Presentation of the mixed Braid group

This question concerns the mixed Braid groups $B_{n,P}$. Suppose $P$ is a partition of $\{1, \ldots, n\}$. Consider the subgroup $S_{n,P}$ of the symmetric group $S_n$ consisting of the permutations which preserve the parts of $P$. There is an epimorphism $\phi: B_n \to S_n$ from the Braid group on $n$ strands $B_n$ to the symmetric group $S_n$. The mixed Braid group for the partition $P$ is defined as $$B_{n,P} = \phi^{-1} S_{n,P}.$$ That is $B_{n,P}$ consists of those braids which preserve the parts of $P$ as a permutation of the endpoints.

A presentation of $B_{n,P}$ is given in S. Manfredini Some subgroups of Artin's braid group (available here). The presentation is given in Theorem 4 where the generators are taken to be $$\sigma_i, \quad 1\leq i < n, \ i \neq h_j, \qquad A_{h_j, h_k+1}, \quad 1 \leq j < k < m.$$

I suspect that there is a typo here and we should have $$\sigma_i, \quad 1\leq i < n, \ i \neq h_j, \qquad A_{h_j, h_k+1}, \quad 1 \leq j \leq k < m.$$ That is we have to allow the elements $A_{h_j, h_j +1}$ to be in the generating set. Otherwise this presentation does not work for the case when $P$ has 2 parts.

Can someone please shed some light on this?

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A series that converges to $2\pi^2 −18$

Inspired by Question 506695, I discovered the following identity $$\sum_{k=1}^{\infty} \frac{414 k^4 + 177 k^3 - 66 k^2 - 49 k - 8}{(1+2k)(1 + 3k)k^2(9k^2 - 1)\binom{3k}{k} } = 2\pi^2 - 18.\tag{1}$$

My questions are:

1. Is the series (1) known?

2. How can it be proven?

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Proof of Alas and Wilson that countably compact hereditarily Lindelof spaces are sequentially compact

In the proof of theorem 2.1 of When is a compact space sequentially compact? by O. T. Alas and R. G. Wilson there are some errors.

First, the transfinite induction only works for limit ordinals. If $\alpha = \beta+1$ we need to take $S_\alpha = S_\beta\setminus U_\beta$. But after doing so this is not a problem anymore.

The second issue with the proof, is that instead of limit points, we need $\omega$-accumulation points

$$A_\omega' = \{x\in X : \text{for all neighbourhoods }U\text{ of }x, U\cap A\text{ is infinite}\}.$$

If $X$ is a $T_1$ space, then $A_\omega' = A'$ and the proof works fine. But we don't want to assume $T_1$ here. We just need to analyze the beginning of the proof, this is of concern here, let me explain the problem.

First we take a sequence without convergent subsequence $(x_n)\subseteq X$ of a countably compact space $X$. We define $S = \{x_n : n\in\omega\}$ to be its image. Now the claim is as follows, every countable closed subset of $X$ is countably compact, so sequentially compact by a simple argument given in here. This implies that if $A\subseteq S$ is infinite, then $A'\setminus S\neq\emptyset$. However we don't want $A'\setminus S\neq\emptyset$! We want $A_\omega'\setminus S\neq\emptyset$.

Can this proof be salvaged, by either proving that $A_\omega'\setminus S$ must be non-empty for infinite $A\subseteq S$, or taking a different $S$ for which this property would hold?

Remark. I was given another proof of this result by A. Bella and P. Nyikos, which I haven't read yet, but I was wondering if this specific proof can be saved.

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Convergence of Maximum Likelihood Estimator

I apologize for the basic question. If $\{p_\theta(x): \theta\in K\subseteq\mathbb{R}\}$ is a smooth family of distributions, then the MLE $\hat{\theta}_n,$ under suitable regularity conditions satisfies $\sqrt{n}(\hat{\theta}_n-\theta)\to\mathcal{N}(0,I(\theta)^{-1})$ where $I(\theta)$ is the Fisher information. We also know from the Cramer-Rao bound that no estimator can outperform this estimator in the sense of having a smaller mean squared error (except on a set of zero Lebesgue measure, the superefficiency phenomenon).

I am interested in knowing similar results for other families of distributions which do not satisfy the above conditions. The canonical examples are: $\{\mathrm{Uniform}[\theta,\theta+1]: \theta\in\mathbb{R}\}$ and $\{\mathrm{Uniform}[0,\theta]: \theta>0\}.$ In these cases, the MLE estimates the parameter to within $O(1/n),$ since in the former case, $\hat{\theta}_n = \min\{X_1,X_2,\ldots, X_n\}$ and $n(\theta-\hat{\theta}_n)\to -Z,$ where $Z\sim\mathrm{exp}(1)$ and in the latter case, $\hat{\theta}_n = \max\{X_1,X_2,\ldots, X_n\}$ and $n(\theta-\hat{\theta}_n)\to W,$ where $W\sim\mathrm{exp}(\frac{1}{\theta}).$

Do we know

(1) If MLE is always an order-optimal estimator in the sense that it estimates $\theta$ to $O(\frac{1}{n^\alpha})$ where $\alpha$ is the best possible over all estimators?

(2) If MLE is not just order-optimal but optimal in a suitable sense similar to that of the Cramer-Rao lower bound, in that, any other estimator can outperform MLE only on a set of Lebesgue measure 0?

(3) Any general results on the distribution of convergence of $n^\alpha(\theta-\hat{\theta}_n)$ where $O\left(\frac{1}{n^\alpha}\right)$ is the rate of the convergence of the estimator?

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Is it true that in any coloring there are two red points at unit distance and one more red point between them?

Is there an $n$ such that for every $100$-coloring of $\mathbb R^n$ there are three collinear, identically colored points, $a,b,c$, such that the distance of $a$ and $c$ is exactly $1$, and $b$ is between them?

Remarks. $100=r$, just I don't like too many parameters.
If we fix any $3$-point configuration that we want to find, then there is such an $n$ if and only if the $3$ points are not collinear.
Such configurations are called Ramsey, and this follows from works by Erdős, Graham, Montgomery, Rothschild, Spencer, Straus, and by Frankl, Rödl.
However, I don't think that their methods apply if we want to avoid all (or just infinitely many) configurations at the same time.
Nevertheless, this particular question might be elementary.

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Determining the vector space for application of Cauchy Schwarz

In the paper "On the Erdős distinct distance problem in the plane" by Larry Guth and Nets Hawk Katz,

https://arxiv.org/PS_cache/arxiv/pdf/1011/1011.4105v1.pdf

they define the functions $d(P)$, (distinct distances between $N$ points), and $Q(P)$, (quadruples on $N$ points where the distances between the first two points of the quadruple are equal to the second two points). They then apply Cauchy Schwarz to obtain

$|d(P)|\ge\frac{N^4}{|Q(P)|}$

However, I don't see how we can consider the functions $d$ and $Q$ to be functions in the same Vector Space, since one yields a set of distances, where the other yields a set of quadruples. If I can't interpret them as elements of a Vector Space, I don't see how Cauchy Schwarz can be applied.

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Homology of smooth fibrations over the punctured affine line

Let $T \subset \mathbb{A}^1_{\mathbb{C}}$ be a non-empty (Zariski) open subset and $\pi:\mathcal{X} \to T$ be a smooth, projective morphism. Suppose that the relative dimension of $\pi$ is $n$. I am a looking for references which gives the generators of the $(n+1)$-th homology group $H_{n+1}(\mathcal{X},\mathbb{Q})$. I was thinking the following: $T$ is homotopy equivalent to a bouquet of circles with say $o \in T$ a common point. Denote by $\mathcal{X}'$ the fibration over the bouquet of circles. We know $H_{n+1}(\mathcal{X}')=H_{n+1}(\mathcal{X})$. My guess is that the genrators of $H_{n+1}(\mathcal{X}')$ are elements coming from $H_{n+1}(\mathcal{X}_o)$ (i.e., the homology of the fiber over the common point $o$) and the parallel transport of $H_{n}(\mathcal{X}_o)^{\gamma_i}$, where $\gamma_i$ ranges over all the loops in the bouquet of circles. My question is: Can we restrict to the generators of the fundamental group of $T$ or do we need $\gamma_i$ to range through all the loops of the fundamental group? More precisely, if $\mathbb{A}^1_{\mathbb{C}}\backslash T=p_1,...,p_m$ and $\gamma_1,...,\gamma_m$ are the loops that go around only $p_i$ exactly once, then is the natural map: $$H_{n+1}(\mathcal{X}_o) \oplus (\oplus_{i=1}^m H_n(\mathcal{X}_o)^{\gamma_i}) \xrightarrow{(i_*,\tau)} H_{n+1}(\mathcal{X}') \mbox{ surjective?}$$ where $\tau$ is the map that translates a monodromy invariant homology class in $H_n(\mathcal{X}_o)$. Any reference that studies the homology group of fibrations over punctured affine line will also be very much appreciated.

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distribution of non-solvable group orders

let $M$ be the set of natural numbers such that there is a group of this order, which is not solvable. what is the minimal distance $D$ of two numbers in $M$?

the examples $660$ and $672$ show $D \leq 12$. the famous theorem of feit-thompson implies $D>1$.

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Is the symmetric inequality $abc > \operatorname{rad}(abc)^3$ a sharper bound for ABC triples?

1. Empirical Data Analysis

Using the dataset from ABC@home (Leiden University) for $c < 2^{63}$, we observe a significant filtering effect when switching to the symmetric form:

• Standard Filter ($c > \operatorname{rad}(abc)$): There are 23,827,716 exception triples.
• Symmetric Filter ($abc > \operatorname{rad}(abc)^3$): There are only 4,452,223 exception triples.

Observation on Filter Strength:

Theoretically, since $a+b=c$ implies $a, b < c$, we have $abc < c^3$. Therefore, the condition $c > \operatorname{rad}(abc)$ does not directly imply $abc > \operatorname{rad}(abc)^3$. However, the data shows that the symmetric condition acts as a much stronger filter. It eliminates 19,375,493 "noise" cases (a reduction of 81.31%), retaining only about 18.69% of the original exceptions. This suggests the structure $abc$ vs $\operatorname{rad}^3$ might align better with the true distribution of these triples.

2. Structural Advantages

Besides the empirical tightening, this form offers structural benefits:

• Symmetry: It treats $a, b, c$ equally.
• Self-Balancing:
  • If $a=1$ (extreme asymmetry), $abc \approx c^2$. The condition $c^2 > \operatorname{rad}^3$ is extremely hard to satisfy, naturally filtering out trivial cases.
  • If $a \approx b$ (balance), $abc \approx c^3/4$. The coefficient $1/4$ provides a natural safety margin against the bound $\operatorname{rad}^3$.

3. Proposed Formulation

For every $\epsilon > 0$, there exist only finitely many triples of coprime positive integers $(a, b, c)$ with $a + b = c$ such that: $abc > \operatorname{rad}(abc)^{3+\epsilon}$

4. Questions

Given the strong empirical filtering and the structural properties discussed, I have two related questions:

Q1. The case for $\epsilon = 0$

Does the symmetric conjecture hold true even with $\epsilon = 0$? Specifically, are there only finitely many coprime triples satisfying the strict inequality $abc > \operatorname{rad}(abc)^3$?

Q2. Asymptotic filtering ratio

Let $N_{std}(X)$ be the number of triples with $c < X$ satisfying $c > \operatorname{rad}(abc)$, and $N_{sym}(X)$ be the number of triples with $c < X$ satisfying $abc > \operatorname{rad}(abc)^3$.Based on the data, the ratio $N_{sym}(X) / N_{std}(X)$ is approximately $0.18$ for $X = 2^{63}$.

What is the expected behavior of this ratio as $X \to \infty$? Does it tend to 0, suggesting that the symmetric bound is asymptotically much stricter than the standard one?

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How is tropicalization like taking the classical limit?

There is a folk — I can't call it a theorem — "fact" that the mathematical relationship between Complex and Tropical geometry is analogous to the physical relationship between Quantum and Classical mechanics. I think I first learned about this years ago on This Week's Finds. I'm wondering if anyone can give me a precise mathematical statement of this "fact". Or to the right introduction to tropical mathematics.

I can do the beginning. In classical mechanics, roughly (lower down I will mention some ways what I am about to say is false), when a system transitions from one configuration to another, it takes the route that minimizes some "action" (this idea dates at least to Maupertuis in 1744, and Wikipedia gives ancient-Greek analogs). Thus, for a system to transition from state $A$ to state $C$ in two seconds, after one second it is in the state $B$ that minimizes the sum of the action to get from $A$ to $B$ plus the action to get from $B$ to $C$. For comparison, quantum mechanics assigns to the pair $A,B$ an "amplitude", and the amplitude to go from $A$ to $C$ in two seconds is the sum over all $B$ of the amplitude to go from $A$ to $B$ times the amplitude to go from $B$ to $C$. (This is the basic principle of Heisenberg's matrix mechanics.) Anyway, we can understand both situations within the same language by considering the a matrix, indexed by states, filled with either the actions or the amplitudes to transition. In the quantum case, the matrix multiplication is the one inherited from the usual $(\times, +)$ arithmetic on $\mathbb{C}$. In the classical case, it is the $(+, \min)$ arithmetic of the tropical ring $\mathbb{T}$.

Let's be more precise. To any path through the configuration space of your system, Hamilton defines an action $\operatorname{Action}(\mathrm{path})$. The classically allowed trajectories are the critical paths of the action (rel boundary values), whereas if you believe in the path integral, the quantum amplitude is $$\int \exp\left(\frac{i}{\hbar} \operatorname{Action}(\mathrm{path}) \right) \mathrm{d}(\mathrm{path}),$$ where the integral ranges over all paths with prescribed boundary values and the measure $\mathrm{d}(\mathrm{path})$ doesn't exist (I said "if"). Here $\hbar$ is the reduced Planck constant, and the stationary-phase approximation makes it clear that as $\hbar$ goes to $0$, the integral is supported along classically allowed trajectories.

Of course, the path integral doesn't exist, so I will describe one third (and more rigorous) example, this time in statistical, not quantum, mechanics. Let $X$ be the space of possible configurations of your system, and let's say that $X$ has a natural measure $\mathrm{d}x$. Let $E : X \to \mathbb{R}$ be the energy of a configuration. Then at temperature $T$, the probability that the system is in state $x$ is (unnormalized) $\exp(-(kT)^{-1} E(x))$, by which I mean if $f : X \to \mathbb{R}$ is any function, the expected value of $f$ is (ignoring convergence issues; let's say $X$ is compact, or $E$ grows quickly and $f$ does not, or...):

$$\langle f \rangle_T = \frac{\int_X \exp(-(kT)^{-1} E(x)) f(x) \mathrm{d}x}{\int_X \exp(-(kT)^{-1} E(x)) \mathrm{d}x}$$

Here $k$ is Boltzmann's constant. It's clear that as $T$ goes to $0$, the above integral is concentrated at the $x$ that minimize $E$. In tropical land, addition and hence integration is just minimization, so in the $T \to 0$ limit, the integral becomes some sort of "tropical" integral.

Here are some of the issues that I'm having:

1. When you slowly cool a system, it doesn't necessarily settle into the state that globally minimizes the energy, just into a locally-minimal state. And good thing too, else there would not be chocolate bars.
2. The problems are worse for mechanics. Classically allowed paths are not necessarily even local minima, just critical points of the Action function, so the analogy between quantum and warm systems isn't perfect. Is there something like $\min$ that finds critical points rather than minima?
3. More generally, the path integral is very attractive, and definitely describes a "matrix", indexed by the configuration space. But for generic systems, connecting any two configurations are many classically-allowed trajectories. So whatever the classical analogue of matrix mechanics is, it is a matrix valued in sets (or sets with functions to the tropical ring), not just in tropical numbers.
4. Other than "take your equations and replace every $+$ with $\min$ and every $\times$ with $+$", I don't really know how to "tropicalize" a mathematical object.

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Does the embedding ???? ↦ ???? ???? ( ???? ) n↦t R(n) in a Recursive-Adic Number Field define a genuine valuation-theoretic realization of recursive depth?

In my manuscript The Recursive-Adic Number Field: Construction, Analysis, and Recursive Depth Transforms, I study a recursively defined depth function R(n) and two related constructions intended to parallel standard non-Archimedean frameworks. I would like to ask two focused questions about the mathematical validity and interpretation of these constructions.

Let α > 1. Define R : ℕ → ℝ≥0 recursively by

R(1) = 0,

R(n) = 1 + min_{1 ≤ k < n} (R(k) + R(n − k)) / α, n ≥ 2.

This recursion is well defined by dynamic programming, and one can show that

lim_{n→∞} R(n) = α / (α − 1).

Extend R to ℤ by symmetry R(−n) = R(n) and set R(0) = +∞.

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Construction A: Additive ultrametric completion

Fix 0 < σ < 1 and define

d_R(a, b) = σ^{R(|a − b|)}, a, b ∈ ℤ.

One can verify that

d_R(a, c) ≤ max{ d_R(a, b), d_R(b, c) },

so (ℤ, d_R) is an ultrametric space, and its completion Ẑ_R exists as a complete topological abelian group under addition.

Question A. Is the metric d_R genuinely induced by a valuation in the standard non-Archimedean sense, or should this construction be viewed as defining only an additive ultrametric group? More specifically, is there a principled obstruction to extending this additive structure to a topological ring or field while preserving the ultrametric?

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Construction B: Valuation via Hahn-type embedding

Define a map φ : ℕ → ℚ((t))× by

φ(n) = t^{R(n)}.

Inside the Hahn (or Laurent) series field ℚ((t)), the ambient valuation satisfies

v(φ(nm)) = v(φ(n)) + v(φ(m)),

even though R(nm) ≠ R(n) + R(m) holds in ℤ.

Question B. Is this embedding best interpreted as defining a genuine valued-field realization of the recursive depth function, or is it merely a formal device that bypasses the failure of multiplicativity of R on integers? In valuation-theoretic terms, does this construction correspond to a recognized method of inducing valuations from non-multiplicative invariants, or does it fall outside the standard framework?

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I am not asking whether the Recursive-Adic Number Field is “new”, but rather:

• whether Construction A admits extension beyond an additive ultrametric, • whether Construction B constitutes a legitimate valuation-theoretic realization, • or whether one can precisely identify which valuation axioms fail and why.

Any clarification regarding the correctness, interpretation, or limitations of these two constructions within non-Archimedean analysis would be greatly appreciated.

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Higher topos theory- Circular argument

Someone made me notice recently that in Lurie's ''Higher topos theory'' there is what seems to be a logical problem. Proposition 4.1.1.8 uses corollary 4.2.4.8, then corollary 4.2.4.8 needs proposition 4.2.3.14, but proposition 4.2.3.14 uses proposition 4.1.1.8, this is circular reasoning.

To the experts out there, to what degree does this constitute an obstruction to the validity of the results ?

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For which sets do Arthur and Nimue have a winning strategy in this game (“communicate a bit”)?

TL;DR: I define a three-player game (Arthur, Nimue, Merlin) where Nimue is shown a hidden bit $b$ chosen by Merlin and tries to communicate it to her ally Arthur, but Arthur must act computably while Merlin acts adversarially, replacing Nimue's chosen symbols $\mathtt{P},\mathtt{Q}$ by actual numbers from fixed sets $P,Q\subseteq\mathbb{N}$. When do Arthur+Nimue have a winning strategy in this game?

In other words, for which pairs $(P,Q)$ can the information contained in a single bit can be transmitted through an adversarial interpreter to a computably bounded receiver?

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THE GAME:

Let $P,Q\subseteq\mathbb{N}$ (we will soon assume $P\cap Q = \varnothing$). Consider the following game played between three players, “Arthur”, “Nimue” and “Merlin”, with Arthur and Nimue allied against Merlin.

(The game is described in the format used by Takayuki Kihara's paper “Lawvere-Tierney topologies for computability theorists”; however I do not assume familiarity with this paper, and I suspect it may not even be particularly relevant for answering the question. But see remarks below for what the question “means” in the context of Kihara's paper.)

Informal description: Merlin initially chooses a bit $b \in \{0,1\}$ and shows it to Nimue but not to Arthur. Nimue's goal is to communicate $b$ to Arthur (i.e., Arthur's goal is to guess the value of $b$), but Nimue isn't allowed to communicate directly to Arthur (they did agree a common strategy beforehand). At each turn, Nimue chooses an element of $\{\mathtt{P},\mathtt{Q}\}$ and shows it to Merlin, who then chooses an element of the corresponding set ($P$ or $Q$) and shows that element to Arthur. (Arthur only sees the elements of $\mathbb{N}$ that Merlin shows him; Merlin and Nimue see everything.) If Arthur can guess $b$ then he declares its value and the game ends. Otherwise, the game continues for a new turn. The crucial limitation is that Arthur must play following a computable strategy (Merlin and Nimue are not limited in this way). Arthur and Nimue win if Arthur makes a correct guess; otherwise, Merlin wins.

More precise description: On turn $0$, Merlin chooses $b \in \{0,1\}$. On each subsequent turn $i\geq 1$:

• Nimue chooses an element of $\{\mathtt{P},\mathtt{Q}\}$.

• Merlin chooses an element of $P$ if Nimue chose $\mathtt{P}$, and an element of $Q$ if Nimue chose $\mathtt{Q}$.

• Arthur (seeing only Merlin's moves on turns $i\geq 1$) either chooses to end the game by declaring a value $a \in \{0,1\}$ or to continue the game.

Arthur+Nimue win the game if Arthur ends the game by declaring $a=b$ (or if Merlin fails to abide the requirement of playing in $P$ or $Q$). Merlin wins if $a\neq b$ or if the game lasts forever.

A Nimue strategy is a function $\sigma \colon \{0,1\} \times \bigcup_{i\geq 0}\mathbb{N}^i \to \{\mathtt{P},\mathtt{Q}\}$ that picks an element in $\{\mathtt{P},\mathtt{Q}\}$ in function of Merlin's previous moves ($b$ and the finite sequence with values in $\mathbb{N}$). An Arthur strategy is a partial computable function $\tau \colon \bigcup_{i\geq 1}\mathbb{N}^i \dashrightarrow \{0,1,\texttt{continue}\}$ that picks an element of $\{0,1,\texttt{continue}\}$ in function of Merlin's visible moves (the finite sequence of length $i$ with values in $\mathbb{N}$, but obviously not $b$). The function is not required to be defined on all possible input values, but if $\tau$ is undefined when it is Arthur's turn to play, this is considered a forfeit by Arthur and Merlin wins (just as when the game lasts indefinitely).

We say that Arthur+Nimue have a winning strategy if there is a (computable!) Arthur strategy $\tau$ and a Nimue strategy $\sigma$ such that, if they play according to them, they win against any valid play by Merlin.

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THE MAIN QUESTION: Can we characterize pairs $(P,Q)$ such that Arthur+Nimue have a winning strategy?

Since the above main question might be too difficult or too general to admit a satisfactory answer (there might not be a characterization meaningfully different from the game itself), I am interested in partial answers as well, or even just interesting examples. To avoid posing a problem that is too open-ended, here are two specific questions that I don't know the answer to:

• Specific question #1: Does there exist $(P,Q)$ such that Arthur+Nimue have a winning strategy but for which it does not hold that there exists a partial computable function defined at least on $(P\times Q) \cup (Q\times P)$ with value $0$ on $P\times Q$ and $1$ on $Q\times P$? (See examples below for the significance of this condition, which gives Nimue a simple strategy.)

• Specific question #2: Do Arthur+Nimue have a winning strategy if $P$ is the set $\{e \in \mathbb{N} : \varphi_e\text{ is total}\}$ of codes of total computable functions, and $Q = \mathbb{N} \setminus P$ is its complement? (A positive answer here would imply a positive one to specific question #1.)

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SPECIAL CASES AND EXAMPLES:

• If $P$ resp. $Q$ is empty, then Arthur and Nimue trivially win by having Nimue play $\mathtt{P}$ resp. $\mathtt{Q}$ so that Merlin cannot make a choice. This is uninteresting so we now assume $P$ and $Q$ are inhabited.

• If $P$ and $Q$ have an element $n$ in common, then Merlin can play $n$ on every turn $i\geq 1$ and clearly Arthur+Nimue cannot have a winning strategy. So we might as well assume $P\cap Q = \varnothing$.

• If ($P\cap Q = \varnothing$ and) $P$ is computable, then Arthur+Nimue have a winning strategy: Nimue chooses $\mathtt{P}$ if $b=0$ and $\mathtt{Q}$ if $b=1$, and Arthur computably tests whether the integer $n$ communicated to him by Merlin is in $P$: if so, he ends the game by declaring $0$, and if not, he ends the game by declaring $1$. The same strategy works, mutatis mutandis, if $Q$ is computable.

• If ($P\cap Q = \varnothing$ and) $P$ is merely computably enumerable, then Arthur+Nimue also have a winning strategy, it is just slightly more complicated. On turn $1$, Nimue chooses $\mathtt{P}$ if $b=0$ and $\mathtt{Q}$ if $b=1$ (and Arthur continues the game), and on turn $2$, Nimue chooses the opposite of what she chose on turn $1$. On turn $1$, Arthur chooses to continue the game. On turn $2$, he runs runs a program enumerating $P$ while searching for either integer $n_1,n_2$ communicated to him by Merlin: if he finds $n_1$ in $P$ then he declares $0$, and if he finds $n_2$ in $P$ then he declares $1$, and one of the two will necessarily occur because of how Nimue played. The same strategy works, mutatis mutandis, if $Q$ is computably enumerable.

• If ($P\cap Q = \varnothing$ and) $P$ or $Q$ is co-computably enumerable, Arthur+Nimue still have a winning strategy where this time Arthur enumerates the complement of whichever set is assumed to be co-c.e. until he can rule out one of the elements presented by Merlin.

• If $z \in \mathbb{R}$ is any irrational number and, fixing a coding of rationals by integers, $P = \{r \in \mathbb{Q} : r < z\}$ and $Q = \{r \in \mathbb{Q} : r > z\}$, then Arthur+Nimue have a winning strategy: again, on turn $1$, Nimue chooses $\mathtt{P}$ if $b=0$ and $\mathtt{Q}$ if $b=1$ (and Arthur continues the game), and on turn $2$, Nimue chooses the opposite of what she chose on turn $1$. Arthur now wins by simply comparing the two rationals received from Merlin. Since $P$ and $Q$ have the same Turing degree as $z$, this example shows that Arthur+Nimue can have a winning strategy with $P,Q$ (complement to each other and) of arbitrarily Turing degree.

• If $P \subseteq P'$ and $Q \subseteq Q'$ and Arthur+Nimue have a winning strategy for $(P',Q')$, then they obviously also have one for $(P,Q)$.

• If $P$ is “sufficiently generic” and $Q = \mathbb{N}\setminus P$ is its complement, then Arthur+Nimue do not have a winning strategy. This is not completely obvious (or at least it wasn't to me…), so I give a proof of this proposition at the bottom for completeness of this question.

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FURTHER REMARKS:

• The case where $P$ and $Q$ are complement to each other seems most interesting. I don't know if this is a genuine simplification, however.

• In all my examples where Arthur and Nimue win, Nimue's strategy is very simple: she ignores what Merlin does except for the initial $b$, and simply plays $\mathtt{P},\mathtt{Q}$ or $\mathtt{Q},\mathtt{P}$ according to the value of $b$ and Arthur uses a computable function that separates $P\times Q$ from $Q\times P$ inside their union (so the game is over in two turns). This leads to the specific question #1: are there $(P,Q)$ for which Arthur+Nimue have a winnning strategy but not one of this kind?

• A trivial point that might be worth stating explicitly because it confused me: the game need not be determined in the usual sense. It is obvious that Merlin cannot have a winning strategy because Arthur could always guess the right $b$ by accident; but this does not imply that Arthur+Nimue have one.

• Relation to the effective topos (and motivation for the question): for those who know what this is about, $(P,Q)$ defines a subobject $X \hookrightarrow \nabla 2$ in the effective topos or equivalently, its characteristic function $\chi\colon\nabla 2 \to \Omega$ (taking $0$ to $P$ and $1$ to $Q$ with the obvious abuse of language). If I didn't mess up my translation of the formalism, Arthur+Nimue have a winning strategy exactly when the smallest Lawvere-Tierney topology $j$ making $X \hookrightarrow \nabla 2$ be $j$-dense (viꝫ. which satisfies $\forall b:\nabla 2.\,(j(\chi(b)))$) coincides with double negation $(\neg\neg)$ topology. (The double negation topology is itself the smallest L-T topology such for which the subobject $2 \hookrightarrow \nabla 2$ defined by $\{0\}$ and $\{1\}$ is dense.)

• For secondary motivation: the game is somewhat reminiscent of the indistinguishability game used to define the IND-CPA security level of in cryptography.

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Proposition: There exists $P$ such that (in fact, for all “sufficiently generic” $P$) if $Q = \mathbb{N}\setminus P$ then Arthur+Nimue do not have a winning strategy.

Proof. Endow the powerset $\mathcal{P}(\mathbb{N})$, identified with $\{0,1\}^{\mathbb{N}}$, with the product topology. We wish to show that there is a countable intersection of dense open sets $G$ (so $G$ is dense by Baire's theorem) such that, if $P$ is taken in $G$, then Arthur+Nimue do not have a winning strategy.

Since there are countably many Arthur strategies $\tau$ (since they are supposed to be computable), it is enough to show that, for any given Arthur strategy $\tau$, there is a countable intersection of dense open sets $G_\tau$ such that if $P$ is taken in $G_\tau$ then Nimue does not have a winning strategy along with Arthur playing $\tau$.

So, fix an Arthur strategy $\tau$: it takes as input a finite sequence $s$ with values in $\mathbb{N}$ and (computably) returns an element of $\{0,1,\texttt{continue}\}$. Define the “game tree” $T$ (associated to $\tau$) to be the (countable) set of finite sequences $s$ of natural numbers such that $\tau$ returns $\texttt{continue}$ for every subsequence obtained by removing some of the final elements of $s$ (by convention, we agree that the empty sequence is in $T$ and $\tau$ returns $\texttt{continue}$ on it).

Note that we are now dealing with a perfect information game between Nimue and Merlin.

Let us now label each node $s\in T$ with either “$0$” or “$1$” or both or none, and we also define sets $K_{s,v}$, as follows. First, if $\tau$ returns $c\in\{0,1\}$ on $s$, then we label $s$ by $c$; and if $\tau$ fails to terminate on $s$, then we label it both $0$ and $1$ (this won't really matter). Next, recursively, if the set of $m \in \mathbb{N}$ such that $s^\frown m$ (the concatenation of $s$ and $m$) is labeled $c$, is infinite, then we also label $s$ itself by $c$, and we let $K_{s,c}$ be the (infinite) set of such $m$. Finally, if $s$ has remained unlabeled after all of this, then there must be infinitely many $m$ such that $s^\frown m$ is unlabeled (because at most finitely many are labeled $0$ and at most finitely many are labeled $1$), and in this case we let $K_{s,\bot}$ be the (infinite) set of such $m$.

Now note that if $K$ is an infinite subset of $\mathbb{N}$, then the set $W_K \subseteq \mathcal{P}(\mathbb{N})$ of the $E\subseteq\mathbb{N}$ such that both $K \cap E \neq \varnothing$ and $K \setminus E \neq \varnothing$ hold, is open and dense in $\mathcal{P}(\mathbb{N})$. So intersection $G_\tau$ of the $W_{K_{s,v}}$ for $s\in T$ and $v \in \{0,1,\bot\}$ such that $K_{s,v}$ has been defined (always an infinite set) in the previous paragraph, is a countable intersection of dense open sets.

I claim that if $P$ is in $G_\tau$ and $Q$ is its complement, then Merlin has a winning strategy against Arthur playing $\tau$ (whatever Nimue does). Indeed:

• If the empty sequence $s$ is labeled $c\in\{0,1\}$ (if it is labeled both, pick one arbitrarily), then Merlin plays $b := 1-c$. Whether Nimue plays $\mathtt{P}$ or $\mathtt{Q}$, since $P$ and $Q$ both meet $K_{s,c}$, Merlin can play an $m \in K_{s,c}$, so that $s^\frown m$ is also labeled $c$. Continuing this way, two things can happen: either Arthur eventually declares $c$ and loses since $c\neq b$, or the play continues indefinitely (including when $\tau$ doesn't terminate), so Arthur also never wins. So in all cases, Merlin wins.

• If the empty sequence $s$ is unlabeled, then Merlin plays an arbitrary $b$. Whether Nimue plays $\mathtt{P}$ or $\mathtt{Q}$, since $P$ and $Q$ both meet $K_{s,\bot}$, Merlin can play an $m \in K_{s,\bot}$, so that $s^\frown m$ is also unlabeled. Continuing this way, the play continues indefinitely, and Arthur also never wins. So again, Merlin wins.

So in all cases, if $P$ is in $G_\tau$ then Arthur playing $\tau$ cannot give Arthur+Nimue a winning strategy. So if $P$ is in $G := \bigcap_\tau G_\tau$, then Arthur+Nimue do not have a winning strategy. ∎

(The same reasoning shows that for all “sufficiently random” $P$ the same conclusion holds: just note that $W_K$ has full measure on top of being open dense.)

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A question on integral functionals in Sobolev spaces

Let $r>0$ and $$ X:=\{u\in H^1_0(0,1) : \int_0^1|u'(t)|^2dt\leq r^2\}. $$ Find a continuous function $f:{\bf R}\to {\bf R}$, with $f(0)\neq 0$, having the following property: for every $u\in X$ there exists $v\in X$ such that $$ \int_0^1f(u(t))f(v(t))dt\leq 0 $$ In general, characterize such functions.

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$C^\infty$ Quotient of polynomials in $\mathbb{R}[x, y]$ with isolated zeroes at the origin

(Somewhat of a cross post of this question on MSE: https://math.stackexchange.com/questions/5115316/is-the-quotient-of-two-coprime-polynomials-in-mathbbrx-y-with-isolated-ze )

Suppose $p(x, y), q(x, y) \in \mathbb{R}[x, y]$ have isolated zeroes at $(0,0)$ in the sense that $p(0,0) = q(0,0) = 0$ and there is a disk punctured at the origin on which $p(x, y)$ and $q(x, y)$ are nonzero. Suppose also that the function $p(x, y)/q(x, y)$ extends to a function $f(x, y)$ which is $C^\infty$ at the origin with $f(0,0) = 1$. Does this imply that $p(x, y)$ and $q(x, y)$ have a common factor?

Example: $p(x, y) = x^2 + y^2 - x^3$. I have not been able to find a polynomial $q(x, y)$ which is both coprime to $p$ and makes the quotient extend to a $C^\infty$ function. But I also don't see how to prove that such a quotient can never extend to a $C^\infty$ function.

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Bytedance AIME set answer wrong? [closed]

Problem: The Bank of Pittsburgh issues coins that have a heads side and a tails side. Vera
has a row of 7873 such coins alternately tails-up and heads-up, with the leftmost
coin tails-up.
In a move, Vera may flip over one of the coins in the row, subject to the following
rules:
• On the first move, Vera may flip over any of the 7873 coins.
• On all subsequent moves, Vera may only flip over a coin adjacent to the coin
she flipped on the previous move. (We do not consider a coin to be adjacent
to itself.)
Determine the smallest possible number of moves Vera can make to reach a state in
which every coin is heads-up.

https://huggingface.co/datasets/ByteDance-Seed/BeyondAIME (the rows are ordered by answer)

It's reporting 15744,but as there are 3937 initial tails, so it seems like the answer has to be odd. Or am I missing something?

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Symmetric (under the swapping) recursions for Stirling numbers of both kinds

This question was cross-posted and resolved on math.se

• Let ${n \brack k}$ be the (unsigned) Stirling numbers of the first kind.
• Let ${n \brace k}$ be the Stirling numbers of the second kind.

I conjecture that for $1 \leqslant k < n$ we have $$ {n \brack k} = \frac1{n-k} \sum_{j=2}^{n-k+1} \binom{-k}j{n \brack k+j-1}(-1)^j, \\ {n \brack k} = \frac1{n-k} \sum_{j=2}^{n-k+1} (j-2)!\binom nj{n-j+1 \brack k}. $$ After swapping $(-1)^j$ and $(j-2)!$, I also conjecture that for $1 \leqslant k < n$ we have $$ {n \brace k} = \frac1{n-k} \sum_{j=2}^{n-k+1} (j-2)!\binom{-k}j{n \brace k+j-1}, \\ {n \brace k} = \frac1{n-k} \sum_{j=2}^{n-k+1} \binom nj{n-j+1 \brace k}(-1)^j. $$ Note that first part of second conjecture has already been proven here, but it would be nice to have a more general proof covering all the above cases.

Is there a way to prove it?

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Twisted cochain as a model for universal cover

Let $X$ be a pointed connected cw-complex and $C_{\ast}(X)$ the singular chain complex associated to $X$. Let denote $G=\pi_{1}(X)$ and $\tilde{X}$ the universal covering space for $X$. As far as I understand there is a twist $\tau: C_{\ast}(X)\rightarrow \mathbb{Z}G[1]$ i.e. a chain map (of degree $-1$) from $C_{\ast}(X)$ to $\mathbb{Z}G$ verifying the Maurer–Cartan equation such that the twisted chain complex $C_{\ast}(X)\otimes_{\tau}\mathbb{Z}G$ is quasi-isomorphic to $C_{\ast}(\tilde{X})$. I was wondering if there is a concrete description of the twist $\tau$? Moreover can we construct it in a natural (functorial) way?

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Conditions for continuity: eigenfunctions of the Dirichlet Laplacian divided by a function approaching 0 on the boundary

Suppose $\eta_1, \eta_2, \dots$ are the eigenfunctions of the Dirichlet Laplacian on a Lipschitz domain $\Omega$. For fixed $J$ and for some function $\phi \in C_0(\overline{\Omega})$, I would like $\frac{\eta_j}{\phi}$ to be continuous on $\Omega$ for $j \leq J$. Furthermore, defining $f = \frac{\eta_j}{\phi}$ on the interior of $\Omega$, and $f_j(y_0) = \lim_{y \to y_0}\frac{\eta_j(y)}{\phi(y)}$ for $y_0 \in \partial \Omega$, I would like $f$ to be continuous on $\overline{\Omega}$. My question is: what conditions are sufficient (and perhaps also necessary) on $\phi$ for $f_j$ to be continuous on $\overline{\Omega}$ for all $j \leq J$?

If we set $\phi = \eta_1$, then this is clearly true for some domains where we know the eigenfunctions explicitly, such as circular and rectangular domains in $\mathbb{R}^2$. What might also be critically relevant are $C^1$ bounds on the eigenfunctions of the Laplacian. I'm aware of the Hassell-Tao bounds that give bounds on the $L^2(\partial \Omega)$ norm of the derivative of the eigenfunctions in terms of the eigenvalues, but this doesn't quite give the conditions I need.

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A bidegree-$(2,2)$ equation

This is a refinement of my earlier question.

Consider the equation

$$\displaystyle (a_1 x_1 + a_2 x_2)^2(b_1 y_1 + b_2 y_2)^2 - 4x_1 x_2 y_1 y_2 = k^2,$$

where $a_1, a_2, b_1, b_1, k$ are (fixed) non-zero integers. Suppose that there is at least one integer tuple $(x_1, x_2; y_1, y_2)$ which solves this equation. Is there a way to parametrize all other integer solutions?

One can compare this to the analogous question of

$$\displaystyle f(x,y) = x^2 + bxy + cy^2 = k^2$$

for $f$ an indefinite and irreducible binary quadratic form and $k$ a non-zero integer. In this case, one identifies certain fundamental solutions that correspond to the arithmetic of the number field associated with $f$, and each fundamental solution corresponds to an infinite family corresponding to the units of the associated number field.

Is there an analogous story for equations of the above type?

It would already be interesting to me if one solves an explicit equation of this form, say an example with $\max\{|a_1|, |a_2|, |b_1|, |b_2|, |k| \} \leq 3$.

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