One form of de Franchis theorem for algebraic curves is the following: let $X$ be an algebraic curve (defined over $\mathbb{C}$ say) with genus $g > 1$. Then there are only finitely many (isomorphism classes of) curves $Y$ with genus $g' > 1$ such that there is a non-constant map $f : X \rightarrow Y$.

My question is the following: suppose that $X, X^\prime$ are two curves of the same genus $g > 1$ admitting a dominant map to the same curve $Y$ of genus $h > 1$. What can be concluded about $X,X^\prime$? Can there be infinitely many isomorphism classes of such $X$, say?

Let $p$ be an odd prime. It is well-known that $$\det[i^{j-1}]_{1\le i,j\le p-1}=\prod_{1\le i<j\le p-1}(j-i)\not\equiv0\pmod p.$$ I'm curious about the behavior of the permanent $\text{per}[i^{j-1}]_{1\le i,j\le p-1}$ modulo powers of $p$. This leads me to formulate the following conjecture.

**Conjecture**. Let $p>3$ be a prime. Then
$$\text{per}[i^{j-1}]_{1\le i,j\le p-1}\equiv\begin{cases}\frac{p-1}2!\ p\pmod{p^2}&\text{if}\ p\equiv1\pmod4,\\0\pmod{p^2}&\text{if}\ p\equiv 3\pmod4.\end{cases}$$

I have verified this conjecture for $p=5,7,11,13,17,19,23$. Note also that $\text{per}[i^{j-1}]_{1\le i,j\le 3-1}=3$.

QUESTION: Is the above conjecture true?

Now I show that $\text{per}[i^{j-1}]_{1\le i,j\le p-1}\equiv0\pmod p$ for any odd prime $p$. Let $g$ be a primitive root modulo $p$. Then $$\text{per}[i^{j-1}]_{1\le i,j\le p-1}\equiv\sum_{\sigma\in S_{p-1}}\prod_{i=1}^{p-1}g^{i(\sigma(i)-1)}=\prod_{i=1}^{p-1}g^{-i}\times\text{per}[g^{ij}]_{1\le i,j\le p-1}\pmod p.$$ Thus we may appy the argument in the related question 316836 to get that $\text{per}[i^{j-1}]_{1\le i,j\le p-1}]\equiv0\pmod p$ since the order of $g$ is the even number $p-1$.

Your comments are welcome!

What is the value of $c$ such that $$\lim_{n\rightarrow\infty}\sum_{k=1}^{n-1}\frac{1}{(n-k)c+\log(n!)-\log(k!)}=1?$$ Numerically, it seems that the answer is $c=\log 2$. But I'd like to see a reason why.

Motivated by Question 316142 of mine, I consider the new sum $$S(n):=\sum_{\pi\in S_{n}}e^{2\pi i\sum_{k=1}^{n}k\pi(k)/n}$$ for any positive integer $n$, where $S_n$ is the symmetric group of all the permutations of $\{1,\ldots,n\}$. Note that $S(n)$ is the permanent of the matrix $[e^{2\pi ijk/n}]_{1\le j,k\le n}$. As $S(n)\equiv\det[e^{2\pi ijk/n}]_{1\le j,k\le n}\pmod2$, it is easy to see that $S(n)\equiv n\pmod2$ in the ring of all algebraic integers.

Suppose that $S(n)\not=0$. Then the coefficient of $x_1^{n-1}\ldots x_n^{n-1}$ in the polynomial $$P(x_1,\ldots,x_n):=\prod_{1\le j<k\le n}(x_k-x_j)\left(e^{2\pi ik/n}x_k-e^{2\pi ij/n}x_j\right)$$ is $\text{per}[(e^{2\pi ij/n})^{k-1}]\not=0$. Applying Alon's Combinatorial Nullstellensatz to the subset $A=\{z\in\mathbb C:\ z^n=1\}$ of the complex field $\mathbb C$, we see that there is a permutation $\sigma\in S_n$ such that $j+\sigma(j)\not\equiv k+\sigma(k)\pmod n$ for all $1\le j<k\le n$. Thus $$\sum_{k=1}^n(k+\sigma(k))\equiv\sum_{j=1}^n j\pmod n$$ and hence $\sum_{k=1}^nk=n(n+1)/2\equiv0\pmod n$. This shows that $n$ must be odd.

Via a computer I find that \begin{gather}S(1)=1,\ S(3)=-3,\ S(5)=-5,\ S(7)=-105,\ S(9)=81,、 \\S(11)=6765=3\cdot5\cdot11\cdot41,\ S(13)=175747=11\cdot13\cdot1229, \\ S(15)=30375=3^5\cdot 5^3,\ S(17)=25219857=3\cdot13\cdot17\cdot38039.\end{gather} Thus it is natural for me to formulate the following conjecture.

**Conjecture**. (i) For each $n=1,3,5,\ldots$, the sum $S(n)$ is an integer divisible by $n$.

(ii) For any odd prime $p$, we have $S(p)\equiv-p\pmod{p^2}$.

Any ideas towards the solution?

Often the most difficult part of venturing into a field as a researcher is to come up with an appropriate definition. Sometimes definitions suggest themselves very naturally, as when you solve a problem and then ask, ‘What if I generalize this a bit?’

Other times they arise only after struggling with a subject and realizing you were looking at it from the wrong angle. An appropriate definition can then make all the difference, by reorganizing the thought and sheding light into the problems, somehow making them sharper and more focused.

I would like to collect evidences and instances of this idea. An answer should be a story of how someone came up with a good definition and how this was crucial to their understanding of a topic. If you talk about someone else, then ideally provide a reference.

(My interest in this is mainly psychological, namely, how the action of naming something somehow brings it into existence and organizes the world around it.)

editIt was suggested that this question is a duplicate of (Examples of advance via good definitions), which asks about good definitions. It was pointed out there, and also here, that basicaly all notions that stood the test of time qualify as good definitions.

I was not asking, although it seems many people construed it this way, for a collection of good definitions, but for a collection of **stories** that showed how a proper definition actually **changed** the perception about a field.

A typical such story would have someone saying "Wait a moment! I should not be dealing with [concept A] at all! That's the wrong way to approach this. Instead I should define this other guy, [concept B], and then everything will make a lot more sense!"

I realize this is hard to make precise, so I understand if the question gets closed.

Let's denote a sentence $P$ as "** weak Godel sentence of theory $T$**", if and only if
$$[\neg (T \vdash P) \wedge \neg (T\vdash \neg P)] \wedge [Con(T)=Con(T+P) \wedge Con(T)=Con(T+ \neg P)] $$

In English this is: $P$ is independent of $T$ and the addtion of $P$ or $\neg P$ to $T$ doesn't prove the consistency of $T$.

Let's denote a sentence as ** complex** if it has a proper subformula of it that is a sentence, or when de-prenexed it results in a sentence that has a proper subformula of it that is a sentence. A sentence is

Let's fix the language of $T$ to a classical first order logic language that doesn't contain any constants in its signature. By sentence it is meant the usual meaning of a fully quantified formula (i.e. has no free variables).

Definition: $$T \text{ is complete for simple sentences below } Con(T) \iff \forall P (P \text { is a weak Godel sentence of }T \to P \text { is complex})$$

*In other words:* all sentences if the addition of them or their negation to $T$ doesn't result in a theory that can prove the consistency of $T$, that are simple, then those sentences are decidable by $T$.

Question: is it possible to have a theory that meets Godel's incompleteness criteria and yet is complete for simple sentences below its consistency level?

I am looking for a cool and simple (but not worn-out) application of Bochner--Weitzenböck type formulas which could be used as a motivation.

What is your favourite example?

**P.S.** Here is one which I like, but want more:
Assume two discs $\Delta_1$ and $\Delta_2$ with common boundary $\gamma$ bound a convex set in a positively curved three-dimensional manifold $M$.
Then
$\int_{\Delta_1}k_1\cdot k_2$
is small if the maximal angle between the discs on $\gamma$ is small;
here $k_i$ denote the principle curvatures of $\Delta_1$.

Let $(Y,g_y)$ be a closed Riemannian manifold, $Z^T=[-T,T]\times Y$ with the standard product metric.

When we consider the Neunmann boundary condition, i.e. $$\begin{cases} \Delta u=f\\ \frac{\partial u}{\partial\nu}=0,~\partial Z^T, \end{cases}$$ if we let $\int_{Z^T}f=0$, we know that we have the estimate $$\|u\|_{L^p_{k+2}}\leq C (\|f||_{L^p_k}+\|u\|_{L^p_k}).$$

**Q**
How to show that the constant $C$ does not depend on $T$.

Dirichlet's Arithmetic Progression Theorem states that:

Given $a, b\in\mathbb{Z^+}$ with $(a,b)=1$, then $a+kb$ is prime for an infinite number of $k\in\mathbb{Z^+}.$

For any given $a$ and $b$ let $K_{a,b}=\{k\mid a+kb \text{ is prime}\}$.

Also consider another Dirichlet-Valid AP $c+jd$. Restrict $j$ to $j_k\in K_{a,b}$.

Is $c+j_k d$ prime an infinite number of times?

I am trying to do some work on some math conjecture. I am testing the conjecture numbers using very large math numbers (1000+ digits ). I am currently using python to test these numbers.

In the conjecture's calculation, I quickly need to multiply and divide a large number several times. Python can do this quickly(~3000 calculations per minute). Is there any other faster software to achieve this.

Let say I have two different sites $(\mathcal{C},I)$ and $(\mathcal{D},J)$ for an ordinary topos $\mathcal{T}$. I.e.

$$Sh(\mathcal{C},I) \simeq \mathcal{T} \simeq Sh(\mathcal{D},J)$$

And we want to know if one also have an equivalence of the $\infty$-topos of sheaves of spaces on these sites:

$$ Sh_{\infty}(\mathcal{C},I) \overset{?}{\simeq} Sh_{\infty}(\mathcal{D},J)$$

This question is about finding an explicit counter example where this is not the case. But I'll give a bit more context.

If my understanding is correct, there are two cases where one can say something:

**1)** *If $ Sh_{\infty}(\mathcal{C},I)$ and $Sh_{\infty}(\mathcal{D},J)$ are hypercomplete*, or more generally if one only care about their hypercompletion. Indeed, one always has:

$$ Sh_{\infty}(\mathcal{C},I)^{\wedge} \simeq \ Sh_{\infty}(\mathcal{D},J)^{\wedge}$$

(the $^{\wedge}$ denotes hypercompletion) this is because one can show that they are both equivalent to the $\infty$-category attached to category of simplicial objects of $\mathcal{T}$ with the Jardine-Joyal model structure on simplicial sheaves. And the equivalences of this model structure only depends on the underlying ordinary topos.

**2)** *If $\mathcal{C}$ and $\mathcal{D}$ have finite limits*, then one gets the equivalence:

$$ Sh_{\infty}(\mathcal{C},I) \simeq Sh_{\infty}(\mathcal{D},J)$$

This follows from J.Lurie Lemma 6.4.5.6 in Higher topos theory. This lemma assert that under these assumptions, there is a natural equivalence between

$$ Geom( \mathcal{Y}, Sh_{\infty}(\mathcal{C},I)) \simeq Geom( \tau_0 \mathcal{Y} , Sh(\mathcal{C},I) ) $$

Where $\mathcal{Y}$ is any $\infty$-topos, $Geom$ denotes the spaces of geometric morphsisms, either of $\infty$-topos or ordinary toposes, and $\tau_0 \mathcal{Y}$ is the ordinary topos of homotopy sets of $\mathcal{Y}$. THis in particular gives a universal property to $Sh_{\infty}(\mathcal{C},I)$ which only depends on the $Sh(\mathcal{C},I)$ and so this implies the isomorphisms above.

So what about the general case ? I have quite often heard that this was not true in general, and I'm willing to believe it. But I would really like to see a counter-example !

In some of the places where I have seen asserted that this is not true in general, people were pointing out to the examples where $Sh_{\infty}(C,J)$ is not hypercomplete, and often these examples fall under the assumption of the second case.

In a comment on this old closely related question of mine, David Carchedi mention a counter-example due to Jacob Lurie which indeed avoids both situation... But I havn't been able to understand how it works, and it does not seem to appears in print. If someone can figure it out I'll be very interested to see the details.

Also a counter example to this lemma 6.4.5.6 mentioned above (without the assumption that the site has finite limit) would probably produce an answer immediately. Moreover (assuming such a counterexample exists) there must exists one where the topology is trivial, so I guess there has to be some relatively simple counter examples.

I'm a statistician working on a research project dealing with metrics on SPD matrices, specifically the log-Euclidean $d_{LE}(X, Y) = \|\log(X) - \log(Y)\|$ and the Riemannian metric $d_{R}(X, Y) = \|\log(Y^{-1/2}XY^{-1/2})\|$, where the norm is the Frobenius norm.

I understand that the two metrics are closely related, and that the LE metric is something like a linearization of the Riemannian metric, but I haven't been able to find a reference that makes this precise. Any comments or suggestions are appreciated.

For some references, [1] is a good intro to different metrics for SPD matrices, and [2] gives some more details on the context I'm working in.

I am familiar with Barnette's Lower Bound Theorem on the number of facets a $d$-dimensional simplicial convex polytope with $n$ vertices can have. Is there a similar result for a general (i.e. not necessarily simplicial) $d$-polytope with $n$ vertices?

Let $Y$ be an integral affine variety over $\mathbb{C}$ and $K$ be its function field. How to find a sufficiently small Zariski open set of $Y$ such that it is isomorphic to $K(\pi,1)$? Here $\pi$ is the absolute Galois group of $K$.

This statement is from the Galois Cohomology of SGA 4.5.

On sait qu’il existe des ouverts de Zariski arbitrairement petits qui pour la topologie classique sont des $K(\pi, 1)$. On ne sera donc pas surpris si l’on considère Spec(K) lui-même comme un $K(\pi, 1)$, étant le groupe fondamental (au sens algébrique) de X, autrement dit le groupe de Galois de $\overline{K} /K$, où $\overline{K}$ est clôture séparable de $K$.

Consider the Weil group $W$ of $\mathbb{Q}_p$, that is, the subgroup of those elements of $\mathrm{Gal}(\overline{\mathbb{Q}}_p/\mathbb{Q}_p)$ mapping to an integer power of Frobenius. Class field theory tells us that there is an isomorphism $$ \mathbb{Q}_p^\times \stackrel{\sim}{\longrightarrow} W^{\text{ab}}. $$

There are several ways to normalise it, say I pick the one that sends the uniformiser $p$ to a lifting of *geometric* Frobenius.

**Question**: what is then the image of $-1 \in \mathbb{Q}_p^\times$ under this map?

SGA 4 VI Discusses finiteness conditions one can impose on topoi to make limits behave correctly. I am not that familiar with SGA but it is my impression that this expose only discusses abelian sheaves.

Ideally I would like to find written in the literature an analogue of the statement 8.7.7 but in a "non-abelian context" that proves the same thing for $H^1$. Something like the formula: $H^1(\lim F_i, G_i)=\lim H^1(F_i, G_i)$ as pointed sets.

I suppose one could do a careful read of this expose and verify that this holds true, this statement is probably both true and known. I am just wondering if it is written somewhere I can reference.

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.)