12:03 AM
It's been a while since I've been here. Looks like this chat isn't as happening a place as it used to be. I sure hope somebody can answer my question though

12:18 AM
Hi there, Ted! Long time no see

12:48 AM
hi
how do I get to display the math here?

1:01 AM
@justanotheruser go here: math.meta.stackexchange.com/questions/1088/… and follow the instructions

@ALannister thank you!

@justanotheruser you're welcome :)

Hello just anotherusers
Can you please point out how is that form I used ambiguous so I can improve In the future how I convey information?
How can I write a message to a user so he gets a notification I am talking to him?

Oh nice :D

1:06 AM
@MadSpaceMemer well, more experienced users are here, people with a lot more experience and knowledge.

@justanotheruser I am not sure I understand you.
@justanotheruser which other people?

@MadSpaceMemer you are in a chat room, with other people.

No doubt there are a lot of people that are smarter/more experienced and have more knowledge. But what do you mean by stating that? *sorry English is not my mother tounge
@justanotheruser I am heading to bed. It was nice conversing with you and thank you for your help on the question of <>. Good night.

1:44 AM
If anybody is an expert in using Matlab, I have no doubt you could help me.

2:16 AM
It's too bad there isn't a Matlab stackexchange lol

2:49 AM
HI @ALannister. My brief time working with Matlab was highly frustrating and unsuccessful. I went back to Mathematica.

3:43 AM
@TedShifrin you should use Coq :D
It's better than all those, except isn't a CAS
Everything in math has to be proven formally anyway, so who cares about vis

4:34 AM
@TedShifrin my experience was starting with IDL programming, which is like Matlab but even less intuitive
compared to that Matlab is pleasant
(it's all relative)

5 hours later…
9:15 AM
Posted this question on philosophy stack exhange:
https://philosophy.stackexchange.com/questions/65486/why-are-you-doing-something-in-a-certain-situation/65497#65497
Went pretty bad, if anyone here is interested, I would like to know your opinion on this since we're all mathematicians here.
Question is understable if you have a driving license or you attended driving school.

Morning all

gm

Morning @ÍgjøgnumMeg
When does the semester begin in Heidelberg?

Hey @Alessandro
I think October the 13th

That's a Sunday so probably the 14th :P

9:29 AM
yeah I just saw that
lol

(I know only because we start on the 7th)

oh nise
Mathein told me last night that Greenberg is giving a short course on iwasawa theory a week before uni starts

I'm afraid I don't know who Greenberg is
(or what Iwasawa theory is)

he was one of Iwasawa's doctoral students
and Iwasawa theory is (in my limited knowledge) a way of getting information on the p-torsion in ideal class groups
or smth

Oh ok, so we're talking about algebraic number theory

9:32 AM
yis, of course
when do I talk about anything else
lol

Fair enough :P

you have a tower $\Bbb Q(\zeta_p) \subset \Bbb Q(\zeta_{p^2}) \subset \cdots \subset \Bbb Q(\zeta_{p^\infty}) = \bigcup \Bbb Q(\zeta_{p^n})$ where each intermediate extension has group $\Bbb Z/p^n\Bbb Z$ so the galois group of the big extension is $\varprojlim \Bbb Z/p^n\Bbb Z \cong \Bbb Z_p$
and then you look at the p-parts of the ideal class groups of each intermediate extension and you get another inverse system over those guys, which gives you a module over $\Bbb Z_p[[\operatorname{Gal}(\Bbb Q(\zeta_{p^\infty})/\Bbb Q]]$
and then there's just a big structure theory for modules over that guy
lol

Uhm
That's a bit over my head, I didn't do much ant :P

lol fair, well the motivation was apparently that p-torsion in the ideal class group of $\Bbb Q(\zeta_p)$ was an obstruction to Kummer's original proof (recall that his proof only works for regular primes, i.e. primes for which $\operatorname{Cl}(\Bbb Q(\zeta_p))$ has no p-torsion)
if that makes any sense to you

Yep, it does

9:44 AM
Cool, then given the exact power $p^{e_n}$ dividing the order of the class group of $\Bbb Q(\zeta_{p^n})$ Iwasawa theory tells you asymptotically that $e_n = \lambda n + \mu p^n + \nu$, for some structural invariants $\lambda, \mu, \nu$
whose derivation is unknown to me lol

10:11 AM
welcome @RScrlli

Hello everyone!
I would like to ask, what might be a trivial question in measure theory. Let's say we have a set $\mathcal{G} \subset \mathcal{P}(X)$ and we have a sequence of sets $(G_j)_{j\in \mathbb{N}}\subset \mathcal{G}$, and $\bigcup_{j\in \mathbb{N}} G_j=X$
then, is $\mathcal{G}=X$?
or at least, $\mathcal{G}$ contains all the elements in $X$

Yes. Since $G_j\subseteq G$ for all $j$ we have $\bigcup G_j\subseteq G$, so $X\subseteq G$, the other inclusion should be clear

The converse inclusion holds since $\mathcal{G}$ is part of the Power set of $X$
In this case I don't really understand the idea behind this theorem, this is related with my previous question
1

I think the textbook wants to talk about a generating set (base) for the $\sigma$-algebra $\mathcal{A}$ and so in particular $\mathcal{G} \subseteq \mathcal{A} \subseteq \mathscr{P}(X)$. That's about all we know.

Which theorem?

This is a theorem about the uniqueness of the measures,

10:23 AM
You don't state any theorem in the question though

It states that any two measures that coincide in $\mathcal{G}$ and are finite, must coincide for all the elements of the sigma algebra generated by $\mathcal{G}$
But I don't understand why the author introduces the set $\mathcal{G}$
Since we already "showed" that $\mathcal{G}=X$ why should be consider $\mathcal{G}$ at all?

Because $\mathcal G$ is usually much smaller than $\sigma(\mathcal G)$, so this theorem makes testing for equality of measures easier
@AlessandroCodenotti Ah, no, wait, I see what went wrong, you have $\mathcal G\subseteq \mathcal P(X)$, not $\mathcal G\in\mathcal P(X)$, so this is wrong

ohh so my original thought is not valid? ( the fact that $G=X$)

No, look at $G$ as the collection of open intervals in $\Bbb R$, which generates the Borel $\sigma$-algebra

10:39 AM
What confuses my is the fact that there is an exhausting sequence in $\mathcal{G}$ whose "upper-limit" is $X$

Look at the open intervals again

I'll try to work with it and I'll let know, thanks again for the patience with this pretends-to-be-a-mathematician economist!

any pretender is welcome to try to be a contender in this room :-)

every dual basis is dual of some basis
is this true

11:06 AM
@AlessandroCodenotti So let's say I take $X=\mathbb{R}$, $\mathcal{G}=\mathcal{O}$, then the sigma algebra generated by $\mathcal{G}$ is the Borel sigma algebra. Then we have an exhausting sequence $(G_j)_{j\in\mathbb{N}}\subset \mathacal{G}$ whose limit is $X$. And then what you told me, the fact that $\mathcal{G}$ is much smaller than the sigma algebra generated by it makes absolute sense. An we have that $X\neq \mathcal{G}$. Is that ok?

11:21 AM
Yes, but note that $X$ is not even a subset of $\mathcal P(X)$, it is only an element

4 hours later…
2:59 PM
anything in math where you have a graph and the vertices encode geometric objects?
spatial graph maybe?
A spatial network (sometimes also geometric graph) is a graph in which the vertices or edges are spatial elements associated with geometric objects, i.e. the nodes are located in a space equipped with a certain metric. The simplest mathematical realization is a lattice or a random geometric graph, where nodes are distributed uniformly at random over a two-dimensional plane; a pair of nodes are connected if the Euclidean distance is smaller than a given neighborhood radius. Transportation and mobility networks, Internet, mobile phone networks, power grids, social and contact networks and neural...

a quiver is a digraph with R-modules at the vertices and R-linear maps as the edges
alg. geo

@ÍgjøgnumMeg thank you
I'm looking for planar spatial graphs that have degree four vertices as algebraic numbers
the obvious one is a lattice

3:14 PM
Is it possible to write $\text{proj}_U(\vec{v})=\vec{v} \iff \vec{v} \in U$ 1-3 line proof for this?
It seems very trivial

do diffeomorphisms of a mesh care about spatial locations
like if you take a diffeomorphism on a mesh could you possibly identify the geometric location of the vertices after the transformation

3:39 PM
@ÍgjøgnumMeg that's a quiver representation
the quiver is just the digraph

$$\sum_p 2^{-p}$$

4:00 PM
It certainly converges. It still will if 2 is replaced by any -1<x<1, even (ratio test)

yeah

(removed)
:-)

hey @Semiclassical Isn't there shorter proof for the problem you helped me yesterday?
specially the case "$\impliedby$"

@Tuki "hey"?
is for horses

I don’t know about whether $\sum_p z^{-p}$ can be analytically continued outside the unit circle, though

4:03 PM

Couldn't you just say $\text{perp}_U(v) = v-v = 0 \implies v \in U$?

why would one want to analytically continue it outside the circle? please

For instance, $\sum_{n=1} x^{-2^n}$ converges in the unit circle but has $|z|=1$ as natural boundary of analytic continuation
It’s a pretty typical question, since it gets you into the realm of complex analysis and all the fun that brings

is $x$ meant to be $z$?

Yeah. I’m on mobile
@Tuki perp_U?

4:08 PM
equivalent to $v = u+u_\perp$
I think we set $u=P_u(v)$
Then the premise tells that $P_u(v)=v$

Hello guys, I was thinking, is it correct to say that $\mathbb{R}^n \in \mathcal{\mathbb{R}^n }$? (Where $\mathcal{B}$ is the Borel sigma algebra)
typo
$\mathbb{R}\in \mathcal{B}$

As in, $\text{perp}_U(u+u_\perp)=u_\perp$?

Well the idea is that if $u + u_\perp = 0$ it means that the projection doesn't "move" the vector in any direction

I am asking about the use of $\in$

I actually didn't know that hey is rude @skullpetrol . (Obviously you have to nice when asking for help, but I never knew hey is considered rude).

4:13 PM
I don’t think it’s necessarily rude either

I don't think this is rude aswell
Isn't "hey" short for "hello"?

My issue is that you haven’t actually said what perp_U is

yes sorry let me get the definition from our material
$\text{perp}_U(v)=v-\text{proj}_U(v)$
now if you take what the premise says about projection $\text{proj}_U(v)=v$
then we get $\text{perp}_U(v)=v-v = 0$

Sure. But you still need to justify why having zero perpendicular component means you’re in the subspace

This is exactly what I find "hard". This seems so trivial but It doesn't prove it to be exact.

4:20 PM
0

Is anything known about this sum? I'm summing over the primes, $p.$ $$\sum_p \frac{1}{2^p}$$ I've calculated that it converges, but I can't determine whether the following related sum can be analytically continued outside the unit circle: $$\sum_p z^{-p}$$

anyone in the usa?

Ted recommended me the book "How to think like a Mathematician" by Kevin Houston, maybe something like this helps indeed with this.

Hey tuki

yes?

are you studying linear algebra?

I'm taking a course called "Linear algebra 2" right now so yes.

4:25 PM
@ultradark btw, the one comment about z^p vs z^-p is correct: I should have said the former
The first converges inside the unit circle, the latter outside
If you insert z=e^(-x) into the former case (so $\sum_p e^{-p x}$ then that goes over to convergence in some half of the complex plane (let me think of which one)
Oh, right half-plane (positive real part)
And the question becomes whether you can analytically continue that to the other half of the complex plane

One reason why this question is interesting, btw:

I don't understand
it looks so complex
@Tuki I'm asking because I'm bored

Well, my above comments (after the z^p bit) were intended to clarify the appearance of f(x) in reuns answer

oh
okay

4:35 PM
That said, the details from then on are indeed involved

i'm going to peruse the lacunary functions page on wiki

Heh, that’s what I was heading towards
The example I gave earlier: $\sum_{n=0}z^{2^n}$
The reason it can’t be analytically continued outside the unit circle is because the gaps between subsequent exponents keeps getting bigger
By contrast, the gaps between powers in the geometric series are just 1,1,1,...

yeah

And that one can be analytically continued
For the primes, you’ve got an interesting variation on that

you have a asymptotic relation on the exponents

4:41 PM
The gaps keep changing, and I think that the gaps can be arbitrarily large if you go far enough. On the other hand, if the twin prime conjecture is true, then gaps as small as 2 will occur infinitely often

mhm
$g_n$ has been extensively studied
like Cramer's conjecture

So there’s probably an intimate—if not easily accessible!—relation between prime gaps and analytic continuation outside the obvious region of convergence

$g_n$ is the gaps between successive primes
I know an ANT

Hence why reuns answer is necessarily hard: the problem is interesting enough to be difficult but not intractable

analytic number theorist (ANT)

4:45 PM
@Ultradark What are you learning yourself?

In number theory, Cramér's conjecture, formulated by the Swedish mathematician Harald Cramér in 1936, is an estimate for the size of gaps between consecutive prime numbers: intuitively, that gaps between consecutive primes are always small, and the conjecture quantifies asymptotically just how small they must be. It states that p n + 1 − p n = O ( ( log ⁡ p...
$\ln^2(p)=p_{n+1}-p_n$

lol the weaker version assumes the Riemann hypothesis.

@J.Doe I'm learning Abstract Algebra and Geometry
geometry from a transformational approach (using mainly abstract algebra)

Nice. What book(s) are you using?

A First Course in Abstract Algebra, John B. Fraleigh 3rd Edition
and I'm not sure about the textbook yet for Geometry
something weird is going on with cramer's conjecture

2 hours later…
6:58 PM
Hello! ncatlab.org/nlab/show/Introduction+to+Topology+--+2 I don't get the first sentence of the proof of Lemma 2.6. Don't we need that the open neighborhood is in the image?

Linear algebra question of mine: Suppose $A,B$ are $r$-dimensional linear subspaces of $V=\mathbb{R}^n$ where $0<r<n$. Then as usual one has $V=A\oplus A^\perp=B\oplus B\^\perp$ where $A^\perp,B^\perp$ are the $(n-r)$-dimensional orthogonal complements of $A,B$ respectively. We know $A\cap A^\perp=B\cap B^\perp=\{0\}.$
Does this imply: $$A\cap B^\perp =\{0\}\iff A^\perp \cap B =\{0\}\iff V=A\oplus B^\perp=A^\perp \oplus B$$

@KonformistLiberal yes whats written there is not correct

Bah, missed the typo in my Latex before

the steps carried out do contain the details of the proof but the words around it are wrong and and misleading

I think the first equivalence is true. Not at all sure about the other.

7:07 PM
@Semiclassical $A\cap B^\perp =\{0\}\iff V= A\oplus B^\perp$ definitely is true

Nice. That seemed intuitively right

thats just from comparing dimensions
$A\oplus B^\perp$ has dimension $r+(n-r)=n$ and must be $V$

Oh, of course. Makes sense

the other equivalences are correct I think

Yeah, I think so too
I guess I should’ve left in the equivalence $V=A \oplus B^\perp \iff V=B\oplus A^\perp$
Since it seems like that should be easy to prove under the assumptions

7:24 PM
@s.harp, is it because the last set we find which is open around (e,e) is actually in the image?

@Semiclassic: It's all correct. Dimension is all you need.
(For the last part)
For the first part, you can use $V\subset W \iff W^\perp\subset V^\perp$, I believe.
Oh, as usual, @s.harp beat me to it.
<--- resigns :P

@KonformistLiberal the set they find is "$(E\times_X E)\cap (U_{p(e)}\times U_{p(e)})$" which is notational abuse for $(E \times_X E)\cap ( [U_{p(e)}\times \{e\}] \times [U_{p(e)}\times \{e\}])$

Nice

@KonformistLiberal in particular that open set is in the image of the diagonal map (so yes to what you said)

@s.harp thanks!

7:39 PM
16

Having had my formative years in New Zealand, I was born in South Africa. I vaguely recall when I was VERY young having someone tell me when I said "hey" that "hay is what horses eat". I got that then it was more when people would yell "hey" across a room to attract attention, and that was co...

since we're not really in any "country" this doesn't strictly apply :P

7:59 PM
0

This is related to a remark in Iitaka's algebraic geometry sec 1.12. "...It should be noted that sheaves of germs of differentiable functions are by no means coherent. These facts seem to suggest coherence is linked with the property of being algebraic or analytic." $\textbf{Q1:}$ What is the ...

whats the structure sheaf for $C^\infty (M)$ in order for this question to make sense?

8:29 PM
@s.harp: I presume Iitaka is starting with an analytic or algebraic variety, so it's the "usual" structure sheaf. You obviously can't get a presentation of continuous or smooth functions out of holomorphic/algebraic functions.

@TedShifrin that makes the most sense to me, but also makes the statement itself kind of lame, if you are interested in $C^\infty$ you shouldn't use polynomial quotients for the structure ring
i guess the statement is the author is making is precisely that: $C^\infty$ is not coherent over the sheaf of algebraic functions, so dont try to use that structure sheaf to do anything smooth

1 hour later…
10:04 PM
@s.harp: Yeah. I mean, if we use the smooth structure sheaf, then it's certainly coherent over itself :P

$$\int_0^1 \ln(\ln(x))-\ln(\ln(1-x))dx$$
trying to integrate this
I think it might be $0$?
due to symmetry?

@TedShifrin apparently a sheaf of rings being coherent over itself is not always true, what one needs is that every $O$-module map $O^n\to O$ has finitely generated kernel. For $C^0(-\epsilon,\epsilon)$ for example im pretty sure the map $f\mapsto f(0)$ doesn't have finitely generated kernel (but for $C^\infty$ it does, being generated by the function $x$, which can be seen from the morse lemma)

yeah it is $0$

1 hour later…
11:35 PM
@s.harp: Or Malgrange preparation theorem, or somethin'.