Recent MathOverflow Questions

Does the cyclic group $\Bbb Z/4 \Bbb Z$ acts freely on $S^{2k} \times \Bbb CP^n$?

Math Overflow Recent Questions - Sun, 01/06/2019 - 01:29

I was wondering whether the cyclic group $\mathbb Z/4\Bbb Z$ acts freely on $S^{2k} \times \Bbb CP^n$ where $n>1$? It seems to me that it does not act freely. In case it acts freely then the induced action on cohomology must be non-trivial as the Euler characteristic is non-zero. I was trying to prove using Lefschetz fixed point theorem. But I could not able to derive any contradiction.

Thank you so much for your help.

finite dimensional representation of corona algebra

Math Overflow Recent Questions - Sun, 01/06/2019 - 00:29

If $I=\bigoplus M_n(\Bbb C)$,the multiplier algebra $M(I)$ of $I$ is $\prod_nM_n(\Bbb C)$.I wonder whether corona algebra $M(I)/I$ has a finite dimensional representation?

Is there any work related to Hypertype Theory?

Math Overflow Recent Questions - Sat, 01/05/2019 - 17:46

First, let me state that I am not a formally-trained mathematician, therefore please forgive my approximations and help me correct my errors. Second, I am looking for prior work related to what I would like to call Hypertype Theory. What follows is a short introduction to the topic. But before that, here are a few specific questions that will hopefully make this question less open-ended.


  1. Where can we find prior work on the subject?
  2. Does $\mathbb{H}_n$ converge toward $\mathbb{R}$?
  3. What would be a simple way of establishing such a convergence?
  4. What are examples of important numbers that appear in $\mathbb{H}_{n\ge4}$?
  5. Could the proposed notation be improved upon?  



The basic idea is this: a group is roughly defined with a triplet of two operators and an identity term, while a field is made of two such triplets plus a distributivity law. But what do we get with a third triplet? And what if an unlimited number of such triplets could be added in a recursive manner? Would such an object exhibit any valuable properties?

Interestingly, there is one such object, and it can be constructed in a quite straightforward fashion by using Albert Bennett’s commutative hyperoperations. Most importantly, this object is isomorphic to a strict superset of $\mathbb{Q}$ and to a strict subset of $\mathbb{R}$, while possibly converging toward $\mathbb{R}$.


Commutative Operators

$a\overset{1}{\oplus}b = e ^ {ln(a)} + e ^ {ln(b)} = a + b$

$a\overset{2}{\oplus}b = e ^ {ln(a)} × e ^ {ln(b)} = a × b$

$a\overset{3}{\oplus}b = e ^ {ln(a) × ln(b)} = a ^ {ln(b)} = b ^ {ln(a)}$

$a\overset{4}{\oplus}b = e ^ {\displaystyle e ^ {ln(ln(a)) × ln(ln(b))}}$


$\displaystyle a\overset{n+1}{\oplus}b = e ^ {ln(a)\overset{n}{\oplus}ln(b)}$


Non-Commutative Operators

$a\overset{1}{\ominus}b = e ^ {ln(a)} - e ^ {ln(b)} = a - b$

$a\overset{2}{\ominus}b = e ^ {ln(a)} ÷ e ^ {ln(b)} = a ÷ b$

$a\overset{3}{\ominus}b = e ^ {ln(a) ÷ ln(b)} = a ^ {1 ÷ ln(b)}$

$a\overset{4}{\ominus}b = e ^ {\displaystyle e ^ {ln(ln(a)) ÷ ln(1 ÷ ln(b))}}$


$a\overset{n+1}{\ominus}b = e ^ {ln(a)\overset{n}{\ominus}ln(b)}$


Identity Terms

$i_1 = 0$

$i_2 = 1$

$i_3 = e$

$i_4 = e ^ e$


$i_{n+1} = e ^ {i_n} = e \upuparrows (n - 1) \quad (n > 1)$


Core Identities

From these definitions, we can recursively prove the following core identities:

$a\overset{n}{\ominus}i_n = i_n$

$a = b \Longleftrightarrow a\overset{n}{\ominus}b = i_n$

$a\overset{n}{\ominus}(b\overset{n}{\ominus}c) = c\overset{n}{\ominus}(b\overset{n}{\ominus}a)$

$(a\overset{n}{\oplus}b)\overset{n}{\ominus}c = a\overset{n}{\ominus}(c\overset{n}{\ominus}b)$

In general, these core identities can be used as axiomatic definitions, but it just so happens that we can prove them when the operators are defined using commutative hyperoperators. This ensures that we are not dealing with purely-theoretical objects.


Group Properties

From the core identities stated above, we can prove the following properties:


$\overset{n}{\ominus}$ Anticommutativity

$a\overset{n}{\ominus}b = i_n\overset{n}{\ominus}(b\overset{n}{\ominus}a)$


$\overset{n}{\oplus}$ Left and Right Identity

$i_n\overset{n}{\oplus}a = a\overset{n}{\oplus}i_n = a$


$\overset{n}{\oplus}$ Left and Right Identity

$i_n\overset{n}{\oplus}a = a\overset{n}{\oplus}i_n = a$


$\overset{n}{\oplus}$ Commutativity

$a\overset{n}{\oplus}b = b\overset{n}{\oplus}a$


Additional Properties

Along the way, the following properties can be established as well:


Double Reverse

$i_n\overset{n}{\ominus}(i_n\overset{n}{\ominus}a) = a$


Associative Commutativity

$(a\overset{n}{\ominus}b)\overset{n}{\ominus}c = (a\overset{n}{\ominus}c)\overset{n}{\ominus}b$


Affine Equivalence

$a\overset{n}{\ominus}b = c \Longleftrightarrow a\overset{n}{\ominus}c = b$


Dual Substitution

$a\overset{n}{\oplus}b = a\overset{n}{\ominus}(i_n\overset{n}{\ominus}b)$


Dual Identities

$(a\overset{n}{\ominus}b)\overset{n}{\oplus}b = a$

$(a\overset{n}{\oplus}b)\overset{n}{\ominus}b = a$



We define a sequence of hypertypes by granting the hypertype $\mathbb{H}_n$ the triplet $(\overset{n}{\oplus}, \overset{n}{\ominus}, i_n)$ and all the triplets granted to its predecessors. This makes $\mathbb{H}_1$ isomorphic to $\mathbb{Z}$ and $\mathbb{H}_2$ isomorphic to $\mathbb{Q}$, while $\mathbb{H}_3$ is isomorphic to a strict subset of $\mathbb{R}$, which we call exponential numbers $\mathbb{E}$.

We have yet to study distributivity properties on hypertypes for $n > 2$.

We call our structure Hypertypes instead of Hypergroups for three main reasons: first, Hypergroups refer to an already-existing collection of objects that have nothing to do with Hypertypes; second, we are interested in developing the Hypertype Theory in the context of Type Theory, without relying on any of the axioms of Set Theory; third, we are interested in treating commutative operators separately from their non-commutative duals.

The third point is of critical importance: in order to properly handle real-world measures like temperatures that are non-additive (physicists and statisticians call them intensive), we must be able to distinguish them from measures that are additive (extensive). Therefore, our hierarchy of types should add one operator at a time, always starting with the non-commutative operator. Therefore, for every hypertype $\mathbb{H}_n$, we will have an intensive sub-hypertype $\mathbb{H}_{\bar{n}}$ defined with the non-commutative operator but without its commutative dual.

This approach is nicely supported by the fact that the Core Properties defined above are written with affine-style identites focused on the non-commutative operator. Therefore, we can properly define the commutative operator from its non-commutative dual. This approach also reduces the number of axiomatic definitions that need to be stated when defining the two operators.



The notation introduced above is especially useful for $n = 3$. Therefore, we suggest that the index $n$ can be ommitted whenever it is equal to $3$. Doing so, we would benefit from a single-symbol notation for the commutative power law and its inverse.

$a \oplus b = a\overset{3}{\oplus}b = e ^ {ln(a) × ln(b)} = a ^ {ln(b)} = b ^ {ln(a)}$

$a \ominus b = a\overset{3}{\ominus}b = e ^ {\frac{ln(a)}{ln(b)}} = a ^ {\frac{1}{ln(b)}}$

Alternatively, if we were to limit ourselves to symbols found on most keyboards, we would suggest:

$a \# b = a\overset{3}{\oplus}b = e ^ {ln(a) × ln(b)} = a ^ {ln(b)} = b ^ {ln(a)}$

$a \backslash b = a\overset{3}{\ominus}b = e ^ {\frac{ln(a)}{ln(b)}} = a ^ {\frac{1}{ln(b)}}$



If any of this is of any value, please let me know, because I really have no clue.



Many thanks to the following people for their contributions:

  • @Henry for having found the triplet for $\mathbb{H}_3$.
  • Prof N J Wildberger for being such an inspiration to young contrarian mathematicians.

Anisotropic perimeter and regularity of anisotropic minimal surfaces

Math Overflow Recent Questions - Sat, 01/05/2019 - 15:59
1. Introduction.

By-now classical results assert that minimal surfaces (in $\mathbb R^n$) are generically "smooth" out of a "small" set.

Question. What are the known regularity results for anisotropic minimal surfaces?

2. A more detailed version of the question

For instance, reading a 1972 paper by Bombieri and Giusti I found the following definition:

Definition. A set $A \subset \mathbb R^n$ has an oriented boundary of least area if $\chi_A \in BV(\mathbb R^n)$ and for every $g \in BV(\mathbb R^n)$ with compact support $K$ we have $$\tag{1} P(A) := \vert D \chi_A\vert(K) \le \vert D(\chi_A + g)\vert(K) $$ in the sense of measures, being $\chi_A$ the characteristic function of the set $A$ and $P(A)$ the Euclidean perimeter of $A$.

The authors say - right after the definition - that "It is known that if $A$ has oriented boundary of least area then [..] the boundary is an analytic hypersurface, except possibly for a closed set whose Hausdorff dimension does not exceed $n-8$".

I would like to investigate the analogue of this problem in the anisotropic case. Let me clarify what I mean: let $f \colon \mathbb R^n \to [0,+\infty)$ be some good function (say non-negative, convex and positively 1-homogeneous, as usual in Calculus of Variations). We define the anisotropic perimeter of a set $A$ (of finite perimeter) by $$ P_f(A) := \int_{\partial^e A} f(\nu_A(x))\, d\mathcal H^{n-1}(x) $$ where $\nu_A$ is the measure theoretic outer unit normal, $\partial^e A$ is the essential boundary and $\mathcal H^{n-1}$ is the Hausdorff measure.

What is the analogue of (1) in this case? And are there known regularity results for the solutions to this minimum problem? Can you point out some reference to the literature investigating this problem? Thanks.

Serre spectral sequence for de Rham cohomology

Math Overflow Recent Questions - Sat, 01/05/2019 - 15:58

Suppose we a given a fibration of manifolds $p\colon E\to M$ with a path connected fiber $F$ and simply connected $M$, then we have the Serre spectral sequence with

$$ E_2^{p,q} = H^p(M,\underline{H^q(F)}) $$

The standard proof of its convergence to $H^n(E)$ is purely topological and goes for singular or cellular cohomology. Can one give the proof in terms of de Rham cohomology?

In fact, I'm even more interested in de Rham cohomology with compact support. Now we have some difficulties in defining the local system as cohomology with compact support are no longer homotopy invariant but I hope these problems can be overcome.

EDIT. The proof for cohomology is given in Bott, Tu as V. Zaccaro told. In my answer I try to mimic the proof given there for cohomology with compact support but not completely successfully. I write the first sheet of the spectral sequence modulo two conjectures (I'd be very surprised if they're not true). I'd be grateful if you prove them and write the second sheet.

In the proof I use the notion of "quasi-sheaf with compact support" which I made up some days ago. Am I the first to introduce such a notion? If not, could you give me a reference on it?

R. Godement in "Topologie algébrique et théorie des faisceaux" uses another approach. He doesn't change the category ($Sh(X)$) but introduce another functor $\Gamma_{\Phi}$ where ${\Phi}$ is a family of supports, a family of compact sets in our case. Perhaps my proof can be retold in his language.

Tate algebras and fundamental theorem of algebra

Math Overflow Recent Questions - Sat, 01/05/2019 - 14:01

Let $\mathbb K$ be an algebraically-closed complete non-archimedean field whose absolute value is non-trivial. Consider the Tate algebra $T_n=\mathbb K\langle X_1,\dots, X_n \rangle$ and fix $f\in T_n$.

Then $f(x_1,\dots,x_n)$ converges for every $x=(x_1,\dots,x_n)\in B(\mathbb K)$ where $B(\mathbb K)$ denotes the unit ball consisting of $x$ with $|x_i|\le 1$.

Question: Suppose $f(x)=0$ for any $x$ with $|x_i|=1$, then can we conclude $f\equiv 0$ as an element of $T_n$?

For a polynomial this is obviously true due to the fundamental theorem of algebra. I am just curious about whether there is something similar in the non-archimedean world. More generally, let's assume $f=0$ on some subset $E$ of $T_n$. What conditions for $E$ can ensure $f\equiv 0$ in $T_n$?

Limit circle/point of an ODE with finite eigenvalues

Math Overflow Recent Questions - Sat, 01/05/2019 - 11:56

Consider the following Sturm–Liouville (SL) eigenvalue problem defined in $(-\infty,0]$ or $[0,\infty)$ or $(-\infty,+\infty)$ $$(py')'-qy=-\lambda^2wy,$$ in which $p(x)=x^2$, $w(x)=1$, and $q(x)=(x/2+a)^2+a$ with parameter $a>0$. It has a regular singularity $x=0$. We basically hope for something like homogeneous Dirichlet b.c.

It is solved by making the substitution $y(x)=e^{x/2}x^{-\frac{1}{2}+\sqrt{(a+\frac{1}{2})^2-\lambda^2}}u(x)$, leading to Kummer's equation with two independent solutions (1st & 2nd kind) $$xu''(x)+(\gamma-x)u'(x)-\alpha u(x)=0,$$ in which $\alpha=\sqrt{(a+\frac{1}{2})^2-\lambda^2}-a+\frac{1}{2}$ and $\gamma=1+2\sqrt{(a+\frac{1}{2})^2-\lambda^2}$.
Let's then follow the ubiquitous argument.

Requiring nondivergence at $x=0$, the 2nd kind solution is dropped. Requiring nondivergence at infinity, the 1st kind is reduced to a polynomial when $-\alpha$ is a non-negative integer and eigenvalue $\lambda^2$ is attained. Seen from this condition for $\alpha$, we only have a bounded and finite sequence of eigenvalues. Therefore, one probably also expects an additional continuous spectrum. But I'm not sure whether it starts from the largest eigenvalue.


Is it possible to know the limit circle/point classification of this ODE near $0,\pm\infty$? Is the above solution useful for this purpose? How should one proceed? I also found page 13 of this paper has a limit point/circle classification of the Kummer equation I used. Not sure if relevant.

Given B,C incomplete, incomparable r.e. sets must C compute low r.e. set avoiding cone below B? (ADDED: Uniformly?)

Math Overflow Recent Questions - Sat, 01/05/2019 - 01:53

I feel like there must be a classical result answering this question (or easily modified to do so) but a quick flip through Soare didn't produce anything so rather than waste time I figured I'd just ask.

Given incomplete r.e.sets $C \nleq_T B$ must there exist a low r.e. set $A <_T C$ with $A \nleq_T B$?

I'm guessing the answer is no and somehow you can put together Robinson low splitting with one of the non-splitting/non-bounding theorems to show this but I'm not seeing it right away.

EDIT: Great answer to the original question below (thanks Ted!) but I realized I should have specified that I wanted the set produced uniformly for the application I wanted.

Numbers that can be written as a sum of three cubes in exactly one way (a^3 + b^3 + c^3)

Math Overflow Recent Questions - Fri, 01/04/2019 - 21:14

Based on online info, it seems that most of these numbers have many solutions. Are there any that have only 1 known solution or only a few solutions?

Closed form of integration of modified Bessel function composed with trigonometric function times a linear term

Math Overflow Recent Questions - Thu, 01/03/2019 - 11:36

Assuming one draws two points from a von Mises distribution on a circle, I am looking for the expected distance between two such points.

Given the pdf of a centered von Mises distribution $$f_X(t \mid \kappa) = \frac{e^{\kappa \cos(t)}}{2\pi I_0(\kappa)} \cdot 1_{[-\pi, \pi]}(t) \; ,$$ one can calculate the pdf that describes the distance of two points by $$f_{\Delta}(t \mid \kappa) = \frac{I_0 \left( 2\kappa \cos{\frac{t}{2}} \right)}{\pi I^2_0(\kappa)} \cdot 1_{[0, \pi]}(t)\; ,$$ where $\Delta := \min{\big\{ |X_1 - X_2|, \, 2\pi - |X_1 - X_2| \big\}} = \pi - \big||X_1 - X_2| - \pi \big|$. Now, the expected distance corresponds to the expression $$\mathbb{E}[\Delta] = \frac{1}{\pi I^2_0(\kappa)}\int_0^{\pi} t \cdot I_0 \left( 2\kappa \cos{\frac{t}{2}} \right) dt ,$$ which I was not able to solve in a meaningful way even though there are some nice known integrals involving Bessel functions like $\int x I_0(x)\, dx = xI_1(x)$.

Trying to integrate it on its power series yields $$\int_0^{\pi} t \cdot I_0 \left( 2\kappa \cos{\frac{t}{2}} \right) dt = \sum_{m=0}^{\infty} \left(\frac{\kappa^m}{m!}\right)^2 \int_0^{\pi} t\cos^{2m}{\left(\frac{t}{2}\right)} dt,$$ which seemed equally hard to solve.

Another desperate attempt to tackle this problem with integration by parts led to this question.


Is there a closed form for the integral $$\frac{1}{\pi I^2_0(\kappa)}\int_0^{\pi} t \cdot I_0 \left( 2\kappa \cos{\frac{t}{2}} \right) dt \quad ?$$

An operad-like structure, is there a name for it?

Math Overflow Recent Questions - Thu, 01/03/2019 - 09:13

Here is an example which I like to have a name for.

Let $P$ be a compact smooth manifold of dimension $p$, possibly with non-empty boundary.

Define $E(k,P)$ to be the space of smooth (codimension zero) embeddings $$ \coprod_{k} D^p \to P \, , $$ that is the space of embeddings of $k$ disjoint $p$-disks in $P$, where the image of each such embedding lies in the interior.

In particular, when $P = D^p$, the spaces $\{E(k,D^p)\}_{k\ge 0}$ form an operad.

There is an evident "action" map $$ E(\ell,P) \times (E(k_1,D^p) \times \cdots \times E(k_\ell,D^p)) \to E(k_1 + \cdots +k_\ell,P) $$ given by insertion.

Question: What is this action an example of? (Does it have a name?)

Independence of $\ell$ in $\ell$-adic cohomology

Math Overflow Recent Questions - Thu, 01/03/2019 - 08:50

Let $X$ be the base change of a smooth projective variety over a finite field, to a separable closure $k$ of the ground field.

Do we expect for any endomorphism $f$ of $X$ to have an effect $f^* : H^n(X,\mathbf{Q}_{\ell})\to H^n(X,\mathbf{Q}_{\ell})$ whose characteristic polynomial has coefficients in $\mathbf{Q}$? or at least independent of $\ell$ in some suitable sense?

Is there an example of an $f$ such that its characteristic polynomial on $\ell$-adic cohomology does depend on $\ell$?

Link between homotopy equivalence of simplicial sets and categorical equivalences

Math Overflow Recent Questions - Thu, 01/03/2019 - 08:35

In Higher Topos Theory, a map $f: S \rightarrow T$ of simplicial set is a categorical equivalence if after applying the functor $\mathfrak{C}[-]$ we have a equivalence of simplicial categories.

In the proof of Theorem, we see for example that any trivial fibration in the Kan model structure on simplicial sets is a categorical equivalence.

My question is the following : under what condition can we say that a Kan weak equivalence is a categorical equivalence?

I am asking this question because when Lurie finally proves Propisition in section 2.4.5 he claims (I think) at some point that a weak equivalence between two Kan complexes is a categorical equivalences. (This is to use Corollary

If more details are needed I will gladly add them.

Is the following analysis correct? [on hold]

Math Overflow Recent Questions - Thu, 01/03/2019 - 06:08

Given three infinite divergent series:

$Series_1: (\sum_{n=1}^{\infty}\frac{1}{F(n)^p} +\frac{1}{G(n)^p}) = S_1(n)$

$Series_a: \sum_{n=1}^{\infty}\frac{1}{F(n)^p}$

$Series_b: \sum_{n=1}^{\infty}\frac{1}{G(n)^p} = S_b(n)$

(here $F(n)$, $G(n)$, $S_1(n)$, $S_b(n)$ represents some functions on $n$.)

$Series_1$ and $Series_b$ are equal to functions $S_1(n)$ and $S_b(n)$ respectively. Since the series are divergent, both $S_1(n)$ and $S_b(n)$ (by definition) will also be some divergent functions over $n$. Now, can we (using above equations and their functional equals) say that:

$Series_a: \sum_{n=1}^{\infty}\frac{1}{F(n)^p} = S_1(n) - S_b(n)$ ?

Note: For a convergent series the sums $S_1(n)$ and $S_1(n)$ will evaluate to a finite value (that are independent of variable $n$), but for divergent series this is not possible.

($p$ here is some real/complex variable)

Homomorphisms from BV

Math Overflow Recent Questions - Thu, 01/03/2019 - 05:08

Denote by $\mathsf{BV}(\mathbb T)$ the Banach space of functions on the circle with bounded variation which is a Banach algebra under the pointwise product. Is there a surjective homomorphism from $\mathsf{BV}(\mathbb T)$ onto $\ell_1(\mathbb Z)$ (with convolution)?

My motivation comes from the fact that Fourier coefficients of functions in $\mathsf{BV}(\mathbb T)$ vanish at infinity however I am not sure if this is directedly related.

University year 1, chapter 7 (estimation) [on hold]

Math Overflow Recent Questions - Thu, 01/03/2019 - 04:11

A random sample of 25 female students is chosen from students at higher education establishments in a particular area of a country, and it is found that their mean height is 163 centimeters with a sample variance of 64.

a. Assuming that the distribution of the heights of the students may be regarded as normally distributed, calculate a 97% confidence interval for the mean height of female students.

b. You are asked to obtain a 97% confidence interval for that mean height of width 2 centimeters. What sample size would be needed in order to achieve that degree of accuracy?

c. Suppose that a sample of 15 had been obtained from a single student hostel, in which there were a large number of female students (still with a distribution of heights which was well approximated by the normal distribution). The mean height is found to be 155 centimeters. Calculate a 99% confidence interval for the mean height of female students in the hostel. How do their heights compare with the group in 4(a)(i) ? [(159.3, 166.7), 339, (148.9, 161.1)]

Hi! Are you able to solve this qns?

Limit of sequence of vectors in $\ell^2$ with coefficients approaching $0$

Math Overflow Recent Questions - Thu, 01/03/2019 - 04:08

Let $\{v_m\}_{m \in \mathbb{N}} \subset \ell^2$ be a sequence in $\ell^2$ over the complex plane $\mathbb{C}$ such that: $\{v_m\}_{m \in \mathbb{N}}$ is linearly independend and $v_m \to v$

Let $V= \operatorname{span} \{v_m\}_{m \in \mathbb{N}}$

Let $\{u_p\}_{p \in \mathbb{N}} \subset V$ be a sequence in $V$ such that $u_p \to u$ so we have $$ \forall p \in \mathbb{N}: u_p = \sum_{m=1}^\infty \left( a_{p,m} \cdot v_m \right) $$ with $a_{m,p} \in \mathbb{C}$ and for each fixed $p \in \mathbb{N}$ there are only finitely many $m$ with $a_{p,m} \neq 0$

Further, we have for each fixed $m \in \mathbb{N}$ $$ \lim_{p \to \infty} a_{p,m} =0 $$ My question is if it is true that: $$ \lim_{p \to \infty} u_p = a \cdot v $$ with $a \in \mathbb{C}$


Explicit examples of Azumaya algebras

Math Overflow Recent Questions - Thu, 01/03/2019 - 03:48

I'm trying to understand the Brauer group of a scheme better. I know how to compute $\text{Br}(X)$ as an abstract group in some cases, but don't have a good idea of what the individual Azumaya algebras actually look like (in terms of an explicit trivialisation on an etale open cover). For instance:

  • What do the Azumaya algebras on $\mathbb{P}^n_k$ look like? (Since $\text{Br}(\mathbb{P}^n_k)=\text{Br}(k)$, I imagine there's an explicit way to get an Azumaya algbera on $\mathbb{P}^n$ from a central simple algebra).
  • What do the Azumaya algebras on a smooth curve in $\mathbb{P}^2$ given by $f=0$ look like? (For some $k$ not algberaically closed or finite, like a number field, since otherwise $\text{Br}(X)=0$).
  • Are there any other examples like this?

I'm mainly interested in this to compute examples of the Brauer-Manin obstruction, i.e. taking Azumaya algebra $A$ and point $p\in X(k_\nu)$ and computing $$\text{inv}_\nu(A_{p})$$ where $A_{p}$ is the image of Azumaya algebra $A$ under $\text{Br}(X)\to \text{Br}(k_\nu)$ arising from the inclusion of $p\hookrightarrow X$, and $\text{inv}_\nu$ is the Hasse invariants map. I'd be interested in seeing this computed in the examples above.

My confusion might just arise from not knowing how to explicitly compute the map $\text{inv}_\nu:\text{Br}(k_\nu)\to \mathbb{Q}/\mathbb{Z}$.

Construct example satisfying some inequlaities

Math Overflow Recent Questions - Thu, 01/03/2019 - 01:55

How do I construct two vectors $a,b\in \mathbb{R}^{n}$, $a=(a_1,a_2,\ldots a_n)^{T}$ and $b=(b_1,b_2,\ldots b_n)^{T}$ which satisfy in the following conditions ‎\begin{align} & a_ib_i\geq 1,a_ib_j<1, i\neq j, \quad i,j=1,2,\ldots n,\\ & or\\ & a_ib_i> 1,a_ib_j\leq1, i\neq j, \quad i,j=1,2,\ldots n. ‎\end{align}

Your professional $\LaTeX$ experiences that saves your time in typesetting

Math Overflow Recent Questions - Wed, 01/02/2019 - 23:08

In $\LaTeX$ typesetting, when we repeat a long and complex formula in long documents, it is appropriate to create a new command that just by calling this new command we get the desired output. For example, I have used the following math expression in my previous document frequently: $$\{a^1,a^2,\ldots,a^n\}$$ For doing this in usual way, we need to press 22 keys on keyboard (and think about $(\frac{\partial}{\partial x^1}, \cdots,\frac{\partial}{\partial x^n})$ and other terrible formulas). Of course we can do this by copy and paste from similar one in the text. it is much better to define the following new command on preamble

\newcommand{\set}[1]{\setaux#1\relax} \def\setaux#1#2#3\relax{% \{ {#1}#2 1, \ifnum\pdfstrcmp{#3}{3}=0 {#1}#2 2 \else \ldots \fi , {#1}#2{#3} \} }

and just by typing \set{a^n} in our text we get the same output.

Question: What are your professional $\LaTeX$ experiences that saves your time in long document typesetting?


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