On a relaxed form of Goldbach's conjecture proposed by Erdős

The Goldbach's conjecture says that:

"Every even integer greater than $2$ is the sum of two prime numbers".

Let $\varphi$ denote the Euler's totient function. I remember that a long time ago I read that Erdős, motivated by the identity $\varphi(p) = p - 1$ for all prime numbers $p$, asked if the following weaker form of Goldbach's conjecture is true:

"Every positive even integer is the sum of two totients $\varphi(x)$ and $\varphi(y)$ ($x$ and $y$ positive integers)".

Unfortunately, I do not remember any reference.

Anyway, recently this problem has come to my mind, so I am curious: Has Erdős' question been settled?

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Why is $A_5$ the unique non-commutative simple group whose order is not divisible by 8

I am looking for a short (and elementary) proof of the following known

Fact. The alternating group $A_5$ is the unique non-commutative finite simple group whose order is not divisible by $8$.

By elementary I understand a proof that does not use the Classification of Finite Simple Groups. But the Feit-Thompson Theorem on the solvability of groups of odd order can be used.

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Intersection of 6 quadratics define genus 1 curve with given 8 rational points

We found a genus 1 curve over the rationals with given 8 rational points and the defining equations are 6 quadratics using very crude opportunistic construction with linear algebra.

Q1 Why the genus is 1? (We expect significantly larger genus).

Q2 Can this approach be used to construct low genus curves with many given points?

We are aware that we can find given 9 points on bivariate cubic.

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How the curve was found.

Set $8$ points $P_i \in \mathbb{Q}^7$ where each point is consecutive primes starting from $2$.

Working over $\mathbb{Q}[x_0,x_1,...x_6]$ we generate all monomials $m_i$ up to degree $D=2$, there are $T=\binom{7+2}{2}$ of them.

Then we make a matrix $M$: For each point $P_i$, the $i$-th row of $M$ is $m_j(P_i)$ for $0 \le j < T$. To get solution of the linear system compute echelon basis of the right kernel $k_i$ of $M$. Take $k_0,k_1,..,k_5$ and $f_i=\sum_j(k_j m_i)$ are defining polynomials of the given curve.

Some observations: It is very tempting to use linear equations or linearly dependent $P_i$, but we couldn't get better result. Taking fewer points give reducible curves or higher genus, which is counterintuitive. Computing the genus is slow for us.

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The curve in Magma format.

//Author: Georgi Guninski, 2025-12-28
K<x0,x1,x2,x3,x4,x5,x6>:=AffineSpace(Rationals(),7);
C:=Curve(K,[-2931660220757389/157134357590956018*x0^2 - 3667151920315521/157134357590956018*x0*x1 - 767360019563525/157134357590956018*x0*x2 - 1435108441048859/33080917387569688*x0*x3 + 74849216875776849/628537430363824072*x0*x4 + 15104741939290599/628537430363824072*x0*x5 - 4763865690345441/89791061480546296*x0*x6 - 123349000253628393/157134357590956018*x0 + 1, -696867356184847/1726751182318198*x0^2 - 192895633848475/1726751182318198*x0*x1 - 308964494777883/1726751182318198*x0*x2 - 201043073424537/363526564698568*x0*x3 + 10977160900688171/6907004729272792*x0*x4 + 1915764829672221/6907004729272792*x0*x5 - 4300570647046965/6907004729272792*x0*x6 - 22611461693911881/1726751182318198*x0 + x6, -1204830892294570861/157134357590956018*x0^2 + 580852504952248215/157134357590956018*x0*x1 - 813699710468856285/157134357590956018*x0*x2 - 206200710084444891/33080917387569688*x0*x3 + 12442880354469220017/628537430363824072*x0*x4 + 1492089576059044647/628537430363824072*x0*x5 - 703479992442465441/89791061480546296*x0*x6 + x6^2 - 29252820561327576753/157134357590956018*x0, -57274738219385249/157134357590956018*x0^2 + 6641160271884007/157134357590956018*x0*x1 - 29478617950566289/157134357590956018*x0*x2 - 11652971051545699/33080917387569688*x0*x3 + 667629956639099401/628537430363824072*x0*x4 + 96277192057344023/628537430363824072*x0*x5 - 31966335526489225/89791061480546296*x0*x6 - 1589777724532206687/157134357590956018*x0 + x5, -1078586274226104017/157134357590956018*x0^2 + 862272808289562063/157134357590956018*x0*x1 - 793036057530751861/157134357590956018*x0*x2 - 110660706587088547/33080917387569688*x0*x3 + 7635275108226960393/628537430363824072*x0*x4 - 274112126213548121/628537430363824072*x0*x5 - 267887356418295377/89791061480546296*x0*x6 + x5*x6 - 22284836892528486873/157134357590956018*x0, -965090317156350709/157134357590956018*x0^2 + 1031081615052252735/157134357590956018*x0*x1 - 764167612302321741/157134357590956018*x0*x2 - 44200055340445539/33080917387569688*x0*x3 + 4257314504189521161/628537430363824072*x0*x4 - 1682793117141967473/628537430363824072*x0*x5 + x5^2 + 58161259818701703/89791061480546296*x0*x6 - 17119262261114582001/157134357590956018*x0]);
Genus(C);

pts:=[[2, 3, 5, 7, 11, 13, 17], [19, 23, 29, 31, 37, 41, 43], [47, 53, 59, 61, 67, 71, 73], [79, 83, 89, 97, 101, 103, 107], [109, 113, 127, 131, 137, 139, 149], [151, 157, 163, 167, 173, 179, 181], [191, 193, 197, 199, 211, 223, 227], [229, 233, 239, 241, 251, 257, 263]];
//len(pts)=8

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Amenable inclusion of von Neumann algebras

Let us call an inclusion $A\subseteq B$ of von Neumann algebras amenable if ${}_A L^2(A)_A$ is weakly contained in ${}_A L^2(B)_A$ as $A$-$A$-correspondences.

A nice and deep result by Bannon, Marrakchi and Ozawa $(*)$ proves that $A\subseteq B$ is amenable if and only if there exists a (not necessarily normal) conditional expectation $E: B \to A$.

Suppose now that $A\subseteq B\subseteq C$ are inclusions of von Neumann algebras such that $A\subseteq C$ is amenable. Then it follows trivially from $(*)$ that $A\subseteq B$ is amenable as well. Is there a more elementary proof that $A\subseteq B$ is amenable, without making use of the result $(*)$?

$(*)$ Jon Bannon, Amine Marrakchi, and Narutaka Ozawa. "Full factors and co-amenable inclusions." Communications in mathematical physics 378.2 (2020): 1107-1121.

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Pre-image complete $T_2$-spaces

Let us call a topological space $(X,\tau)$ pre-image complete if for all $U, V \in (\tau\setminus \{\varnothing\})$ there is a continuous map $f:X\to X$ such that $f^{-1}(V) = U$.

For example, $\mathbb{R}$ with the Euclidean topology is not pre-image complete as $U = (0,1) $ can never be a continuous pre-image of $V = (0,1) \cup (2,3)$, because $V$ is not connected. Also, discrete spaces with more than $1$ point are not pre-image complete, as Henrik Rüping noted in the comments below (the previous version of this post contained the incorrect statement that discrete spaces are pre-image complete).

Question. Is there a pre-image complete $T_2$-space with more than $1$ point?

(Comment on this)

Inequality with Euler's totient

Using the first 100,000 values of $\varphi(n)$ it seems that the following is true.

Let $\mathcal A$ be a finite subset of $\mathbb{N}$, $\forall n\in \mathbb{N} \setminus \mathcal{A}$, $\displaystyle \frac{1}{\varphi(2n)} - \frac{1}{\varphi(2n+1)} \geqslant \frac{1}{2n\ln (2n)} $.

Is this true? Is there a stronger lower bound?

P.S.: I looked at Handbook of Number Theory I by Mitrinović and Sándor which has a lot of info about $\varphi (n)$ but it doesn't appear there.

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Equivariant fundamental groups of general $G$-varieties

I am reading the book "Perverse Sheaves and Applications to Representation Theory" by P. Achar. In proposition 6.2.13, it is written that, if $G$ is an algebraic group (over $\mathbb{C}$) and $X$ is a principal $G$-variety, then $$\operatorname{Loc}_G(X)[\dim(X)] \simeq \operatorname{Perv}_G(X)$$ namely, all equivariant perverse sheaves are shifted equivariant local systems and there is an equivalence $$\operatorname{Loc}_G(X) \simeq \operatorname{Mod}_{\mathbb{Q}[(G^x)/(G^x)^{\circ}]}$$ to category of $\mathbb{Q}[(G^x)/(G^x)^{\circ}]$-modules, where $x \in X$, $G^x$ the stabilizer, $(-)^{\circ}$ connected component of identity. Then $\pi_1^G(X,x) = (G^x)/(G^x)^{\circ}$ is called the equivariant fundamental group of $X$.

In general, I wonder what happens if we do not require $X$ to be a principal $G$-variety. I can indeed define $$\operatorname{Loc}_G(X)[\dim(X)] = \left \{L \in \operatorname{Perv}_G(X) \mid \operatorname{Res}^G(L) \in \operatorname{Loc}(X)[\dim(X)] \right \}$$ where $\operatorname{Res}^G \colon \operatorname{Perv}_G(X) \longrightarrow \operatorname{Perv}(X)$ is the forgetful functor. I claim that this category is still a neutral Tannakian category with fiber functor is the composition $$\operatorname{Loc}_G(X) \longrightarrow \operatorname{Loc}(X) \simeq \operatorname{Rep}(\pi_1(X)) \longrightarrow \operatorname{Vect}_{\mathbb{Q}}.$$ Indeed, the only nontrivial thing that we have to check is that every object in $\operatorname{Loc}_G(X)$ admits a strong dual. This follows from the fact that local systems can be described as strongly dualisable objects (see for instance, Lemma 1.2.9 here), since forgetful functor $\operatorname{Loc}_G(X) \longrightarrow \operatorname{Loc}(X)$ is conservative, exact, commutes with $\otimes, \underline{\operatorname{Hom}}$, it must reflect strong dual.

My question. How does the dual group of $\operatorname{Loc}_G(X)$ look like when $X$ is not a principal $G$-variety?

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Optimal rule for multiple stopping times for defect finding

Suppose a quality inspector is inspecting $b$ black amongst which $d_B$ are known to be defective and $w$ white gadgets amongst which $d_W$ are known to be defective. The gadgets come down along an assembly line one by one in uniformly distributed random order. As each gadget passes, the inspector observes its color, and he chooses to let the gadget pass or use a device to detect whether the gadget is defective. But he can only use the device a total of $n$ times. What is the optimal stopping rule to use the inspecting device to maximize the expected number of defective gadgets found.

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Edit: As fedja points out below, there is a difference between detecting a defect by the device and deducing for sure by logic a defect. Both are legitimate objectives. The solution to the first is easier than that to the second. I for now choose the first definition, i.e., to define finding a defective gadget as only indicated by the device and use up one quota of using the device even if the quality inspector is sure by logic alone the gadget under inspection is defective.

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Suppose at a pass the number of black gadgets already inspected by the device is $i_B$, amongst them $f_B$ are detected to be defective. Suppose the current passing gadget is black and yet to be inspected. Then the probability of this black gadget being defective is $p_B=\frac{d_B-f_B}{b-i_B}$. Symmetrical probability holds for if the current passing gadget is white.

I have a conjecture for the explicit solution, which is a greedy algorithm, as follows and am seeking a proof.

At a pass of the gadget, without loss of generality, suppose the current gadget which has not yet been inspected is black. Suppose there are $n_B$ black including the current one and $n_w$ white gadgets left. Suppose $i_B$ black and $i_W$ white gadgets have been inspected, amongst which $f_B$ black and $f_W$ have been found to be defective. If $p_B\ge p_W$ or $n_W=0$ the inspector inspect the current black gadget with the device. Otherwise, the inspector let the current black gadget pass without inspection.

I have set up the dynamic programming formulation but fail to see either the proof or a counterexample to my conjecture.

(Comment on this)

old notation for equations of 4-variate quartics

I am reading an old book on the topic, Jessop, C. M. Quartic surfaces with singular points, 1916, CUP, where notation like $(a(\!\!)A,B,C)^2=0$ means, most probably, something like $A^2+B^2+C^2=0$, with $A,B,C\in\mathbb{C}[x_0,\dots,x_3]$. This notation is not explained in the text. I wonder where I can check its meaning.

(Comment on this)

Has this been studied before? [closed]

Note: English is not my first language (I'm Polish). I used AI assistance to translate and structure this writeup, but the mathematical concepts and exploration are my own work.

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Theory of Orders: A Framework for Exponential Hierarchies

Abstract

We propose a systematic study of exponential hierarchies through a property called order, which extends the concept of iterated logarithm to negative and complex domains. While the base formulation r = log_b(log_b(n)) is known, we provide a comprehensive framework including:

• Characterization of numbers with integer orders
• Extension to negative numbers and complex domains
• Inverse formula n = b^(b^r)
• Investigation of algebraic properties

This is an exploratory work seeking feedback on potential applications and connections to existing theory.

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1. Definition

For a given base b ∈ ℝ (b > 0, b ≠ 1) and number n ∈ ℝ₊, we define the order of n as:

r = order_b(n) = log_b(log_b(n))

Inverse formula:

n = b^(b^r)

Special cases:

• Order 0: Numbers whose b-th root is not an integer
• Positive order: Numbers greater than b
• Negative order: Numbers between 0 and b (0 < n < b)

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2. Computing Orders

2.1 For Real Positive Numbers (n > 0, b > 0)

Formula:

r = log_b(log_b(n)) = ln(ln(n)/ln(b)) / ln(b)

Primitive method (for integers):
Extract the b-th root repeatedly until reaching a number whose b-th root is not an integer. The number of steps is the order.

Example (base 2):

• 256 → √256 = 16 → √16 = 4 → √4 = 2 → order = 3
• √2 → (√2)² = 2 → order = -1

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2.2 Extension to Negative Numbers and Complex Domain

For negative n or b, the order becomes complex-valued.

General formula using complex logarithm:

r = log_b(log_b(n))

where:

log_b(z) = ln(z) / ln(b)
ln(z) = ln|z| + i·arg(z)

with principal branch arg(z) ∈ (-π, π].

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3. Numbers with Integer Orders

Numbers with integer order k form a special sequence:

n = b^(b^k)  for k ∈ ℤ

Examples:

Base 2: {2, 4, 16, 256, 65536, ...} = {2^(2^k) : k ≥ 0}
Base 3: {3, 9, 81, 6561, ...} = {3^(3^k) : k ≥ 0}

These numbers define exponential layers in the hierarchy.

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4. Examples

Observation: The same number has different orders in different bases.

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5. Properties (Under Investigation)

The following properties are currently being investigated:

5.1 Behavior under arithmetic operations:

• order_b(n^k) = ?
• order_b(n₁ · n₂) = ?
• order_b(n₁ + n₂) = ?

Preliminary observation: For numbers with integer orders in base b:

order_b(n²) = order_b(n) + 1

5.2 Base conversion:

How does order_b₁(n) relate to order_b₂(n)?

5.3 Limits:

• lim_{n→∞} order_b(n) = ?
• lim_{n→b⁺} order_b(n) = ?

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6. Open Questions

1. Has this framework been studied systematically before under a different name?
2. Are there natural applications in:
   • Computational complexity theory?
   • Information theory?
   • Number theory?
3. What are the complete algebraic properties of the order function?
4. Is there a meaningful interpretation of complex-valued orders?

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7. References and Prior Art

I'm aware this builds on the well-known iterated logarithm. I'm seeking references to:

• Similar frameworks in existing literature
• Potential applications I haven't considered
• Related work on exponential hierarchies

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Feedback Welcome

This is exploratory work by someone without formal mathematical training. Constructive criticism, references, and suggestions for improvement are greatly appreciated.

(Comment on this)

Fourier series of smooth functions in infinitely many variables

Let $J$ be a set (usually countable). Let $t_j$, $j\in J$, be variables in ${\mathbb R}/2\pi i{\mathbb Z}.$ Put $u_j=\exp(it_j),$ $j\in J.$ Introduce the following semi-norms on the space of Fourier polynomials ${\mathbb C}[u_j^{\pm 1}|j\in J]:$ for $N\geq 0,$ put $$||f||_N=\max_{t\in (S^1)^J} \sum _{|\alpha|=N} |\partial ^{[\alpha ]} f(t)|$$ where the sum is taken over multi-indices $\alpha=(\alpha_j|j\in J)$; $\alpha_j\geq 0;$ $|\alpha|=\sum_{j\in J}\alpha_j;$ and $\partial ^{[\alpha ]}=\prod_{j\in J} \frac{1}{\alpha_j !}(\frac{\partial}{\partial t_j})^{\alpha_j}$.

Let $C^\infty ((S^1)^J)$ be the completion of ${\mathbb C}[u_j^{\pm 1}|j\in J]$ in the topology defined by these semi-norms.

Question. How to describe $C^\infty((S^1)^J)$ in terms of the Fourier coefficients $a_n$?

(Comment on this)

Conjecture on the existence of non-trivial integer solutions to $A^4+B^4+C^4=2 n^2 D^4$

I have been investigating the existence of non-trivial integer solutions to the Diophantine equation: $A^{4}+B^{4}+C^{4}=2n^{2}D^{4}$ where $n$ is a square-free positive integer.

Through extensive computational search (covering $n$ up to $1013$), I have found that this equation possesses extremely large primitive integer solutions for many values of $n$. For instance, when $n=101$, the smallest solution involves integers with over 40 digits, and I have found even larger solutions (hundreds of digits) for other values of $n$.

Local Considerations:

It is easy to show via infinite descent that no non-trivial solutions exist if $n$ is divisible by $2,5,$ or $29$.

For $p=29$, the set of 4th powers modulo 29 is $\{0,1,7,16,20,23,24,25\}$. The only combination of three elements that sums to $0\quad (\mod 29)$ is $(0,0,0)$. Similar local obstructions exist for $p=2$ and $p=5$.

The Conjecture:

Based on numerical evidence for $n=33,41,47,51,59,69,77,83,89,101,119, 123, 137, 141, 149, 159, 161, 173, 179,$ $\quad\quad 187, 197, 209, 213, 227, 233, 383, 389, 393, 1013$, I would like to propose the following: 

Conjecture: Let $n$ be a square-free integer such that $\gcd (n,290)=1$. Then the equation $A^{4}+B^{4}+C^{4}=2n^{2}D^{4}$ always has a non-trivial integer solution.

My Questions:

(1) Is there any known counterexample to this conjecture for $n$ satisfying $ \gcd (n,290)=1 $?

(2) Does this specific form of the equation relate to a known family of surfaces (e.g., K3 surfaces) for which the Brauer-Manin obstruction is known to vanish?Given the extreme size of the fundamental solutions (often exceeding $10^{100}$ or $10^{200}$, is there a geometric explanation for why the "height" of the rational points on the associated curves/surfaces grows so rapidly with $n$?

The details of the huge solutions I found are documented (in Japanese) here:

 https://www.kaynet.or.jp/~kay/misc/de46-to1000.html

I would appreciate any theoretical insights or references to similar problems.

(Comment on this)

Concern with corollary $2.4.4.4$ of HTT

Let me restate the corollary in question, for convenience:

Corollary $2.4.4.4$:

Suppose we have functors of $\infty$-categories: $$\mathcal{C}\overset{p}{\longrightarrow}\mathcal{D}\overset{q}{\longrightarrow}\mathcal{E}$$such that both $q$ and $q\circ p$ are locally Cartesian fibrations, and that $p$ maps locally $(q\circ p)$-Cartesian morphisms to locally $q$-Cartesian morphisms.

If for every object $Z\in\mathcal{E}$, $p$ induces a categorical equivalence of the fibres over $Z$: $$p_Z:\mathcal{C}_Z\overset{\sim}{\longrightarrow}\mathcal{D}_Z$$then $p$ is itself a categorical equivalence.

I am quite concerned: $(1)$ with its terse proof, and: $(2)$ with its level of generality. It is not even assumed that $p$ is an inner fibration. I suspect the proof is wrong, and have heard from various people that it's "well-known" to experts, in a folkloric way, that there are a fair few minor mistakes in HTT - not conceptual ones, though. Since I've been unable to find this result in the Kerodon, and this corollary sees quite a lot of use through the book, I would like to know if it really is correct (perhaps under stronger hypothese) and, if so, see a more detailed proof.

The proof essentially says: [from the hypothesis on fibres over $\mathcal{E}$, $p$ is essentially surjective (agreed)] and by proposition $2.4.4.2$ $p$ is fully faithful. That is all that is said, and it's the last claim I dispute.

Proposition $2.4.4.2$:

Let $p:\mathcal{C}\to\mathcal{D}$ be an inner fibration of $\infty$-categories. Then for any pair of objects $X,Y\in\mathcal{C}$ and morphism $e':p(X)\to p(Y)$ which admits a locally $p$-Cartesian lift over $Y$, there is an homotopy fibre sequence: $$\operatorname{Map}_{\mathcal{C}\times_{\mathcal{D}}\{p(X)\}}(X,X')\to\operatorname{Map}_{\mathcal{C}}(X,Y)\overset{p_{X,Y}}{\longrightarrow}\operatorname{Map}_{\mathcal{D}}(p(X),p(Y))\\\text{over the point $e'$}$$where $X'\in\mathcal{C}$ is any object which is the domain of a locally $p$-Cartesian lift $e:X'\to Y$ of $e'$.

I believe that $2.4.4.2$ neither applies nor gives a strong enough conclusion.

The conclusion isn't strong enough since, although if we define $Z:=q(p(X))$ and $\mathcal{C}_{p(X)}\to\{\ast\}$ is a strict pullback of the categorical equivalence $\mathcal{C}_Z\to\mathcal{D}_Z$, that does not imply - since this categorical equivalence is not necessarily a Joyal fibration, unless I am missing something, so the strict pullback is not an homotopy pullback - that $\mathcal{C}_{p(X)}\sim\ast$ is an homotopically trivial category; which is exactly what is wanted, to conclude from the fibre sequence of $2.4.4.2$ that $p$ is indeed fully faithful.

I believe the proposition does not apply since, as mentioned, I do not know why $p$ ought to be an inner fibration in this context and because I cannot actually construct a (locally) $p$-Cartesian lift of an arbitrary morphism $p(X)\to p(Y)$ in $\mathcal{D}$.

Let us grant, for the moment, that $p$ is an inner fibration and then try to proceed. Let $e'$ be such a morphism. We can consider $a:=q(e')$, a morphism $Z\to q(p(Y))$ in $\mathcal{E}$. For simplicity, assume both $q$ and $q\circ p$ are 'globally' Cartesian fibrations.

Choose a $(q\circ p)$-Cartesian lift $f:X'\to Y$ of $a$. By assumption on $p$, $g:=p(f):p(X')\to p(Y)$ is also a $q$-Cartesian lift of $a$, and testing $g$'s own lifting property against $e'$ we can find a $2$-simplex $\sigma\in\mathcal{D}_2$ witnessing $p(f)\cdot h\simeq e'$, for some $h:p(X)\to p(X')$ a morphism in the fibre $\mathcal{D}_Z\sim\mathcal{C}_Z$; we can choose some $h':X\to X'$ in $\mathcal{C}_Z$ with $p(h')$ being homotopic with $h$ over $Z$. Thus, there is a potentially different $2$-simplex $\sigma'\in\mathcal{D}_2$ witnessing $p(f)\cdot p(h')\simeq e'$. Inside the $\infty$-category $\mathcal{C}$ we can choose a morphism $e:X\to Y$ which then lifts $e'$ up to homotopy. When we ultimately come to take fibres over the point $e'$ of the mapping space, that $e$ and $e'$ represent the same morphism in $h\mathcal{D}$ means we may as well assume $p(e)=e'$ on the nose.

By proposition $2.4.1.3$, we know $f$ is also a $p$-Cartesian morphism. However, as $h'$ is not at all guaranteed to itself be an equivalence or more generally a $p$-Cartesian morphism itself, we cannot infer that $e$ is a $p$-Cartesian lift of $e'$. To me, this attempt seems natural and about the most you can do, while following your nose, without doing serious, extra work, but it's insufficient and I cannot think of another way of trying to construct a Cartesian lift of $e'$.

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If $ab^2$ is a sum of three squares, then so is $a$. How to see it quickly?

Here $a, b$ are positive integers, and the squares are the squares of integers. This follows from Legendre's three squares theorem, but is there a direct way?

(Comment on this)

How to find roots of xcos(x)-sin(x) [closed]

As stated; my professor was unsure on analytically finding the second solution of xcos(x) - sin(x) = 0 within [0,2π]. More generally, how would you go about finding the general form for roots in (-inf, inf)?

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Average distance between numbers of the form $2^{a}3^{b}$

I want to order all numbers of the form $2^a3^b$. I need to find the average distance between a random consecutive pair.

For example, in case of a random consecutive pair $2^{n'}$ and $2^{n'+1}$, the distance is $2^{n'}$. Similarly, whats the distance between $2^{a_1}3^{b_1}$ and $2^{a_2}3^{b_2}$ such that there is no number of the form $2^{a}3^{b}$ between them.

Seems like an interesting problem.

(Comment on this)

Alternating sum of powers

Fix a positive integer $k$. Is it true that for large enough $n$ there exist signs such that $$\pm 1^k\pm 2 ^k\pm \dots \pm n^k=(n(n+1)/2\, \text{mod}\, 2)? $$

It is not hard to show that such sums may be made bounded (since $2^{k+1}$ consecutive signs may be chosen so that the contribution of corresponding summands equals to 0.) But I do not see immediate reasons why it can not be made so ultimately small.

(Comment on this)

Direct summands of a module $M_R$ and those of $E:=\text{End}(M)$

Let $M_R$ be a right $R$-module with $E=\text{End}(M)$. Let $\bigoplus_{\lambda \in \Lambda} e_\lambda M$ be a summand of $M_R$ where $e_\lambda ^2 = e_\lambda \in E$. Without getting involved in the details (which we don't need by the way), it is true that the sum $\bigoplus_{\lambda \in \Lambda} e_\lambda E \subseteq E$ is direct.

My question is: Is it true that $\bigoplus_{\lambda \in \Lambda} e_\lambda M \subseteq^\oplus M \Longrightarrow \bigoplus_{\lambda \in \Lambda} e_\lambda E \subseteq^\oplus E_E$. If the statement is in general false, under what conditions is it true?.

Edit: Is the following argument correct?

As $\bigoplus_{\lambda \in \Lambda} e_\lambda M \subseteq^\oplus M$, for each $\lambda\in \Lambda$ let $\xi_\lambda:M\to M$ be the canonical projection onto $e_\lambda M$. Hence, $\xi^2_\lambda = \xi_\lambda \in E$, $\xi_\lambda M = e_\lambda M$ and $\xi_\lambda \xi_{\lambda'}=0$ for $\lambda \neq \lambda'$. Note that $\xi_\lambda E = e_\lambda E$ for each $\lambda$.

We show that $\bigoplus_{\lambda\in \Lambda} e_\lambda E = \bigoplus_{\lambda\in \Lambda} \xi_\lambda E \subseteq^\oplus E_E$. As $\bigoplus_{\lambda \in \Lambda} \xi_\lambda M \subseteq^\oplus M$, we have $\bigoplus_{\lambda \in \Lambda} \xi_\lambda M = \alpha M$ for some $\alpha^2=\alpha \in E$. We claim that $\bigoplus_{\lambda \in \Lambda} \xi_\lambda E = \alpha E$. Let $\lambda\in \Lambda$ and $x\in M$ be arbitrary. As $\xi_\lambda M \subseteq \alpha M$, then $\xi_\lambda(x)=\alpha(y)$ for some $y\in M$. So, $\alpha \xi_\lambda(x)=\alpha^2(y)=\alpha(y)=\xi_\lambda(x)$. Hence, $\xi_\lambda = \alpha \xi_\lambda$ on $M$ which proves that $\xi_\lambda \in \alpha E$ for each $\lambda$. Thus, $\xi_\lambda E \subseteq \alpha E$, so $\bigoplus_{\lambda \in \Lambda} \xi_\lambda E \subseteq \alpha E$. Conversely, let $x\in M$ be arbitrary. We have, as $\alpha M \subseteq \bigoplus_{\lambda \in \Lambda} \xi_\lambda M$, $\alpha(x)=\xi_{\lambda_1}(y_1)+\ldots+\xi_{\lambda_k}(y_k)$ where $\lambda_{\ell} \in \Lambda$ and $y_\ell \in M$ for each $\ell=1,\ldots,k$. By the orthogonality of the family $\{ \xi_\lambda \}_{\lambda \in \Lambda}$, for each $\ell$, $\xi_{\lambda_\ell} \alpha(x)=\xi_{\lambda_\ell}(y_\ell)$. So, $\alpha(x)= \xi_{\lambda_1}\alpha(x)+\ldots+\xi_{\lambda_k}\alpha(x)$ which implies, as $x\in M$ is arbitrary, $\alpha= \xi_{\lambda_1}\alpha+\ldots+\xi_{\lambda_k}\alpha \in \bigoplus_{\lambda\in \Lambda} \xi_\lambda E$. Consequently, $\alpha E \subseteq \bigoplus_{\lambda\in \Lambda} \xi_\lambda E$. Hence, $\bigoplus_{\lambda\in \Lambda} e_\lambda E = \bigoplus_{\lambda\in \Lambda} \xi_\lambda E = \alpha E$, i.e. $\bigoplus_{\lambda\in \Lambda} e_\lambda E \subseteq^\oplus E_E$.

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"Three great cocycles" in Complex Analysis as cohomology generators

In his lecture notes, C. McMullen discusses "the three great cocycles" in Complex Analysis: the derivative $$f\mapsto\log f',$$ the non-linearity $$f\mapsto (\log f')'dz$$ and the Schwarzian derivative $$f\mapsto \left(\frac{f'''}{f'}-\frac32\left(\frac{f''}{f'}\right)^2\right)(dz)^2$$ I would like to understand what "cocycle" means here precisely, and in which sense these maps are "the only" cocycles.

For diffeomorphisms of the circle, this is very clearly explained, e. g., in the book by Ovsienko and Tabachnikov. Namely, $\mathrm{Diff}(S^1)$ is a group acting naturally on each of the spaces $F_\lambda(S^1)$ of tensor densities of degree $\lambda$. The corresponding (group) 1-cohomology is trivial unless $\lambda$ equals $0$, $1$, or $2$, where it is one-dimensional and generated by the above cocycles.

Question: is there an equally simply formulated statement in the context of complex variable, e. g. treating the above cocycles as generators of a group (or perhaps groupoid?) cohomology? This MO answer by user8248 to Is there an underlying explanation for the magical powers of the Schwarzian derivative? suggests that the group is $\operatorname{Aut}$(meromorphic functions), but, at least naïvely, this seems to coincide with a subgroup of the Möbius group.

Update: Let me restate the question in simpler terms. The maps above satisfy the cocycle equation $$ D(g\circ f)(z)=(f'(z))^\lambda D(g)(f(z))+D(f(z)), $$ for $\lambda=0,1,2$, respectively. For a tensor density $\psi$, one can define the coboundary $$ D_\psi:f\mapsto (f')^\lambda(\psi\circ f)-\psi, $$ satisfying the same equation. So, the question is: what are the explicit conditions guaranteeing that the three maps above are the only, up to coboundaries, solutions to the cocycle equation?

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What are dessins d'enfants?

There was an observation that any algebraic curve over Q can be rationally mapped to P^1 without three points and this led Grothendieck to define a special class of these mappings, called the Children's Drawings, or, in French, Dessins d'Enfants (his quote was something like "things as simple as the drawings...").

I'm not an expert in this field, so could somebody please write more about those dessins, and what things they are related to? What's their importance? How does the cartographic group act on these?

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