Can one prove by elementary means (such as Paul Erdös' proof of Bertrand's Postulate) that every prime greater than 5 is less than the sum of the two primes immediately preceding it?

This fact holds true in absolute geometry, and I would like to see an elementary synthetic proof not using the classification of absolute planes (Euclidean and hyperbolic planes) and specific models. Actually I know such a proof (from Skopets - Zharov book), but it uses the third dimension which has to be justified itself.

Do you recall having seen the attached pages in a textbook once? If so, would you be so kind as to share its bibliographic record (or the main items in it) with me below?

A teacher provided us xerox copies of these exercises *maaany* years ago, but I did not have the presence of mind at the moment to ask him where he had gotten them from.

Let me thank you in advance for your attention and for allowing me to pose this question in spite of the fact that it may be a wee bit borderline for the site...

Let $M_{g,n}$ be the moduli space of $n$-pointed curves, and $M_g[m]$ the moduli space of (unpointed) curves with $m$-level structure.

Fix $m>0$. Is it true that for $n$ large enough, there is a dominant rational map $M_{g,n}\to M_g[m]$?

I am not aware of the existence of a "natural" (i.e. functorial) map $M_{g,n}\to M_g[m]$. For my purposes, any dominant rational map would work, but I understand that it is probably difficult to construct a "non-canonical" one.

Assume $\pi$ is a tempered representation of $GL_n(\mathbb{Q}_p)$. $N_n$ is a maximal unipotent radical of $GL_n$, and $\xi$ is a non-degenerate character of $N_n(\mathbb{Q}_p)$. Let $\Pi$ be the Whittaker model of $\pi$ associated to $N_n$ and $\xi$. Assume $W \in \Pi$.

Then the question is whether $W\otimes \overline{W} \in \mathcal{C}^{\omega}(N_n\times N_n \backslash GL_n \times GL_n, \xi \otimes \xi^{-1}),$ where $\mathcal{C}^{\omega}$ is a Harish-Chandra Schwartz space.

Thank you very much.

Take natural numbers $A_1,B_1,A_2,B_2$ random pairwise coprime in $[n,2n]$ for $n$ large enough and consider the space of solutions to $A_1a+B_1b=0$ and $A_1A_2a+A_1B_2b+B_1A_2c+B_1B_2d=0$ spanned by $1\times 2$ and $3\times 4$ matrices $$N_1=\begin{bmatrix} -B_1&A_1 \end{bmatrix}$$ and $$N_2=\begin{bmatrix} -B_2&A_2&0&0\\ -B_1&0&A_1&0\\ 0&0&-B_2&A_2 \end{bmatrix}$$ respectively.

Denote the $\Bbb Q$-linear space spanned by rows of $N_1$ by $T_{A_1,B_1}\subseteq\Bbb Q^2$ and $N_2$ by $T_{A_1,B_1,A_2,B_2}\subseteq\Bbb Q^4$ and denote the set of non-zero $\Bbb Z$ vectors in $T_{A_1,B_1}$ by $T_{A_1,B_1}[\Bbb Z]^\star$ and in $T_{A_1,B_1,A_2,B_2}$ by $T_{A_1,B_1,A_2,B_2}[\Bbb Z]^\star$.

Consider the quantities $$\mu_1((A_1,B_1))=\min_{v\in T_{A_1,B_1}[\Bbb Z]^\star}\|v\|_\infty$$ $$\mu_2((A_1,B_1)\otimes(A_2,B_2))=\min_{v\in T_{A_1,B_1,A_2,B_2}[\Bbb Z]^\star}\|v\|_\infty$$ where $\|v\|_\infty$ is largest coordinate by magnitude of vector $v$. One can show with probability $1-o(1)$ value of $\mu((A_1,B_1))$ is at least $c_1\cdot n$ for a constant $c_1>0$ and of $\mu_2((A_1,B_1)\otimes(A_2,B_2))$ is at least $c_2\cdot n^{2/3}$ for a constant $c_2>0$.

Similarly define $$\mu_k((A_1,B_1)\otimes(A_2,B_2)\otimes\dots\otimes(A_k,B_k))$$ where $$A_1,B_1,A_2,B_2,\dots,A_k,B_k$$ are random pairwise coprime in $[n,2n]$ for $n$ large enough.

It is possible to show that the typical value of $\mu_k((A_1,B_1)\otimes(A_2,B_2)\otimes\dots\otimes(A_k,B_k))$ is at least $$c_k\cdot n^{k/(2^k-1)}$$ for some $c_k>0$ with probability $1-o(1)$ at every given $k\in\mathbb N$.

However what is the minimum $\gamma_k$ in $$\max(A_1,\dots,A_k,B_1,\dots,B_k)-\min(A_1,\dots,A_k,B_1,\dots,B_k)\leq n^{\gamma_k}$$ to get typical value of $\mu_k((A_1,B_1)\otimes(A_2,B_2)\otimes\dots\otimes(A_k,B_k))$ to be at least $$\Omega(n^{k/(2^k-1)})$$ with probability $1-o(1)$ at every given $k\in\mathbb N$? Is $n^{\gamma_k}$ as low as $n^{k/(2^k-1)}$ or does at least $2^k\gamma_k=2^{o(k)}$ hold (at $k=1$ we have $\gamma_1\rightarrow0$)?

.

Consider an undirected graph $K(n,k,i)$, with the all $k$-element subsets of $\{1,\dots,n\}$ as vertices, and two vertices connected by an edge if their sets intersect in less than $i$ elements.

This paper claims on page 74 (Theorem 5.1) that the chromatic number of $K(n,k,i)$ is at least $n-2k+2i$. However, their is a place in the proof of the theorem which I don't understand: They say that

Otherwise we would have two $k$-subsets of color $j$ such that each of them has at least $k - i + 1$ elements in one of two disjoint hemispheres, so their intersection has at most $i-1$ elements which is impossible.

Since the intersection can be of size up to $i-1$ in each hemisphere, don't we only get that it has size at most $2i-2$ in total?

By a *good writer*, I mean someone who is clear and motivates their approach. By *modern*, I mean someone who is currently active in the mathematical community.

Let $K$ be a cyclic cubic number field with conductor $f$ and ring of integers $\mathcal{O}_K$.

Define $K$ to be *blue* if and only if $$\operatorname{Norm}_{K/\mathbb{Q}}(w) = \operatorname{Norm}_{K/\mathbb{Q}}(1-w) = -1\quad\text{for some $w\in K$}.$$
Define $K$ to be *green* if and only if $$\operatorname{Norm}_{K/\mathbb{Q}}(w) = \operatorname{Norm}_{K/\mathbb{Q}}(1-w) = -1\quad\text{for some $w\in \mathcal{O}_K$}.$$ (So green implies blue).

**Question 1:** Are all cyclic cubic number fields blue?

**Question 2:** What is the density of green number fields restricted to blue number fields? That is, defining $$B_N:=\{K:K\text{ is a blue (cyclic cubic) number field of conductor }<N\},$$ $$G_N:=\{K:K\text{ is a green (cyclic cubic) number field of conductor }<N\},$$ what is
$$
\lim_{N\to\infty} \frac{\#G_N}{\#B_N}?
$$
(and does the limit exist?)

**Question 3:** Define $$\mathcal{G}:=\{f: K \text{ is green, where $K$ is a cyclic cubic number field of conductor $f$}\}.$$ What is $\mathcal{G}$ explicitly?

**Remarks:** I wrote some magma code that proved that $K$ is blue for all of the 1822 cubic cyclic number fields given from LMFDB (http://www.lmfdb.org/NumberField/start=0°ree=3&galois_group=C3&count=20). The code also explicitly gives the minimal polynomial of $w$. Here are the first few examples.

\begin{align*} f=7, \quad & t^3 - 2t^2 - t + 1 \\ f=9, \quad & t^3 - 3t + 1 \\ f=13, \quad & t^3 + t^2 - 4t + 1 \\ f=19, \quad & t^3 - 5t^2 + 2t + 1 \\ f=31, \quad & t^3 - (5/2)t^2 - (1/2)t + 1 \end{align*}

The polynomials above prove that $\{7,9,13,19\}\subseteq\mathcal{G}$. Notice that for $f=31$, this polynomial implies $K$ of conductor $31$ is blue, but it may or may not be green.

Suppose $X$ is a connected separable metric space, and each point of $X$ is a cut point.

KP Hart's answer to my last question Does every cut-point space embed into the plane? shows that $X$ may not embed into the plane. But what if $X$ is locally connected and/or locally compact? Then $X$ more closely resembles a dendrite (defined in the linked post), and dendrites embed into the plane. Is there a positive result here?

Let $\alpha$ be a standard contact form on $\mathbb{R}^{2n+1}$. We say that a map $f:\mathbb{R}^k\to\mathbb{R}^{2n+1}$ *contact* if $f^*\alpha=0$.

- Is it true that a $C^1$-contact immersion can be approximated (locally in the supremum norm) by $C^\infty$-contact immersion?
- Is it true that a $C^1$-contact map $f:\mathbb{R}^{k}\to\mathbb{R}^{2n+1}$ can be approximated (locally in the supremum norm) by $C^\infty$-contact maps?

In the case of question 2 I do not pose any restrictions on how large $k$ is. I expect that the answer to question 2 is in the negative and I have an idea how to construct a counterexample so I would like to know if it would be something new or not.

I am also interested in other variants of the questions stated above.

The following is a conjecture due to Littlewood

For any set of non-zero integers $n_1,\cdots,n_k$ the inequality $$\int_0^{2\pi}|1+e^{in_1x}+\cdots+e^{in_kx}|dx\geq C\log k$$ holds. Is this proven to be true(or false)?

Here is a triangular region $T$ whose two curved edges are complicated analytic curves that I know only numerically, but can compute to any desired precision:

And here is a simpler region $H$ whose one curved side is a circular arc:

Both regions have vertex angles of $\pi/5$, $\pi/4$, and $\pi/2$; so there is a unique conformal map $f\colon T\to H$ that takes each vertex of $T$ to the corresponding vertex of $H$. And the map $f$ is holomorphic, not only on the interior of $T$, but also on its boundary.

I have managed to approximate $f$ numerically by a degree-13 polynomial $p(z)$ of a complex variable $z$ whose coefficients are real numbers. So $p$ maps the real axis precisely to itself. And $p$ maps the two curved boundaries of $T$ to analytic curves that are never farther than 0.00003 from the corresponding boundary of $H$. I have thus approximated $f$ to roughly four digits.

But I would like to approximate $f$ much more precisely --- to at least ten digits, maybe twenty. Can anyone suggest a technique by which I could achieve this? Replacing $p$ by a polynomial of higher degree isn't an attractive strategy, since the precision achieved is likely to grow only slowly with increasing degree, while solving for the coefficients of a high-degree polynomial is numerically challenging.

There are Fortran libraries for numerically approximating conformal maps. But they typically use a fixed precision, either single or double; and they deal with conformal maps either to or from the upper half-plane. I am hoping to implement something in Mathematica, so that I can specify a high working precision. And approximating $f$ directly, rather than detouring through the upper half-plane, seems like it should be easier, since $f$ is holomorphic also on the boundary of $T$.

I will append a few remarks about how this problem arose, but I won't give all the details, since the full story is rather complicated.

Suppose that we identify labeled equilateral pentagons $ABCDE$ in the Euclidean plane that differ by any combination of translation, magnification, and rotation --- but that we don't identify a pentagon with its reflection. The resulting moduli space $M$ is known, topologically, to be a compact, orientable, smooth 2-manifold of genus 4. The region $T$ is the triangular subset of $M$ in which the pentagons are convex, are traversed counterclockwise, and have $A$ and $B$ as their narrowest and widest vertices. The region $T$ turns out to be $1/240^\text{th}$ of $M$. (Along the bottom boundary, the widest vertex $B$ straightens out; along the upper-left boundary, the vertices $B$ and $E$ are tied for widest; and along the right-hand curved boundary, the vertices $A$ and $C$ are tied for narrowest.)

A metric on $M$ needs to assign some length to each tangent vector, where a tangent vector at a pentagon $P$ is a recipe for altering the shape of $P$ as function of time. Extend all pairs of the edges of $P$ until they intersect. We can then define the length of that tangent vector as the Euclidean norm of the vector in $\mathbb{R}^{10}$ whose coordinates are the rates of change of the ten resulting angles, say in radians per time step. Call that the *ten-pairs metric*. (We use all ten intersections, rather than just the five vertex angles, to achieve greater symmetry.)

Equipping the moduli space $M$ with this ten-pairs metric gives us a Riemannian manifold with an isometry group of order 240, a manifold that is partitioned into 240 regions, each isometric to $T$. The Gaussian curvature of this manifold is everywhere negative, but is not constant.

Pulling the metric of the hyperbolic plane back onto $M$ through the conformal map $f$ will allow us to view $M$, with the conformal structure of the ten-pairs metric, as the quotient of the hyperbolic plane by a Fuchsian group, an index-240 subgroup of the (2,4,5)-triangle group.

The opening illustration of $T$ is drawn using a coordinate system that is isothermal for the ten-pairs metric. I compute my isothermal coordinate system using Gauss's technique for reducing the Beltrami partial differential equation, in the real-analytic case, into an ordinary differential equation over the complex numbers. That ODE has the form $q'(t)=F(t,q(t))$ where the function $F$ is algebraic, but is so complicated that solving the ODE in closed form seems quite unlikely. Instead, I have Mathematica solve it numerically.

Thus, while I don't know the curved boundaries of $T$ in closed form, I could easily compute thousands of points along each of them, each point to any desired precision (within reason). Note also that I don't need to refine my approximation to $f$ efficiently; I'll be happy to spend weeks of CPU cycles, is that's what it takes to get $f$ to twenty digits.

The sample pentagons in the drawing of $H$ are positioned using $p$, my degree-13 polynomial approximation to $f$. I would like to approximate $f$ more accurately since I would like to know, for example, just where the two pentagons $P_\text{adj}$ and $P_\text{nadj}$ end up along the boundary of $H$, those being the equilateral pentagons with two right angles, either adjacent or nonadjacent.

How can I calculate the remainder of divide (1!+2!+3!...+2018!) by 18? I have already tried different things but nothing looked like to work.

I'm studying the Ramsey numbers, especially $R(3,6) = 18$

I understand that the proof using the theorem $R(m,n) <R(m-1,n)+R(m,n-1)$ can only prove that $R(3,6)<20$. However by Cariolaro's "On the Ramsey number $R(3,6) = 18$" I understand the proof for $R(3,6)<19$.

Now I try to understand the proof for $R(3,6)>17$, but the graph there is built with some program. I read that it is possible to build a graph to test $R(3,6)>17$ without programs.

In "thesis (Ph. D.)--University of Waterloo, 1966, Chromatic Graphs and Ramsey's Theorem" by J. G. Kalbfleisch this result supposedly exists, but unfortunately I can not find the document. Can someone give the idea of the construction or do you know of any document or article where you do it?

This is of course a very-well known problem, but still let me ask the questions my way. Let $L(s)$ be a "motivic" $L$-function, whatever that means: in particular, it has an Euler product (including at "bad" primes), and a (possibly conjectural) functional equation of standard type with an arithmetic conductor $N$ occurring as $N^{s/2}$ in the factor at infinity. Let $p$ be a prime, denote by $L_p(s)$ the denominator of the Euler factor at $p$, by $d_p$ the degree of $L_p(s)$ in $p^{-s}$, and by $v_p(N)$ the valuation of $N$ at $p$. Finally let $d$ be the degree of the $L$-function, equal both to the number of $\Gamma_{\mathbb R}(s)$ factors and to the generic degree in $p^{-s}$ of $L_p(s)$.

I define a prime to be good if it does not divide $N$ (note that if $L$ comes from a variety with bad reduction at $p$, it may still be a good prime in this sense). For the following statements, please tell me if they are true, or only sometimes true, sometimes false (and in that case possibly some extra info):

1) If $p$ is good then $d_p=d$.

2) If $d_p=d$ then $p$ is good.

3) $v_p\ge d-d_p$

4) I know that one can have $v_p>d-d_p$, but can one give a reasonably simple criterion for equality ?

Let $\mathscr{A},\mathscr{B}$ be abelian categories, the first with enough projectives, together with a right-exact functor $F\colon \mathscr{A}\to\mathscr{B}$ (in my example, it is a tensor product, but I am interested in the general setting). Let $\mathcal{D}^-(\mathscr{A}),\mathcal{D}^-(\mathscr{B})$ be their (bounded-above) derived categories and $\mathbb{L}F$ the left-derived functor of $F$. Consider a chain complex $X_\bullet\in\textbf{comp}(\mathscr{A})$ together with a Cartan–Eilenberg projective resolution $Q_{\bullet\bullet}\to X_\bullet$. Define $P_\bullet=\mathrm{Tot}(Q_{\bullet,\bullet})$ for the total complex (assume all needed boundness, existence of sums, etc.): it gives rise to a quasi-isomorphism $\epsilon\colon P_\bullet\to X_\bullet$. I can then perform two operations, and I'd like to show that they coincide.

The first is to consider the composition of (the extension to complexes of) $F$ with $q_\mathscr{B}\colon \textbf{comp}(\mathscr{B})\to\mathcal{D}^-(\mathscr{B})$: applying this to $\epsilon\colon P_\bullet \to X_\bullet$ I obtain $(q_\mathscr{B}\circ F )(\epsilon)\colon F(P_\bullet)\longrightarrow F(X_\bullet)$ (as a morphism in the derived category $\mathcal{D}^-(\mathscr{B})$). By inverting signs, in order to turn chain complexes into cochain ones, and applying the cohomological functor $H^n\colon\mathcal{D}^+(\mathscr{B})\to\text{Ab}$, I end up with a map $$ \alpha_1=(H^n\circ q_\mathscr{B}\circ F)(\epsilon)\colon H^n(F(P^\bullet))=(\mathbb{L}^nF)(X^\bullet)\longrightarrow H^n(F(X^\bullet)). $$

The second is the edge-map in the corresponding spectral sequence, built as follows. By applying $F$ to the Cartan–Eilenberg resolution we obtain a double complex $F(Q_{\bullet\bullet})$: by construction, this is a second quadrant double complex, but since I prefer to do cohomology rather than homology, I invert signs and consider $F(Q_{-\bullet,-\bullet})$: it is a fourth quadrant double complex. This we can filter by its "second degree" to obtain a fourth quadrant spectral sequence $E_2^{p,q}=H^p(F(Q^{\bullet q}))$ which converges to the hypercohomology of $X^\bullet$, namely $$ E_2^{p,q}=H^p(F(Q^{\bullet,q}))\Longrightarrow \mathbb{L}^{p+q}(X^\bullet). $$ Corresponding to this spectral sequence there is a edge morphism obtained as follows. Since the spectral sequence is concentrated in the fourth quadrant, all differentials landing in the term $E_{2}^{n,0}$ are $0$: it follows that $E_r^{n ,0}$ is a subobject of $E_2^{n,0}$ for all $r\geq 2$ (it is the kernel of the differential $d_{r-1}^{n,0}$) and since the sequence stabilizes we obtain a map $\beta\colon E_\infty^{n,0}\to E_2^{n,0}$. On the other hand, convergency means that $\mathbb{L}^{p+q}(X^\bullet)$ has a dicreasing filtration $\mathrm{Fil}^j(\mathbb{L}^{n}F)(X^\bullet)$ with associated grading $E_\infty^{j ,n-j}$: in particular, $\mathrm{Fil}^{n}(\mathbb{L}^{n}F)(X^\bullet)=\mathbb{L}^{n}(X^\bullet)$ and $E_\infty^{n,0}=(\mathbb{L}^{n}F)(X^\bullet)/\mathrm{Fil}^{n+1}\mathbb{L}^{n}(X^\bullet)=(\mathbb{L}^{n}F)(X^\bullet)$ is a quotient of $(\mathbb{L}^{n}F)(X^\bullet)$. Composing this quotient map with the map $\beta$ constructed above gives us the edge map \begin{equation} \alpha_2\colon (\mathbb{L}^{n}F)(X^\bullet)\longrightarrow E_\infty^{n,0}\longrightarrow E_2^{n,0}=H^n(F(Q^{\bullet,0})=H^n(F(X^\bullet)) \end{equation}

Does anyone know a reference, or an easy proof, that $\alpha_1=\alpha_2$?

Suppose $X$ is a subset of $\{1, \cdots, n\}$ such that the equation $ax_i+bx_j=cx_k+dx_{\ell}$ where $a+b=c+d,$ $1 \leq a,b,c,d \leq O(n^{1/4 })$ and $x_i, x_j, x_k, x_{\ell} \in X,$ has only trivial solution. A solution is trivial if $x_i=x_j=x_{k}=x_{\ell}.$

What can we say about the size of $X?$ Is this possible that $|X|\geq n^{1-o(1)}$?

I think the answer is related to Sidon Sets, but I could not find any references. Any help is greatly appreciated.

Suppose I have a diagonal matrix $D$ whose entries are bounded in absolute value. I also have a matrix $A$ that is positive (entry-wise, so $A_{ij} > 0\ \forall\ i,j$): one can assume that the entries of $A$ are all bounded away from 0 by $a_0$. I would like to understand what the eigenvalues of the matrix $A^T D A$ look like, and in particular how they compare to the eigenvalues of $D$. Can one say anything about $x^TA^T D Ax$ compared to $x^T D x$?

EDIT: If it helps, one can also assume the $A$ is PD.

I'm interested in the quantitative isoperimetric inequality. I am currently reading https://arxiv.org/pdf/1007.3899.pdf and I have some questions regarding lemma 3.5 which states a certain situation in which a minimizer of a certain functional is a strong $\Lambda$-minimizer .

I have several questions regarding the proof.

In the very beginging, why can we w.l.o.g. assume $\alpha(E_j)=\frac{|E_j\triangle B|}{|B|}$? Here $\alpha$ is the Fraenkel assymetry, $B$ is the open unit ball in $\mathbb{R}^n$ and $|B|$ denotes the Lebesgue measure of $B$. My guess is that since $E_j$ is a minimizer for $Q_j$, one has $|E_j\triangle B|\to 0$ for $j\to \infty$. Thus, for $j$ large enough, $\alpha$ becomes smaller so that one can assume that $B$ is a optimal ball ( i.e. the minimum is attained at $x=0$). Is that correct?

The second inequality in (42), i.e. $\frac{9}{4}(\alpha(F)-\alpha(E_j))(\alpha(F)+\alpha(E_j)-2\alpha(W_j))\le C_1|E_j\triangle F|$. By lemma 3.4. it is $|\alpha(F)-\alpha(E_j)|\le\frac{2^{n+2}}{(2^n-1)\omega_n}|E_j\triangle F|$, but I don't know how to estimate the $\alpha(F)+\alpha(E_j)-2\alpha(W_j)$ term using $\alpha(E_j)\le \frac{3}{2}\alpha(W_j)$.

The inequalities in (44) from "Observe now that.." Here I have no idea how to obtain this.

The first inequality on the next page, I don't see why $Q(F)\le \frac{P(B)\alpha(E_j)^2}{P(B_F)\alpha(F)^2}Q(E_j)+(\frac{P(B)}{P(B_F)}-1)\frac{1}{\alpha(F)^2}$. Since $Q(E_j)=\frac{\delta P(E_j)}{\alpha(E_j)^2}$ and the same for $Q(F)$, the inequality is equivalent to $(\delta P(E_j)+1)\frac{P(B)}{P(B_F)}\ge \delta P(F)+1$, but I don't see which "previous estimates" are used to explain this inequality.

And lastly, the next one, how to conclude $Q(F)\le 2Q(E_j)+2$ for j large enough? Maybe $\frac{P(B)\alpha(E_j)^2}{P(B_F)\alpha(F)^2}\le 2$ and $(\frac{P(B)}{P(B_F)}-1)\frac{1}{\alpha(F)^2}\le 2$ for $j$ lare enough.. However, I don't know how to obtain this with (44).

Sorry for posting such basic questions here, but the setting is quite specific and, therefore, I hope it's ok to ask here. Thank you