In homotopy theory, we can construct the $\infty$-category of spaces from the ordinary category of manifolds $\rm Man$, by freely co-completing it, imposing gluing relations, and homotopy invariance. This sort of procedure is described model-categorically by Dugger in https://arxiv.org/abs/math/0007070.

My question is: Is it possible to perform an analog of this construction, where instead of homotopy invariance, we only impose that $\mathbb R^1$ is invertible? That is, we freely make $\mathbb R^1 \times -$ an equivalence.

Is there a source for this sort of localization of multiplication by objects in a symmetric monoidal category? It is analogous to the process of constructing spectra.

If this sort of localization exists, what do we get in this case? I suppose it's possible that we just get the ordinary homotopy category again. Alternately, we could get something related to the proper homotopy category.

Given $n$ points in general position in the plane, let $P_n$ be the maximum proportion of the $\binom{n}{3}$ triangles with three acute angles. What is the limit $\lim\limits_{n \rightarrow \infty} P_n$?

An upper bound of $\frac{7}{10}$ is shown in the proof of the sixth question of the 1970 edition of the International Mathematical Olympiad.

In the opposite direction, we can establish a trivial lower bound of $\frac{2}{9}$ by splitting the points into three equal clusters of $\frac{n}{3}$ points, with one cluster at each vertex of an equilateral triangle.

Despite the previous edit, we can establish a trivial lower bound of $5/9$ by spreading the points at $A$ so they are on a circle, centre $B$; the points at $B$ so they are on a circle, centre $C$, and the points at $C$ are on a circle centre $A$. If the points at $A$ are on an arc of width $N^{-13}$, those at $B$ on an arc of width $N{-9}$ and those at $C$ on an arc of width $N^{-11}$; and further the angle at $B$ is $\pi/2-N^{-7}$, then all triangles $A_iA_jB_k, B_iB_jC_k$ and $C_iC_jA_k$ are acute.

I would be very grateful for a reference to the following results (which are, I think, true, though I never saw it in the literature).

Let $G\subset GL(n,{\Bbb C})$ be $U(n)$, abd $A\in GL(2n,{\Bbb R})$ an endomorphism which satisfies $AGA^{-1}=G$. Then $A\in {\Bbb R}^* \times U(n)$.

Let $G\subset GL(n,{\Bbb C})$ be the group $Sp(n)$ of quaternionic Hermitian matrices, and $A\in GL(2n,{\Bbb R})$ an endomorphism which satisfies $AGA^{-1}=G$. Then $A\in {\Bbb R}^* \times Sp(n)\times Sp(1)$.

Many thanks in advance.

Let $F_n$ be the $n$-th Fibonacci number, started with $F_0=0,F_1=1$, and consider the matrices
$$M_n=\pmatrix{F_{n+3} & F_{n+1} \\ F_{n+2} & F_{n}}.$$

Let
$$\pmatrix{\alpha_n & \beta_n \\ \gamma_n & \delta_n}=M_1\cdot M_2\cdot \ldots \cdot M_n .$$

It is easy to see by computer, that the quotients $\frac{\alpha_n}{\gamma_n},\frac{\beta_n}{\gamma_n},\frac{\delta_n}{\gamma_n}$ are converging. I found that $$ \lim_{n\to\infty} \frac{\delta_n}{\gamma_n}={\phi^2}$$ where $\phi=\frac{\sqrt{5}-1}{2}$. Unfortunately I can't find the other two limit, but numerically it seems to be that $$ \lim_{n\to\infty} \frac{\alpha_n}{\gamma_n}\approx 1.3876267558043602953$$ $$ \lim_{n\to\infty} \frac{\beta_n}{\gamma_n}\approx 0.53002625701851519880$$ Can anyone give me a "nice" description of these numbers? Alternatively, it would be enough, if someone can decide whether these numbers are algebraic or not.

(I remark that the limit of $\alpha_n / \gamma_n$ is the most interesting for me, since it has a continued fraction expansion: $[1;2,1,1,2,1,1,1,2,...]$ and so on, the number of ones between twos increasing by one. It is a non-periodic, badly approximable number.)

Let $C_{j,k}^l$ ,usually called class structure constants, eg Jansen and Boon and/or JQ Chen, be the number of times the class $l$ is generated from the product of classes $j,k$ and $c_j=c_{-j}$ (a fronting minus sign just means the inverse class which is the same class for ambivalent classes) is the order or number of elements in class $j$. It is given (i) $c_j c_k=\sum C_{j,k}^l c_l$ , (ii) $C_{j,k}^l=C_{(-j),(-k)}^{-l}$ both of which I can understand. It is also stated in Jansen and Boon (iii) $c_l C_{j,k}^l=c_j C_{k(-l)}^{-j}$ which I can't prove in the general case. I can reconcile it in the special identity case $l=1$ , say, where $c_l=c_{-l}=1$ meaning the class of the single identity element in which case the only non zero result in (iii) is $j=-k$ and $C_{j,-j}^1=c_j , C_{j,-1}^j=C_{j,1}^j=1$. Years ago I think I did reconcile the general case but now cannot and have tried in vain to search for the proof which has to be given in some texts likely quite a few decades ago with no luck. I can reconcile, eg by $g_k$ I mean a group element of group G from class $k$, that for some triplet of elements $g_j, g_k,g_l$ lhs. of (iii) says $g_j g_k=g_l$ which means $g_k g_l^{-1}=g_j^{-1}$ which corresponds to rhs. But this isn't saying much nor any quantitative one to one correspondance. I could go on and on in a round about way with examples etc. showing it is true but need a direct concise proof which seems should not be all that difficult.

Greenberg conjectured that given $E/K$, there always exists $E^\prime/K$ such that $E'$ is isogenous to $E$ and $\mu(E^\prime)=0$. Michael Drinen has shown that for an elliptic curve $E/K$, it is possible that there are no isogenous elliptic curves with $\mu=0$. He says that it is possible when $E$ satisfies some very restrictive hypothesis (which can happen only at finitely many primes, if at all). This was a counter-example to Greenberg's original conjecture. The conjecture is still open for $K=\mathbb{Q}$.

Is there any work which indicates that hypothesis N (as Drinen calls it) is the only obstruction to the conjecture? (ie, if $E/K$ does not satisfy hypothesis N, then Greenberg's conjecture should hold?)

Gromov conjectured in 1985 and LLarull proved in 1998 that: If $g > g_0$ on the sphere, then there exists some point p on the sphere with $Sc(p) < Sc_0(p)$. Here $g, g_0$ are Riemannian metrics and $g_0$ is the standard metric on the sphere. $Sc$ is the scalar curvature. His proof is by contradiction and uses some non-vanishing index on the sphere.

Is the theorem also true for sectional curvature? Or is it the case that if one Riemannian metric is bigger than the other one on the sphere, then there must exist some point such that the scalar curvature at that point is smaller than the other one?

In general, is it true that *the bigger the metric, the smaller the curvature* (sectional, Ricci or scalar curvaure), at least in the pointwise setting?

For a dynamical project, I need to bound uniformly a quantity defined from a metric on the segment $I=[0,1]$ over *all* metrics inducing the usual topology. The details do not matter too much, I wonder whether any non-completely-trivial information is known about such metrics, whether there is some sort of rough classification. I moreover have some monotony property that may help.

So, given a topological space $X$, let $\mathcal{D}(X)$ be the set of all metrics $d:X\times X\to\mathbb{R}_+$ that moreover induce the topology of $X$. Let say that a subset $\mathcal{M}\subset \mathcal{D}(X)$ is *minorizing* if for all $d'\in\mathcal{D}(X)$, there exist some $d\in\mathcal{M}$ and some $C>0$ such that $d\le C d'$.

My question would be

Can we describe some (all ?) inclusion-minimal minorizing subsets of $\mathcal{D}(I)$? Or at least, a small minorizing subset of $\mathcal{D}(I)$?

but I have honestly not yet given enough thought to it yet; my real question is rather whether any information on this sort of questions is already available in the literature. It is very difficult to find keywords that would extract relevant references without being flooded with automatic classification research.

**Edit:** to make things more precise (but I do not want to reduce the scope of relevant reference: any one remotely related to the question would be nice), let me give a more focused question:

Is the set of metrics that are conjugate to the Euclidean metric by a homeomorphism minorizing?

Let $H$ denote the irreducible component of $\text{Hilb}^{3t+1}\mathbb{P}^3$ whose general member corresponds to a non-singular twisted cubic. Let $C$ be a subscheme lying in the boundary of $H$ and assume it lies in a surface $S \subseteq \mathbb{P}^3$.

Then why is it possible that we can find families $C_R, S_R \subseteq \mathbb{P}^3_R$ over a DVR $R$ with fraction field $K$ such that

1) $C_R \subseteq S_R$

2) The generic fiber $C_K$ is a non-singular twisted cubic

3) $C \subseteq S$ are the closed fibers of the family.

The authors in "Hilbert Scheme Compactification of the Space of Twisted Cubics": https://www.uio.no/studier/emner/matnat/math/MAT4230/h10/undervisningsmateriale/Hilbertscheme.pdf make the claim on page 4 (pg 763), line 7 of the proof. Although they are studying embedded points, this claim seems to be something more general about flat limits?

More generally, is it true that if something on the boundary of my component in a Hilbert scheme lied in a hypersurface, then I could find a family over a DVR like above?

I was working on some mathematics of Wasserstein GAN and found out a seemingly interesting research problem but *I am not quite sure whether it has already been studied in some recent literature of Optimal Transport Theory* (As far as I know, it hardly is)

The observation is as follows (which has been validated with some toy simulations).

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**Problem (not rigorously stated)**

Let $\mu, \nu$ probabilistic measures on *regular* manifolds $\mathcal{M}, \mathcal{N}$, $C^{\infty}(\mathcal{M}, \mathcal{N})$ the set of continuous mapping from $\mathcal{M}$ to $\mathcal{N}$, and $\Pi(\mu,\nu)$ the set of measures on $\mathcal{M}\times\mathcal{N}$ s.t. its marginal distributions are respectively $\mu, \nu$.

Consider the following optimization problem

$$ (*) = \min_{T\in{C^{\infty}(\mathcal{M}, \mathcal{N})}} \inf_{\gamma\in\Pi(\mu,\nu)} \int d^{2}(T(p), q) d\gamma(p,q) $$ where the cost function can be considered as the $L_2$ distance on $\mathcal{N}$'s total space as a real vector space.

My question is that whether $(*) \propto h(\chi(\mathcal{M}), \chi(\mathcal{N}))$ where $h$ is certain metric function and $\chi(\cdot)$ denotes *Euler characteristic*. **Generally, would it be possible that the minimum cost of the Wasserstein game is deeply related with the difference between some topological invariants of underlying manifolds'?**

==================================================================

Look forward to any feedbacks and welcome discussions and potential references :D. I am willing to provide details of my toy experiments if one is interested in this problem.

It is well-known that any positive rational number can be written as the sum of finitely many distinct unit fractions. This is easy since $$\frac1n=\frac1{n+1}+\frac1{n(n+1)}\quad\text{for all}\ n=1,2,3,\ldots.$$ In 2015 I thought that this easy fact should have a further refinement which is somewhat sophisticated. Note that the series $\sum_p\frac1{p-1}$ and $\sum_p\frac1{p+1}$ (with $p$ prime) diverge just like the harmonic series $\sum_{n=1}^\infty\frac1n$. Also, for any positive integer m there are infinitely many primes $p$ congruent to $1$ (or $-1$) modulo $m$ (by Dirichlet's theorem). Motivated by this, I made the following conjecture in Sept. 2015.

**Conjecture**. For any rational number $r>0$, there are finite sets $P_r^-$ and $P_r^+$ of primes such that
$$r=\sum_{p\in P_r^-}\frac1{p-1}=\sum_{p\in P_r^+}\frac1{p+1}.$$

This appeared as Conjecture 4.1 of this published paper of mine. For example, $$2=\frac1{2-1}+\frac1{3-1}+\frac1{5-1}+\frac1{7-1}+\frac1{13-1}$$ with $2,3,5,7,13$ all prime, and $$1=\frac1{2+1}+\frac1{3+1}+\frac1{5+1}+\frac1{7+1}+\frac1{11+1}+\frac1{23+1}$$ with $2,3,5,7,11,23$ all prime. Also, \begin{align*}\frac{10}{11}=&\frac1{3-1}+\frac1{5-1}+\frac1{13-1}+\frac1{19-1}+\frac1{67-1}+\frac1{199-1} \\=&\frac1{2+1}+\frac1{3+1}+\frac1{5+1}+\frac1{7+1}+\frac1{43+1}+\frac1{131+1}+\frac1{263+1} \end{align*} with $2,3,5,7,13,19,43,67,131,199,263$ all prime. The reader may see more numerical data in my detailed introduction to this conjecture.

After learning this conjecture from me, Prof. Qing-Hu Hou and Guo-Niu Han checked my above conjecture seriously and their computational results support my conjecture. For example, in 2018 Prof. Han found 2065 distinct primes $p_1<\ldots<p_{2065}$ with $p_{2065}\approx 4.7\times10^{218}$ such that $$\frac1{p_1+1}+\ldots+\frac1{p_{2065}+1}=2.$$

My question is whether the above conjecture is true. I would like to offer 500 US dollars as the prize for the first correct solution.

**Remark**. Let $r$ be any positive rational number, and let $\varepsilon\in\{\pm1\}$. As the series $\sum_p\frac1{p+\varepsilon}$ (with $p$ prime) diverges, there is a unique prime $q$ such that $$\sum_{p<q}\frac{1}{p+\varepsilon}\le r<\sum_{p\le q}\frac1{p+\varepsilon}.$$ Thus
$$0\le r_0:=r-\sum_{p<q}\frac1{p+\varepsilon}<\frac1{q+\varepsilon}\le1.$$
If $r_0=\sum_{j=1}^k\frac1{p_j+\varepsilon}$ with $p_1,\ldots,p_k$ distinct primes, then $p_1,\ldots,p_k$ are all greater than $q$, and
$$r=\sum_{p<q}\frac1{p+\varepsilon}+\sum_{j=1}^k\frac1{p_j+\varepsilon}.$$ Therefore it suffices to consider the conjecture only for $r<1$.

Recently I was playing several rounds of the game of pairs with my children. I was surprised that almost every time, one matching pair was adjacent (either next to each other in a row, or vertically). This led to the following question.

Let $n$ be a positive integer. Consider the set $$C_n = \{1,\ldots, 2n^2\}\times\{0,1\}.$$ We say that $(k,0)$ and $(k,1)$ for $k\in \{1,\ldots,2n^2\}$ is a *matching pair* in $C_n$.

Let $G_n = \{1,\ldots, 2n\} \times \{1,\ldots,2n\}\subseteq \mathbb{R}^2$ be thought of the "gaming grid" where we place the numbers. Formally: shuffling and distributing the cards corresponds to a bijection $\varphi: C_n \to G_n$. So we can define the *minimum distance of a matching pair* by $$M_n = \min\big\{||\varphi(k,0), \varphi(k,1)||: k\in \{1,\ldots,2n^2\}\big\}.$$

By $||\cdot||$ we denote the Euclidean distance.

$M_n$ is a random variable, so we can calculate its expected value $E(M_n)$.

**Question.** In plain English: if we play the game of pairs on an ever growing quadratic play-field, can we expect to find some matching pair in a reasonably close distance? Or more formally: Do we have $\lim_{n\to\infty} E(M_n) < \infty$?

Let $G$ be the free pro-p group with $n$ generators; we can assume $n=2$ and generators are $x,y$ at first. Let $G_0=G$ and $G_{n+1}=[G,G_n]$ be the group generated by certain commutators. Then the quotient group $\Delta_n=G/G_n$ is a nilpotent group of class $n$.

Now I want to know more about the structure of the groups $\Delta_n$. For example, $\Delta_1=\mathbb{Z}_px\bigoplus\mathbb{Z}_py$ and $\Delta_2$ is isomorphic to the group of upper triangular matrix with diagonal equal to 1 and coefficients belonging to $\mathbb{Z}_p$.

Moreover, I guess that $\Delta_{n+1}/\Delta_n$ (it is abelian) can be viewed as a free $\mathbb{Z}_p$-module, am I right? (If so, then the structure of the associated graded Lie algebra is known, c.f. Serre "Lie Algebras And Lie Groups".)

Any ideas or references will be welcome. Thanks!

Consider function $f: (\mathbb{R}^{d})^{n} \rightarrow \mathbb{R}$ with spatial invariance property of the form : $f(x_1,x_2,...,x_n) = f(x_1 + \zeta, x_2 + \zeta,..., x_n + \zeta)$ for $\zeta \in \mathbb{R}^{d} $.

I want to show that $\int \limits_{(\mathbb{R}^{d})^{n}} \exp(\,f(x_1,...,x_n) \,) dx_1..dx_n \,=\, \infty $

I would like some help if possible.

My only idea is to fix ball $B(0,r) \subseteq \mathbb{R}^{d}$ and $\zeta \in \mathbb{R}^{d}$ and then compute $\int \limits_{{B(0,r)}^{n}} \exp(\,f(x_1,...,x_n) \,) dx_1..dx_n \, = \int \limits_{{B(\zeta,r)}^{n}} \exp(\,f(y_1 - \zeta,...,y_n - \zeta) \,) dy_1..dy_n = \int \limits_{{B(\zeta,r)}^{n}} \exp(\,f(y_1,...,y_n) \,) dy_1..dy_n \, $ where the first equality comes from setting $x_i = y_i - \zeta $ and the second from the spatial invariance property.

So we get that the integral of this strictly positive f over any ball is the same. But does it imply that the integral over $(\mathbb{R}^{d})^{n}$ is $\infty$ ?

I'm trying to solve below probability question ;

fuel station has $5$ filling points of which $3$ are use to fill petrol and remaining $2$ are use to fill diesel for different kind of vehicles.It was noticed that an average $3$ minutes to fill fuel to any vehicle (diesel/petrol).number of petrol vehicles to enter the fuel station within an hour estimated is $24$ and the diesel vehicle estimated 36 within an hour.

want to find below probabilities :

- all petrol filling points are idle
- all diesel filling points are busy
- all petrol filling points are busy
- all filling points are busy

Let $G$ be a torsion free group. Let $\alpha$ be an element in $\mathbb CG$, the group algebra of $G$, with $\|\alpha\|_1=1$ and assume that

- $\{1,\alpha,\alpha^2,\dotsc\}$ is linearly independent,
- $(\alpha^n)_{n\in\mathbb N}$ converges to 0 in strong operator topology, so in particular $\lim_n\|\alpha^n\|_2=0$.

Assume that $K$ is the closed linear span of $\{1,\alpha,\alpha^2,\dotsc\}$ in $\ell^2(G)$. Is $\{1,\alpha,\alpha^2,\dotsc\}$ a Schauder basis for $K$?

Let's suppose we have an objective function $\max_\limits{x} \sum_\limits{i} f_i(x_i)$ with the constraint that $\ x_i \geq 0, \sum_\limits{i} x_i = 1$.

Each function $f_i$ is continuous and differentiable within the boundary, with the following properties: $f_i(x_i) \geq 0$, $f_i(0) = 0$, $f'_i(x) > 0$, $f'_i(1) > 0$, $f_i''(x_i) < 0$.

By adding a Lagrange term and taking the partial derivative, we get

$\frac{\partial \sum_i f_i(x_i) + \lambda (1 - \sum_i x_i)}{\partial x_i} = f'_i(x_i) - \lambda$.

By setting the above expression to zero, moving $\lambda$ to the right side of the equation and multiplying an $x_i$ with each side, we would come up with a fixed-point update $x_i^\mbox{new} \propto x_i \cdot f'_i(x_i)$.

Would this iterative algorithm monotonically increase the concave objective and is it guaranteed to converge?

I asked this question at MSE now I repeat it at MO:

Let $A,B,C$ be $3$ unital $C^*$ algebras. Assume that we have the following short exact sequence of $C^*$-algebras:

$$0\to A\to C\to B\to 0$$

Assume that $A,B$ are generated by their projections. Is $C$ necessarily generated by its projections, too?

Assume that $A,B$ are von Neumann algebras, is $C$ necessarily a von Neumann algebra, too?

Does the last question has an obvious answer when $A,B$ (hence $C$) are commutative algebras?

Let $k$ be a field of characteristic $0$, and let $\mathcal{C}= \mathbf{Vect}_k^{\leq 0}$ be the $\infty$-category of vector spaces concentrated in degrees $\leq 0$. Consider the category $\mathbf{Pr}(\mathcal{C}):= \operatorname{Fun}(\mathbf{CAlg}(\mathcal{C}), \mathcal{S})$ of prestacks over $k$, where $\mathcal{S}$ is the $\infty$-category of spaces or $\infty$-groupoids.

Suppose we have a grouplike prestack $G \in \mathbf{Pr}(\mathcal{C})$. That is, a functor $G: \mathbf{CAlg}(\mathcal{C}) \to \mathbf{Sp}^{\text{cn}}$, where $\mathbf{Sp}^{\text{cn}}$ is the $\infty$-category of connective spectra, thought of as a functor to spaces by composing with the forgetful functor $\mathbf{Sp}^{\text{cn}} \to \mathcal{S}$. We can then form the iterated classifying spaces $B^nG$.

Suppose we have a nice enough stack $X \in \mathbf{Pr}(\mathcal{C})$ (e.g. a perfect stack). When will the category $\mathbf{QCoh}(\text{Map}(X,B^nG))$ of quasicoherent sheaves on the mapping stack be compactly generated? Is the assumption that $X$ be perfect enough? Do we have to make any assumptions on $G$?

Sage 50 Error solution Ireland 353-766-803-988. The main cause of this error 1628 is:-

Corrupt download file of sage. An invalid process of installation. Incomplete installation of sage in the system. Windows may be corrupt due to frequent installation. When the .NET framework is expired. Missing company files corrupt software. Some programs are delete Virus and malware corrupt the file of sage Let Us discuss how to Troubleshoot the Error as follows:-

SOLUTION 1:-

Start system [Windows, Mac, iPhone, iPad]. Insert the installation CD. Install the .NET file in the system. Install the setup file. SOLUTION 2:-

Run Microsoft easy fix utility. SOLUTION 3:-

Repair all windows Run system scan > remove the virus, malware, and threats. Delete junk and temporary files Check notification > run updates. Undo recent changes. Uninstall sage 50 software Re-install accounting software Run windows system Clean system installation SOLUTION 4:-

Rename the InstallShield folder > My computer > Click InstallShield folder > Click Rename > Press Enter SOLUTION 5:-

Install latest window installer > Install the .exe file > Click install or download.