Mathematics - 2020-04-09 (page 1 of 2)

Masterphile

12:01 AM

by that argument bijection also exists

Thorgott

yes

I'm not sure what you're trying to get at

to construct one :)

a bijection?

yes

you can do the usual trick

fix a copy of $\mathbb{N}$ in there and shift that by one, leave the rest the same

I'm not sure why you'd want to construct such a bijection though

geocalc33

12:06 AM

math.stackexchange.com/questions/3614831/… very interesting q

Masterphile

very very

geocalc33

because apparently you can compactify $\Bbb R^2$ and $\Bbb R^n$

it's incredible

it's not homeomorphic to $\Bbb RP^n$ I believe

Masterphile

won´t do that, i am focused on open sets :P

geocalc33

Nice what are open sets

Masterphile

i am just considering those in $\mathbb R^n$, you know what they are :P

Open set

In mathematics, particularly in topology, an open set is an abstract concept generalizing the idea of an open interval in the real line. The simplest example is in metric spaces, where open sets can be defined as those sets which contain a ball around each of their points (or, equivalently, a set is open if it doesn't contain any of its boundary points); however, an open set, in general, can be very abstract: any collection of sets can be called open, as long as the union of an arbitrary number of open sets is open, the intersection of a finite number of open sets is open, and the space itself...

geocalc33

12:15 AM

nice lol

Thorgott

in fact, it appears that an open subset of $\mathbb{R}^n$ is homeomorphic to $\mathbb{R}^n$ if and only if it is connected and simply connected at infinity (source: MSE)

geocalc33

yes exactly. what I've done is transferred the structure of $\Bbb R^n$ to the $n-$cube

and then mapped the $n-$cube to the $n-$ torus

Masterphile

now map n-torus to n-koch snowflake :D

geocalc33

no @Masterphile lol :)

Mike Miller

@Thorgott Does that hold for all $n>2$ or $n>4$? (For $n<3$ you just want simply connected, not simply connected at infinity.)

Thorgott

12:34 AM

$n\ge3$, according to this answer

should've specified that

geocalc33

I'm using a metric that restricts contractibility

Mike Miller

12:51 AM

@Thorgott Yeah, I just didn't know the bottom two cases. I guessed 4 was Freedman but forgot how to do 3.

Thanks for the link, explains what I needed.

Thorgott

np

Simple

2:05 AM

in Probability and Statistics, 24 hours ago, by Simple

Let $\{S_n\,:\,n\geq0\}$ be a simple random walk with $p$ be the probability of stepping up and $q=1-p$ be the probability of stepping down. Let $T_k=\min\{n\geq1\,;\,S_n=k\}$. If $p<1/2$, is $E(T_1\,|\,T_1<\infty)<\infty$?

Currently I have the probability generating function, $G(s)=\frac{1-\sqrt{1-4pqs^2}}{2qs}$ Not sure how to proceed,

Simple

5:05 AM

I got it.

Astyx

8:50 AM

I know the irreduible representations of $SL_2$ (one for each natural integer $m$, by acting on polynomials of degree $m$ - let's denote thtat representation $V_m$) and I want to deduce they are irr rep of $SU_2$ as well. I am asked to prove that a if $W\subset V_m$ is stable under the action of $SU_2$, then $W = \oplus_{i\in I}e_i$ for some finite set of indices I, where $e_i = X^iY^{m-i}$

I have done this already but I fail to remember the proof of this argument.

Astyx

9:10 AM

Nevermind, I got it

By looking at the eigenspaces of diagonal matrices

Bob Oakley

10:31 AM

Hi Can I ask a question?

I try to simplify the following singular integral:

>$$\int_{R^2}\hat{p}(\xi)\hat{q}(\xi)|\xi|^{-2}d\xi= A\int_{R^2}\int_{R^2}p(x) q(y) (-\log|x-y|)dxdy$$
where $p, q\in C_c^{\infty}$ and $\int q=0$ and $\hat{p}(\xi):= (1/(2\pi))\int e^{-ix\xi}p(x)dx$ which is the Fourier transform, and $A$ is a constant.

1. My way is LHS=
$$\int_{R^2\setminus B(\epsilon)}\hat{p}(\xi)\hat{q}(\xi)|\xi|^{-2}d\xi+\int_{B(\epsilon)}\hat{p}(\xi)\hat{q}(\xi)|\xi|^{-2}d\xi$$
But why the RHS appears $(-\log|x-y|)$ which is the log potential of Poisson equation.

Masterphile

10:56 AM

Is the book written in English and entitled "Set Theory" by Hausdorff actually translation of his work written originally in German, and entitled "Grundzüge der Mengenlehre"?

Leaky Nun

@Masterphile yes, according to proofwiki.org/wiki/Book:Felix_Hausdorff/Set_Theory

Masterphile

thanks, although it seems it is a translation of not the first German edition but of some later one

MathStudent

If I got a matrix $A = X^T D X$ with $D$ being a diagonal matrix, are the diagonal entries in $D$ the eigenvalues of $A$?

Tobias Kildetoft

11:12 AM

@MathStudent Not unless $X$ is orthogonal

MathStudent

@TobiasKildetoft Thanks. That's unfortunate.

Is there any other easy way to find out whether such a matrix would be positive definite? Or only by calculating the leading minors because it is symmetric which is not necessarily easy either, right?

Masterphile

@LeakyNun yes, it is not the translation of the first (which has the largest number of pages!) edition:
https://en.wikipedia.org/wiki/Grundz%C3%BCge_der_Mengenlehre

why, on Earth, wasn´t the first edition translated, instead of third?

does anyone has some clues?

Thorgott

11:34 AM

isn't it standard practice to just translate the newest edition

Masterphile

yes, but the third has serious omissions, and relatively much lesser number of pages than the first, i am thinking whether the first survived the happenings of WW2?

learn german

they did

Lohnt sich

according to Wikipedia it appears the first edition had sections on measure theory and topology, which were then considered set theory, that got removed in the later ones

Here's the first edition archive.org/details/grundzgedermen00hausuoft/page/n13/mode/1up

Compare with books.google.de/books/about/…

It seems that the measure theory and topology chapters were removed, but the other chapters actually increased in content

Considering that measure theory and topology became even larger independent topics during the times between the first and third edition, this makes sense

Tobias Kildetoft

11:44 AM

@MathStudent What do you mean "because it is symmetric"? You did not mention any of these being symmetric

Masterphile

@Thorgott you can also read the first edition, written in German?

Thorgott

Yes

Masterphile

you haven´t read it all never before?

Thorgott

no, I haven't read it

TechnoKnight

Hi

Alessandro Codenotti

11:56 AM

In the entrance of the main building of the math department in Bonn there's a few framed posters telling the story of Hausdorff and his set theory book

TechnoKnight

Guys, I need some help with something I didn't understand.

Can you please help me with it?

Masterphile

@AlessandroCodenotti you were there?

Astyx

Not if you don't tell us what it is @TechnoKnight

TechnoKnight

Thanks.

Alessandro Codenotti

I'm still a student there even though I haven't been in Bonn physically for almost two months because of the coronavirus

TechnoKnight

11:57 AM

The equation of wave is:

A = Amax*sin(wt + f)

My question is, how we figured out that Amax is Amax?

Astyx

What does that mean ?

TechnoKnight

I mean, let's write this equation: A = Bsin(wt + f)

How did we know that B is Amax?

Astyx

Because sin only takes values in $[-1;1]$

TechnoKnight

As we know, B is the max positive value of A...right?

Yes. But did we set wt + f to 90 or 0 or 180 or which one?

I'm really sorry for my bad English.

Astyx

It's the maximum as t varies I guess

TechnoKnight

12:00 PM

Yes, but which part of that equation is responsible to control the angle?

Masterphile

ah, he was a professor in Bonn

apparently

Astyx

@TechnoKnight I'm not sure what you mean by that

Alessandro Codenotti

He was

MathStudent

@TobiasKildetoft I meant the matrix A is symmetric. I'm pretty sure that a matrix of such form is always symmetric, I am double checking to make sure now though

TechnoKnight

@Astyx I will try to write in a different way.

sin(wt + f) = 1 means that wt + f = pi/2

Tobias Kildetoft

12:02 PM

@MathStudent Ahh, right, it is. And then you just need the eigenvalues, which will be those on the diagonal, once you pick $X$ to be orthogonal

TechnoKnight

Right?

Astyx

no, just that wt+f = pi/2 + 2 k pi

TechnoKnight

Then that means wt + f is the responsible for setting the angle for wave, right?

MathStudent

@TobiasKildetoft Unfortunately, I can't "pick" X to be orthogonal. It's arbitrary I think.

TechnoKnight

All what I mean, is that when wt + f = pi/2 + 2kpi, it is when A = Amax...right?

Tobias Kildetoft

12:06 PM

@MathStudent Right, but you can pick a different one which is, since the matrix will be orthogonally diagonalizable

Astyx

Yes

TechnoKnight

Another question

It's the final one

What does sin(wt + f + pi/2) mean?

Does it equal to 1 or something else?

Astyx

it's a function

Masterphile

@TechnoKnight it can take "much" values, 1 being only one of the possible values, try to find some info about geometric definition of sine function, with triangles and circle

MathStudent

@TobiasKildetoft Ah. I think I see what you mean. Do you know if the diagonal matrix would not stay the same then though if X isn't already orthogonal though, right?

Tobias Kildetoft

12:10 PM

@MathStudent I am trying to think of whether we are in fact guaranteed that $X$ must be orthogonal, but I am not seeing it

TechnoKnight

@Masterphile I know. I was just confused by someone writing the same equation and finding it equal to 1

Tobias Kildetoft

@MathStudent No idea what that means :)

Masterphile

the equation can be set to be equal to 1, but the function takes many values, you can set, for example A(sin(wt+r))=1 and try to solve that @TechnoKnight

Tobias Kildetoft

Though it might be that there is a hidden assumption that $X$ is in fact orthogonal

Mike Miller

An operator is positive definite it $\langle Qv, v\rangle > 0$ for nonzero $v$. Let's show that this is true for $Q = X^T DX$ so long as $D$ has all positive eigenvalues (or more generally, is positive definite itself) and X is invertible.

MathStudent

12:14 PM

@TobiasKildetoft I sent the message too early by accident sorry. I meant to add that I found out what the fisher information matrix looks like from here web.stanford.edu/class/archive/stats/stats200/stats200.1172/… but i guess if you dont have any idea what it means then it probably doesnt help either

Mike Miller

We have $$\langle Qv, v\rangle = \langle X^T DX v, v\rangle = \langle DXv, Xv\rangle.$$

Now $X$ is assumed invertible, so if $v$ is nonzero, then $Xv$ is nonzero. Now we get that $$\langle Qv, v\rangle = \langle D(Xv), (Xv) \rangle > 0,$$ as desired.

MathStudent

@TobiasKildetoft Anyway, I'm pretty sure it is not assumed that it is orthogonal. Do you have any idea whether it would be easier to try to find out what D would look like with an orthogonal matrix or to look at the leading minors?

Tobias Kildetoft

@MathStudent See what Mike wrote above

Mike Miller

I just proved for you that you only need to know that X is invertible and D is positive

MathStudent

Oh, I'm sorry, I didn't see those messages. Thanks a lot, @MikeMiller ! However, unfortunately, I don't think I can assume X is invertible either.

I do know that D is positive though

Tobias Kildetoft

12:20 PM

If $X$ is not invertible, then $A$ is at most positive semidefinite

VJ123

@Thorgott So would I be right in saying that if a function is not absolutely convergent, it is one of these three cases? (1) Integrals of both the positive and negative parts of the function are infinite. It is not HK integrable on the region either. The two iterated integrals are different, and the double integral does not exist. e.g. f(x,y) = (x^2-y^2)/(x^2+y^2)^2 integrated with x between 0 and 1 and y between 0 and 1.

(2) Integrals of both the positive and negative parts of the function are infinite. It IS HK integrable on the region, so the iterated integrals are equal and the double integral does exist, e.g. f(x,y) = sinx/x integrated with x between 0 and infinity and y between 0 and 1. (3) Integral of positive part of function is infinite and that of negative part is finite, or vice versa. Iterated integrals are equal to +/- infinity, and the double integral does exist and is also +/- infinity.

Edward Evans

12:32 PM

https://math.stackexchange.com/questions/3617190/minkowski-theory-an-isomorphism-of-bbb-r-vector-spaces-induces-a-scalar-prod

longboi for a dumb question

Astyx

If $H\triangleleft G$, what's the link between $L^2(G/H)$, $L^2(G)$ and $L^2(H)$ ?

If there is one

MathStudent

@TobiasKildetoft Because then there exists a $v$ such that $Xv = 0$ and looking at Mike's proof that gives positive semidefinite. Very interesting, thanks a lot, you two @MikeMiller @TobiasKildetoft !!!

Rafael Nadal

please explain this solution

Leaky Nun

@EdwardEvans when you said long boi I thought you meant long exact sequence

Astyx

$L^2(G/H)$ are the elements of $L^2(G)$ that are compatible with the quotient.

Edward Evans

12:34 PM

@Leaky hahaha

Rafael Nadal

i can't understand 2nd line of soln "if dt/dx = p"

Thorgott

Hmm, I think it may be possible to have the function non-integrable, but with equal iterated integrals

Edward Evans

@Leaky I think I'm confusing a couple of definitions but I can't pinpoint where lol

Leaky Nun

what definitions?

Rafael Nadal

@LeakyNun please help if you're free.

Edward Evans

12:36 PM

Well, I think I'm confusing what I'm trying to show; I think I'm trying to show that $\langle \psi(z_\tau), \psi(z_\tau^\prime)\rangle_c = \langle z_\tau, z_\tau^\prime\rangle_M$ in the notation given in the question

VJ123

@Thorgott oh even non HK integrable?

MathStudent

@MikeMiller I got another question about this after all.. How do you know that it would be $>0$ at the end and not just that it is nonzero?

ah wait a second..

Thorgott

I'm not sure, I'll have to think about it later

Mike Miller

Because $D$ is positive, we have that $\langle Dv, v\rangle > 0$ for all nonzero $v$ --- including $Xv$

More precisely, if $c$ is the smallest positive eigenvalue of $D$, you will see by hand that $\langle Dv, v \rangle \geq c |v|^2$

which is positive if $v$ is nonzero

MathStudent

I didn't think of the definition of the standard scalar product and thus didnt get to it sorry, silly mistake and thanks again!

Mike Miller

12:53 PM

No worries

Masterphile

the definition of limit superior an limit inferior confuse me, they are given without any examples of the sequences

without any concrete examples

any! :D

i think they are equal to the usual limit if they are equal

MathStudent

@Masterphile I was looking at those recently again as well. It is simply the limit if the limit exists. One example that helped me understand it for when the limit doesn't exist is $sin(n)$. the limsup is 1 and the liminf is -1. Just reading the wiki page for it helped me: en.wikipedia.org/wiki/Limit_superior_and_limit_inferior

Masterphile

@MathStudent so they are unique?

they seem to be lowest upper and greatest lower bound of the set of all accumulation points of a sequence

@Thorgott are they that?

MathStudent

@Masterphile Yes I think

@Masterphile the wiki article also gives a definition of them being inf sup and sup inf so that would fit to what you are saying there I think and it at least fits in the example of the sinus function

Masterphile

1:08 PM

yes, it seems it fits

Thorgott

yes, that's one characterization

VJ123

1:31 PM

@Thorgott Okay thanks please do let me know.

user131753

1:42 PM

1

Q: What conditions on the topological spaces are necessary and sufficient to ensure the existence of the following kind of functions?

Suppose that $(X,\tau_X)$ and $(Y,\tau_Y)$ are two topological spaces where neither is given the discreet (or indiscreet) topology. Does there always exists a nonconstant function $f:X\to Y$ is such that for all $Z\subseteq X$ if $\operatorname{Cl}_Y{\left(f(Z)\right)}\subseteq f(\operatorname...

general-topology

MathStudent

2:00 PM

@Masterphile I just thought of an example where this is not the case I think. Say we have a sequence that is defined as $x_n = 1/n$ for all $n$ except for $n = 5$ with $x_5 = 100$. Then the lowest upper bound is 100. However the limsup is still 0 because the definition isnt just supremum but lim sup with n going to infinity or infimum over $n \geq 0$ and supremum taken over $m \geq n$.

Masterphile

that lowest upper bound is not from the set of accumulation points

58 mins ago, by Masterphile

they seem to be lowest upper and greatest lower bound of the set of all accumulation points of a sequence

MathStudent

Ah okay. That could be it then.

Astyx

How do I find the decomposition of the representation of $SO_3$ on $L^2(SO_3/SO_2) = L^2(\Bbb S^2)$ if I know $L^2(SO_3) = \bigoplus_m \left(W_{2m+1}\right)^{2m+1}$ ?

Where $W_{2m+1}$ is the irreducible rep of dimension $2m+1$ of $SO_3$

Tobias Kildetoft

@Astyx What does it mean for a function in that space to be compatible with the quotient?

Astyx

I think it has to only depend on $z$

But it seems weird

Tobias Kildetoft

2:12 PM

what is $z$?

Ahh, so functions that are constant on the cosets?

Astyx

One of the canonical variables of $\Bbb R^3$

I think I'm confusing a lot of things

Tobias Kildetoft

I don't think you can use the information about the big decomposition directly without knowing something about how it comes about

Masterphile

@MathStudent if it isn´t that then it should be

Tobias Kildetoft

But I have not looked at anything like this for a long time (and never very much)

user8718165

Could someone please have a look at this question I posted. I also mentioned what I tried but the question has received very little attention. I'd be very grateful if someone could have a look. Thanks.

0

Q: Geometry, prove that $E$ bisects $\overline{HI}$

Here in the figure, we have: $\bullet$ The radii of both the circles are equal $\bullet$ $E$ bisects both $\overline{AC}$ and $\overline{FG}$ $\bullet$ $\overline{HI}$ is a line passing through $E$ We have to prove that $E$ bisects $\overline{HI}$ as well. (I think it is really obvious fro...

geometry analytic-geometry

Anonymous

2:23 PM

Is there an elegant method to see that alternating group $A_5$ can be written as the disjoint union of its Sylow subgroups and the trivial element? I mean, without having to explicitly write the subgroups in cycle notation.

Tobias Kildetoft

@SanchayanDutta The two ingredients are that the Sylow subgroups intersect trivially (even for same prime), and all elements have prime power order.

Anonymous

@TobiasKildetoft How do you prove trivial intersection for the same prime $p$?

Anonymous

(That is, without explicit decomposition...)

Tobias Kildetoft

For that part you do need to consider the prime 2 explicitly

Anonymous

@TobiasKildetoft And not the primes 3 and 5 too? Could you explain why?

Tobias Kildetoft

2:35 PM

because those only divide the order once

Anonymous

@TobiasKildetoft (Please bear with me here a bit, I'm new to this.) Let's take the 3-Sylow subgroup for instance. It has 10 conjugate subgroups. So wouldn't we also have to show that all the conjugate subgroups also intersect trivially with each other?

Tobias Kildetoft

@SanchayanDutta Yes, but that is clear just from the order

Anonymous

@TobiasKildetoft How so? I'm apparently missing that argument. Does that follow from the Sylow theorems?

Tobias Kildetoft

no, from Lagrange

Anonymous

@TobiasKildetoft Alright, how exactly? I'm probably being silly here, but Lagrange's just says that the order of every subgroup $H_i$ of a certain group $G$ divides the order of the group. Well, it is also true that the cosets of the subgroup $H_i$ partitions the whole group.

Tobias Kildetoft

2:44 PM

@SanchayanDutta So what can the intersection be if both the subgroups have prime order?

user193319

0

Q: Injective Morphism of Commutative $C^*$-algebras

Let $X$ and $Y$ be compact Hausdorff spaces, and let $\varphi : C(X) \to C(Y)$ an injective morphism of $C^*$-algebras. The book I'm reading through claims that there exists a continuous surjection $\alpha : Y \to X$ such that $$\varphi (f) = f \circ \alpha$$ for every $f \in C(X)$. But I am havi...

general-topology compactness operator-algebras c-star-algebras

Anonymous

@TobiasKildetoft If they have the same prime order? Let's label two conjugate Sylow 3-subgroups of $A_5$ as $H_1$ and $H_2$. Both $H_1$ and $H_2$ have prime order 3. But can we directly conclude from there that $H_1 \cap H_2 = \{1\}$?

Tobias Kildetoft

@SanchayanDutta Same or different prime order. Same conclusion

Because what does Lagrange say about the order of the intersection?

Anonymous

@TobiasKildetoft I'm not sure what Lagrange's says about the order of intersection...

Anonymous

This is probably what I'm missing though

Tobias Kildetoft

2:49 PM

What does Lagrange say about a subgroup?

Anonymous

@TobiasKildetoft That its order divides the order of the whole group...and its cosets partition the whole group...

abenthy

Given a geodesic line $L$ in the Poincaré disc and a point $P \in L$, how can one construct a point on $L$ that has a given distance $D > 0$ from $P$?

Astyx

Hey @ShaVuklia

Sha Vuklia

ola @Astyx !

how ya doin'

Astyx

Good and you ?

Tobias Kildetoft

2:52 PM

@SanchayanDutta Right. So use the first part here

Sha Vuklia

ye, pretty good (given the circumstances x))

Thorgott

think about $H_1\cap H_2$ as subgroup of $H_1$ (or $H_2$, doesn't really matter)

Sha Vuklia

also question: say we have a one-parameter subgroup $\gamma\colon\mathbb R\to G$ of a Lie group G. How do we know the push-forward $\gamma_*(d/dt)$ exists? (treating $d/dt$ as a left-invariant vector field on $\mathbb R$)

Astyx

What do you do nowadays ?

Sha Vuklia

I'm finishing my bachelor

basically

and you?

Astyx

2:53 PM

Finishing my year

Sha Vuklia

nice, what was it again you were doing?

Astyx

Engineering school, but mostly pure maths

Sha Vuklia

oh right ye

Anonymous

@Thorgott Aha! So the intersection of (say) two Sylow 3-subgroups $H_1$ and $H_2$, i.e., $H_1 \cap H_2$ necessarily divides both $|H_1|=3$ and $|H_2|=3$. So $|H_1 \cap H_2|$ is either $1$ or $3$?

Thorgott

yes

Anonymous

2:56 PM

(But it doesn't make much sense for $|H_1 \cap H_2|$ to be 3 because then $H_1$ and $H_2$ would be the same. So it's necessarily 1.)

Anonymous

Interesting. So now I need to show that the Sylow 2-subgroups all intersect trivially. How do I go about this?

Anonymous

24 mins ago, by Tobias Kildetoft

For that part you do need to consider the prime 2 explicitly

abenthy

3:09 PM

I have now posted my question on SE: math.stackexchange.com/questions/3617405/…

Mad Spaces

Hello! i plotted a function in mathematica and i want to make the gridlines look such, that it isbetween each point and point like a graph paper

Or also called millimetric paper

What ist the command ? i am suffering since an hour with this

Masterphile

i think i found three statements such that 1) and 2) imply 3) and 1) and 3) imply 2) and 3) and 2) imply 1), is this unusual?

Edward Evans

Hey @abenthy, you wouldnt happen to know your way around Minkowski theory would you? :)

Alessandro Codenotti

Hi @Edward

Edward Evans

Hey @Alessandro

Masterphile

3:16 PM

yeah, for analysis it´s unusual

classical elementary analysis

Mad Spaces

nvm got it

Masterphile

i did once transform the expression of Riemann zeta to some other function H such that, i think, H(1/2+epsilon)=H(1/2-epsilon), and i think i only needed to show epsilon=0, but i do not know where are those papers, i think i threw them away when cleaning a room, i do not tend to collect all that i write, so i lose some of my work without being published, that was when i was better in transforming functions so they take another form, i forgot what i exactly did

that wasn´t so much close to resolving RH but the expression was nice to work with

i remember i introduced some new functions there

they also had some nice properties

although i could try to recall the procedure exactly

and now i think of open sets!

what a downfall

abenthy

@EdwardEvans What do you mean? Minkowski space time?

Edward Evans

@abenthy I mean Minkowski theory within algebraic number theory hehe

abenthy

I've used some of this stuff, but am far from an expert. You have a question on it?

Mad Spaces

3:30 PM

No sorry i did not get it. If anyone knows how to create a grid that is millimitric in mathematica for plotting without making it look like utter eye Pain please mark me in your reply i was aiming for something like this

Edward Evans

I have a question posted on the main site, it's about the isomorphism between the Minkowski space and $\Bbb R^{r + 2s}$ and the scalar product it induces

0

Q: Minkowski theory: an isomorphism of $\Bbb R$-vector spaces induces a scalar product on $\Bbb R^{r + 2s}$

I'm following Neukirch's algebraic number theory. The situation is as follows: Let $K$ be a number field of degree $n$. Then $n = r + 2s$, where $r$ is the number of real embeddings $\rho : K \to \Bbb C$ (i.e. those embeddings $\rho$ such that $\rho(K) \subseteq \Bbb R$) and $s$ is the number of...

number-theory algebraic-number-theory geometry-of-numbers

I think I'm just being a doofus but if you have 5 mins at some point then I'd be grateful if you took a look :P No rush

Leaky Nun

@EdwardEvans what if I don't have 5 mins at any point

Alessandro Codenotti

Then don't take a look

Edward Evans

I'll only be grateful if you spend exactly 5 mins looking at it

Alessandro Codenotti

So if I opened it, read "Neukirch" and closed it immediately I won't be receiving any gratitude?

Edward Evans

3:43 PM

You won't, and you'll also have to pay royalties to Neukirch's grave for having seen his name

Alessandro Codenotti

Meh

Leaky Nun

you mean to the grave of he-who-must-not-be-named?

Edward Evans

nah that's Voldemort you're thinking of mate

Leaky Nun

gasps

Edward Evans

oh fu- is immediately killed by death eaters

user714630

4:03 PM

@EdwardEvans What should be happening is that $z\bar w + \bar z w = 2\text{Re}(\bar z w) = 2(x_1x_2+y_1y_2)$ where $z=x_1+iy_1$ and $w=x_2+iy_2$.

feynhat

4:17 PM

Hello.

feynhat

4:47 PM

$$ \begin{tikzcd}
& \Omega^*(U) \arrow{dr} & \\
\Omega^*(U \cup V) \arrow{dr} \arrow{ur}& & \Omega^*(U \cap V) \\
& \Omega^*(V) \arrow{ur} & \\
\end{tikzcd}
$$

ugh... How do I commutative diagram?

Thorgott

you google "commutative diagram latex", then copypaste the first result and try to adjust it without breaking it

the second part sounds easier than it is

Alessandro Codenotti

tikz is not supported in this chat or MSE

feynhat

5:02 PM

@Thorgott How do I not break it? I mean what I posted earlier is a valid latex code, but mathjax doesn't render it for some reason.

@AlessandroCodenotti oh... For that reason.

Tobias Kildetoft

@feynhat No, it is not valid basic LaTeX

It uses a package

feynhat

I see.

Tobias Kildetoft

(also, MathJax is only a subset of LaTeX, so there are occasionally other stuff that also doesn't quite work)

Alessandro Codenotti

There is a weird commutative diagram package usable on MSE and here but I forgot what it's called or how it works. It's also rather limited

feynhat

5:14 PM

Is this a pullback?

All arrows are induced by inclusion.

Suppose there is a module $M$, with maps $\alpha: M \to \Omega^k(U)$ and $\beta: M \to \Omega^k(V)$. I have to come up with a map $M \to \Omega^k(U \cup V)$.

$\alpha(m)$ is a k-form in $U$, $\beta(m)$ in $V$.

Do I have to invoke partitions of unity?

Of course, $U$, $V$ are open sets such that $U \cup V = N$ (a smooth manifold).

by $M$ being a module I meant, $M$ is an $\Bbb R$ vector space.

choose a partition of unity for $\{U, V\}$. Define $\gamma : M \to \Omega^k(N)$ as $\gamma(m) = \rho_U\alpha(m) + \rho_V\beta(m)$.

(Sorry for the poor choice of symbols, I should have chosen M for the manifold)

Does this $\gamma$ work?

anakhro

5:41 PM

@AlessandroCodenotti how best to describe the weak convergence on $L^\infty(\mathbb R)$?

Weak-* convergence is pretty easy to describe as $\int x_ny\,dx\to \int xy\,dx$ for all $y\in L^1(\mathbb R)$.

Alessandro Codenotti

6:26 PM

Hm I'm not sure

The dual of $L^\infty$ is some space of measures, right?

Archer

Why do we need p values when we have rejection regions? or vice versa?

Masterphile

6:45 PM

is some branch of topology specially suited for research of $\mathbb R^n$?

Mike Miller

@feynhat Yes, that works. On the overlaps $\alpha(m) = \beta(m)$ and $\rho_U(m) + \rho_V(m) = 1$, so that $\gamma(m) = \alpha(m) = \beta(m)$. On $U \setminus V$ the desired equality is automatic.

Similarly with $V \setminus U$. I wouldn't have done this with partitions of unity.

Masterphile

for example, this theorem seems to be specially suited for $\mathbb R^n$:

https://en.wikipedia.org/wiki/Invariance_of_domain

Mike Miller

I would have just said $\gamma(m) = \alpha(m)$ on $U$ and $\beta(m)$ on $V$, with no ambiguity on the overlap by assumption.

Masterphile

it seems "algebraic topology" could be that branch

Mike Miller

Invariance of domain is proved by algebraic topology. Most theorems of algebraic topology about $\Bbb R^n$ also extend to the context of manifolds much more generally, for instance invariance of domain.

Algebraic topology is not about Euclidean space in particular.

Masterphile

6:51 PM

i think there is also "geometric topology", whatever that branch is about, is it specially for $\mathbb R^n$

Ted Shifrin

does \begin{CD}\end{CD} work in chat?

Astyx

Not for me

Masterphile

sort of, as an introduction to the subject, i would like to do topology in $\mathbb R^n$ mostly

eventually also in the case $n= + \infty$

Ted Shifrin

salut, @Astyx

Astyx

Salut

Masterphile

6:57 PM

ah, doesn˙t matter much, i think i can find a way to handle even more general cases

when some efforts to prove NT result which would be an improvement failed, i started to do topology

Ted Shifrin

"do"?

Masterphile

yes, meaning "to research"

Ted Shifrin

sounds like the first step should be to study ...

Leaky Nun

@TedShifrin what's the possessive of "Tom and I"?

Masterphile

yes, i am learning basics, but i am always research-oriented, much of results can be improved, some slightly but much of them can, although i do not know how much i am going to be skilled in some particular branch

Ted Shifrin

7:03 PM

@Leaky: "Tom's and my ..." most likely.

Leaky Nun

hmm

English is weird

Ted Shifrin

I can also work around ... "that belongs to Tom and me."

Mike Miller

@Masterphile Just study. Research means you just keep asking questions while being curious. Don't worry about the forefront of research. It is very far.

4

Ted Shifrin

Well put, @MikeM.

Masterphile

@MikeMiller yes, but i am more researching than proving actually, either known theorems, or some new ones (of not-so-great importance), i know that that is usual, but i am worried about that, and i shouldn´t be, since i am not student nor a professor, i do it as a hobby, so, in some senses, i won´t perish if i do not publish, most probably not even if i publish, if you understand me well

Mike Miller

7:18 PM

You seem to be saying you're immortal.

Masterphile

no :D, i just have no complete understanding of the word "perish", i tried much to understand that word but i cannot fully understand its meaning, English is not my native so some words to me are very difficult to fully grasp

i know some synonyms, but they are not equivalent fully to that word, it is a very interesting word

Mike Miller

To perish means to die. The phrase "publish or perish" does not use it literally, but as an analogy: to lose your job is like to die. The reality is that you publish or you won't keep your job.

I understood you meant "I won't lose my job, whether or not I publish something unrelated". But I was taking you literally (as in to die), as a joke.

Masterphile

nah, "to perish" is different than "to die", it is one of the most interesting words that i know of, but i do not know much about genesis of English language, nor of its evolution to be competent as much as i would like to

NoName

It does mean "to die".

Khallil

7:33 PM

Who knew I walked into english.stackexchange.com? xD

Ted Shifrin

It used to be french/german ... too :)

Leaky Nun

yeah, it's the Wurst

Ted Shifrin

Knockwurst

NoName

Reminders of mortality everywhere now. Me don't like it.

Everybody say wurst keep it going

Masterphile

did some of you read Elements, or at least, for example approx. 50-100 pages of it?

Khallil

7:39 PM

I actually wrote it, @Masterphile

NoName

lol

Masterphile

@MikeMiller so we now know who is immortal here

Khallil

Immortality jokes xD

Masterphile

well, i have read relatively much of it, why not, it is a classic

Ted Shifrin

I actually do have Euclid here (in English translation), but I only read pieces of it when it's relevant to do so. :P

Khallil

7:41 PM

Isn't it all available online?

anakhro

I took a history of math course where we went over it.

@Khallil half a billion different translations, too.

Ted Shifrin

A former student gave me the paperback book years ago to thank me ...

Masterphile

@anakhro yes, i actually wanted to ask Ted which one he has

NoName

I thought they sold that type of geometry ages ago and exchanged it for some algebra.

Khallil

The devil was very happy with that deal, @NoName lol

Ted Shifrin

7:43 PM

Mine is translated by Sir Thomas L. Heath.

Érico Melo Silva

@TedShifrin the best english scholar of greek math

Ted Shifrin

Glad to hear that, @Eric !

I am woefully ignorant in such matters.

Érico Melo Silva

his two volumes on the history of greek mathematics are very very good

Masterphile

who decides if a mathematician will get the title "Sir"?

Khallil

I think it's the queen for England and the head of state elsewhere

Masterphile

7:46 PM

@Khallil i observe that as not a joke

could be, Newton was "Sir" also, so it is relatively an old title

NoName

I don't know how much it meant back then but now it hardly means anything.

Sir Harry Kane lol

CaptainAmerica16

does anyone know how to put $m \times n$ into a subscript with mathjax? I can't figure it out.

Masterphile

i think Heath-Brown was (or is) a "Sir"

Khallil

\text{stuff}_{m \times n} gives $\text{stuff}_{m \times n}$ @CaptainAmerica16

CaptainAmerica16

@Khallil Thanks!

Khallil

7:52 PM

You're most welcome :)

Masterphile

it seems i thought wrong, or there is no information on Wikipedia, i really thought Heath-Brown is a "Sir"

Khallil

This is a fairly basic guide I used when I was first getting into typesetting, @CaptainAmerica16

Astyx

How can I make $$\int_{\Bbb R^d}{\overline{\nabla u} \cdot xu(x)\over|x|^2}dx$$ look nice ?

Ted Shifrin

Start with \displaystyle

Khallil

What does it mean to look nice, @Astyx?

CaptainAmerica16

7:58 PM

@Khallil Cool :D I was using another one, but it didn

have all of the necessary info

Astyx

As in, how can I solve it :P

Ted Shifrin

You also need $u(x)$ at the front of the numerator or parentheses.

Oh.

Khallil

I always use \text{ d}x for my differential forms but I'm slightly strange

Ted Shifrin

I never bother, Khallil.

Khallil

Oh x2 combo

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$Mathematics$ Mathematics