I wish to cite that the automorphism groups of $X_1(N)$ have already been completely calculated, and what they are, but I am having difficulty finding this calculation in the literature.

I have found in multiple [22] papers [29] on related topics a reference to "F. Momose, Automorphism groups of the modular curves X1(N). Preprint."

I cannot find it. I am wondering if this paper ever made it to publication, or where a copy of it might be? If not, is there a published reference on the topic?

Galaz-Garcia and Guijarro proved the geometrization of closed (compact, boundaryless) Alexandrov 3-spaces. Part of the strategy was to use the so-called ramified double cover $\tilde{X}$ of the space $X$. This ramified cover is a smooth $3$-manifold. Being this the case, the space $X$ would be isometric to a Riemannian $3$-orbifold.

I don't quite follow why then, it's not immediate that the geometrization of $X$ follows from the geometrization of $3$-orbifolds?

Jenny is traveling at a rate that allows him to go 34 miles in 39 minutes. What is her average speed, in miles per hour?

Consider the optimization problem \begin{array} \ \min\text{ / } \max &f(x_1,\cdots,x_n) \\ \text{subj. }& g(x_1,\cdots,x_n)=\text{constant} ,\end{array} where $g$ is a symmetric function, i.e. $g(x_1,\cdots,x_n)=g(\sigma(x_1,\cdots,x_n))$ where $\sigma\in\text{Symm}\{x_1,\cdots,x_n\}$. There is a principle (sometimes called Purkiss principle see here) which states, under some mild assumptions, that if $f$ is also symmetric, then the optimal solution is of form $x_1=\cdots=x_n$(see, also other post).

**Q.** Are there other versions of Purkiss principle that address aforementioned problem, meaning not necessarily symm. objective function but symmetric constraints?

Let X be a smooth projective connected curve over $\mathbb{C}$ and let $n>1$ be an integer. Let $Y= Sym^n_X$ be the $n$-th symmetric product of $X$.

Is there, for every $i$, a nice formula for the Hodge decomposition of $H^i(Y,\mathbb{C})$?

If not, what part of the Hodge diamond can be described easily?

Let $A$ be a quiver algebra with an acyclic quiver and primitive idempotents $e_i$. The Cartan matrix $C_A$ of $A$ is defined as the matrix with entries $dim(e_i A e_j)$ and the Coxeter matrix $\phi_A$ of $A$ is defined as $\phi_A=-C_A^{-1} C_A^T$. The Coxeter polynomial of $A$ is defined as characteristic polynomial of $\phi_A$. The Coxeter polynomial is a derived invariant and thus derived equivalent algebras share the same Coxeter polynomial.

Is the following true:

A is derived equivalent to a path algebra $KQ$ of Dynkin type if and only if it has the same Coxeter polynomial as $KQ$?

In case the answer is no, is this true in case $A$ is additionally a Nakayama algebra with a linear quiver?

I think this is at least true when $A$ is a Nakayama algebra with a linear quiver (corresponding to a Dyck path) and there is computational way using trivial extensions and representation-finiteness of those trivial extensions to test it, but it gets very ugly when $Q$ is of type $D_n$ ($E_n$ can be done with the computer and indeed, for $E_6$ it is true. It is also true for Dynkin type $A_n$ and $n \leq 8$ and Dynkin type $D_n$ for $n \leq 6$).

Let $F$ be a continuously differenable function over $\mathbb{R}$. Let $\Omega$ be a bounded subset of $\mathbb{R}^2$. Assume that for every $w\in L^2(\Omega)$ then $v(x)=F(w(x))$, $x\in \Omega$, is also in $L^2(\Omega)$. Can we say that $$\|F(w_1(\cdot)+\eta w_2(\cdot))\|_{L^2(\Omega)}$$ is continuous in $\eta$?

I am trying to find a closed form solution for the generating function $x+x^2+x^4+x^8+...=\sum_{i=1}^\infty x^{2^i}$. I began with the fact that the positive integers can be neatly partitioned into sets $2^i*(2\mathbb{Z}^+-1)$ for i from 0 to /infty along with some other summation manipulation to get from $$x+x+x^2+x^3+...=\sum_{i=0}^\infty\sum_{j=0}^\infty(x^{2^i*(2j+1)})$$ To $$x+x^2+x^4+x^8+...=\frac{x}{1-x}+\sum_{i=0}^\infty\frac{x^{3*2^i}}{x^{2*2^i}-1}$$ But I am unsure of how to proceed from there. To clarify, a closed form solution would be a finite summation or product.

Consider a motivating example:

Let $E\in \mathbb{Q}[y][x]$ be of degree $n=2$ (in $x$) and separable when viewed as a member of $\mathbb{Q}[x,y]$. Therefore we can calculate it's roots in $\mathbb{Q}[y][\sqrt[n]{D_E(y)}]$[1] where $D_E(y)\in\mathbb{Q}[y]$ is the discriminant of $E$

If we look for rational solutions $(x,y)$ for $E$ we must have $D_E(y)=z^n$ for some rational $z$ and therefore we can see that every such solution gives rise to a rational point $(y,z)$ on the curve $E': z^n=D_E(y)$.

Now this is not a bijection on the rational points of the curves, so we might as well make it one:

We can express $E'$ as a function of $y$ and $x(y,z)$ only (due to the fact that $D_E$ is a symmetric function of the roots) and writing $w=x(y,z)$ we can get a new curve $E''(y,w)$ with the same rational points as $E$.

[1] We can do all of the above in degrees $3, 4$ as well, only we would have to repeat the construction a number of times. We only need the solvaibilty of $S_n$ for $n\leq 4$ for the construction.

Example:

Let $E$ denote $x^2+3xy^2+2y^4+4$

$\Rightarrow D(y)=y^4-16$

$E(F)=\{(\frac{-3y^2+\sqrt{D(y)}}{2}, y) \cup (\frac{-3y^2-\sqrt{D(y)}}{2}, y): y\in F\}$ for every field $F/\mathbb{Q}(y)(\sqrt{D(y)})$

Now $(x,y)\in \mathbb{Q}^2 \iff (y, \sqrt{D(y)})\in\mathbb{Q}^2$

And we can now write $z^2 = y^4 - 16$ and we want different $z$s that give rise to the same $x$ to yield the same solution. We can see that if $x(y,z) = x(y, z')$ then we must have either $z=z'$ or $z'=-z$ and therefore we care about solutions to $z^2-y^4+16$ up to the sign of $z$ which we know how to solve (this is a pet example as we could solve both equations without doing this either way).

Now I want this this to lift this operation to an endomorphism on some subspace of curves with $E \mapsto E''$ (that preserves all rational points), the problem is that we seem to not have much control on the degree of $E''$ (in fact it seems to increase so this surely can't work in the present form).

All the above construction is just repeated application of basic Galois theory on the ground field $\mathbb{Q}(x_1,...,x_n)$

**Question:** Can this be replicated in some other ground field to make the
following a well defined operation on some subspace of curves (and still preserve points in some "field of definition", or at least say these points lie over each other, if this field of definition is not the same between $E$ and $E''$)?

This comes from playing around with some examples, and I might be missing something basic that would not make this possible, so the answer to the question is probably "only trivial classes", but maybe there is interesting theory behind this.

I think this can only apply in some generalization of "theory of solvability by radicals" but am not sure what nature should this have.

Suppose $v\in R^n$ is a constant unit vector. $P_l$ is a random projection matrix to an $l$ dimensional subspace of $R^n$ which is uniformly sampled from $G(l,R^n)$ which is the collection of all $l$-dimensional subspace in $R^n$. What is the upper bound of the following: $$\mathbb{P}(P_lv\leq\delta)$$ What is the order of the above probability as $\delta\to0$?

Also, when $l=n$, the above probability is the indicator function $1_{\{\delta\geq1\}}$. Can we derive an upper bound of the above probability for $l<n$ such that when $l\to n$, the upper bound tends to $1_{\{\delta\geq1\}}$?

The simple formula (x - y) / (x + y) has me scratching my head, where is it used?

I don't remember seeing it during my engineering studies, it looks a bit familiar but not sure if I should know it or if it is a bit of a rare thing.

I found mention of it in a specification of a Russian *analogue computational device*. Many such devices might have been used for military or space programs to calculate trajectories or something that needed fast solutions to targeting systems.

It is from the family of vintage CRT storage devices and has the model designation "ЛФ2" but not very much information is to be found on the internet.

These devices were variously used to do scan conversions (between TV standards), static pattern generation, polar to Cartesian conversions (for radar), static echo background cancellation (for radar), image integration (for vector displays, film printers and scanners). The other more popular use was for digital storage memory for the earlier computers.

So I am wondering if this tube would have some interesting modern application or is it just something that a microcontroller will work out in a microsecond these days.

My first question on MO and I cannot think of any more tags so please add some if you can figure something appropriate.

A private facebook group can only be joined if someone from the group invites you. Originally, the group has two members : Adam and Eve. Occasionally a random person from the group invites a new member. Someone is part of Adam's clique if they are Adam themselves or if someone from Adam's clique invited them.

a) What is the probability of Adam's clique having k members when the Facebook grouo has 4 members? ( k= 1,2,3)

b) Let X be the number of people in Eve's cliquewhen Facebook has n memberd. Give the distribution of x

I need help, for a) is it okay if i say for $K = 1$ probability is 1/4 For k=2 then P=1/2 and When k=3 then P= 1/4

Then for part b) finding the distribution, I dont know how to solve that, How can i use mathematical induction to solve it

Please help

Denote the set of prime numbers by $P$. Let $u,v,m,n \in \mathbb{N}-\{0\}$ satisfy: $m \leq n$, $\gcd(u,m)=1$ and $\gcd(v,n)=1$.

Is it possible to find $L \in \mathbb{N}$, such that $u+Lm \in P$ and $v+Ln \in P$? (different primes, probably).

Of course, by Dirichlet's theorem on arithmetic progressions, we can find $L$ such that one of $\{u+Lm,v+Ln\}$ is prime. But are those $L$'s 'dense enough' to guarantee that both $\{u+Lm,v+Ln\}$ are primes?

A similar, hopefully a less difficult question, is as follows:

Is it possible to find $L \in \mathbb{N}$, such that:

(i) $u+Lm < v+Ln$;

(ii) $u+Lm = p \in P$;

(iii) $p$ does not divide $v+Ln$?

From the answer to this question, it is clear that there exists $L \in \mathbb{N}$ such that: (i) $u+Lm < v+Ln$ and (ii) $u+Lm = p \in P$. The problem is how to guarantee that $p$ will not divide $v+Ln$?

Thank you very much!

**A remark about the answer:** Let us concentrate on the special case mentioned in the answer: $m=n=1$, $u=1$, $v=3$, so $A:=1+L$ and $B:=3+L$. There is a major difference between asking: 'Are there infinitely many $L$'s such that $A,B \in P$' (very very difficult question) and 'Does there exist $L$ such that $A,B \in P$',
which is a very easy question that has a positive answer, for example $L=2$ yields $(A,B)=(3,5) \in P^2$. Therefore, I expect that my (first) question has a positive answer, or maybe I am missing something, and even finding only one such $L$ is a difficult task? (I do not require that $A$ and $B$ will be greater than a given number).

Let $A\in\mathbb{R}^{n\times n}$ be a diagonalizable matrix with real and strictly positive eigenvalues (note that $A$ is not required to be symmetric).

**My question.** Do there exist an orthogonal matrix $T\in\mathbb{R}^{n\times n}$ and a symmetric positive definite matrix $P\in\mathbb{R}^{n\times n}$ such that
$$
TAPT^\top = D+S,
$$
where $D\in\mathbb{R}^{n\times n}$ is a diagonal matrix with *positive* diagonal entries and $S\in\mathbb{R}^{n\times n}$ is a skew-symmetric matrix?

Of course, if the diagonal entries of $D$ are not required to be positive then the answer is in the affirmative (see, e.g., this related question).

Let's define the n-th degree Chebyshev polynomials by

$$ T_{n} (x)=\cos(n\arccos(x)).$$

Find a polynomial $P$ such that

$$\mid y- P (x) \mid$$

is minimal, using the first three Chebyshev polynomials for the following data:

$$ \begin{bmatrix} x & -1 & -0.5 & 0 & 0.5 & 1 \\ y & 0.6346 & 0.6565 & 1 & 1.5230 & 1.5756 \end{bmatrix}. $$

How could we manage to approach such a problem?

This question is actually from *MSE*. I had to post it here due to the lack of response there even after placing a bounty. Here goes the question

Let tangents be drawn to the curve $y=\sin x$ from the origin. Let the points of contact of these tangents with the curve be $(x_k,y_k)$ where $x_k\gt 0 ;(k\ge 1)$ such that $x_k\in (\pi k, (k+1)\pi)$ and $$a_k=\sqrt {x_k^2+y_k^2}$$ (Which is basically the distance between the corresponding point of contact and the origin i.e. the length of tangent from origin) .

I wanted to know the value of

$$\sum_{k=1}^{\infty} \frac {1}{a_k ^2}$$

Now this question has just popped out in my brain and is not copied from any assignment or any book so I don't know whether it will finally reach a conclusion or not.

I tried writing the equation of tangent to this curve from origin and then finding the points of contact but did not get a proper result which just that the $x$ coordinates of the points of contact will be the positive solutions of the equation $\tan x=x$

On searching internet for sometime about the solutions of $\tan x=x$ I got two important properties of this equation. If $(\lambda _n)_{n\in N}$ denote the roots of this equation then

$$1)\sum_n^{\infty} \lambda _n \to \infty$$ $$2)\sum_n^{\infty} \frac {1}{\lambda _n^2} =\frac {1}{10}$$

But were not of much help.

I also tried writing the points in polar coordinates to see if that could be of some help but I still failed miserably.

On trying a bit more using some coordinate geometry I found that the locus of the points of contact is $$x^2-y^2=x^2y^2$$

Hence for third sum we just need to find $$\sum_{k=1}^{\infty} \frac {\lambda _k ^2 +1}{\lambda _k ^2 (\lambda _k ^2 +2)}=\sum_{k=1}^{\infty} \frac {1}{\lambda _k ^2} -\sum_{k=1}^{\infty} \frac {1}{\lambda _k ^2 (\lambda _k ^2 +2)}=\frac {1}{10} -\sum_{k=1}^{\infty} \frac {1}{\lambda _k ^2 (\lambda _k ^2 +2)}=\frac {1}{10} -\sum_{k=1}^{\infty} \frac {1}{2\lambda _k ^2} +\sum_{k=1}^{\infty} \frac {1}{2(\lambda _k ^2 +2)} =\frac {1}{20}+\frac {1}{2}\sum_{k=1}^{\infty} \frac {1}{\lambda _k ^2 +2} $$

Now for the second summation I did think about it to form a series but for the roots to be $\lambda _k^2 +2$ we just need to substitute $x\to \sqrt {x−2}$ in power series of $\frac {\sin x-x\cos x}{x^3}$ and then get the result but it was still a lot confusing for me.

Using $x\to\sqrt {x-2}$ in the above power series and using *Wolfram Alpha* I have got a series. So we need ratio of coefficient of $x$ to the constant term so is the value of second summation equal to
$$\frac {5\sqrt 2\sinh(\sqrt 2)−6\cosh(\sqrt 2)}{4(2\cosh(\sqrt 2)−\sqrt 2\sinh(\sqrt 2))}?$$

Is this value correct or did I do it wrong?

I would also like to know if there is some other method to solve this problem

Call two Dyck paths $D_1$ and $D_2$ derived equivalent in case their corresponding Nakayama algebras are derived equivalent (The Dyck path of a Nakayama algebra with a linear quiver is just the top boundary of its Auslander-Reiten quiver, so we can identify Nakayama algebras with a linear quiver with Dyck paths). Derived categories of Nakayama algebras appeared in the literature (see for example https://www.sciencedirect.com/science/article/pii/S0001870813000182 ) but it seems to be wide open to classify Dyck paths with respect to derived equivalences (or?).

Call a Nakayama algebra (or the corresponding Dyck path) bouncing in case the Kupisch series is of the form $[a_1+1,a_1,...,3,2,a_2+1,a_2,...,2,...,a_r+1,a_r,...,3,2,1]$.

The truth of the conjecture in What are the periodic Dyck paths? would imply the following: In case a Dyck path $D_1$ is derived equivalent to a bouncing Dyck path $D_2$, then also $D_1$ is bouncing. Thus, it would give that being bouncing is a derived invariant.

Question 1:

Do we know whether it is true that being bouncing is a derived invariant? That is: In case a Nakayama algebra $D_1$ is derived equivalent to a bouncing Nakayama algebra $D_2$, then also $D_1$ is bouncing?

Question 2:

Is it known which bouncing Nakayama algebras with a fixed number of simple modules are derived equivalent to eachother? How many equivalence classes (up to derived equivalence) exist of bouncing Nakayama algebras?

Question 3:

Can a quiver algebra with acyclic quiver be derived equivalent to a quiver algebra with non-acyclic quiver? Can a Nakayama algebra with linear quiver be derived equivalent to a Nakayama algebra with a non-linear quiver?

edit: In case the conjecture of What are the periodic Dyck paths? is true and question 1 of this thread together with the guess that there might be only one equivalence class (answering question 2), one would have the following nice equivalences for a Dyck path $D$ of length $n$:

1.$D$ is bouncing (combinatorial characterisation).

2.$D$ is iterated tilted of Dynkin type $\mathcal{A}$ (homological characterisation).

3.The Coxeter matrix of $D$ has period $n+1$ (linear algebraic characterisation).

4.The trivial extension of $D$ is a Brauer tree algebra (representation-theoretic characterisation).

5.The coxeter polynomial is equal to $\sum\limits_{k=0}^{n}{x^k}$ (polynomial characterisation).

The equivalence of 2. and 4. is well known by a result of Happel .

Question 1 and 2 had an easy positive answer using classification results from the literature. But maybe there is a more direct proof? Question 3 remains open.

Let $X$ be a curve (proper, smooth, ...) over a finite field $\mathbb F_q$ where $q$. Suppose also that $\mathbb F_q$ contains the $p$-th roots of unity, in this case we have the following (unique) chain of field extensions: $$\mathbb F_q \subset \mathbb F_{q^p} \subset \dots \subset \mathbb F_q^{(p)}$$ where each step in the extension is degree $p$ and the union of all these extensions is defined to be $\mathbb F_q^{(p)}$.

I am interested in the $\ell$ (possibly equal to $p$) torsion of the degree $0$ Picard group of $X_{\mathbb F_q^{(p)}}$. If we consider the algebraic closure $X_{\overline{\mathbb F_q}}$, then the $\ell$ torsion here looks like $P = (\mathbb Q_\ell/\mathbb Z_\ell)^{2\lambda}$ for $\lambda \leq g(X)$, the genus.

One strategy to get information about the Picard group of $X_{\mathbb F_q^{(p)}}$ would be to consider the Galois group $\Gamma = \mathrm{Gal}(\overline{\mathbb F_q}/\mathbb F_q^{(p)}) \cong \prod_{p' \neq \ell}\mathbb Z_{p'}$ and it's action on $P$.

Can this strategy prove the following:

1) If $\ell = p$, then the $\ell$-torsion in the Picard group over $\mathbb F_q^{(p)}$ is of the form $(\mathbb Q_\ell/\mathbb Z_\ell)^r$ for $r \leq 2\lambda$.

2) If $\ell \neq p$, then the $\ell$-torsion in the Picard group over $\mathbb F_q^{(p)}$ is finite.

Ideally, I would like to avoid any knowledge of the zeta function of curves.

The proposition 1.9 from "Duality, Trace and Transfer" by Dold and Puppe states that:

Given a commutative ring $R$, a chain complex of $R$-modules is strongly dualizable in $Ho(Ch(R))$, the homotopy category of chain complexes, iff it has the same homotopy type of a strongly dualizable object in $Ch(R)$.

However, they omit the proof (an I can't find the references cited there). I was wondering how can one prove such a thing.

Let $P(M,G)$ be a principal bundle. Giving a connection on $P(M,G)$ means two equivalent things. One as an assignment of subspace of $T_pP$ for each $p\in P$ and another as a $\mathfrak{g}$ valued $1$ form.

Given connection as a $\mathfrak{g}$ valued $1$ form $\omega$ on $P(M,G)$ we have associated a $2$ form (differential of $\omega$ and modifying it little bit) which we called the curvature form for the connection, denoted by $\Omega$.

Given connection as a distribution $p\mapsto H_pP\subseteq T_pP$, we know what it means to say a curve is

**horizontal curve**. We write $v\sim u$ to mean that $v$ and $u$ are related by**horizontal curve**. Given $u\in P$ we have defined what is called a Holonomy bundle $P(u)=\{v\in P:v\sim u\}$ based at $u$ and what is called a Holonomy group $\Phi(u)=\{a\in G:u\sim ua\}$ based at $u$.

Given $p\in P$ we have $\Omega(p)(X(p),Y(p))\in \mathfrak{g}$ for all $X(p),Y(p)\in T_pP$.

As $\Phi(u)$ is a Lie subgroup of $G$, we have, Lie algebra of $\Phi(u)$ to be subalgebra of $\mathfrak{g}$.

So, from curvature form, we get a collection that is a subset of $\mathfrak{g}$ and from holonomy group/holonomy bundle we get a collection that is a subset of $\mathfrak{g}$ and considering suitable restrictions, it is natural to expect these two sets to be same. **Ambrose-Singer Holonomy theorem** what is the precise relation between these. The set
$\Omega(p)(X(p),Y(p))$ where $p\in P(u)$ and $X(p),Y(p)\in H_pP$ generated the Lie algebra of $\Phi(u)$.

More precisely, we have the following.

Let $P(M,G)$ be a Principal bundle where $M$ is connected and paracompact. Let $\Gamma$ be a connection in $P$, $\Omega$ the curvature form, $\Phi(u)$ the holonomy group with reference point $u\in P$ and $P(u)$ the holonomy bundle through $u$ of $\Gamma$. Then the Lie algebra of holonomy group $\Phi(u)$ is equal to the subspace of $\mathfrak{g}$, Lie algebra of $G$, generated by all elements of the form $\Omega_v(X,Y)$ where $v\in P(u)$ and $X$ and $Y$ are arbitrary horizontal vectors at $v$.

This is very natural thing one can expect but the proof is not that simple (I am not saying I could have guessed the statement, I am saying that if some one show me this statement and ask me to guess if that is true or not I would have said that it is mostly true). The proof given in Kobayashi involves some thing called involutive distribution. I want to know if there is any alternative proof for this result.

We have to prove $\mathfrak{g}'$ which is Lie algebra of $\Phi(u)$ is spanned by elements of the form $\Omega(p)(X(p),Y(p))$ where $p\in P(u)$.

Let $v\in \mathfrak{g}'$ i.e., there exists a smooth curve $\alpha:[-1,1]\rightarrow \Phi(u)$ such that $\alpha(0)=e$ and $\alpha'(0)=v$. Then, I want to prove $v=\sum a_i \Omega(p_i)(X(p_i),Y(p_i))$. So, I have to look for some elements $p_i\in P(u)$ and then some horizontal vectors $X(p_i),Y(p_i)$. It is not clear what choice of $p_i$ should one make. I am not very sure if this approach works or not. Any suggestion on making this approach work are welcome. Any other alternative proof is also welcome. An exposition of proof given in Kobayashi and Nomizu is also welcome.