User jim belk - MathOverflow

How do we recognize a Markov partition?

2013-04-28T20:51:48Z

I'm looking for theorems that can be used to show that a topological partition for a given expanding map is Markov. Here are the relevant definitions:

Let $\phi\colon\mathbb{R}^m\to\mathbb{R}^m$ be a $C^1$ map, let $J\subseteq\mathbb{R}^m$ be a compact set, and suppose that $\phi(J) = J$.
We say that $\phi$ is expanding on $J$ if there exists an $n\in\mathbb{N}$ so that $\|D[\phi^n]_x(v)\|>\|v\|$ for all $x\in J$ and all nonzero $v\in\mathbb{R}^n$.
A topological partition of $J$ is a finite collection $U_1,\ldots,U_k$ of (relatively) open, disjoint subsets of $J$ whose closures cover $J$.
A topological partition $U_1,\ldots,U_k$ of $J$ is called a Markov partition if it satisfies the following conditions:
1. For each $i$ and $j$, either $\phi(U_i)\cap U_j = \emptyset$ or $U_j\subseteq \phi(U_i)$
2. For every sequence $i_0,i_1,i_2,\ldots$, the intersection $\bigcap_{k=0}^\infty \phi^{-k}\bigl(\,\overline{U_{i_k}}\,\bigr)$ contains at most one point.
(These conditions guarantee that $\phi$ is semi-conjugate to a subshift of finite type.)

My question is, how can we tell that a given topological partition $U_1,\ldots,U_k$ is Markov? For example, suppose that a partition satisfies condition (1) above for a Markov partition, and that $\phi$ is expanding on $J$ and one-to-one on each $\overline{U_i}$. Does it follow that $U_1,\ldots,U_k$ satisfies condition (2) above? If not, what extra hypotheses are required?

If it helps, the $J$ that I am interested in is the connected Julia set for a hyperbolic rational map on the Riemann sphere, and each $U_i$ is connected. I have a specific Markov partition that I want to prove is Markov, and I would prefer to simply cite some theorem.

Answer by Jim Belk for Actions of Thompson group F

2013-05-05T16:25:48Z

Well, every action of $F$ corresponds to a subgroup $H\leq F$ in the standard way. Specifically, the "standard" action on the interval corresponds to the stabilizers of various points in the interval. The Schreier graphs of these actions have been studied here.

If you want other actions of Thompson's group $F$, you need to look at subgroups that aren't just the stabilizer of a point. If you want to avoid a binary tree in the Schreier graph, you want this subgroup to be "large" enough so that every conjugate intersects the $\langle x_0,x_1\rangle$ monoid. However, if you want the action to be faithful, the subgroup cannot be so large that it contains the commutator subgroup.

Of course, it's not clear how to make this concept of "large" precise. I suppose one possible definition is "does not stabilize any point in $(0,1)$".

With that in mind, here are some relatively "large" subgroups of Thompson's group $F$. I have no idea whether their Schreier graphs contain binary trees.

Given any subset of the dyadics (or indeed any subset of the interval), one could consider the consider certain Cantor sets inside the interval, such as the set of points whose binary expansion subgroup of elements of $F$ that stabilize that subset. One interesting subset to look at might be $\{\ldots,\frac{1}{16},\frac18,\frac14,\frac12,\frac34,\frac78,\frac{15}{16},\ldots\}$. Another might be the set of dyadics of the form $k/4^n$, where $k$ is an odd integer. Finally, one can consider certain Cantor sets inside of $[0,1]$, such as the set of points whose binary expansion has a "$0$" in every odd-numbered position.
There is a copy of $F_3$ inside of $F$, which can be described as follows:
- Start by labeling the interval $[0,1]$ with the letter $A$.
- Now, whenever you subdivide an $A$ interval, label the left half $A$ and the right half $B$
- Finally, whenever you subdivide a $B$ interval, label both halves $A$.
- Let $H$ be the subgroup of elements of $F$ that map linearly between the intervals of two dyadic subdivisions in a label-preserving way. Then $H$ is isomorphic to $F_3$.
This construction can be generalized to give a copy of $F_n$ for any $n$.
More generally, it is possible to find copies of many different diagram groups inside of $F$ by using different labelings. The copy of $F_3$ above corresponds to the diagram group for the monoid presentation $\langle A,B \mid A=AB, B=A^2\rangle$. See this paper by Guba and Sapir for some further examples.
Another source of "large" subgroups comes from various laminations of the unit disk. Given any lamination, one can consider the subgroup consisting of all elements of $F$ that preserve the lamination. These groups have not been studied at all, though in a recent preprint Bradley Forrest and I have considered an analogous subgroup of Thompson's group $T$.

Can you tell whether a space is Banach from the unit ball?

2011-02-28T17:35:35Z

Let $V$ be a real vector space. It is well known that a subset $B\subset V$ is the unit ball for some norm on $V$ if and only if $B$ satisfies the following conditions:

$B$ is convex, i.e. if $v,w\in B$ and $\lambda\in[0,1]$ then $\lambda v+(1-\lambda)w \in B$.
$B$ is balanced, i.e. $\lambda B \subset B$ for all $\lambda \in [-1,1]$.
$\displaystyle\bigcup_{\lambda > 0} \lambda B = V$ and $\displaystyle\bigcap_{\lambda>0} \lambda B = \{0\}$.

My question is: is there some simple way to determine from $B$ whether the resulting norm on $V$ will be complete? Keep in mind that $V$ does not yet have a topology.

Edit: I guess the word "simple" is a bit misleading. What I'm looking for is some geometric insight into how the shape of $B$ affects whether the result is a Banach space. When $V$ is finite dimensional, all sets $B$ satisfying conditions (1) - (3) give equivalent norms, so all $B$'s are somehow roughly the same shape. In what way do the shapes vary when $V$ is infinite-dimensional, and how does this affect the completeness of the resulting norm?

Answer by Jim Belk for Approximating fractal curves

2012-06-20T19:19:00Z

I'm certainly not an expert on this subject, but I do have a few naive suggestions.

The function $f(x,y) = \sum_{n,m=-N}^N t_{mn}e^{inx+imy}$ is just a finite Fourier sum defined on a square. It is well-known how to find a finite Fourier sum that "best" approximates a given continuous function on a square.

Therefore, one possible approach would be the following:

Choose a continuous function $F(x,y)$ on the square whose zero set is the given fractal.
Find the Fourier sum $f$ which best approximates the function $F$.

For step (1), one potential choice would be to use standard distance function to a compact set, i.e. $$ F(p) \;=\; \min\{ d(p,q) \mid q\in\text{the fractal set} \} $$ where $d$ denotes Euclidean distance in the plane. The problem with this choice is that, if you perturb the function $F$ slightly, the zero set might disappear entirely.

It would be better to start with a continuous function $F$ whose graph intersects the $xy$-plane transversely along the fractal curve. For something like the the Koch snowflake, you could use a continuous function $F$ which is positive outside of the snowflake and negative inside, e.g. use the distance function for the outside and the negative of the distance function on the inside.

Also, practically speaking, there's no reason why the function $F$ really needs to be continuous. For example, if you start with a piecewise function $F$ which is $1$ outside of the Koch snowflake and $-1$ inside the Koch snowflake, then the Fourier approximations for $F$ might work fairly well.

Edit: I tried this in Mathematica, and it seems to work. Here is a plot of the zero set for a Fourier series with $N=50$:

For $F$, I used a function which is $1$ outside the third iterate for the Koch snowflake, and $-1$ inside. I used Green's Theorem to compute the double integrals for the coefficients.

Answer by Jim Belk for Abstract Commensurator Group of $\mathbb{Z}^n$ $Comm(\mathbb{Z}^n)\cong GL(n,\mathbb{Q})$?

2012-05-23T15:32:46Z

First, it is easy to see that $\mathrm{Comm}(\mathbb{Z}^n)$ must be isomorphic to a subgroup of $\mathrm{GL}(n,\mathbb{Q})$. In particular, every finite-index subgroup of $\mathbb{Z}^n$ is an $n$-dimensional lattice, so any isomorphism between two such subgroups extends uniquely to an isomorphism $\mathbb{Q}^n\to\mathbb{Q}^n$. Note that two isomorphisms of $\mathbb{Q}^n$ that agree on an $n$-dimensional lattice must be equal, so it really does work to think of elements of $\mathrm{Comm}(\mathbb{Z}^n)$ as matrices.

All that remains is to show that every element of $\mathrm{GL}(n,\mathbb{Q})$ corresponds to some element of $\mathrm{Comm}(\mathbb{Z}^n)$. This is fairly easy: if $A\in\mathrm{GL}(n,\mathbb{Q})$, then there exists a positive integer $k$ so that the $n\times n$ matrix for $kA$ has integer entries. In this case, $A$ maps the finite-index subgroup $k\,\mathbb{Z}^n$ (of index $k^n$) isomorphically to another finite-index subgroup of $\mathbb{Z}^n$. The image subgroup has finite index since it spans $\mathbb{Q}^n$, and is therefore an $n$-dimensional lattice.

Edit: As Mark points out, this is essentially the same answer given in the comments above.

Answer by Jim Belk for Conformal structure does not see conical singularities

2012-02-25T04:14:17Z

Consider the following pair of surfaces:

$P\;$ is the plane with the origin removed.
$C$ is the cone $z = \sqrt{x^2+y^2}$ in $\mathbb{R}^3$, with the origin removed.

There are several possible structures we could put on these surfaces. For example, both of the surfaces support a differentiable structure. As differentiable manifolds, $P\;$ and $C$ are diffeomorphic. Thus the differentiable structure cannot "see" the conical singularity at the tip of the cone.

Each of these spaces also has a Riemannian metric. Both Riemannian metrics are flat (i.e. Euclidean), but the two surfaces are not isometric. In particular, a loop on $C$ that surrounds the origin will have a holonomy of $(2-\sqrt{2})\pi$ radians, while a loop on $P\;$ that surrounds the puncture has zero holonomy. Thus the metric can "see" the conical singularity.

You do not need the full metric structure to detect the conical singularity. For example, if we give each surface a Euclidean similarity structure, then it is still possible to define the holonomy of a closed curve, and we can detect the difference between $P\;$ and $C$.

However, it turns out that $P\;$ and $C$ are equivalent as conformal surfaces. For example, the map $P\to C$ that maps the point in the plane with polar coordinates $(r,\theta)$ to the point on the cone with cylindrical coordinates $\bigl(r^{1/\sqrt{2}},\theta,r^{1/\sqrt{2}}\bigr)$ is a conformal equivalence. Thus the conformal structures cannot "see" the conical singularity.

Answer by Jim Belk for Delaunay triangulations and convex hulls

2012-01-23T07:06:42Z

It certainly appears in this survey paper by Franz Aurenhammer (see Figure 11). The paper cites Klee, On the complexity of d-dimensional Voronoi diagrams as well as K.Q. Brown's Ph.D. thesis, Geometric transforms for fast geometric algorithms.

Answer by Jim Belk for How is the Julia set of $fg$ related to the Julia set of $gf$?

2012-01-04T05:32:47Z

I'm not sure if this is helpful, but here is an example. The following picture shows the filled Julia set for $z^6 - 1$.

and the following picture shows the filled Julia set for $(z^2-1)^3$:

This is the case where $f(z) = z^2 - 1$ and $g(z) = z^3$. Note that the bottom image is a double cover of the top, while the top image is a triple cover of the bottom.

(These images were produced using Mathematica.)

Answer by Jim Belk for Are there any "related rates" calculus problems that don't feel contrived?

2011-12-10T17:39:48Z

The skills that students are practicing in related rates problems are:

Differentiating a known equation implicitly with respect to time.
Interpreting the time derivative of a quantity as a rate of change.

The main reason that related rates problems feel so contrived is that calculus books do not want to assume that the students are familiar with any of the equations of science or economics. Every related rates problem inherently involves differentiating a known equation, and the only equations that the calculus book assumes are the equations of geometry.

Thus, you can find related rates problems involving various area and volume formulas, related rates problems involving the Pythagorean Theorem or similar triangles, related rates problems involving triangle trigonometry, and so forth. A few of these problems are compelling -- for example, computing the speed of an airplane based on ground observations of its altitude and apparent angular velocity -- but most of them do feel a bit contrived.

The reality, of course, is that students are familiar with many of the basic equations and concepts of science and economics, and there's no rule against using these in problems. For example, you can make up all sorts of compelling related rates problems by starting with any physics or chemistry equation and imagining a situation where you might want to take its derivative:

The kinetic energy of an object is $K = \frac{1}{2}mv^2$. If the object is accelerating at a rate of $9.8 \text{m}/\text{s}^2$, how fast is the kinetic energy increasing when the speed is $30 \;\text{m}/\text{s}$?
An ideal gas satisfies $PV = nRT$, where $n$ is the number of moles and $R \approx 8.314\;\; \text{J}\; \text{mol}^{-1} \text{K}^{-1}$. Give the rate at which the temperature and volume of the gas are increasing, and then ask about the rate of change in pressure when the volume and temperature reach certain amounts.
The total energy stored in a capacitor is $\frac{1}{2} Q^2 / C$, where $Q$ is the amount of charge stored in the capacitor and $C$ is the capacitance. Give the value of $C$ and the rate at which $Q$ is decreasing, and ask about the rate at which the capacitor is losing energy when the energy is a certain amount.
In astronomy, the absolute magnitude $M$ of a star is related to its luminosity $L$ by the formula $$ M \;=\; M_{\text{sun}} -\; 2.5\; \log_{10}(L/L_{\text{sun}}). $$ where $M_{\text{sun}} = 4.75$ and $L_{\text{sun}} = 3.839 \times 10^{26} \text{watts}$. (Note that, by convention, brighter stars have lower magnitude.) If the absolute magnitude of a variable star is decreasing at a rate of $0.09 / \text{week}$, how quickly is the luminosity of the star increasing when the magnitude is $3.8$?

It's easy to make these up: just think of any equation in science or economics whose derivative might be interesting. Wikipedia and/or science textbooks can be helpful for finding equations from a wide variety of fields.

Expanding Measurable Sets

2011-06-09T20:10:03Z

Let $S,T \subset \mathbb{R}^n$ be measurable sets, and suppose that there exists a measurable bijection $f\colon S\to T$ so that $$ \|f(x)-f(y)\| \;\geq\; \|x-y\| $$ for all $x,y \in S$. Does it follow that $\mu(S) \leq \mu(T)$?

Answer by Jim Belk for Orthogonal transformations fixing a subspace (setwise)

2011-06-08T18:17:22Z

I posted this answer over at Math Stack Exchange, but since I'm not too sure about it I thought I should post it here as well. Hopefully someone who knows more about this than I do can look it over and check whether it's right.

I claim that the following statements hold for any subspace $W$ of $V$ (not just totally isotropic spaces). In the totally isotropic case, $W = R$ and $W^\perp = S$, which makes the solution slightly simpler.

Let $W$ be an arbitrary subspace of $V$, and let $G \leq O(V)$ be the group of orthogonal transformations that leave $W$ invariant. Define $$ R \;=\; W \cap W^\perp \qquad\text{and}\qquad S \;=\; W + W^\perp. $$ Note that $W^\perp$ is invariant under the action of $G$, and therefore $R$ and $S$ are also invariant.

The structure of $G$ is as follows. First, there is a short exact sequence $$ 1 \;\;\to\;\; A \;\;\to\;\; G \;\;\to\;\; O(S) \;\;\to\;\; 1 $$ where $A$ is an abelian group isomorphic to $K^d$ for some value of $d$. The homomorphism $G \to O(S)$ is surjective because of Witt's theorem.

The group $O(S)$ is a semidirect product. Specifically, $$ O(S) \;\cong\; \bigl( \text{Lin}(R,W/R) \times \text{Lin}(R,W^\perp/R) \bigr) \;\rtimes\; \bigl(O(R) \times O(W/R) \times O(W^\perp/R) \bigr). $$ Here $\text{Lin}(R,W/R)$ is the additive abelian group of all linear functions $R\to W/R$, and $\text{Lin}(R,W^\perp/R)$ is similarly an additive abelian group. Since $Q$ restricts to the null quadratic form on $R$, the orthogonal group $O(R)$ is the same as $GL(R)$. Moreover, since $Q$ is null on $R$, the quotients $W/R$ and $W^\perp/R$ are quadratic spaces, and $O(W/R)$ and $O(W^\perp/R)$ are the corresponding orthogonal groups. Note also that the quadratic forms on $W/R$ and $W^\perp/R$ are nondegenerate.

Edit: Here is a bit more information on the kernel $A$ of the epimorphism $G \to O(S)$. Since $Q|_{R\times S} = 0$, the quadratic form $Q$ defines a bilinear map $B \colon R \times (V/S) \to K$, and it is not hard to show that $B$ is a perfect pairing. It follows that the action of an element $g\in G$ on $V/S$ is entirely determined by the action of $g$ on $R$. In particular, every element of $A$ acts trivially on $V/S$. Therefore, every element $g\in A$ has the form $$ g(v) \;=\; v + \varphi(\pi(v)) $$ where $\pi\colon V \to V/S$ is the quotient map, and $\varphi\colon V/S \to S$ is a linear map. Thus $A$ is isomorphic to some subgroup of the abelian group $\text{Lin}(V/S,S)$.

To be specific, $A$ is isomorphic to the group of all linear maps $\varphi\colon V/S \to S$ that satisfy the following conditions:

The range of $\varphi$ lies in $R$.
The map $\varphi$ is "antisymmetric" with respect to $B$ in the sense that $$ B(\varphi(u),v) + B(\varphi(v),u) = 0 $$ for all $u,v \in V/S$.

In particular, $A$ is isomorphic to the additive group of all $m\times m$ antisymmetric matrices over $K$, where $m = \dim(R)$.

Answer by Jim Belk for Can non-homeomorphic spaces have homeomorphic squares?

2011-05-26T02:16:22Z

Yes. Let $M$ be the Whitehead Manifold, which has the property that $M \not\cong \mathbb{R}^3$, but $M\times\mathbb{R}^3 \cong \mathbb{R}^6$. (In fact $M\times\mathbb{R} \cong \mathbb{R}^4$.) Let $$ X \;=\; \mathbb{R}^3 \:\uplus\: M \:\uplus\: M \:\uplus\: M \:\uplus\: M \:\uplus\: \cdots $$ and $$ Y \;=\; \mathbb{R}^3 \:\uplus\: \mathbb{R}^3 \:\uplus\: M \:\uplus\: M \:\uplus\: M \:\uplus\: \cdots\text{,} $$ where $\uplus$ denotes the disjoint union of topological spaces. Then $X$ and $Y$ are not homeomorphic, but $$ X^2 \;\cong\; Y^2 \;\cong\; (\mathbb{R}^6 \:\uplus\: \mathbb{R}^6 \:\uplus\: \cdots) \:\uplus\: (M^2 \:\uplus\: M^2 \:\uplus\: \cdots). $$

Answer by Jim Belk for Bases for infinitely generated free groups

2011-05-24T17:25:58Z

If I understand the question correctly, you are concerned that there might be hidden relations between the elements of $T$. Moreover, since the images of $T$ are linearly independent in the abelianization, these would have to be commutator relations.

This kind of thing can't happen: if there were a nontrivial relation involving elements of $T$, then said relation would involve only finitely many elements of $T$, so it would be a relation on the free subgroup generated by that finite set of elements. Essentially, condition (2) guarantees that the elements of any finite subset of $T$ are algebraically independent, and therefore all of the elements of $T$ are algebraically independent.

Codimension of Measurable Sets

2011-02-13T08:20:48Z

I am currently teaching an advanced undergraduate analysis class, and the following question came up.

Intuition suggests that "most" subsets of $[0,1]$ are not Lebesgue measurable. However, the power set $\mathcal{P}([0,1])$ has the same cardinality as the collection of measurable sets, so it is not clear how to make this statement precise.

One method is to view $\mathcal{P}([0,1])$ as a vector space over $\mathbb{Z}_2$, with addition corresponding to symmetric difference of sets. Then the measurable sets $\mathcal{M}$ form a subspace, and the quotient $\mathcal{P}([0,1])/\mathcal{M}$ is clearly uncountable.

So my question is: what is the cardinality of $\mathcal{P}([0,1])/\mathcal{M}$? It seems like it should be $2^c$, just like $\mathcal{P}([0,1])$, but I don't know how to prove it.

Also, in what other senses are "most" subsets of $[0,1]$ non-measurable?

Answer by Jim Belk for If a result is apparently provable with AC, is actually independent of ZF?

2011-02-05T19:44:53Z

The paper "Division by Three" by Doyle and Conway (link to PDF) gives a proof of the following result without appeal to the axiom of choice:

Let $A$ and $B$ be sets, and let $3$ denote a three-element set. If there exists a bijection from $3\times A$ to $3\times B$, then there exists a bijection from $A$ to $B$.

(This result is not due to Doyle and Conway -- it was first obtained by Lindenbaum and Tarski in 1926.)

Answer by Jim Belk for Distinguishing pro-finite completions

2010-09-25T22:21:18Z

It is known that in a topologically finitely-generated profinite group, every subgroup of finite index is open. (See this paper.)

If $G$ is a finitely-generated residually finite group, then the profinite completion $\hat{G}$ contains $G$ as a dense subgroup, and is therefore topologically finitely generated. If $F$ is a finite-index subgroup of $\hat{G}$, it follows that $F$ is open, so every coset of $F$ contains elements from $G$, and therefore $F \cap G$ is a finite-index subgroup of $G$ with the same index. This defines an isomorphism between the the lattice of finite-index subgroups of $\hat{G}$ and the lattice of finite-index subgroups of $G$.

As long as $G$ and $H$ are finitely generated, this gives a rather strong invariant that can be used to distinguish $\hat{G}$ from $\hat{H}$. Specifically, $\hat{G}$ and $\hat{H}$ can only be isomorphic if the lattices of finite-index subgroups of $G$ and $H$ are isomorphic. Moreover, the isomorphism between these lattices must preserve the permutation action on the cosets of each finitely-generated subgroup. In particular, the lattices of finite-index normal subgroups of $G$ and $H$ must also be isomorphic, in a way that preserves the isomorphism types of the finite quotient groups.

Answer by Jim Belk for Distinguishing finite-orbit permutation groups by action on tuples

2010-09-04T08:03:50Z

Here's a case where $G$ and $H$ can be conjugate. First some notation: given a sequence $\{k_n\}$ of positive integers, let $[k_1,k_2,\ldots]$ denote the permutation

$$(1,\ldots,k_1)(k_1+1,\ldots,k_1+k_2)(k_1+k_2+1,\ldots,k_1+k_2+k_3)\cdots$$

with cycles of size $k_1,k_2,k_3\ldots$. For example, $[1,1,1,1,\ldots]$ denotes the identity, $[2,2,2,2,\ldots]$ denotes $(1,2)(3,4)(5,6)(7,8)\cdots$, and $[2,3,2,3\ldots]$ denotes $(1,2)(3,4,5)(6,7)(8,9,10)\cdots$.

Let $$g = [1,2,\;\;1,2,4,\;\;1,2,4,8,\;\;\ldots],$$ let $$h = [1,1,1,\;\;1,1,1,2,2,\;\;1,1,1,2,2,4,4,\;\;\ldots],$$ and let $G$ and $H$ be the cyclic subgroups generated by these elements. Since $g$ and $h$ have the same cycle structure, they are conjuagte in $Sym(\mathbb{N})$, so $G$ and $H$ are conjugate subgroups. However, for sufficiently large $n$, the orbit of $(\pi(1),\pi(2),\ldots,\pi(n))$ under $G$ will be precisely twice the size of the orbit under $H$.

Of course, in this example $G$ and $H$ both have infinitely many orbits of size $2^k$ for every $k$, so this does not answer the more restrictive version of the question.

Answer by Jim Belk for Best tablet computer for mathematics

2010-09-02T01:06:27Z

Just about any tablet PC ought to work for this. I used a Thinkpad X61 Tablet for several years, and after it broke I bought a Thinkpad X200 Tablet. A typical Windows tablet comes with software called "Windows Journal" which works well for taking notes, and can print to PDF using PDFCreator. I also purchased a program called PDF Annotator, which lets me hand-write notes on top of any existing PDF file.

When I was first considering buying a tablet, I had trouble deciding between the slate type and the convertibles that come with a keyboard and a swivel screen. I decided on a convertible, and I'm very glad that I made this decision -- my tablet also functions perfectly well as a standard laptop, and I keep it in laptop mode about 90% of the time. A straight-up tablet with no keyboard would be almost as expensive, and would be considerably less useful.

Answer by Jim Belk for How to shuffle a deck by parts?

2010-07-20T19:42:56Z

Though David's answer settles the original question, there is no reason to restrict to the case where the total number of cards is a multiple of $n$. In general, we can ask whether it is possible to completely shuffle a deck with $M$ cards if only $n$-card subdecks are directly shuffleable.

As David points out, this is necessarily impossible if there exists a prime $p$ for which $n < p \leq M$. This means that the smallest case that's still open is $n=3$ and $M=4$. That is, is it possible to completely randomize a 4-card deck if only 2 or 3 cards may be shuffled at a time?

Answer by Jim Belk for Function with range equal to whole reals on every open set

2010-07-16T07:35:53Z

For a non-constructive solution, let $\pi : \mathbb{R} \to \mathbb{R}/\mathbb{Q}$ be the projection homomorphism, and let $g : \mathbb{R}/\mathbb{Q} \to \mathbb{R}$ be a bijection. Then the composition $g\circ \pi$ has the desired property.

Answer by Jim Belk for Counting $(n,k)$-forests of binary trees

2010-07-13T17:22:49Z

The number of $(k,n)$-binary forests is the $(n-1,n-k)$ entry of Catalan's triangle. Thus the formula is: $$ f_{k,n} \:=\: \frac{\:k\:}{n}\binom{2n-k-1}{n-1}. $$ Given this formula, you can use Stirling's approximation to obtain asymptotic estimates.

Answer by Jim Belk for The lie algebra of the orthogonal group of an arbitrary space time metric

2010-07-11T19:09:41Z

(1) There is no difficulty in diagonalizing the quadratic form $g$, regardless of its signature. However, you must be careful: either $g = B^{-1} \eta B$, where $B$ is an orthogonal matrix and $\eta$ is an arbitrary diagonal matrix of signature (3,1), or $g = B^T \eta B$, where $B$ may not be orthogonal, but $\eta$ is the signature-(3,1) identity matrix.

(2) I am somewhat puzzled by your assertion that $g = g^{-1}$. This will not be true for most metrics (although it might be true for the case you care about).

(3) In any case, here's how I think the calculation ought to go. From the formula $A^TgA=g$, we get $$ gA^{-1} g^{-1} = A^T, $$ which, after substituting in $g = B^T \eta B$, becomes $$ B^T \eta B A^{-1} B^{-1} \eta B^{-T} = A^T, $$ which can be rearranged as $$ \eta B A^{-1} B^{-1} \eta = B^{-T} A^T B^T, $$ or $$ \eta (B A B^{-1})^{-1} \eta = (BAB^{-1})^T. $$ As you can see, $A$ is an element of $O(g)$ if and only if $BAB^{-1}$ lies in $O(1,3)$

Answer by Jim Belk for The orthogonal group of a riemannian metric

2010-07-10T18:33:07Z

A matrix $A$ preserves this inner product if $g_{ij} \: A_k^i \: A_l^j \; = \;g_{kl}$ (similar to the $AA^T=I$ definition of orthogonal matrices). One interpretation of this equation is that the columns of $A$ must be orthonormal with respect to the given inner product. Similarly, the rows of $A$ must be orthonormal with respect to the dual inner product $g^{ij}$.
The defining equation for matrices $B$ of the Lie algebra is $g_{ik} \: B_j^k \;+\; g_{kj}\:B^k_i \;=\; 0$.
Assuming the inner product is positive definite, the Lie group and Lie algebra in question are in fact isomorphic to the standard orthogonal group and Lie algebra of antisymmetric matrices, respectively.

Is this what you're looking for? I'm not entirely sure what some parts of your question are referring to. (For example, what is "a formula that gives the parameterized components...of a euclidean metric"?)

Answer by Jim Belk for Constructing 4-manifolds with fundamental group with a given presentation.

2010-07-09T18:55:32Z

Taking the boundary of tubular neighborhood in codimension two doesn't preserve fundamental group -- the codimension must be at least three.

For a simple example, if you take the boundary of a tubular neighborhood of a single point in $R^2$, you get a circle, which has nontrivial fundamental group. Similarly, the boundary of a tubular neighborhood of a 2-plane in $R^4$ is the product of a 2-plane with a circle, which again has nontrivial fundamental group. In general, if you start with a 2-manifold in $R^4$, the boundary of a tubular neighborhood will be a circle bundle over the manifold, and will therefore not have the right fundamental group.

This problem is fixed in $R^5$, because you get a sphere bundle instead of a circle bundle, and the 2-sphere has trivial fundamental group.

Answer by Jim Belk for Two-dimensional random walk

2010-07-09T17:33:53Z

Let $a$ and $b$ be fixed points in the integer lattice, and let $f(p)$ be the probability that a random walk starting at the point $p$ will arrive at $a$ before $b$. Then for every point in the plane other than $a$ and $b$, we have, $$ f(p) = \frac{f(p+i)+f(p-i)+f(p+j)+f(p-j)}{4} $$ where $i$ and $j$ are the basis unit vectors. That is, the value of $f$ at a point is equal to the average of the values of $f$ at the neighboring points.

A function on the square lattice with this property is called harmonic, and satisfies a discrete version of Laplace's equation: $$ \Delta f = 0 $$ where $\Delta$ is the discrete Laplace operator. Unfortunately, the function $f$ is not quite harmonic, since the equation above need not hold when $p=a$ or $p=b$.

In particular, the function $f$ actually satisfies the Poisson equation $$ \Delta f(p) = C_1 \delta_a(p) + C_2 \delta_b(p), $$ where $\delta_a$ is the function which is $1$ at $a$ and zero elsewhere, $\delta_b$ is the same for $b$, and $C_1$ and $C_2$ are unknown constants.

Since Poisonn's equation is linear, it suffices to solve the equations $$ \Delta f(p) = \delta_a(p)\qquad\text{and}\qquad\Delta f(p) = \delta_b(p) $$ independently, and then take an appropriate linear combination of the solutions. Solutions to equations such as these are called lattice Green's functions. For the integer lattice, the lattice Green's functions cannot be written in a closed form, but there are definite integral formulas that can be used to compute the function to arbitrary precision (see here).

Once you know the values of the lattice Green's functions, you ought to be able to solve for the constants $C_1$ and $C_2$ by using the boundary conditions $f(a) = 1$ and $f(b) = 0$.

Answer by Jim Belk for Is Thompson's Group F amenable?

2010-06-02T17:28:33Z

My understanding is that the situation has not changed much. A group of mathematicians at Binghamton University had been investigating Shavgulidze's argument, and they found a flaw which Shavgulidze has not addressed. As of now they are still waiting for Shavgulidze to respond.

I haven't heard anything about the Akhmedov paper recently.

Comment by Jim Belk

2013-05-06T16:36:24Z

@Kate: Using Yves's argument, it follows that all of the actions above are faithful. I have no idea whether the Schreier graphs contain binary trees, although if I had to guess, I would guess the Schreier graph for every faithful action of $F$ contains a binary tree.

Comment by Jim Belk

2012-06-04T03:31:44Z

One point which may be relevant is that the fundamental group of the complement of the Alexander horned ball is not finitely generated. Therefore, the complement of an Alexander horned ball is not homotopy equivalent to a finite CW complex.

Comment by Jim Belk

2012-06-04T03:26:37Z

I'm not entirely sure I understand your first question. What exactly does it mean for the closure of a cell to be a ball "with the Alexander horned sphere as its boundary"? The Alexander horned sphere is topologically a sphere -- it's only the embedding of the sphere in $\mathbb{R}^3$ that makes it special. Is the given CW complex embedded in $\mathbb{R}^3$?

Comment by Jim Belk

2012-05-24T18:35:42Z

@Mark: Oh, I hadn't noticed!

Comment by Jim Belk

2012-01-03T18:10:47Z

What kinds of properties are you interested in? As you point out, each Julia set is a branched cover of the other, so the local structures will be very similar, but the global structures may be very different.

Comment by Jim Belk

2011-06-09T22:33:51Z

Oh, great. Thanks!

Comment by Jim Belk

2011-03-03T04:44:44Z

That's fair. I'll accept this answer and move on. Thanks everybody!

Comment by Jim Belk

2011-03-03T04:38:07Z

Cool, so Bill Johnson's answer solves the case I asked about, and makes me understand that completeness is a much more general concept than I had realized. Thanks!

Comment by Jim Belk

2011-03-03T04:20:02Z

Ah, I see -- every topological vector space has an obvious uniform structure.

Comment by Jim Belk

2011-03-02T21:14:36Z

It is very reasonable to argue that the restrictive nature of the question is wrong. Indeed, the point of view that I'm taking is very different from the usual perspective on such things. This is intentional -- I'm curious what can be said about completeness in this context, without any of the usual tools available. It's certainly possible that nothing very interesting can be said, in which case the question is not successful.

Comment by Jim Belk

2011-03-02T20:58:58Z

I'm a bit confused by this -- $\mathbb{R}^\Omega$ is not metrizable in the product topology, so I don't see how it can be complete.

Comment by Jim Belk

2011-03-02T16:05:04Z

I'm picturing $V$ as something like $\mathbb{R}^\Omega$, where $\Omega$ is an uncountable set. It doesn't come with a topology, other than perhaps the infinite product topology. I find it very puzzling that there can be something so different about different convex sets in a vector space like this, and I'm wondering if there's any way to understand this difference from an external point of view. For example, suppose you are given a set $S\subset\mathbb{R}^\Omega$, and you let $B$ be the convex hull of $S\cup-S$. Under what conditions on $S$ will the norm induced by $B$ be complete?

Comment by Jim Belk

2011-03-02T04:23:55Z

The total convexity condition is certainly interesting, as is the nested intersection condition, though neither of these seems that easy to check. These are good conditions because they manage to avoid discussing the topology induced by $B$, which is something relatively complicated. I guess my question is whether someone living in $V$ can "see" whether a given convex set will lead to a complete norm.

Comment by Jim Belk

2011-02-28T23:11:21Z

These observations are all of course correct, and have made me realize that "simple" isn't quite what I meant. I have edited the question to clarify what I'm looking for.

Comment by Jim Belk

2011-02-13T18:11:57Z

It seems to be right now. Since each $S_J$ has outer measure $1$, and $[0,1] - S_J = S_{I-J}$, every $S_J$ is non-measurable as long as $J\ne\emptyset$ and $J\ne I$. Since $S_J + S_K = S_{J+K}$, it follows that $S_J$ and $S_K$ are distinct modulo $\mathcal{M}$ as long as $J\ne K$ and $J \ne I-K$.