ arranged anticlockwise on the boundary.

Both annuli and quadrilaterals, as defined above, have a conformal invariant, in both cases known as the modulus. Let $A_R$ be a geometrical annulus with inner boundary a circle of radius $1$ and outer boundary a circle of radius $R$. If $A$ can be mapped conformally and bijectively onto $A_R$ we say that $A$ has modulus $\ln{R}/2\pi$. Likewise, let $Q_m$ be a geometrical rectangle with vertices $(0, m, i+m, i)$, we say that $B$ has modulus $m$ if there exists a conformal bijection mapping $B$ onto $Q_m$ with $z_1$,$z_2$,$z_3$, and $z_4$ being mapped onto $0$, $m$, $i+m$, and $i$ respectively. The two concepts are linked by the fact that if we take a geometric annulus centred on the origin with modulus $m$, remove all the points on the positive real line from the annulus, and then take the preimage under the exponential map, we obtain (infinitely many copies of) a geometric rectangle also with modulus $m$.

My questions relate to the degree in which the choice of branch cut is arbitrary.

More particularly:

Let $C \subset \mathbb{C}$ be another annulus and represent $C$ as the union $R_1 \cup R_2$ of two simply connected regions with the intersection $R_1 \cap R_2$ consisting of two smooth simple curves $\gamma_1$ and $\gamma_2$ , disjoint from each other and each of which has one end point, $x_1$ and $x_2$ respectively, on the inner boundary and one end point, $y_1$ and $y_2$ on the outer boundary. Assume we have a continuous function $f_1$ mapping $R_1$ onto $A$, and such that the restriction of $f_1$ to the interior of $R_1$ is a conformal bijection and such that the image of $\gamma_1$ and $\gamma_2$ is a single simple curve from the inner to the outer boundary of $A$, with $f_1(x_1) = f_1(x_2)$ and $f_1(y_1) = f_1(y_2)$. Similarly assume that $f_2$ is a bijection mapping $R_2$ onto $B$, conformal on the interior, with $\gamma_1$ being mapped onto the component of the boundary joining $z_1$ and $z_2$ and $\gamma_2$ being mapped to the component of the boundary joining $z_3$ and $z_4$ . See the diagram below.

My first question is whether the modulus of $C$ is determined entirely by the modulus of $A$ and $B$? (This seems unlikely to me, but no harm in asking)

Secondly, if this is not the case can we give any bound for the modulus of $C$ given the modulus of $A$ and $B$.

Thirdly, are there any conditions we can impose to reduce these bounds? For example specifying that the curves $\gamma_1$ and $\gamma_2$ meet the boundary curves of $C$ orthogonally.

Embedding a Riemann surface in the sphere

2012-05-12T18:23:15Z

Assume we have a Riemann surface, the underlying topological surface of which is a sphere with (possibly uncountably many) points removed. Can we always conformally embed this Riemann surface in the Riemann sphere? If not, can someone suggest a counter example?

Thurston's definition of an orbifold

2012-03-22T21:34:14Z

I'm currently trying to understand the definition of an orbifold as expressed in Thurston's Geometry and topology of three manifolds (The definition is in chapter 13 p300). I'm confused about the following:

Assume we have a surface $X$, the cover of which includes two open sets $U_a$ and $U_b$ with $U_a \subset U_b$. To $U_a$ we associate the set $\tilde{U}_a \subset \mathbb{R}^2$ and let $\Gamma_a$ be an order three rotation acting on $\tilde{U}_a$. Likewise, to $U_b$ we associate the set $\tilde{U}_b$ and let $\Gamma_b$ be an order six rotation acting on $\tilde{U}_b$. Assume also that the fixed point of $\Gamma_a$ and $\Gamma_b$ are associated with the same point $x$ in $X$. It seems to me that the these charts are consistent with the conditions laid out in the definition, but I'm sure this can't be the case as it would lead to an ambiguity as to whether we have an order three or order six cone point at this point. Can anyone tell me where I'm going wrong?

Comment by uncooltoby

2012-11-24T18:16:35Z

Hi Lasse, my understanding is that absolute area zero is not sufficient in general, and the set needs to be of Hausdorff dimension less than 1 to guarantee holomorphic removability. Is this not the case? If absolute area zero is sufficient then this is good news as I'm pretty sure that the other two examples are also of absolute area zero, due to the 'parabolic' points of the limit set being countable.

Comment by uncooltoby

2012-08-13T01:27:42Z

One extra comment though, your definition of extremal length above is wrong. It should include 'sup inf' instead of 'sup sup', perhaps you or someone else with the required privileges would like to change it?

Comment by uncooltoby

2012-08-13T01:23:27Z

Cheers Alex, this might have just been a 'restatement of the obvious proof strategy', but being new to this sort of thing I hadn't in fact considered determining bounds on the extremal length across as opposed to around the annuli. Doing this with the obvious flat metric on the rectangular regions , as you say, gives the required lower bounds. The upper bounds follow from the standard additive property of extremal length - namely that the extremal length around the annuli is always greater than or equal to the sum of the extremal length of its constituent parts.

Comment by uncooltoby

2012-08-03T23:58:54Z

You're right that my definition is a bit non-standard. However, we need this extra condition so that the extremal length of two glued together quadrilaterals is well determined.

Comment by uncooltoby

2012-08-03T22:42:42Z

Hi Lee. You're quite right on the first point, I'll edit that straight away. On the second point, I only mean to imply by the use of "generic" that the parameterisation of the arc does not necessarily induce the linear parameterisation needed to call it a rectangle, as defined above. For a given $k$ the parameterisation of $\gamma$ and $\gamma'$ induces an equivalence relation between quadrilaterals. I'll think about how to express this more clearly in the question.

Comment by uncooltoby

2012-05-16T18:45:44Z

Yes, I believe that in the cases I'm interested in one can say that $f$ is bounded.

Comment by uncooltoby

2012-05-12T19:40:41Z

I think the problem with this is that you are assuming the sphere had a complex structure before the points were removed, which might not be the case. For example assume we take a topological sphere $X$ and remove one point $x$. Then topologically $D = X \setminus \{ x \}$is a disc and we can give it the conformal structure of a disc. However there is no homeomorphism from $X$ to the Riemann sphere which embeds $D$ conformally. This is because the image of $D$ must be the whole of the Riemann sphere minus one point, and thus be conformally equivalent to the plane by the uniformization theorem.

Comment by uncooltoby

2012-03-23T19:57:54Z

Kevin, I think you're right. The problem was that I was ignoring the fact that things have to commute with respect to the inclusion map from $U_a$ to $U_b$ as well. If this is the case then my example no longer satisfies the conditions, as the composite map $\phi_j^{-1} \circ f_{ij} \circ \phi_{ij} \circ \phi_i$ now maps $U_i$ two to one into $U_j$.