# Recent MathOverflow Questions

### A question on a certain family of complete intersection varieties

Math Overflow Recent Questions - Thu, 02/01/2018 - 17:30

Let $k$ be a field of characteristic zero. Given integers $2 \leq s \leq r < n$, define the variety $X_n$ in $P_K^n$ with coordinates $y_0, \cdots, y_n$ and $K=k(u_0, \cdots, u_n)$ ( where $u_i$'s are independent transcendental vairables) by the folloing $n-r$ equatuions:

$$f_{i-r}(y_0, y_1, \cdots, y_n)= \begin{vmatrix} 1 & 1 & \cdots & 1 & 1 \\ u_0& u_1& \cdots & u_r & u_i\\ \vdots & \vdots & \cdots & \vdots & \vdots \\ u_0^r & u_1^r & \cdots & u_r^r& u_i^r\\ y_0^s & y_1^s & \cdots & y_r^s & y_i^s \end{vmatrix} =0, \ (r+1 \leq i \leq n).$$

It is easy to see that for fixed $r$, $s$, and $n > (sr + 1)/(s−1)$, $X_n$ is a smooth complete intersection variety of general type with $\dim(X_n)=r$. For a set of pairwise distinct elements $b=\{b_0, b_1, \cdots, b_n\}$ in $k$, letting $u_i=b_i$, one can get $X_{b, n}$ a smooth variety defined over $k$.

Question: Is it possible to show that $X_{b, n}$ is a Geometric Mordellic variety, i.e., $X_{b, n}$ does not contain subvarieties which are not of general type over $\bar{k}$ the algebraic closure of $k$?

### Orbifolds with maximal diameter

Math Overflow Recent Questions - Thu, 02/01/2018 - 17:12

Orbifolds with positive curvature and maximal diameter are investigated in this article, by J. Borzellino. Theorem 1 of the article states:

Let $\mathcal{O}$ be a complete $n$-dimensional Riemannian orbifold with Ricci curvature satisfying $\mathrm{Ric}_\mathcal{O}\geq n-1$ and diameter $\mathrm{diam}(\mathcal{O})=\pi$. Then $\mathcal{O}$ is a good orbifold.

Recall that a Riemannian orbifold is good when it is a quotient orbifold of a Riemannian manifold by an effective, (proper) discontinuous, isometric action. Also, $\mathrm{Ric}_\mathcal{O}\geq n-1$ means that the local charts $\tilde{U}\to U\cong \tilde{U}/\Gamma$ satisfy $\mathrm{Ric}_{\tilde{U}}\geq n-1$, and the diameter of $\mathcal{O}$ is the diameter of its underlying metric space (with the metric induced by the Riemannian structure).

The article aims for an analog of Cheng's maximal diameter sphere theorem, claiming (Theorem 2) that $\mathcal{O}$ must then be a quotient of the sphere by a finite group of $\mathrm{O}(n+1)$.

My question is: why isn't the $\mathbb{Z}_p$-teardrop orbifold, i.e., the $2$-orbifold with underlying space $\mathbb{S}^2$ and with a single cone singularity of order $p$, a counterexample to Theorem 1?

The proof of Theorem 1 is based on volume comparison and, sincerely, I don't quite understand it, mostly because the notation does not distinguish between $\mathcal{O}$ and its underlying space (that I will denote by $|\mathcal{O}|$). Am I missing something or the correct interpretation of the conclusion in Theorem 1 is actually: "$|\mathcal{O}|$ can be realized as the underlying space of a good Riemannian orbifold"?

### Is there any pseudoprime that pass this test above tested range, or any prime that does not show these ending patterns?

Math Overflow Recent Questions - Thu, 02/01/2018 - 17:08

if the recurrence sequence is defined by the following foormula, $d_{n + 3} = 3d_{n + 2} - d_{n + 1} - 2d_n$ where $d_1 = 1, d_2 = 3$ and $d_3 = 7$, this produce the following complex sequence $$1, 3, 7, 16, 35, 75, 158, 329, 679, 1392, ...$$.

if this sequence is evaluated over finite field, i mean modulo some interger, $p$ $$(d_{n + 3} = 3d_{n + 2} - d_{n + 1} - 2d_n) \text{mod}\,p$$ It follows that, if $p$ is prime number, the last three term if evaluated up to $n = p + 2$, last three terms i mean $d_{p}, d_{p + 1}$ and $d_{p + 2}$ fall in one of the following ending sequence $\{...,1, 3, 7\}$ or $\{...,4, 7, 14\}$, if $p$ is smaller than these ending patterns, then modulo by $p$. the only exception for prime numbers less than $7$

i have tested all numbers up to $100,000$, no pseudo prime exist

Prior Studied Sequences

i studied Lucas numbers for existence of pattern like this, but there existed many pseudo prime below $10,000$ e.g $377$. then i studied recussive sequence defined by the following formula $d_{n+3} = d_{n + 2} + 2d_{n + 1} - d_n$. in this sequence pseudo prime exist above $10^6$, may be this sequence can pull up this boundary

Questions
• is there any pseudo prime above tested range of this sequence
• can anyone explain to me further why this patterns appear to prime only, if mo pseudoprime pass this test
UPDATES ONE DAY AFTER POST Facts about Pseudo primes of this recurence
• all pseudo primes are divisible by primes. $p_1, p_2,..., p_n$, all of these primes are in the form of $10k + 1$. My limitation is computational power, i can not test big numbers
• all prime factors of these pseudo primes have period of $p - 1$. prime numbers of the form $10k - 1$ also have periods of the form $p -1$. no counter example for other primes having this form of period. if any found?
• if a composite number $n$ have all of its prime factors of this form, $p = 10k + 1$. the resulting period of $n$ get reduced by factor of $10$ and sometimes more than that. the gcd of resultant period $r$ and $n - 1$ is equal to $r$. this explain why $nth$ period occurs at $p - 1$ and the number get evaluated as prime number while not

### History of Fargue-Fontaine curve

Math Overflow Recent Questions - Thu, 02/01/2018 - 16:49

In this paper, Pierre Colmez wrote about some history of the Fargue-Fontaine curve. In this schedule of London Number Theory Study Group, Fargues was said to give a talk on November 15th on " Where does the curve and the conjecture come from? What happened in Trieste, Orsay and Berkeley? ". " What happened in Trieste " is well-documented by Colmez. Is there any reference about "what happened in Orsay and Berkeley"?