User john mangual - MathOverflow

Symmetric sums and Representations of SO(3)

2013-05-23T16:47:43Z

I had tried to help someone on math.StackExchange to prove the identity:

$$ (1-Tr(A))^2+\sum_{1\le i\le j\le 3}(a_{ij}-a_{ji})^2=4$$

I guess you could argue the left hand side is independent of basis. Then we can diagonalize. But I couldn't come up with an invariant way of expressing the 2nd term.

Someone came up with $(1-\mathrm{Tr}\,A)^2- \frac{1}{2}\mathrm{Tr}\left[(A-A^T)^2\right]=4$ and prove it.

This look a bit Pythagoras' theorem since $1 - Tr A = 2 \cos \varphi$ so the other term must be $\sum_{1\le i\le j\le 3}(a_{ij}-a_{ji})^2 = 2 \sin \varphi$ which I also couldn't prove.

Is this sum related to a direct sum of representations of SO(3)?

sequences of non-crossing matchings by mutation

2013-03-22T14:33:09Z

In Polynomials, meanders, and paths in the lattice of noncrossing partitions, they talk about sequences of non-crossing matchings related by "flips".

Savitt counts "maximal chains of 2-divisible noncrossing partitions to the set of necklaces" and there seems to be some relation to parking functions.

What do these bijections look like?

How to get 3-manifold, Knots from Number Fields

2013-05-14T21:19:18Z

I'm reading a paper On the Torsion Jacquet-Langlands correspondence by Akshay Venkatesh and Frank Calegari.

Truthfully speaking I have no idea what Jacquet-Landlands is. I'm just trying to understand why there are knots in a paper on algebraic number theory and some of the players involved in that paper.

To keep matters simple, how do we pass between number fields and 3-manifolds and why is this beneficial? Poking around this paper, I find a version of the congruence groups: $$ \Gamma_0(\mathfrak{n}) = \left\{ \left( \begin{array}{cc} a & b \\ c & d\end{array} \right) : \mathfrak{n}\big|\, c\right\} \subset PGL_2(\mathbf{O}_F)$$

where $\mathbf{O}_F$ is an order of a number field. We get a 3-manifold by quotienting hyperbolic 3-space: $\mathbb{H}^3/\Gamma_0(\mathfrak{n})$. Apparently, there's also another similar way to do it with quaternions.

They then proceed to look at look at some group-cohomology invariants of the group and then they use some spectral theory and the rest of paper mostly goes over my head. Well... we do get this:

However, along the way, we took many detours to explore related phenomena, some of which was inspired by the data computed for the ﬁrst author by Nathan Dunﬁeld. In view of the almost complete lack of rigorous understanding of torsion for (nonHermitian) locally symmetric spaces, we have included many of these results, even when what we can prove is rather modest.

The take-home message seems to be that we have constructed a large collection of infinite groups and actions on low-dimensional spaces of interest number theorists. Arithmetic lattices look like they play an important role.

I guess I'm trying to understand better how this connection between knots and number fields works and how Calegari and Venkatesh are using it to get a handle of the many invariants (which I may save for later questions).

Hilbert Matrix and Approximation Theory

2013-03-10T16:53:14Z

I was reading about the Hilbert matrix and Cauchy determinants:

\[ \det \left[ \frac{1}{i+j-1} \right]_{i,j} \]

By guessing where this determinant is 0 or ∞ we can guess the right formula. In Wikipedia, I found this problem:

"Assume that I = [a, b] is a real interval. Is it then possible to find a non-zero polynomial P with integral coefficients, such that the integral $\int_a^b P(x)^2\, dx$ is smaller than any given bound ε > 0, taken arbitrarily small?"

To answer this question, Hilbert derives an exact formula for the determinant of the Hilbert matrices and investigates their asymptotics. He concludes that the answer to his question is positive if the length b − a of the interval is smaller than 4.

I'm asking for a reference / proof to this exercise. I think you can expand $P$ in Legendre polynomials, or use the Gram determinant.

In general, why is this matrix related to approximation theory?

Analysis of Misere Nim?

2011-08-01T15:15:51Z

My friend likes to impress people by playing 3-5-7 which has three piles of counters of sizes 3, 5 and 7. You can remove any number of coins from a single pile, the last player to move loses.

ooo
ooooo
ooooooo

This is a winning position for the first player, but With a solid understanding of the game tree she wins nearly every time playing second. She says, it reduces to knowing a few winning positions.

Two piles of the same size is second player win, in the jargon of Combinatorial Game Theory. Here is the pile (5,5).

ooooo
ooooo

If first player moves to (5,n) for n > 1, second player can imitate on the other pile, moving to (n,n). However, if first player moves to (5,1), second player moves to 1 and wins.

The other winning positions she remembers is (3,4,1) and (4,5,1). She can win once she recognizes these positions. Eventually (after losing many times) I told her that (n, n+1,1) is a losing position for any n...

If our game were played in normal play convention (player moving last wins), but real life Nim is played as a misere game. Probably the analysis is similar to normal-play Nim with some modification.

Recently there was a theory of Misere quotients where each game has a commutative monoid assoicated with it. What does that monoid looks like here? Is it finitely generated?

Examples of Amenable Groups other than Z_n

2011-04-19T17:24:56Z

I'm reading about amenable groups. What are explicit examples of nonabelian discrete amenable groups other than finite groups? Perhaps a group presentation or matrix representation would be useful.

variant of Haar measure

2013-03-25T20:57:14Z

I found a certain trig identity in a discussion on Lie groups

\[ \frac{\prod_{i < j} 2 \sin (\mu_i - \mu_j) \prod_{i< j} 2 \sin (\nu_i - \nu_j) }{\prod_{i, j} 2 \cos (\mu_i - \nu_j) } = \det_{i,j} \frac{1}{2 \cos (\mu_i - \nu_j)} \]

This is the Cauchy determinant formula.

\[ \det_{i,j} \frac{1}{\mu_i - \nu_j} = \frac{\prod_{i < j} \mu_i - \mu_j \prod_{i< j} \nu_i - \nu_j}{\prod_{i, j} \mu_i - \nu_j } \]

It seems to be related to the Haar measure on $U(n)$, which would have a factor:

\[ \prod_{i < j} \sin (\mu_i - \mu_j) \]

I am thrown off by the cosine factor in the denominator.

Could it be the Haar measure of a quotient on $U(n) \times U(n)$?

Answer by John Mangual for Hilbert Matrix and Approximation Theory

2013-03-14T14:54:08Z

In the generically titled, Ein Beitrag zur Theorie des Legendre'schen Polynoms Hilbert says integral $\int_a^b P(x)^2 \, dx $ defines quaratic form over the space of polynomials of degree $\leq n$.

Over $[0,1]$ the determinant in the basis $\{ 1, x, x^2, \dots, x^n \}$ is:

\[ D_{[a,b]} = \int_{[a,b]^n} x_1^{n-1}x_2^{n-2}\dots x_{n-2}^2 x_{n-1} \prod_{i < j}(x_i - x_j)^2 d\mathbf{x} = \left( \frac{b-a}{2} \right)^{n^2} D \]

Usually Legendre polynomials are defined over [0,1], but you can do a change of variables. In that basis the quadratic form is diagonalized:

\[ \int_a^b P(x)^2 \, dx = \frac{2}{2n-1}b_1^2 + \frac{2}{2n-3} b_2^2 + \dots \]

where we expand in Legendre polynomials $P(x) = b_1 p_1(x) + b_2 p_2(x) + \dots $ You are finding the determinant of this change of variables:

\[ \left| \begin{array}{cccc} 1 & \frac{1}{2} & \dots & \frac{1}{n} \\ \frac{1}{2} & \frac{1}{3} & \dots & \frac{1}{n+1} \\ \vdots & & & \vdots \\ \frac{1}{n} & \frac{1}{n+1} & \dots & \frac{1}{2n+1} \end{array}\right| = \frac{[1^{n-1}2^{n-2}\dots (n-2)^2(n-1)]^4}{1^{2n-1}2^{2n-2}\dots (2n-2)^2(2n-1)} \]

The discriminant of this quadratic form to be proportional to $ \left(\frac{b-a}{4}\right)^n$ which tends to 0 for $b-a < 4$.

Polynomials with integer coefficients form an lattice in the space of polynomials. For this discriminant to tend to 0, there must be polynomials with arbitrarily small norm.

I still find this result counterintuitive... We can find explicit examples, when $|a|,|b|<1$

\[ \int_{a}^b x^{2n} \, dx = \frac{b^{2n+1}-a^{2n+1}}{2n+1} \]

but not sure about $a=0,b=2$, for example.

Answer by John Mangual for The unreasonable effectiveness of Pade approximation

2013-02-21T15:30:01Z

Walter Van Assche gives a modern account of Pade Approximation, a variant of these were used in the proof that e is transcendental.

In the case that $f(z)$ is the Cauchy transform of a compactly supported measure $\mu(x)$,

\[ f(z) = \int \frac{1}{z-x} d\mu(x) \]

Then $P_n(x)$ is an orthogonal polynomial with respect to $\mu(x)$ while

\[ Q_{n-1}(z) = \int_a^b \frac{P_n(z) - P_n(x)}{z-x} \, \mu(x) \]

Then we approximate $f(z)$ as a rational function

\[ P_n(z) - f(z) Q_{n-1}(z) = \int_a^b \frac{P_n(x)}{z-x}\, d\mu(x) \]

Then there exists a function $r(z)$ such that the pointwise convergence of the Pade approximants is exponential as we move along the diagonal.

\[ \lim_{n \to \infty} \left| f(z) - \frac{Q_{n-1}(z)}{ P_n(z)}\right|^{1/n} = \frac{1}{r^2} \]

ORIGINAL ANSWER

The coefficients of the series $a_1, a_2, a_3, \dots$ are an infinite amount of data. The radius of convergence is a property of this sequence $1/R = \limsup |a_n|^{1/n}$ using the root test.

The Pade approximation is defined the outside the radius of convergence of the Taylor series
The Pade $[m/n]_f(x)$ and Taylor approximations $[(m+n)/0]_f(x)$ agree up to order $O(x^{m+n+1})$.

A paper by Hubert S. Wall, dating back to 1929, relates Pade approximants the Stieltjes moment problem and continued fractions.

EDIT: My guess is Pade approximant is not always better than Taylor series.

As a counter example let's find $[0/1]_{z+1}$:

\[ z + 1 \approx \frac{1}{1-z} \mod z^2 \]

Taylor series is exact and the 1-1 approximant diverges.

In your example, $\sqrt{\frac{1+\frac{z}{2} }{1+2z}} \to \frac{1}{2}$ for large values and likely the 1,1 approximant does the same, making it a good global fit. The 0,2 or 2,0 approximants will have different global behavior. I was unable to find any precise comparison, overall.

Continued fractions are known to be "best approximations" in a certain sense.

A best rational approximation to a real number x is a rational number d/n, d > 0, that is closer to x than any approximation with a smaller denominator.

Wikipedia's example is to approximate

\[ 0.84375 = \cfrac{1}{1 + \cfrac{1}{5 + \cfrac{1}{2 + \cfrac{1}{2}}}} = [0;1,5,2,2]\]

We can stop in the middle of this process to get the closes fraction given an upper bound on the denominator.

\[ 0.84375 \approx 1,\frac{5}{6}, \frac{ 11}{13 } , \frac{27}{32}\]

The farey fractions up to denominator 6 are

$$0,\frac{1}{6}, \frac{1}{5},\frac{1}{4},\frac{1}{3},\frac{2}{5},\frac{1}{2},\frac{3}{5},\frac{2}{3},\frac{3}{4},\frac{4}{5},\mathbf{\frac{27}{32}},\frac{5}{6} 1$$

Pade approximants are best approximations of functions and can be calculated used a kind of continued fraction.

You get an approximation $\frac{p(x)}{q(x)}$ for given degrees $m = \deg p, n = \deg q$. You are trying to find a polynomial greatest common divisor between your Taylor series and a monomial,

\[ \gcd(T_{m+n}(x), x^{m+n+1} ) \]

You can do this Euclid algorithm doing polynomial long division and taking the remainder at each step:

\[ \frac{p(x)}{q(x)} \equiv T_{m+n}(x) \mod x^{m+n+1} \]

Here is the table for $e^z$ from Wikipedia: (also http://mathoverflow.net/questions/41226/pade-approximant-to-exponential-function)

\[ \begin{array}{c||c|c|c} & 0 & 1 & 2 \\ \hline \hline 0 & \frac{1}{1} & \frac{1}{1-z} & \frac{1 }{1 - z + \frac{1}{2}z^2 } \\ \hline 1 & \frac{1+z}{1} & \frac{1+ \frac{1}{2} z}{ 1- \frac{1}{2} z} & \frac{1 + \frac{1}{3}z }{ 1 - \frac{2}{3}z + \frac{1}{6}z^2 } \\ \hline 2 & \frac{1 - z + \frac{1}{2}z^2 }{1 } & \frac{ 1 - \frac{2}{3}z + \frac{1}{6}z^2 }{1 + \frac{1}{3}z } & \frac{ 1+ \frac{1}{2} z + \frac{1}{12} z^2}{ 1- \frac{1}{2} z + \frac{1}{12} z^2 } \end{array} \]

Let's try to work out the steps for m=2, n=2 (not in Wikipedia). This involves the GCD of the 4th Taylor polynomial $1 + z + \frac{1}{2}z^2 + \frac{1}{6}z^3 + \frac{1}{24}z^4$ and $z^{5}$.

\[ \frac{1}{24} z^5 = (z-4)\left(1 + z + \frac{1}{2}z^2 + \frac{1}{6}z^3 + \frac{1}{24}z^4\right) + \left(4 + 3z + z^2 + \frac{1}{6}z^3 \right) \]

\[ 1 + z + \frac{1}{2}z^2 + \frac{1}{6}z^3 + \frac{1}{24}z^4 = \left( \frac{1}{4}z - \frac{1}{2} \right)\left( 4 + 3z + z^2 + \frac{1}{6}z^3 \right) + \left( 3 + \frac{3}{2}z + \frac{1}{4}z^2 \right) \]

\[ 4 + 3z + z^2 + \frac{1}{6}z^3 = \frac{2}{3}z \left( 3 + \frac{3}{2}z + \frac{1}{4}z^2 \right) + (z+4) \]

\[ 3 + \frac{3}{2}z + \frac{1}{4}z^2 = \left( \frac{1}{4}z + \frac{1}{2}\right)(z+4) + 1\]

Using the 1st two long divisions, we get the (2,2) Pade approximant.

\[ 1 + z + \frac{1}{2}z^2 + \frac{1}{6}z^3 + \frac{1}{24}z^4 \approx \frac{ 3 + \frac{3}{2}z + \frac{1}{4}z^2 }{(z-4) \left( \frac{1}{4}z - \frac{1}{2} \right)+1 } = \frac{ 1+ \frac{1}{2} z + \frac{1}{12} z^2}{ 1- \frac{1}{2} z + \frac{1}{12} z^2 } \]

Alternatively compare coefficients of your rational approximation and polynomial $a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 \approx \frac{p_0 + p_1 x + p_2 x^2}{q_0 + q_1 x + q_2 x^2 } $ Then you can solve the system of equations: \begin{eqnarray*} a_0 &=& p_0 \\ a_1 + a_0 q_1 &=& p_1 \\ a_2 + a_1 q_1 + a_0 q_2 &=& p_2 \\ a_3 + a_2 q_1 + a_1 q_0 &=& 0 \\ a_4 + a_3 q_1 + a_2 q_0 &=& 0 \end{eqnarray*}

Cramer rule gives you the correct fraction at the end:

\[ \frac{ \left|\begin{array}{ccc} a_1 & a_2 & a_3 \\ a_2 & a_3 & a_4 \\ a_0 x^2 & a_0 x + a_1 x^2 & a_0 + a_1 x + a_2 x^2 \end{array} \right|} { \left|\begin{array}{ccc} a_1 & a_2 & a_3 \\ a_2 & a_3 & a_4 \\ x^2 & x & 1 \end{array} \right|} = \frac{ \left|\begin{array}{ccc} 1 & 1/2 & 1/6 \\ 1/2 & 1/6 & 1/24 \\ x^2 & x + x^2 & 1 + x + \frac{1}{2} x^2 \end{array} \right|} { \left|\begin{array}{ccc} 1 & 1/2 & 1/6 \\ 1/2 & 1/6 & 1/24 \\ x^2 & x & 1 \end{array} \right|}\]

which integers take the form x^2 + xy + y^2 ?

2011-10-17T17:03:13Z

I guess one way of putting it, when does the series $\sum_{x,y \in \mathbb{Z}} q^{x^2+xy+y^2}$ have nonzero coefficients?

The analogous answer for $\sum_{x,y \in \mathbb{Z}} q^{x^2+y^2}$ is that $q^n$ appears when $\mathrm{ord}_p(n)$ be even for all primes.

Is there a closed form for either of these with quadratic characters or theta functions or something?n

A q,t-extension of Plancherel Measure thru Yang-Mills Theory ?

2012-10-07T12:06:49Z

Buried in the physics paper by Nekrasov and Okounkov, a strange identity is proven: $$ \prod_{n > 0} (1 - q^n)^{\mu^2-1} = \sum_{\mathbf{k}} q^{|\mathbf{k}|} \prod_{\square \in k} \left( 1 - \frac{\mu^2}{h(\square)^2}\right) $$ where the left side is a q-series and the right side is the sum over all partitions. Ihis was proven by physical considerations, evaluating the Yang-Mills partition function in 2 different ways.

The partitions could index representations of the permutation group $S_n$. We can define measure on partitions, $\mathrm{Irr}(S_n)$ by

$$ \mathbb{P}_{\mu, t} (\mathbf k) = \prod_{n \geq 1} (1-t^n)^{1-\mu^2} q^{|\mathbf{k}|} \prod_{\square \in k} \left( 1 - \frac{\mu^2}{h(\square)^2}\right) $$

In fact, 3 years later Alexei Borodin explains this formula interpolates between uniform and Plancherel measures on partitions.

Can this be extended to a q,t-deformation of uniform measure on the permutation group? Maybe through something similar to Robinson-Schensted correspondence.

Answer by John Mangual for A q,t-extension of Plancherel Measure thru Yang-Mills Theory ?

2013-01-04T14:10:52Z

Recently there is An insertion algorithm associated with q-Whittaker functions

defining a q-weighted form of the Robinson-Schensted algorithm.

The MacDonald processes contain many examples of "integrable" combinatorial processes on Gelfand-Tsetlin patterns (see this talk).

Many of these feel nature from the point of view of representation theory of symmetric groups. Macdonald functions are a q,t-deformation of this.

spectrum of a polygon and zeta function

2012-12-30T21:55:52Z

Let $\Delta(x,y) = 1,0$ according to whether $(x,y)$ is in some polygon (symmetric with respect to the diagonal axis).

E.g. The convex hull of three points (taken from a paper on dominoes)

\[ \Delta = \bigg\{ r(-1,-1)+s(1,0)+t(0,1): r + s + t = 1\bigg\} \]

So I am looking for eigenvalues and eigenfunctions of this Kernel:

$$ f \mapsto \int_{-1}^1 \Delta(x,\;\cdot\;) f(x)\;dx$$

This occurred to me while reading this paper by Noam Elkies on zeta functions, where he studies the spectrum of the convex hull:

$$ \bigg\{ r(0,0)+s(1,0)+t(0,1): r + s + t = 1\bigg\} $$

In that paper he gets a zeta function by taking the trace of this kernel,

$$ \mathrm{tr}(\Delta^n) = \sum_{i \in \mathbb{Z}} \lambda_i^n $$

For Elkies, $\lambda_k = \frac{1}{4k+1}$. See also the paper by Stanley.

He then interprets the trace $\mathrm{tr}(\Delta^n)$ as the volume of a polytope, which I will address separately.

Rotations, Harmonic Oscillators, Gaussians, Ladders

2012-12-26T21:02:36Z

I am trying to understand better the quantization of the Harmonic Oscillator.

Here are three ways of thinking about the Harmonic Oscillator.

Eigenfunctions of the differential operator: $H = -\frac{d^2}{dx^2} + x^2$
Eigenfunctions of the oscillator $H = a a^\dagger+ \frac{1}{2}$
Special orbits of the $U(1)$ action on the complex plane, level sets of the moment map $H = p^2 + x^2$.

Are there any places that explain all three of these on equal footing? Items 1 and 2 have a Wick formula $$ \langle a b c d\rangle = \langle a b \rangle \langle c d\rangle + \langle a c \rangle \langle b d\rangle + \langle a d \rangle \langle bc \rangle$$ Is there an analogue in the symplectic geometry case (item 3)?

I want to understand better why this is a duality

$$ {\tt rotation,}\;e^{it}\in U(1)\leftrightarrow {\tt gaussians,}\;e^{-x^2} \leftrightarrow {\tt eigenstates, }\;|n\rangle$$

Something to that effect, mentioned in these notes. Does any rotation action get quantized this way?

This question involves rotation actions, in a different way than this other MO qustion: http://mathoverflow.net/questions/35054/characterizing-the-harmonic-oscillator-creation-and-annihilation-operators-in-a-r

EDIT Here is another MO post where the Bargmann transform arises in quantization of Harmonic Oscillator: http://mathoverflow.net/questions/94535/representation-of-double-cover-of-un-on-eigenspaces-of-harmonic-oscillator

G-bundles in classical mechanics

2011-07-15T17:30:05Z

The paper Geometry of the Prytz Planimeter described a mechanical instrument whose configuration space is an $S^1$-bundle with an $SU(1,1)$ action. That paper goes on to study the holonomies of various paths in the base space. Motion is described by a connection on this bundle.

I am interested because it's an example of a $G$-bundle appearing in classical mechanics. Are there other explicit classical mechanical systems that engineers study that exhibit the basic concepts of differential geometry so lucidly?

Answer by John Mangual for G-bundles in classical mechanics

2012-12-29T05:06:27Z

Just to record what I heard from another source, Geometric Control Theory has some promise. Many examples from classical mechanics examined from the point of view of connections on their phase space.

The text is Geometric Control of Mechanical Systems by Francesco Bullo and Andrew D. Lewis.

visualizing singularities of maps from sphere to R^2

2012-12-19T18:23:33Z

Is there a classification of singularities from $S^2 \to \mathbb{R}^2$ ? The critical points of the map $(x,y) \mapsto (f_1(x,y),f_2(x,y))$ where the matrix:

\[ \left[\begin{array}{cc}\frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y}\\ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} \end{array} \right] \]

has less than full rank. Locally, can we draw a picture of what singularities look like?

In the case of maps $S^2 \to \mathbb{R}^1$, we just get the critical points where $f(x,y) = f(x_0,y_0) + (x-x_0,y-y_0)^T (D^2 f )(x-x_0,y-y_0)$

Pairs of Permutations up to Simultaneous Conjugation

2011-10-10T05:29:51Z

The conjugacy classes of $S_n$ are the cycle types since if $\tau = (\dots)(\dots)\dots(\dots)$, the conjugation $\tau \mapsto \sigma \tau \sigma^{-1}$ permutes the labels in the cycles of $\tau$.

Has anyone studied pairs of permutations up to simultaneous conjugation $(\tau_1,\tau_2) \mapsto (\sigma \tau_1 \sigma^{-1}, \sigma \tau_2 \sigma^{-1})$?

These are related to branched covers of a once-punctured torus since $\pi_1(\mathbb{T}-\{ pt\}) = \langle a,b| \text{ no relations }\rangle = \mathbb{F}_2$ we need two generators, $\tau_1, \tau_2$.

Bohr sets, Coin-flip sets and Roth's theorem

2012-11-23T21:11:58Z

I have been learning about Roth's theorem, trying to understand how Fourier series and dynamical systems (or even graph theory and binary sequences)are involved in counting arithmetic sequences in sets.

Any integer set of positive upper density has infinitely many arithmetic arithmetic sequences of length 3. $$ \bar{\delta}(A) = \limsup_{N \to \infty} \frac{|A \cap [-N,N]|}{2N+1} $$

These is a dichotomy between structure and randomness

Bohr sets $A = \{ n \in \mathbb{Z} : ||\alpha n - \theta|| < \delta/2 \}$. (Also, nil-Bohr sets).
"Coin" flip sets
- Flip a coin heads with probability $\delta$, get $\omega \in \{ 0,1\}^\mathbb{Z}$.
- $A = { n\in \mathbb{Z}: \omega(n) = head}$ is Fourier random almost surely

In both cases, the density can be found exactly $\bar{\delta}(A) = \delta$.

After some logical simplifications, the problem boils down to computing correlations between 3 copies of the set $A$

$$ \mathbb{E}[1_A 1_A 1_A] = \sum_{n,r \in \mathbb{Z}} 1_A(n)1_A(n+r)1_A(n+2r) $$

These count arithmetic sequences of all possible lengths and starting points. For Bohr sets and coin-flip sets these terms can be computed exactly.

What are the known asymptotics (if any) for the number of arithmetic sequences of a given difference $r(\bar{\delta})$ as a function of the upper density?

I am just trying to understand what is happening in the proof of Roth's theorem. Maybe it is possible to get an "explicit" proof of Roth theorem at least in some cases.

q-deformation of the permutation group?

2012-05-13T16:22:25Z

The only definition of a quantum group I know of involves q-deforming the relation $EF-FE=H$ or for SL(2): \[ \left[ \left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right), \left( \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array} \right) \right] = \left( \begin{array}{cr} 1 & 0 \\ 0 & -1 \end{array} \right) \]

All the axioms I have seen are very confusing and don't help me with much. I also get the sense, these should be called 'quantum lie algebras' rather than quantum groups. And I never understood the point of co-commutativity.

For now, what does a q-deformation of the permutation group look like? Or the dihedral group?

Answer by John Mangual for Geometric / physical / probabilistic interpretations of Riemann zeta(n>1)?

2012-11-12T05:48:07Z

Bourgade, Fujita & Yor shows to get Zeta functions from Cauchy Random Variables for even values and the $\chi_4$ L-functions for odd values. For some reason they always come in this pair.

This proof is simplified by Luigi Pace for $\zeta(2)$. The Cauchy Random variable is $$ p_X (x) = \frac{2}{1+x^2}$$

when we look at the ration of two such random variables $Y = X/X'$. $$ p_Y(y) = \frac{4}{\pi^2} \frac{\log y}{y^2-1}$$ Then observe $\mathbb{P}(Y \geq 1) = \mathbb{P}(X < X') = \frac{1}{2}$. So they compute $$ \sum_{k=0}^\infty \frac{1}{(2k+1)^2}= \int_0^1 \frac{-\log y}{1 - y^2} = \mathbb{P}(Y \geq 1)= \frac{\pi^2}{8}$$

I learned through a blog a proof using 2D Brownian motion at least for the case $\zeta(2)$.

Suppose that $f: \mathbb{C} \to \mathbb{C}$ is an analytic function on the neighbourhood of the unit disk. This
function maps the unit disk to with boundary where . A two dimensional brownian motion started at $f(0)$ takes on average time $$ \mathbb{E}[\tau] = \sum_{k \geq 1} |a_k|^2 $$ to exit domain $f(\mathbb{D})$ where $f(z) = \sum_{k \geq 0} a_k z^k$ and $\tau = \inf \{ t > 0: B_t \in \partial f(\mathbb{D}) \}$ is the hitting time of the boundary .

You can get $\zeta(2)$ by considering Brownian motion on the strip $\{ x+iy: |x| < \pi/2 \}$ and evaluating the left and right sides. The Brownian motion exit time is $\tau = \pi^2/4$ and $$f(z) = \log(\frac{1-z}{1+z}) = -2\left(z + \frac{z^3}{3} + \frac{z^5}{5} + \dots \right)$$ maps the strip to the unit disk.

This style is traced to the arXiv article by Greg Markowsky.

Also check out this paper by Noam Elkies who relates them to Alternating permutations. One can show:

\begin{eqnarray*} \sum_{k=0}^\infty \frac{1}{(2k+1)^2} &=& \sum_{k= 0}^\infty \int_0^1 \int_0^1 (xy)^{2k}dx\, dy \\ &=& \int_0^1 \int_0^1 \left( \sum_{k= 0}^\infty(xy)^{2k} \right)dx \, dy = \int_0^1 \int_0^1 \frac{ dx \, dy}{1 - (xy)^2} \end{eqnarray*} Then he does the strange Calabi substitution: \[ x = \frac{\sin u}{\cos v} ,y = \frac{\sin v }{\cos u} \]

and recovers a calculus identity: \[ \int_0^1 \int_0^1 \frac{ dx \, dy}{1 - (xy)^2} = \int_{u+v < \pi/2} 1 \, du \, dv = \frac{\pi^2}{8} \]

This proof is extended to higher dimensions in Elkies' paper.

You can then study the transform $T: L^2[0,\pi/2] \to L^2[0,\pi/2]$, the characteristic function of a triangle.

\[ (Tf)(x)=\int_0^{\pi/2 -x} f(t) \, dt \]

and ask when does $Tf = \lambda f$. The spectrum of this operator is

\[ \lambda = \frac{1}{4k+1} , f_\lambda(x) = cos (4k+1)u \]

Then one can take the trace of $T^n$ and compare to the volume of a polytope:

\begin{eqnarray} \sum_{k=-\infty}^\infty \frac{1}{(4k+1)^k}&=& \sum_\lambda \langle f |T^n | f \rangle \\ & =& \mathrm{Vol}\bigg(\{0 < x_1 > x_2 < x_3 > \dots < x_{n-1} > x_n > \frac{\pi}{2}\}\bigg) \end{eqnarray} The volume of this polytope can be expressed in terms of alternating permutations.

I first learned of this iterated integral idea in Stanley's survey on Alternating Permutations, but also in some papers by Chebikin on Parking Functions, this seems to be an example of a chain polytope.

What other L-functions can take neat values like $L(k) \in \mathbb{Q}\pi^k$ where $k \in \mathbb{Z}$ ? Possibly need an algebraic extension $K / \mathbb{Q}$.

Answer by John Mangual for Sperner's lemma and paths from one side to the opposite one in a grid

2012-11-12T05:39:10Z

If you know the path exists, you can pick a vertex at random (from a side) and find all the vertices reachable from it pretty quickly. If you don't reach the other side, just color those vertices (and whatever part you closed off).

Then try again. This process will terminate after finitely many steps.

Then to show such a path exists in the first place. In the dual graph you get a lamination of the disk. Contract all the inner "stuff" to a point. The boundary is divided into two arcs colored black or white and the an the innermost region must touch two points of the same color. The only exception is when there's a from one pt on the black-white boundary to the other.

I can try to draw a graphic of this... the connected regions must have interesting shapes within the graph of diagonals as well.

This is very similar to how you prove that Hex has a winning strategy and it depends on the lattice.

Polygonal billards programs

2010-08-03T03:23:54Z

I'm looking for software that will give billiard trajectories in arbitrary plane polygons. After much work I was able to produce this figure.

It was a good exercise, but at this point I wonder if scripts already exist.

I've heard if the angles are rational multiples of pi you can unfold the polygon so the billiard flow in the polygon becomes the geodesic flow on a translation surface.

Is there a topograph for Pythagorean triples?

2012-10-29T02:48:15Z

I have reading Allen Hatcher's notes on quadratic forms. Naturally, draw a pictures encoding all the values of a quadratic forms in a topographs. These are build by iterating the parallelogram identity:

$$ 2 Q(\vec{v})+2Q(\vec{w}) = Q(\vec{v}+\vec{w}) + Q(\vec{v}-\vec{w})$$

These can be found in Ch 1 of The Sensual (Quadratic) Form by John H Conway.

They are many interesting related to Farey fractions, circle packings, Voronoi tesselations and Kleinian groups.

I am interested points $x,y,z \in \mathbb{Z}^3$ in the quadratic form $Q(x,y,z) = x^2 + y^2 - z^2$ vanishes. It's known such triples exhibit a ternary tree structure. One can multiply vector $(x,y,z)$ by any of

$$ \left[\begin{array}{ccc}1 &-2 & 2 \\ 2& -1& 2\\ 2& -2 & 3 \end{array} \right] \text{ or } \left[\begin{array}{ccc}1 &2 & 2 \\ 2& 1& 2\\ 2& 2 & 3 \end{array} \right] \text{ or } \left[\begin{array}{ccc}-1 &2 & 2 \\ -2& 1& 2\\ -2& 2 & 3 \end{array} \right] $$ and get another Pythagorean triple. The result is an $\Gamma(2)$ action on the Pythagorean triples.

If I had to guess, the topograph would be somehow dual to the hyperbolic tessellation associated to the congruence group. The vertices of the "topograph" would be (similar to) the Farey fractions and the relation would involve 6 numbers instead of 4. I wonder what it could be.

Having a topograph for solutions of quadratic forms rather than values is not without precent. It's been done for Appolonian circle packings and Markoff triples. The topograph itself, has extension to ground fields other than $\mathbb{Q}$.

Branches of the Fibonacci Word Tree

2012-02-17T21:33:33Z

The Fibonacci word starts from $0$ subject to the rules $0 \mapsto 1, 1 \mapsto 01$ (or some variant thereof). The come from cutting sequences of the torus of a line of golden ratio slope. It is a 1D version of the Penrose Tiling.

...1010110101101101011010110110101101101011010110110101101...

The Fibonacci word is has minimal complexity above a periodic word -- there are n+1 subwords of length n, making it a Sturmian word.

5 subwords of length 4: "0101", "0110", "1010", "1011", "1101"
8 subwords of length 7: "0101101", "0110101", "0110110", "1010110", "1011010", "1011011", "1101011", "1101101"

As an experiment, I sorted the subwords of length n alphabetically The words are arranged in an infinite tree, where each word is descendant of its subword.

As a shorthand, I only placed the last letter of each word on each diagonal. Edges represent inclusion... drawn so any path from the top left corner "." appears in the Fibonacci word. Each Fibonacci subword corresponds to a path.

What is the structure of this tree? Is there any regularity to the location of the branches?

The complement of the tree becomes a tesselation of the Euclidean plane by "ribbon $\infty$-ominos", which is amusing.

.-1-1-0-1-1-0-1-0-1-1-0
| |     |              
0 0-1-1 0-1-1-0-1-1-0
|   | |         |      
1-1 0 0-1-1-0-1 0-1
| | |   |     |        
0 0 1-1 0-1-1 0-1      
| |   |     |          
1 1-1 0-1-1 0-1         
| | |     |             
1 0 0-1-1 0-1
| |   | |               
0 1-1 0 1-1
|   | |
1-1 0 1-1
| | |
0 0 1-1
| | |
1 1 0
| |
1 0
|
0

q-deformed group characters

2012-10-24T16:11:55Z

In a paper by Yuji Tachikawa, I found a q-deformed "2d Yang-Mills paritition function for a cylinder". Here it is (adapted):

$$ Z(q, x_L, x_R) = \mu(q, x_L)^{-1/2} \langle x_L | \bigg[ \sum_{R \in \mathrm{Irr}(G)} | R \rangle e^{- aC_2(R) } \langle R | \bigg] |x_R \rangle \mu(q, x_R)^{-1/2}$$

Here's some stuff to help you interpret:

$G$ is a compact lie group and the Irreducible representations should be indexed by the root lattice.
conjugacy classes are indexed by elements of maximal torus $\vec{x} \in \mathbb{T}^n \subset G$
$C_2(R)$ is the quadratic Casimir of the representation.
In my notation, borrowed from quantum mechanics $\langle R|x \rangle = \chi_R(x), \langle x|R \rangle=\overline{\chi_R(x)}$.
$\displaystyle \mu(q, X) = \exp \left[ \sum_{n=1}^\infty \frac{-2q^n}{1-q^n}\chi_{\mathrm{adj}}(x^n) \right]$
The partition function depends on the area $a$ of the cylinder.

In fact, let's turn this into a statement about the Laplacian: The $q$-dependence is hidden:

$$ e^{- a \Delta} =
\sum_{R \in \mathrm{Irr}(G)} | R \rangle e^{- aC_2(R) } \langle R | $$

Let's set the area to $0$. From the last line, we should get the identity matrix. However,

$$
\sum_{R \in \mathrm{Irr}(G)} \langle x_L | R \rangle \overline{ \langle x_R | R \rangle } = \mu(q, x_L) \delta(x_L = x_R)$$

This really looks like orthogonality of characters for compact groups, except the right side should be the identity.

What are these characters $\langle x | R \rangle$ ?

Originally, I wanted to ask about an analogue for finite $G$, but I don't even have a point of reference.

Answer by John Mangual for q-deformed group characters

2012-10-24T20:21:41Z

They are proportional to the Schur polynomials. For representation $\lambda \in \mathrm{Irr}(G), \chi \in \mathbb{T}^n \subset G$:

$$ \langle \lambda | x \rangle = \mu(q,x) \chi^\lambda(x) $$

This weight is is independent of the representation so we can do $\displaystyle \sum_\lambda$ no problem!

Here the Schur polynomial is defined by $\displaystyle \chi^\lambda(a) = \frac{ \det a_i^{\lambda_j + k - j}}{\det a_i^{k-j}}$

The number $\mu(q,x)$ is called superconformal index denoted $\mathcal{I}_q^V(a)$ in Section 6 of Gauge Theories and Macdonald Polynomials by Gadde, Rastelli, Razamat & Yan.

This superconformal index is the trace over representations over a certain superalgebra. It can also be considered a matrix integral over Haar measure. In the case of the cylinder

$$ \mu(a,b) = \Delta(a) \mathcal{I}^V(a) \delta(a, b^{-1})$$

at least, in the zero area limit.

In special cases, they find functions $f^\alpha(a)$ (in our case $=\langle \lambda|x \rangle$) orthogonal with respect to "propagator measure":

$$ \oint [da] \Delta(a)\mathcal{I}^V(a) f^\alpha(a) f^\beta (a^{-1}) = \delta^{\alpha \beta}$$

Then they can define a new metric and structure constants

\begin{eqnarray} \mathcal{I}(a,b,c) &=& \sum_{\alpha, \beta, \gamma} C_{\alpha\beta\gamma}f^\alpha(a)f^\beta(b)f^{\gamma}(c)\ \mu^{\alpha\beta} &=& \oint [da]\oint [db] \mu(a,b) f^\alpha(a)f^\beta(b)\ \end{eqnarray}

These generalize the orthogonality of characters relations (at least for compact Lie groups).

This "superconformal index" for a surface with punctures has an expression in terms of these generalized group characters.

$$ \mathcal{I}_{g,s} (a_1, a_2, \dots, a_n) = \sum_\alpha (C_{\alpha\alpha\alpha})^{2g-2+s} \prod_{i=1}^s f^\alpha(a_i) $$

"Aztec Diamond" analogue for Square-Octagon graph.

2012-10-20T16:56:46Z

I have been reading David Speyer's Perfect Matchings and the Octahedron Recurrence, trying to carry out his "cross-wrenches" generalization of the Aztec diamond. In what follows, I'm asking for a construction of $G_{n_0,i_0,j_0}$ in the case where the infinite graph $\mathcal{G}$ gives square-octagon tiling.

I found Speyer's notation very difficult. Maybe section 4 of Perfect Matchings and Perfect Powers by Mihai Ciucu will be easier to use for this special case.

In section 1.2 the "Aztec Diamond" theorem is stated $f(n_0, i_0, j_0) = \sum m(M)$

$f(n_0,i_0,j_0)$ is the solution to the octahedron recurrence. $$ f(n,i,j)f(n-2,i,j)= f(n-1,i-1,j)f(n-1,i+1,j)-f(n-1,i,j-1)f(n-1,i,j+1)$$
The sum over matchings of generalized Aztec diamond graphs $G(n_0, i_0, j_0)$ is called $\sum m(M)$. These graphs are embeeded in an infinite grid $\mathcal{G}$.
For now, every matching is counted with unit weight: $m(M)=1$.
$n_0+i_0+j_0 \equiv 0 \mod 2$

What is the sequence of shapes corresponding to Speyer's "crosses+wrenches" construction for the square-octagon lattice? Section 3.7 some relevant info.

The faces of his lattices are indexed by pairs of integers $(i,j) \in \mathbb{Z}^2$. He defines a "level" ( I really want to avoid using his word "height", which has another meaning in terms of dominos.) $$ h(i,j) = \left\{ \begin{array}{rc} 0 & \text{if }(i,j) \equiv (0,0) \mod 2\mathbb{Z}\times 2 \mathbb{Z} \\ 0 & \text{if }(i,j) \equiv (1,1) \mod 2\mathbb{Z}\times 2 \mathbb{Z} \\ 1 & \text{if }(i,j) \equiv (0,1) \mod 2\mathbb{Z}\times 2 \mathbb{Z} \\ -1 & \text{if }(i,j) \equiv (1,0) \mod 2\mathbb{Z}\times 2 \mathbb{Z}
\end{array} \right.$$ The Octahedron recurrence has initial conditions $f(h(i,j),i,j)=1$ and specializes here to powers of 5: \begin{array}{rlc} f(2n,i,j) & = 5^{n^2} & \\ f(2n+1,i,j) & = 5^{n^2+n} & \text{if } i \equiv n \mod 2\\ f(2n+1,i,j) & = 2 \cdot 5^{n^2+n} & \text{if } j \equiv n \mod 2 \end{array} I'm not even sure this list of possibilities is exhaustive. For $(2n+1,i,j)$ only one of $i,j$ can be odd.

The shapes $G(n_0, i_0, j_0)\subset \mathcal{G}$ are planar graphs with "open" and "closed" faces. He defines the "lattice", "edges", and "faces": \begin{eqnarray} \mathcal{L} &=& \{ (n,i,j) \in \mathbb{Z}^2 : n = i + j \mod 2\} \\ \mathcal{E} &=& \{ (i,j) \in \mathbb{Z}^2: i + j \equiv 1 \mod 2\} \times \{ a,b,c,d\} \\ \mathcal{F} &=& \mathbb{Z}^2 \end{eqnarray} The faces are indexed by pairs of integers. The edges are labelled a,b,c,d.

In section 2.1 Speyer defined some cones in $\mathbb{Z}^3$: \begin{eqnarray} p_{(n_0,i_0,j_0)} &=& n_0 - |i - i_0| - |j - j_0| \\ C_{(n_0,i_0,j_0)} &=& \{ (n,i,j) \in \mathcal{L}: n \leq n_0 - |i - i_0| - |j - j_0| \} \\ \mathring{C}_{(n_0,i_0,j_0)} &=& \{ (n,i,j) \in \mathcal{L}: n < n_0 - |i - i_0| - |j - j_0| \} \\ \partial C_{(n_0,i_0,j_0)} &=& \{ (n,i,j) \in \mathcal{L}: n = n_0 - |i - i_0| - |j - j_0| \} \\ \mathcal{I} &=& \{ (i,j, n) \in \mathcal{L}: n = h(i,j) \} \\ \mathcal{U} &=& \{ (i,j, n) \in \mathcal{L}: n > h(i,j) \} \end{eqnarray}

I have been unable to sort out definition of $G_{n_0,i_0,j_0}$ in section 3.3

It would be really amazing if one could show how the dominos actually "shuffle".

Answer by John Mangual for "Aztec Diamond" analogue for Square-Octagon graph.

2012-10-21T20:09:16Z

The faces are indexed by $\mathbb{Z}^2$, but $ \mathring{C}_{(n_0,i_0,j_0)} \cap \mathcal{I} \in \mathbb{Z}^3$. The closed faces of $G = G_{(n_0,i_0,j_0)}$ centered at $(n_0,i_0,j_0)$ satisfy an inequality:

$$ \mathring{C}_{(n_0,i_0,j_0)} \cap \mathcal{I} = \{ (i,j,n): n = h(i,j) < n_0 - |i - i_0| - |j - j_0| \}$$

How do we get a planar graph? Speyer defines a projection map that drops the last coordinate:

$$\alpha(\mathring{C}_{(n_0,i_0,j_0)} \cap \mathcal{I}) = \{ (i,j): h(i,j) + |i - i_0| + |j - j_0|< n_0 \}$$

This number $h(i,j) + |i - i_0| + |j - j_0|$ seems to be important for building the graph.

One simple ``height function" is an alternating pattern of 0's and 1's. We then overlay taxicab distances from various points to get the height function. The Aztec Diamond patterns emerge

                      4
010101      3        444
101010     333      44244
010101    33133    4422244
101010   3311133  442202244
010101    33133    4422244
101010     333      44244
            3        444
                      4

Let's try this procedure for square octagon lattice.

We get a clear general sense of the shape. The Mathematica code is:

h[x_, y_] := Switch[{Mod[x, 2], Mod[y, 2]}, 
             {0, 0}, 0, {1, 1}, 0, {1, 0}, 1, {0, 1}, -1]
cut[x_, y_] = If[ x < y, x, "."]
b = Table[
             cut[Abs[k] + Abs[l] + h[k + 1, l], 6],
             {k, -6, 6}, {l, -6, 6}
    ];
MatrixForm[b]

It seems to be possible to take off the corners and still be an acceptable "diamond".

The analogue of "domino" shuffling in these case would be an "inductive" way of moving from one Aztec Diamond to the next. ~~Not sure how to do this at the moment~~. There seems to be more than one way

Answer by John Mangual for Lost soul: loneliness in pursing math. Advice needed.

2012-10-20T16:29:17Z

I don't think a "my condolences" or "I wish you the best" will cut it. Your question is well-thought out and is of general interest to aspiring and active Mathematicians.

Questions about professional development in Mathematics have a place here on MathOverflow - at least until a better place for it comes along. But have you tried quora? Maybe this is a Philosophy of Mathematics question.

Mathematics, in its logical rigor, is a counter-intuitive way of thinking. Most people are not that precise. How does that fit into the professional world?

Comment by John Mangual

2013-05-23T15:02:25Z

Maria, please try math.stackexchange.com same question

Comment by John Mangual

2013-05-17T20:30:51Z

Mr Sausage, or Mr Roll. The typo is fixed. Thank you.

Comment by John Mangual

2013-05-14T22:08:06Z

clearly that is the book on this topic. I have to find it in a library or buy it... I don't know if my question is clear or specific enough to get an answer other than "goto Machlachlan-Reid"

Comment by John Mangual

2013-03-14T17:02:48Z

they are not equivalent. the second one converges to f, while the first one is a convex function related to f

Comment by John Mangual

2013-03-14T15:29:24Z

the first one might be wrong en.wikipedia.org/wiki/Bernstein_polynomial

Comment by John Mangual

2013-02-22T15:44:57Z

I found some papers relating on Pade approximation and the Stieltjes moment problem. Maybe these clarify the sense in which they are the "best approximation". Also, I provided a counter example where Pade approxmation is worse -- but it's off diagonal.

Comment by John Mangual

2013-02-21T19:13:21Z

What is so exact about the Taylor series? How are we comparing approximations? $T_{m+n}$ and the Pade approximant agree up to $O(x^{m+n+1})$. As long as you fix the degree of the numerator and denominator, one can be recovered from the other..

Comment by John Mangual

2013-02-21T18:21:29Z

What do you mean "better" approximation? Do you mean pointwise, $L^1, L^2$ etc ?

Comment by John Mangual

2013-02-21T18:20:29Z

I am explaining the "unreasonable effectiveness" of the Pade approximation by comparing it to continued fractions.

Comment by John Mangual

2012-11-19T15:56:42Z

It can still depend on a,b even though it only takes values +/- 0,1,2,4,8. Think of Mobius function $\mu(n) = 0, \pm 1$.

Comment by John Mangual

2012-10-29T04:20:34Z

Perhaps this question is moot? The topograph is gives a binary tree structure to the values of a quadratic form. For Pythagorean triples we exhibit a ternary tree with a $\Gamma(2)$ action. It remains to overlay the Pythagorean triples in the faces of the $\Gamma(2)$ Poincare disk tiling.

Comment by John Mangual

2012-10-24T21:07:41Z

Tachikawa extended this to nonzero area in the same paper as in the question.

Comment by John Mangual

2012-10-24T16:27:58Z

Is this really about elephants for you? That would be cool!

Comment by John Mangual

2012-10-21T15:45:18Z

The only issue here is to find the graphs that Ciucu and Speyer construct. I would appreciate anyone's help here.

Comment by John Mangual

2012-10-20T20:38:07Z

I guess there are a few separate issues: Flora's question itself, where it should be posted and what constitutes and answer. Flora seems afraid if she posts to a general academic audience, she won't get an answer from mathematicians. To me there's a much broader issue of professional development (or lack thereof) at the undergrad and graduate levels.