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## Is a separable compact Hausdorff space already metrizable?It is a known fact that a 2nd countable compact Hausdorff space is metrizable. What if we weaken the 2nd countable to separable only - is the space still metrizable? | |

## Why arithmetic Langlands?In trying to understand the import of Akshay Venkatesh' most recent work I found myself wondering anew about that old gnawing mystery: Even after decades of its formulation and spectacularly successful resolution in some cases - even more so for its geometric and $p$-adic analogues - my understanding of the literature is that the essential mystery remains in that beyond the equality of L-functions we don't have much of an inkling, at least in print, of a more direct relationship between the two sides, e.g., an action of (entities on) one side on the other, a direct map from one to the other, or both sides springing from a common source. Of course, my reading of the literature may be wrong. It is certainly very incomplete. The proofs of special cases may well have provided the practitioners with insights into why these two sides seemingly so far apart should be so intimately related. If so, have they shared them publicly in a manner that I can access them? If not (perhaps because the insights are too speculative or far from formalizable to put into writing), this seemed to me to be a good - informal, widely shared and relatively anonymous - forum where to learn some of them. Finally, to bring it back to where I had begun: is the recent work of Venkatesh and co-workers (e.g., https://arxiv.org/abs/1609.06370) a step in the direction of showing a more direct relationship? And related: among the myriad (if not quite thirteen) ways of looking at a modular form, are some more promising than others towards such an understanding in your opinion? | |

## I want help to test my prime theoremI want to share to everyone how to calculate primes and really want to ask who can test it and if you find anything I missed or mistakes to tell me? p 301 =1993=11×33+196×10 p301=1993=11×33+196×10 p 51 =233=3×11+2×100 p51=233=3×11+2×100 19122013=31876×1000+1×3+1×10 19122013=31876×1000+1×3+1×10 311=11×1+3×100 311=11×1+3×100 31=11×1+2×10 31=11×1+2×10 29=3×3+2×10 29=3×3+2×10 17=7×1+1×10 17=7×1+1×10 13=3×1+1×10 13=3×1+1×10 11=1×10+1×1 11=1×10+1×1 19=3×3+1×10 19=3×3+1×10 23=2×10+1×3 23=2×10+1×3 p 314 =2083=43×43+10×10 p314=2083=43×43+10×10 How to help me with tests? If we have two primes, x
x
and y
y
and both of them are two different sides of two geometric squares, x 2 x 2 +y 2 +xy+xy, x2+y2+xy+xy, we also can use 2(x+y)+x 2 +y 2 the last prime that we can use before a zillion digits are 1946. After 1991 they reach a zillion digits and are close to infinity. Thank you all for the attention. | |

## Change of variable for the Stokes equationsI asked this question to the Mathematics community but had no response (https://math.stackexchange.com/q/2885217/521741). Let $\Omega$ be domain of $\mathbb{R}^n$ and $\Phi : \Omega \to \Phi(\Omega)$ a deformation. Consider the Stokes equations written in the deformed configuration \begin{align} - 2\mu \operatorname{div}(D(u)) + \nabla p &= f, \quad \text{in } \Phi(\Omega) \\ \operatorname{div}(u) &= 0, \quad \text{in } \Phi(\Omega) \\ \end{align} where $u$ is the velocity of the fluid, $p$ the pressure, $\mu>0$ is the constant viscosity, $f$ is an external force and $D$ is the operator defined by $$ D(u) = \frac{1}{2} (\nabla u + \nabla u^T). $$ How can the Stokes equations be written in the domain $\Omega$ using a change of variable ? | |

## Example: Accessible category without colimitsI am looking for intuitive examples of the way(s) that colimits may fail to exist in the category of (Set-valued) models for a limit/colimit sketch. Bonus points if the sketch and/or the colimit diagram is finite. | |

## Central extension of Tarski monstersSuppose $G$ is a group with the following properties. $G/Z(G)$ is a Tarski $p$-group or another simple finitely generated infinite group in which all proper subgroups are abelian, and $Z(G)$ is a direct sum of two cyclic groups $\langle c_1\rangle$ and $\langle c_2\rangle$ of order $p$, a prime. Clearly setting $c_1\mapsto c_1c_2$ and $c_2\mapsto c_2$ will give us an automorphism of $Z(G)$. Is it possible, in these or similar circumstances to extend this automorphism to an automorphism of the whole group $G$? Are there some sufficient conditions to do it, or one can only hope for some magic? | |

## Roots of lacunary polynomials over a finite fieldIf $P$ is a polynomial over the field $\mathbb F_q$ of degree at most $q-2$ with $k$ nonzero coefficients, then $P$ has at most $(1-1/k)(q-1)$ distinct nonzero roots. Does this fact have any standard name / reference / proof / refinements / extensions? | |

## Is this sum of cycles invertible in QSn?I am interested the following element of the group algebra $\mathbb{Q}S_n$: \begin{align} \phi_n=2e+(12)+(123)+\dots+(1\dots n) \end{align} where $e$ is the identity permutation. My question is whether $\phi_n$ is a unit. For small $n$ I can see numerically that $\phi_n$ is a unit, but I have no idea how to prove it for general $n$. As a linear map from $\mathbb{Q}S_n$ to itself, for $n<10$, $\phi_n$ seems to be diagonalisable with positive integer eigenvalues, and I have no idea why this should be. | |

## Linear dependence of solution?Consider the function $f_k(c):=\sum_{n=0}^{\infty} c^{n^k}$ where $k\ge 1$ is an integer. This one obviously converges for $\left\lvert c \right\rvert <1.$ In the following we want to study the solution to the equation $$\sum_{n=0}^{\infty} \left(n^k -\frac{1}{\alpha}\right) c^{n^k}=0.$$ This one always exists as long as $\alpha \in (0,1).$ Numerically, I discovered something that I would like to understand: As $\alpha \rightarrow 0$ we have that $c= 1-1/k \gamma \alpha.$ So first the solution $c$ seems to depend in a linear way on $\alpha$ for $\alpha$ small and second, the dependence on $k$ also seems to be just $1/k$. I would like to understand these two observations. Of course, I have studied this more carefully, but just as a quick sanity check, one can study in Mathematica $$\sum_{n=0}^{1000} \left(n^k -\frac{1}{0.01}\right) c^{n^k}=0.$$ For FindRoot[!( *UnderoverscriptBox[([Sum]), (i = 0), (1200)](c^i\ ((i - *FractionBox[(1), (a)])))) == 0, {c, 1}] this produces For
this produces and for
this produces: Does anybody have any insights on this? | |

## Does Hartogs's Theorem for complex-analytic functions hold for real-analytic functions?Recall a very famous theorem due to Hartogs for complex analytic functions of several variables.
My question is does this theorem still hold if $f$ real-analytic? By real-analytic, I mean that $f$ has a power series expansion on $G\setminus K$. | |

## Two model structures of some categoryLet $\mathscr{C}$ be a category and $(W,C,F)=$(weak equivalence,cofibration,fibration) is some model structure of $\mathscr{C}$. Another model structure of $\mathscr{C}$, $(W_{S},C,F_{S})$ are called stronger if $W \subset W_{S}$. $Ex$ and $Ex_{S}$ are fibrant replacement functors in each model structures. When $X\longrightarrow Y$ belongs to $W_{S}$, $Ex_{S}(X)\longrightarrow Ex_{S}(Y)$ belong to $W$. Since $F=RLP(C\cap W)$ and $W \subset W_{S}$, $F_{S}\subset F$. In my situation, $\mathscr{C}$ is the category of pointed or unpointed simplicial sets, $(W,C,F)$ is ordinary model structure and $(W_{S},C,F_{S})$ is its localization by some monomorphism $f\colon A\longrightarrow B$ and $W_{S}=\{f$-local equivalence} and $F_{S}=RLP(C\cap W_{S})$. Q. Is $Ex_{S}(F) \subset F_{S}$ ? In other words, when $X\longrightarrow Y$ is a fibration, is $Ex_{S}(X)\longrightarrow Ex_{S}(Y)$ a stronger fibration? | |

## Chow theorem in $\mathbb{C}^2$I have the following question the answer to which I cannot find in the literature (but it must have been studied): Suppose that $M\subset\mathbb{C}^2$ is a real surface which may locally be written as the zero set of a holomorphic function. What conditions are required on $M$ so that $M$ may be written as the restriction of a projective algebraic curve $C\subset\mathbb{CP}^2$ to $\mathbb{C}^2\subset\mathbb{CP}^2$ (where we identify $\mathbb{C}^2$ with $\{[z,w,1]\in \mathbb{CP}^2| z,w \in \mathbb{C}\}$)? Clearly some conditions are necessary for example by considering $M=\{(z,e^z)\in \mathbb{C}^2|z\in\mathbf{C}\}$. I expect something like finite Euler characteristic, finite density at infinity, is this enough? Is there some standard reference for this? Or is there some sneaky way to use the Chow theorem? | |

## Counting "simultaneous squares' over the Gaussian integersLet $n$ be a square-free integer. Then for a given integer $m$, $m$ is a square modulo $n$ if and only if the sum $$\displaystyle \sum_{d | n} \left(\frac{m}{d}\right) > 0.$$ In fact one can write the indicator function $\Psi(m,n)$, which equals unity if $m$ is a square mod $n$ and zero otherwise, as $$\Psi(m,n) = \frac{1}{2^{\omega(n)}} \sum_{d | n} \left(\frac{m}{d}\right).$$ In both instances $\left(\frac{\cdot}{\cdot}\right)$ denotes the Jacobi symbol. It is a problem of central interest to evaluate the following sum: $$\displaystyle \sum_{\substack{mn \leq X \\ mn \text{ square-free}}} \Psi(m,n) \Psi(n,m).$$ This sum, for example, is essential to computing the average size of the 4-class group of quadratic fields (E. Fouvry, J. Kluners, It turns out that in some cases one has to consider the following related sum. Let $m,n$ be positive integers such that $m^2 + n^2$ is square-free. Define the function $$\displaystyle \Phi(m,n) = \begin{cases} 1 & \text{if } p | m^2 + n^2 \Rightarrow \left(\frac{m}{p}\right) = 1 \\ 0 & \text{otherwise}. \end{cases}$$ How does one evaluate the sum $$\displaystyle \sum_{\substack{m^2 + n^2 \leq X \\ m^2 + n^2 \text{ square-free}}} \Phi(m,n) \Phi(n,m)?$$ | |

## Is there an adaptation of the theory of standard forms and Tomita-Takesaki theory to the $\mathbb{Z}_{2}$-graded case?Let $A$ be a von Neumann algebra acting on a Hilbert space $H$, and suppose that $\Omega \in H$ is a cyclic and separating vector for $A$. Then in Tomita-Takesaki theory one defines an unbounded operator $S$ to be the closure of the operator \begin{equation} a \triangleright \Omega \mapsto a^{*} \triangleright \Omega. \end{equation} It then turns out that if we consider the polar decomposition \begin{equation} S = J \Delta^{1/2} \end{equation} into an anti-linear unitary involution $J$ and an unbounded positive operator $\Delta^{1/2}$ it follows that the map $A \ni a \mapsto Ja^{*}J \in (A')^{\operatorname{opp}}$ is an isomorphism. (Here the superscript opp stands for opposite algebra.) Now, let $A$ be any von Neumann algebra, then a standard form for $A$ is a quadruple $(A, H, J, P)$, where $H$ is a Hilbert space in which $A$ acts, $J$ is an antilinear operator in $H$, and $P$ is a self-dual cone, which satisfy the following properties. - $JAJ = A'$,
- $JaJ = a^{*}$, if $a \in A \cap A'$,
- $Jv = v$ for all $v \in P$,
- $aJaJP \subseteq P$ for all $a \in A$.
- $uau^{*} = a$ for all $a \in A$,
- $J' = uJu^{*}$,
- $P' = uP$.
It turns out that if $\Omega \in H$ is cyclic and separating as above, then the quadruple $(A,H,J, P_{\Omega})$, with $P_{\Omega} = \{ aJaJ\Omega \mid a \in A \}$ is a standard form for $A$. Does there exist an adaptation for the $\mathbb{Z}_{2}$-graded case? What I mean by this is as follows. Suppose that $A$ is a $\mathbb{Z}_{2}$-graded von Neumann algebra acting on a $\mathbb{Z}_{2}$ graded Hilbert space $H$, which is equipped with a cyclic and separating vector $\Omega \in H$. (Suppose that $\Omega$ is even.) Furthermore, assume that the representation of $A$ on $H$ respects the grading. Define an involution on $A$ by $a^{\sharp} = a^{*}$ if $a$ is even and $a^{\sharp} = -ia^{*}$ if $a$ is odd. We then consider the unbounded operator $S^{\sharp}$ defined to be the closure of the operator \begin{equation} a \triangleright \Omega \mapsto a^{\sharp} \triangleright \Omega. \end{equation} Then consider the polar decomposition $S^{\sharp} = J^{\sharp} \Delta^{1/2\sharp}$. Is it true that the map $A \ni a \mapsto J^{\sharp} a^{\sharp} J^{\sharp} \in (A^{\backprime})^{\operatorname{opp}}$ is an isomorphism? (Here $A^{\backprime}$ is the graded commutant of $A$.) Similarly, one can define a graded standard form of $A$, by replacing $A'$ by $A^{\backprime}$ and $*$ by $\sharp$. Does Haagerup's theorem still hold? If so, what circumstances guarantee that $u$ is even? (This question was motivated by the analysis done on pages 57 and 58 of these lecture notes on conformal nets.) | |

## What is a true invariant of $G$-crossed braided fusion categories?
(Spherical) fusion categories have well-known invariants like their global dimension, their rank, the categorical and Frobenius dimensions of their simple objects, or the Frobenius-Schur coefficients. Ribbon fusion categories have some more invariants like the $S$-matrix, the twist eigenvalues, and all the invariants of its symmetric centre. Graded fusion categories have e.g. the size of the group and the dimension as an invariant. An example for something that is A (spherical) $G$-crossed braided fusion category (short: $G\times$-BFC) is a $G$-graded fusion category with a compatible $G$-action and a crossed braiding. This implies e.g. that its trivial degree is a ribbon fusion category. All this gives us access to the invariants I've already mentioned, but I want to know whether a $G\times$-BFC has any
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## Prove $\int_0^{\infty}{\frac{1}{e^{sx}\sqrt{1+s^2}}}ds < \arctan\left(\frac1x\right),\quad\forall x\ge1$The question is to prove: $$ \int_0^{\infty}{\frac{1}{e^{sx}\sqrt{1+s^2}}}ds < \arctan\left(\frac1x\right),\quad\forall x\ge1. $$ Numerically it seems to hold true. So I have made some attempts to prove this analytically but have all failed. I also wonder if there is a systematic approach to solve this kind of problem. Thanks for your help. | |

## Number guessing game with lying oracleYou are probably already familiar with the usual number guessing game. But for concreteness I restate it. The usual gameThe At each turn, you make a guess $g$, and the Oracle tells you whether $n \leq g$ or $n > g$. The game ends when you can determine what $n$ is. It is pretty easy to see that the optimal strategy requires exactly 10 turns. The modified gameIn this game, the game runs almost exactly the same as the usual game. Except that there is a probability $p \in [0,1/2)$ (which is known to you) that the Oracle lies. More precisely, at each turn, the Oracle flips a (biased) coin and determines, independently of previous actions and other things in the game, whether to lie to you or not. When she lies she gives the exact opposite of the correct answer for the comparison $n \leq g$ or $n > g$. The game ends when you can determine, to a previously-agreed-upon confidence level, what is the answer $n$. (For argument sake, say that you can say with 95% probability what the value of $n$ is.) To model this, imagine you starting with 0 knowledge, so that each number between 1 and 1024 is equally likely. At each step you can update the probability distribution using the usual Bayesian updating procedure. The game ends when one of the numbers has probability 95% or higher. QuestionCan we estimate the expected number of guesses before the conclusion of the game, as a function of the lying probability $p$? (... and of the desired confidence level, and the number of bins?) Obviously, if the confidence level is anything above 50%, and if $p = 0$, the game reduces the usual game. As $p \to 1/2$, the expected number of guesses tend to infinity (at $p = 1/2$ the Oracle just responds randomly). So in particular, what are the asymptotics of the number of guesses as $p \to 0$ and as $p \to 1/2$? Numeric DataThe simulations below using the following naive strategy for guesses: At each step, based on the current "prior" probability, guess the number $g$ such that the difference $|P(n \leq g) - P(n > g)|$ is minimized. In the case $p = 0$ this reduces (one can check) to the binary search method. This choice is made to "maximize information gained" from that step. I don't have a proof that this is the optimal strategy. From numerical simulations, with 1024 numbers, 95% confidence interval, and 3000 trials, p = 2^-2 avg = 56.806666666666665 p = 2^-3 avg = 23.315 p = 2^-4 avg = 16.112666666666666 p = 2^-5 avg = 13.047666666666666 p = 2^-6 avg = 11.633 p = 2^-7 avg = 10.604333333333333 p = 2^-8 avg = 10.285333333333334 p = 2^-9 avg = 10.156666666666666 p = 2^-10 avg = 10.069 p = 2^-11 avg = 10.029666666666667 p = 2^-12 avg = 10.011333333333333 p = 2^-13 avg = 10.005666666666666 p = 2^-14 avg = 10.0 p = 2^-15 avg = 10.0which suggests a slope for small $p$ of around 100. Frequentist versus BayesianJames Martin brought up a very good point about frequentist versus Bayesian. Let me illustrate that with data from two runs using the strategy described above, with $p = 2^{-8}$. Starting game with 1024 bins. Running game now with secret answer: 857 Guessing 512 Is secret number (857) bigger than guess? Oracle truthfully responds true Update probabilities Sanity check, total probabilities sum to 1.0 My best guess is that the secret answer is 513 with probability 0.00194549560546875 Guessing 767 Is secret number (857) bigger than guess? Oracle truthfully responds true Update probabilities Sanity check, total probabilities sum to 1.0000000000000002 My best guess is that the secret answer is 768 with probability 0.003875850705270716 Guessing 895 Is secret number (857) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 0.9999999999999999 My best guess is that the secret answer is 768 with probability 0.007721187532297775 Guessing 831 Is secret number (857) bigger than guess? Oracle truthfully responds true Update probabilities Sanity check, total probabilities sum to 1.0000000000000004 My best guess is that the secret answer is 832 with probability 0.015441436261097105 Guessing 863 Is secret number (857) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 0.9999999999999986 My best guess is that the secret answer is 832 with probability 0.030880965812145108 Guessing 847 Is secret number (857) bigger than guess? Oracle truthfully responds true Update probabilities Sanity check, total probabilities sum to 1.0000000000000013 My best guess is that the secret answer is 848 with probability 0.0617618727298829 Guessing 855 Is secret number (857) bigger than guess? Oracle truthfully responds true Update probabilities Sanity check, total probabilities sum to 0.9999999999999987 My best guess is that the secret answer is 856 with probability 0.12256258895055303 Guessing 859 Is secret number (857) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 0.9999999999999982 My best guess is that the secret answer is 856 with probability 0.24704724796624977 Guessing 857 Is secret number (857) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 0.9999999999999997 My best guess is that the secret answer is 856 with probability 0.4903092482458648 Guessing 856 Is secret number (857) bigger than guess? Oracle truthfully responds true Update probabilities Sanity check, total probabilities sum to 0.9999999999999998 My best guess is that the secret answer is 857 with probability 0.9881889766208873 I did it! And it only took me 10 tries; the Oracle lied 0 times.Note that in binary, the number 857 is 1101011001 with more or less even distributions of zeros and ones. For 513 whose distribution is more extreme: Starting game with 1024 bins. Running game now with secret answer: 513 Guessing 512 Is secret number (513) bigger than guess? Oracle truthfully responds true Update probabilities Sanity check, total probabilities sum to 1.0 My best guess is that the secret answer is 513 with probability 0.00194549560546875 Guessing 767 Is secret number (513) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 1.0000000000000033 My best guess is that the secret answer is 513 with probability 0.003875733349545551 Guessing 639 Is secret number (513) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 0.9999999999999948 My best guess is that the secret answer is 513 with probability 0.007721187532297778 Guessing 575 Is secret number (513) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 0.9999999999999996 My best guess is that the secret answer is 513 with probability 0.015323132493622977 Guessing 543 Is secret number (513) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 1.0000000000000004 My best guess is that the secret answer is 513 with probability 0.030180182919776057 Guessing 527 Is secret number (513) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 1.0000000000000004 My best guess is that the secret answer is 513 with probability 0.05857858737666496 Guessing 519 Is secret number (513) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 0.999999999999996 My best guess is that the secret answer is 513 with probability 0.1106260320552282 Guessing 515 Is secret number (513) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 1.0000000000000002 My best guess is that the secret answer is 513 with probability 0.19905831824030176 Guessing 513 Is secret number (513) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 0.9999999999999991 My best guess is that the secret answer is 513 with probability 0.3315926624554764 Guessing 385 Is secret number (513) bigger than guess? Oracle truthfully responds true Update probabilities Sanity check, total probabilities sum to 1.0000000000000002 My best guess is that the secret answer is 513 with probability 0.6614346507956576 Guessing 512 Is secret number (513) bigger than guess? Oracle truthfully responds true Update probabilities Sanity check, total probabilities sum to 0.9999999999999979 My best guess is that the secret answer is 513 with probability 0.9902152355556972 I did it! And it only took me 11 tries; the Oracle lied 0 times.Incidentally, this also explains (partly) the drop between $2^{-13}$ and $2^{-14}$ seen in the numerical results above. When $p = 2^{-13}$, when $n = 513$, after | |

## Are there examples of conjectures supported by heuristic arguments that have been finally disproved?The idea for this comes from the twin prime conjecture, where the heuristic evidence seems just so overwhelming, especially in the light of Zhang's famous result from 2014 about Bounded gaps between primes and its subsequent improvements.
I don't mean heuristic in the sense that some conjecture holds for small numbers, e.g. the fact that $li\ x-\pi (x) $ was thought to be always positive, before Littlewood showed that not only it eventually changes sign, but does so infinitely often later on. So my question is different from the question about eventual counterexamples. Likewise, the fact that the first $10^{15}$ or so zeros of the zeta function "obey" RH does tell us something, but not a whole lot compared with infinity. So again this is not what I mean by heuristic. | |

## Question on a proof of density of periodic orbitsIn page 215 and 216 of the book "Introduction to the Modern Theory of Dynamical Systems" by Anatole Katok, Boris Hasselblatt, there is a theorem stated as following:
In the proof, they stated a fact that: given $\epsilon >0$ there exists $\delta>0$ such that when $p \in \mathbb{D}$ is in a $\delta$-neighborhood of $\partial \mathbb{D}$ then any two geodesics through $p$ of Euclidean length greater than $\epsilon$ have a mutual angle of at most $\pi/4$ After that, they used this fact to say that: "most geodesics through $z$ are entirely contained in $U$", with figure: I could not understand this argument. So I hope everyone will help me! I am new in this subject. This is the full proof:
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## Understanding a group of transformations of the plane $\mathbb{Z} \times \mathbb{Z}$I'm doing some research in visualizing arithmetic sets (resp. properties, resp. sequences of integers). I try to create patterns (in which I hope to observe some symmetries) by injectively mapping $\mathbb{N}$ on $\mathbb{Z}\times \mathbb{Z}$ in different ways and highlighting the numbers that have a given property (belong to a given set, are in a given sequence). You can find a little tool with which I'm playing around here. What I came up with has nothing to do with the initial approach (i.e. with arithmetic sets) but with transformations of $\mathbb{Z}\times \mathbb{Z}$. It's possibly interesting, but I'm missing conceptual tools to understand and explain it. What I observed was that different spirals (i.e. regular bijective mappings $m: \mathbb{N} \rightarrow \mathbb{Z}\times \mathbb{Z}$ with $m(0) = (0,0)$) seem to be related by some kind of "quasi-rotations", combined with some scaling, translation and "disturbance" in the middle of the spiral. These transformations can be watched "in action" and indeed "feel" like rotations, sometimes accompanied by some characteristic intermediate "shrinking" (have a look). I'd like to put these findings into mathematical terms and understand how they relate to each other and to the parameters of the spirals. Let a spiral $S^i_j$ be parametrized by the width and height of its rectangular "eye", i.e. pairs of integers $(i,j) \in \mathbb{Z}\times \mathbb{Z}$. Each transition $T^k_l$ between spirals can be described by a pair of numbers $(k,l) \in \mathbb{Z}\times \mathbb{Z}$ such that $$T^k_l(S^i_j) = S^{i+k}_{j+l}$$ Obviously, the transitions $T^k_l$ form a group. A transition seems to be related to an angle of rotation $\frac{n\ \pi}{2}$ (see $S^i_j \leftrightarrow S^{i\pm 1}_j$, $S^i_j \leftrightarrow S^i_{j\pm 1}$) (depending on what and how?)
the relative size of a "disturbed" zone in the middle of the affected spiral (depending on what and how?)
some scaling and translation of "patterns" outside the disturbed zone that remain "intact" otherwise (depending on what and how?) possibly an isolated translation of the upper-right half plane (see $T^{-2}_1$ which sends $S^3_0$ to $S^1_1$)
My question is: How can I treat these ephemeral quasi-rotations in the context of "hard" group theory? Where in group theory is a place for transformations of the form "rotation + X" with a hard to grasp X? |