Mathematics - 2024-09-28

Jakobian

03:39

I found an example of two uniformities in terms of entourages on $\mathbb{N}$ such that their intersection isn't a uniformity, if anyone is interested

@AlessandroCodenotti perhaps

nickbros123

05:16

suppose I label each vertex of a cube with 1,2,3, dots, 8. Suppose I wanna count the rigid rotations of this cube. it must be enough if I know the position of 1 and 2 right?

i.e if i specify position of 1 and 2, the rotation would be uniquely determined? (or rather all vertex position gets specified too?)

Soumik Mukherjee

05:28

@nickbros123 yes

PM 2Ring

08:20

I'm baffled that anyone capable of doing mathematics would consider that this is an acceptable question...

So you have written your exercise on a piece of paper. You don't even write between the lines and it's written wrongly (is it the root of $2x^2$ or of $2x^2+7$?). Then you take a picture of that, add a text with another spelling mistake, and put it, without any other effort, on the internet, and then you expect total strangers to make your homework. Not sorry to disappoint you, but this is really not the way to go! — Dominique 20 hours ago

Cantor Dust Drachen

I mean, don't want to be that guy, but none of these things are intuitive to me as what would be expected by people of the internet

PM 2Ring

@nickbros123 What @Soumik said. There are 24 rotations in the cube group. We can compose that as 8 vertices × 3 neighbours of the chosen vertex. Or 6 faces for the top × 4 faces for the front.

PM 2Ring

08:41

@CantorDustDrachen We don't expect that people are born knowing this stuff. But we do expect people to explore the site a little before they post, to get an idea of what is acceptable and what's likely to get downvoted / closed. Of course, a lot of the bad stuff gets deleted after a few days, but there are still plenty of examples hanging around.

From meta.stackexchange.com/a/402975/334566

> We expect new members to take the Help Tour, and to spend a little bit of time reading existing questions and answers, so that they get a feel for site culture, and a sense of what kinds of questions do well here, and what kinds tend to get closed or downvoted.

> Admittedly, the Help center pages were originally composed in the early days of Stack Overflow, and although they have been revised in the subsequent years, many of those pages still retain a Stack Overflow focus. However, they're still a reasonable introduction to site culture, IMHO.

Ryder is not Rude.

The Cultural Anthropology of Stack Exchange

Ryder is not Rude.

09:15

@CantorDustDrachen that^ may help as an introduction

psie

10:05

I posted a question on the main site. I'm getting hung up on a small remark that succeeds Carathéodory's extension theorem in Folland's real analysis text. I don't see really what the remark has to do with the theorem, why "the proof of this theorem yields more than the statement." Maybe I'm reading too much into this.

Ryder Rude

hi

Jakobian

@leslietownes hey. I have a question about functional analysis if you're there

Jakobian

11:00

I don't have a question anymore

Soumik Mukherjee

11:11

Who stole your question?

Jakobian

Santa Claus

I've came up with a solution to it

I was going to ask how to calculate dimension of $\mathbb{R}^\mathbb{N}/\ell^\infty$

I needed that for an answer I was writing

Jakobian

11:43

not like it matters because no one cares about it

psie

13:36

Let $F(x)=\mu((-\infty,x])$, where $\mu$ is a finite Borel measure on the real line. I see how $F$ is increasing due to monotinicity of $\mu$. I struggle a bit with the claim that $F$ is right-continuous. We have $(-\infty,x]=\bigcap_{n=1}^{\infty}(-\infty,x_n]$ for any sequence that decreases to $x$, and then, since $\mu$ is finite, we can apply continuity from above and conclude via the sequential definition of right-continuity that $F$ is right-continuous.

Why isn't it left-continuous by a similar argument using continuity from below?

Vladimir Lysikov

13:48

Because if $x_n \to x$, then $\bigcup_{n=1}^\infty (-\infty,x_n]$ is not $(-\infty,x]$, it is $(-\infty,x)$

psie

ah right, makes sense! thanks!

Madder

14:18

Hey Guys i am learning about manifolds, can someone check if i am understanding the notation correctly?

If two neighborhoods overlap, a point $ p \in U_i \cap U_j $ has two coordinates. A function $ F: \mathbb{R}^n \to \mathbb{R}^m $, given by

$$
F(q) = (f_1(q), \ldots, f_m(q)),
$$

is differentiable (of class $ C^k $) if the functions $ f_i: \mathbb{R}^n \to \mathbb{R} $ have $ k $-th partial derivatives.

Let $ (U_i, \varphi_i) $ with $ \varphi_i(p) = (x_1(p), \ldots, x_m(p)) = (x_1, \ldots, x_m) = x $ and $ (U_j, \varphi_j) $ mapping to $ y $ be overlapping charts. The transition

$$
\Psi_{ij} = \varphi_i \circ \varphi_j^{-1}: \mathbb{R}^m \to \mathbb{R}^m

It seems to me that the notation $x_i$ is abused for both the function and both $x_i(p)$

what confuses me is that they even talk about $x_i(y)$ as a transition between coordinates map which makes zero sense to me

Ben Steffan

$x_i(y_i^{-1})$ doesn't make sense; you mean $x_i \circ y_i^{-1}$

what $y_i^{-}$ is supposed to be I have no idea

your definition of $C^k$ is also not quite correct

the notational abuse is common, however

Vladimir Lysikov

14:39

Also $\varphi_i \circ \varphi_j^{-1}$ is not $(x_1 \circ y_1^{-1}, \dots x_m \circ y_m^{-1})$
$y_i$ is not necessarily invertible, only the whole $\varphi_j$ is

But it is common to denote $\Psi_{ij}(y)$ as $x(y)$ and $\Psi_{ji}(x)$ as $y(x)$

Shaun

15:04

Research maths is hard, man. (That is all.)

Pizza

👋👋

Sine of the Time

🤝

Pizza

15:20

@SineoftheTime I was discussing a system as a function of the parameter $\lambda \in \Bbb R$ , can I show you what I did?

Sine of the Time

yes

Pizza

Ok then take a moment while I write

$S : \begin{cases} x + y + (\lambda + 1)z = 3 + \lambda \\ x + 2y + z = 4 \\ 2x + y + \lambda z = 3 + \lambda \end{cases}$

Ok I calculated the determinant of the incomplete matrix and I found $\text{det(A)} = -2\lambda - 2$

So $ = 0 $ when $\lambda = -1$ , otherwise $≠ 0$ when $\lambda ≠ 1$

If the determinant is 0 then I find a minor of order 2 $H = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$

With determinant 1 ≠ 0

So $r(A) = 2$ or $r(A) = 3$ if det(A) ≠ 0

Is this correct up to this point?

Sine of the Time

the reasoning is correct

I did not check if what you found for $\det A$ is correct

Vladimir Lysikov

$\det A$ is correct

Pizza

Ok, now when $\lambda = -1$ , $r(A') = 2$ , so $r(A) = r(A') = 2$

So for $n = 3$ variables and $r(A) = 2$ , 3-2 = 1 , so the system admits $\infty^1$ solutions

$H = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$ then the variable I have to take is $z = t$

$S' = \begin{cases} x + y = 2 \\ x + 2y = 4-t\end{cases}$

$x = \begin{pmatrix} 2 & 1 \\ 4-t & 2 \end{pmatrix} = t \quad y = \begin{pmatrix} 1 & 2 \\ 1 & 4-t \end{pmatrix} = 2-t$

$\sum = \{ (t,2-t,t) : t \in \Bbb R\}$

When $\lambda ≠ -1$ then $\text{det(A)} = -2\lambda -2 ≠ 0$

In this case the system admits a unique solution, right?

Vladimir Lysikov

15:37

$\lambda = -1$ case is correct

And yes, for $\lambda \neq -1$ there is a unique solution

Pizza

@VladimirLysikov Yes because it would be rank 3, right?

Vladimir Lysikov

yes

Pizza

Yes however using Cramer's rule I find $S : (1,1,1)$

Vladimir Lysikov

That is the correct solution

Which you can check by substituting into the original system

Pizza

Yes

Anyway that was all, thanks 👍

Vladimir Lysikov

15:44

Did you not cover Gaussian elimination in class?
I think it is significantly more concise for this problem

Pizza

@VladimirLysikov No, we haven't studied it

Vladimir Lysikov

Then everything is good 👍

Madder

@BenSteffan Yes

@VladimirLysikov _Can you elaborate on that? how is it, that the $\phi_j:= (y_1,\cdots ,y_m)$ is invertible, but not the individual functions

and what would the correct notation be in my example

Well actually you do not need to explain any further why the individual functions are not invertible. Just how the correct notation would look like

Ben Steffan

16:07

@Madder look at the $F$ in your question. Its components are maps $\mathbb{R}^n \to \mathbb{R}$. Unless $n = 1$, there's no chance for these to be invertible.

Vladimir Lysikov

@Madder I don't think there is something easier than $\Psi_{ij}(y) = \varphi_i(\varphi_j^{-1}(y))$ or $\Psi_{ij} = \varphi_i \circ \varphi_j^{-1}$

Madder

@BenSteffan if you are refering to the cardinality, these have the same

@VladimirLysikov If i want to express it in terms of the coordinate functions ?

Ben Steffan

nobody's referring to cardinality anywhere

Vladimir Lysikov

Btw $\Psi_{ij}$ is not $\mathbb{R}^m \to \mathbb{R}^m$, it maps $\varphi_j(U_i \cap U_j)$ to $\varphi_i(U_i \cap U_j)$

Madder

that is true, the thing about F in the question was just an example.it should not be considered part of the functions defined under, it is not even referenced

Ben Steffan

16:10

@Madder Are you aware that there are no diffeomorphisms $\mathbb{R}^n \to \mathbb{R}$?

(when $n > 1$?)

Madder

@BenSteffan No because i did not study differential geometry. I am doing that right now :) But i guess that is a good theorem to know!

Ben Steffan

that's not a diff. geo fact

that's a pretty basic real analysis fact

prove it as an exercise :)

Madder

Alright so then these functions can not be invertible, by this theorem.

Vladimir Lysikov

@Madder THey are actually never invertible in dimension more than 1

A concrete example would be the identity on $\mathbb{R}^2$
obviously both coordinate functions are not invertible

Madder

So if i write $ \phi_i = (x_1,\cdots, x_m) $ and $ \phi_j= (y_1,\cdots, y_m) $ how would i express $ \phi_i \circ \phi_j^- $ in terms of these coordinate functions?

Is it even possible to do so?

Ben Steffan

16:18

you can't, generally speaking

Madder

:(

but but how would i conclude then that the $ x_i $ and $y_i$ need to be smooth if the transition map is smooth?

Ben Steffan

components of a smooth map are smooth

so it suffices to show that the charts themselves are smooth

Madder

Ok but how would i know what the components of $ \phi_j^- $ are? or does this not interest me at all because i know that $ \phi_j \circ \phi_i^- $ is also smooth ?

Ben Steffan

There is no such thing as "components of $\phi_j^{-1}$"

$\phi_j^{-1}$ is a map $\mathbb{R}^m \to M$, where $M$ is your manifold

to show that the charts are smooth, you need to know what a smooth map between manifolds is, and then simply unpack the definition a little

Vladimir Lysikov

$x_i$ maps $U_i$ to $\mathbb{R}$
And $U_i$ is in the manifold, not in a euclidean space in general
So first of all we need a definition of a smooth map defined on an open in a manifold

Madder

16:23

i understand, but they are talking about this point, before they even tell me what a smooth map is, albeit they talk about C^k differentiability

students.iiserkol.ac.in/~mms15ms051/courses/PH4105/…

For example in page 3 (or 10 in the PDF)

If you do not want to open the link.

Ben Steffan

I'm tempted to just say "use a better text"

Madder

literally three books i pulled about this topic for physicists introduce it almost similiar

I have checked other mathematic books and their notation is different.

Ben Steffan

this text seems notationally mostly compatible with Lee

"Introduction to Smooth Manifolds" that is

looking at this, the upshot seems to be that you shouldn't worry to much about the details. their definition of smooth manifold explicitly drops any mention of topological space, and therefore ends up being... at odds with the usual definition

Vladimir Lysikov

Yes, it's a bit strange
But what they say is that (in their notation) for every point $p$ in $U_i \cap U_j$ the transition function $\phi \circ \chi^{-1}$ maps a point $(x^1(p), x^2(p), \dots, x^n(p))$ to $(y^1(p), y^2(p), \dots, y^n(p))$.
And we can consider the coordinate functions *of the transition map*, not of a chart
That would be $y^i \circ \chi^{-1}$ in their notation

Together they give the transition map $\phi \circ \chi^{-1}$, and it is smooth if and only if $y^i \circ \chi^{-1}$ are smooth

Madder

@VladimirLysikov do you mean $ \chi \circ \phi^- $ , because $ \phi(p)= (x_1 (p),\cdots,x_m(p))$

Vladimir Lysikov

16:35

@Madder Yes, sorry

And $y^i \circ \phi^{-1}$

Madder

I got it.

Last question would be, if we can not write $ \chi^-$ in regards to the component functions $ y_1 ,\cdots..,y_m$ is it then more appropriate to introduce another variable to express it? say $ \chi^- = (\xi_1 ,\cdots , \xi_m)$ ?

Ben Steffan

to repeat what I said above: there's no such thing as "components of $\chi^{-1}$" (also please write ${}^{-1}$ instead of ${}^{-}$ in the superscripts)

Madder

Oh alright sorry i misunderstood that comment. I am still thinking in terms of $ \mathbb{R}^n$

Ben Steffan

components are associated to maps into some product space, like $\mathbb{R}^n$

but a general manifold is not of that form :)

Madder

Yea true now i got it. Thank you !

Mats Granvik

17:57

Modular forms at squarefree numbers.

The Langlands Program - Numberphile

►

Ben Steffan

?

Mats Granvik

.

psie

19:09

How would you define sequential right-hand continuity? I can't find anything online. I believe it is not that common to use that definition, or maybe you're somehow clever and just use the regular sequential definition of continuity, but I don't know how. This would be very appreciated. Best would be for $f:A\subset \mathbb R\to\mathbb R$, but if that's a hassle, then simply $f:\mathbb R\to\mathbb R$ would do too.

copper.hat

what would the point of such a definition be?

i mean, why would be any different to the usual definition of continuity from the right?

psie

@copper.hat in measure theory, $F(x)=\mu((-\infty,x])$ is right continuous if $\mu$ is a finite Borel measure on the real line. I don't have any firm definition at hand to check that it is right continuous.

copper.hat

I mean and $x_n \downarrow x$ implies that $F(x_n) \to F(x)$.

Lucky Chouhan

Hi @copper.hat

copper.hat

It is basically continuity on $[x,\infty)$.

Hi @LuckyChouhan.

psie

19:16

@copper.hat does $x_n \downarrow x$ mean that $(x_n)$ must strictly decrease to $x$? I think the meaning of $x_n \downarrow x$ is a bit ambiguous.

copper.hat

no, it just means that $x_n > x$.

psie

what would be wrong about $x_n\geq x$, i.e. having weak inequality rather than strict?

Lucky Chouhan

@copper.hat Would you please tell me how I can be better at proving statements in math? I really search a lot, but I don't want to start with too easy and too hard because in both cases I'll be wasting time. So what to do, how you improved your skills?

copper.hat

$x_n \ge x$ is fine too, but does not add anything in terms of proving continuity.

@LuckyChouhan That is waay to broad a question. Just hard work and trying many things.

Lucky Chouhan

@copper.hat yeah, but from where to start. When I try to prove things in math and get failed, I really feel bad and think I can't do serious math :( But I really want to be better at problems to prove.

copper.hat

19:22

I have no suggestions other than the above, just try things. Start simple.

@psie the main result with measures is if $A_n$ are decreasing (nested) sets and at least one has finite measure then $\mu A_n \downarrow \mu ( \cap_k A_k )$.

psie

yeah, indeed, continuity from below (or above, I can never remember)

copper.hat

from above, since they are decreasing

if $x_n \downarrow x$ you can find a subsequence that is decreasing. Then use the fact that $\mu$ is $\sigma$- additive.

psie

@copper.hat inspired by your remark here, here's my definition of right-hand continuity. Let $f:A\subset\mathbb R\to\mathbb R$ be given. Then $f$ is right continuous at a limit point $c$ of $A^+=\{x\in A:x>c\}$ iff $$\lim_{n\to\infty}f|_{A^+}(x_n)=f|_{A^+}(c),$$ for every sequence $(x_n)$ in $A^+$ such that $x_n\to c$ as $n\to\infty$ and where $f|_{A^+}$ is the restriction of $f$ to $A^+$.

copper.hat

you probably want $x \ge c$, otherwise the restriction is not defined.

at $c$, i mean

psie

true :)

copper.hat

19:32

again, focus on what the point is first, then the nitty gritty.

psie

ok 👍

copper.hat

i see Hilbert's Airbnb is getting a recent mention

Martin Brandenburg

19:48

probably a bit too elementary to ask in the forum, but maybe here in chat: what are the orbits of GL(n,Z) in Z^n ? and for which u,v in Z^n can we know that there is some g in GL(n,Z) with g * u = v ?

copper.hat

out of my range

Ben Steffan

is there a reason to suspect a nice description?

Martin Brandenburg

no reason, but on the other hand we know that for u in Z^n with gcd(u1,...,un) = 1 there is some g in GL(n,Z) whose first column is u, which means g * e_1 = u. so all these lie in the orbit, at least.

in one orbit <-- wanted to write this

The Empty String Photographer

You can use MathJax in chat

It’s one of the room description links

copper.hat

this room seems to have emptied out of the usual cast :-(

Martin Brandenburg

20:01

yes I know but for me the mathjax doesn't render anway for some reason (I see the formulas posted previously, but they don't render). :)

copper.hat

check out the tinyurl link at top right.

Martin Brandenburg

ah yes, works now. thx

ok now in all of its LaTeX glory: what are orbits of $\mathrm{GL}_n(\mathbb{Z})$ in $\mathbb{Z}^n$? and for $u,v \in \mathbb{Z}^n$ when do we know that they lie in the same orbit? for example, they do lie in the same orbit when $\mathrm{gcd}(u_1,\dotsc,u_n) = \mathrm{gcd}(v_1,\dotsc,v_n)$.

ah wait maybe that is also necessary

yes. but I have a second question. maybe harder, I don't know. Let $A,B \subseteq \mathbb{Z}^n$ be subgroups such that $f(A) \subseteq f(B)$ for all homomorphisms $f : \mathbb{Z}^n \to \mathbb{Z}$, does this imply $A \subseteq B$?

Ben Steffan

20:31

the more I think about Lurie's discussion of colored operads the less I understand it

or rather, the better I understand it, which leads me to believe that some of the things he does are... not quite correct

Frieren

21:14

I would like to prove this statement: if $(a_n)$ does not have a subsequence tending to $+\infty$, then $(a_n)$ is bounded from above. My strategy was trying to prove this by contrapositive: assume that $a_n$ is not bounded from above. So, for each $k\in\mathbb{N}$ there exists $n_k \in \mathbb{N}$ such that $a_{n_k}>k$. Let $M>0$ be arbitrary. If $k>M$, then $a_{n_k}>k>T$ and this shows that $\lim_{k \to +\infty} a_{n_k}=+\infty$. Is this correct?

copper.hat

it is correct, but it is easier to write that if $a_n$ is unbounded then there is a subsequence whose limit is $\infty$.

how's grandad?

Ben Steffan

actually, I would probably deduct a point here: How do you know $k \mapsto n_k$ tends to infinity?

that's intuitively obvious, but at this level you should probably argue why

copper.hat

I would have stopped at the "such that $a_{n_k} > k$".

Ben Steffan

I would have stopped after the problem statement. This is clear.

but both of us would have a lost a significant amount of points in the course where stuff like this is taught, at least here :)

Sine of the Time

@copper what's your favourite real analysis book?

copper.hat

21:22

first, i am an engineer, not a mathematician. there are 3 books that would content.

one is komogorov & fomir's real analysis

functional analysis by j=kantorovich & akilov

and the most basic, which i still enjoy skimming is an earlier edition of marsden's elementary classical analysis

Sine of the Time

thank you

copper.hat

fomin, i mean. and i should have capitalised, but am very lazy today

KFRA FAKA MECA there, that makes it right

Rietty

Hi, quick question but when I have an imaginary number i, I know that i^2 = -1. However is my understanding that (-i)^2 = 1 accurate?

Cause when I put into google or wolfram it gives me (-i)^2 = -1

Ben Steffan

The equality is true, about your understanding, well, who can say

$(ab)^n = a^n b^n$ holds even when $a, b$ are complex numbers

copper.hat

$(-i)^2 = -1$.

Ben Steffan

21:33

sorry, missed the sign

copper.hat

that's what they all say

jk

Rietty

Okay thank you

copper.hat

@Rietty think of multiplying by $i$ as a 90 degree anti-clockwise rotation.

so twice is the same as -1. similarly $-i$ is a clockwise rotation, and twice rotated is, again, $-1$.

picture of quaternions near Broom Bridge, of Hamilton fame imgur.com/a/4o69cC5

it shows the irish love of squaring

Ben Steffan

22:07

love to spend the whole evening being confused out of my mind because I misguessed a definition

hubris etc.

copper.hat

my normal state

Ben Steffan

being confused?

name a person in this chat of whom this isn't true :)

but looking at a term you don't know and going "oh I know what this should mean" and then not questioning that for a few hours while you descend into madness is another level of [redacted]

and I could easily have guessed the right thing, too

copper.hat

mathematics is an evil and dangerous siren

4

Ben Steffan

there was a vise in the room today and I decided to put my foot in it for fun, then tightened it and suffered for hours

I could have simply not done that

Frieren

@BenSteffan thanks for the reply. I am not completely following why I have to consider that: is that because I must verify that $(a_{n_k})$ is indeed a subsequence (that is, the indexes are a sequence of integers that tends to infinity)?

Ben Steffan

22:17

@Frieren yes

Frieren

@BenSteffan Ok, got it :D I already proved that in the past by induction while proving that a subsequence of a sequence has the same limit, so in my mind it was not necessary. My bad!

Ben Steffan

it's fine, it's fine :)

I pointed this out because I know that you lose points for this in your first term at uni here

it's a good idea to start out being quite pedantic, and then relaxing this later when you have more experience, etc.

Frieren

@BenSteffan Thanks for the precious advice :D

Ben Steffan

welcome :)

Rietty

23:56

@copper.hat This actually helps me a lot, thank you so much.

Transcript for

$Mathematics$ Mathematics