Mathematics - 2020-10-06 (page 1 of 2)

user714630

00:02

What does the book say, for the record?

Stupid question inc

01:47

@BalarkaSen it is something that you can define one by one

by one by one I mean x_11=1, x_21=2,x_12=3,...

n_11=1,n_12=3,n_13=6...

n_21=2,n_22=5,n_23=9...

n_31=4,n_32=8...

and there seems to be no exact pattern to define this

If I have any mistake I would like to know where I have made a mistake

Stupid question inc

02:18

how is set $\{z\in\Bbb{C}:|z-z_0|<r}\$ an open set?

open set is defined as a set which has every interior point contained in it

and interior point is defined as a point with radius greater than 0 ie a disc

pretty confused with how I apply disc concept in $\{z\in\Bbb{C}:|z-z_0|<r}\$

what if a disc was large enough that interior point becomes boundary point

the radius of interior point isn ot even defined

Thorgott

this doesn't make sense

what's the "radius of a point"

Stupid question inc

sorry was in hurry

I will rewrite it clearly

Sophie

02:34

I'm having a brainfart. Someone remind me, is there a way of calculating primitive roots other than trial and error?

Hardik Rajpal

hi so i am certain that this isn't the right place for my doubt but every other chatroom is dead

Stupid question inc

define disc as $B_r(z_0)=\{z\in\Bbb{C}:|z-z_0|<r\}$ and for circle it's same but $|z-z_0|=r$

Hardik Rajpal

pls bear with me. where can i post a question about some technical issue with my laptop

like probably a system bug or a virus

i need an expert's opinion to be certain

Joe Shmo

does anybody know if there are two different definitions of the Fourier transform on $L^1$? (That inevitably agree)

Hardik Rajpal

@Sophie, we can check for prime factors of the constant term in the polynomial equation

Sophie

02:37

I don't know what polynomial equation you're talking about

Stupid question inc

interior point is defined as Let $E\in\Bbb{C}$. A point $z_0$ is an interior point of $E$ if there is some $r>0$ such that $\B_r(z_0)\subset E$

Joe Shmo

$f(\xi) = \int_{\mathbb{R}} f(x) e^{-2 \pi i x \xi} dx$ is one, what's the other?

Hardik Rajpal

when you said primitive roots, we are discussing polynomial roots, yes?

Stupid question inc

so now problem is why $\{z\in\Bbb{C}:|z-z_0|<r}\$ is open set?

it's just a disc

Hardik Rajpal

nvm

looked up primitive roots. i am outta my depth in number theory

so i am out.

Stupid question inc

02:38

and interior point is defined as $B_r(z_0)\subset E$

Hardik Rajpal

@Stupidquestioninc I don't think AJAX works anymore on the site

Joe Shmo

for example, for $L^2$, the Fourier transform is the unique isometric extension from $\mathcal{S}(\mathbb{R})$ to $L^2(\mathbb{R})$

Hardik Rajpal

or is it a thing with my computer

the aforementioned bug or virus?

Stupid question inc

@Thorgott I think I have made it clear there

Joe Shmo

I.e. $\hat{\cdot}$ is the unique operator such that $\|f\|_2 = \|\hat{f}\|_2$

anything corresponding to this in $L^1$?

Stupid question inc

02:41

$\{z\in\Bbb{C}:|z-z_0|<r}\$ just contain point but I can't see interior point

Thorgott

try drawing a picture

Stupid question inc

ok let's say $z_1$ and $z_2$ is element of $\{z\in\Bbb{C}:|z-z_0|<r}\$

what i mean is even though $z_1$ and $z_2$ is in set $\{z\in\Bbb{C}:|z-z_0|<r}\$

if the radius is not constrained then it can't be a interior point

book hasn't defined interior point radius and directly saying it contain all interior point

Thorgott

you need to look at the definition of an interior point again

the definition says that there exists some $r>0$ such that $B_r(z_1)\subset E$

in geometrical words, can you draw some ball around $z_1$ that is entirely contained in the larger ball?

Stupid question inc

yes of course

ok I just found i asked a dumb question lmao

sorry for wasting your time

Stupid question inc

03:31

But why is $\Bbb{C}$ both open and closed ?

alright not having boundary point we can say C contains all boundary point

then I can even say open sets have no boundary point so they contain all

Thorgott

open sets can have boundary points, though

Stupid question inc

would be so embarassing if i didn't chat anonymously

Sophie

04:17

How can I find the publications of mathematician Kevin G Brown? I tried google scholar and the arxiv, but when I try mathscinet it says it's a subscription service

Sayan Chattopadhyay

04:45

@Sophie Are you affiliated with some School/University? You can check if they have a subscription or not to mathscinet. Most math departments usually tend to have a subscription

Ted Shifrin

@Stupidquestioninc: It's clearly open, and the complement, being the empty set, is open, so the set is closed (that way I don't have to ask you what your definition of closed sets is).

Sophie

05:02

@SayanChattopadhyay I'm not

Ted Shifrin

So, what do you know about this mathematician?

Sophie

He writes this website mathpages.com/home/kmath411.htm

I got curious

Ted Shifrin

Where did you even find his name?

Sophie

by googling the site name, it has a "list of pages" thing

Ted Shifrin

I found a source that says that is Kevin S. Brown.

Who knows ...

The only Kevin G. Brown I find did applied math and then disappeared.

Sophie

05:14

Yeah I'm not really sure the initial is and there are lots of Kevin Browns

His website links to a couple of his physics books, so I thought it was the applied math guy lulu.com/en/us/shop/kevin-brown/reflections-on-relativity/…

This is the page with his name btw mathpages.com/home/contents.htm

Ted Shifrin

I'm going to guess this is not someone with research publications in mathematics.

Perhaps after his thesis at Univ Central Florida, if it's the guy with the G. initial.

Sophie

it's the S middle name, plus he likes sums of cubes and G H Hardy

Ted Shifrin

OK, so not related to G.

You should probably remove that, BTW.

Sophie

that's public information

it's the whois information for the domain

Ted Shifrin

OK.

Sophie

05:20

You can tell that he's anticipating people looking into it by the obviously fake address lol

Ted Shifrin

Yeah, true. The phone number doesn't look fake.

Stupid question inc

${z\in\Bbb{C}:1<|z|<2}$ is donut right ?

Ted Shifrin

Oh, wrong area code, in fact.

No, @Stupid, it's called a washer or an annulus. Doughnut is a closed surface in $\Bbb R^3$.

Stupid question inc

ok so this annulus doesn't necessarily need to be centered at (0,0) right?

Ted Shifrin

Nor does it need to have those inner and outer radii.

Stupid question inc

05:24

so what you mean is a shape with boundary and a hole in it is enough?

a plane figure consisting of the area between a pair of concentric circles so this means it does need inner and outer radii

Ted Shifrin

I meant the numbers of the radii could be different, of course.

I have no idea what the context for this is.

Stupid question inc

well today I am quite confused by everything that exists

Ted Shifrin

Well, OK.

Stupid question inc

05:47

$\{z\in\Bbb{C}:|z|<1\}\cup \{z\in\Bbb{C}:|z-2|<1\}\$

so this should be circle centered to origin with radii 1 and the other centered at 2 with radii 1

why centered at 2

why can't I think properly today

AlgebraicGeometryStudent

06:05

What do you call the "owner / parent set" of an element?

I.e. its the element's _____ (blank)?

type?

Stupid question inc

@AlgebraicGeometryStudent I have never heard of owner of an element

AlgebraicGeometryStudent

from object import Object

class Element(Object):
def __init__(self, inhab:Object):
super().__init__()
self._inhabited = inhab

@property
def inhabited_object(self):
return self._inhabited

What I have so far

I'm trying to code a categorical type system

from scratch

Stupid question inc

ok I am really confused by everything that exist so I must go for a walk before I get insane

Sayan Chattopadhyay

08:07

@Stupidquestioninc Those aren't circles. Those are open balls in C

Stupid question inc

08:20

yes open balls

and z is less than 1 distance from 2 so it is centered there

stupid mistake

ok now I have question

in real line we go to plus and inus infinity

but in complex plane we got only infinity ? and not minus infinity?

so this confused me

Nate

Not sure about the rules on the chat here, but I was asked a question today for which I couldn't immediately answer. Is it possible for a function to be differentiable at the origin even though it's paritals would tend to infinty when approaching the origin?

Intuitively this would be no, but I couldn't come up with an example...

Broadsword93

08:39

Hi, everyone.

I have the following functions that I need to rank in increasing order of Big-O complexity:

$(log(n))^3$, $10\sqrt(n)$, $nlog(n)$, $n\sqrt(n)$, $n^4 + n^3$, $(2.1)^n. n^2$, $3^n$, $2^n . n^3$, $n! + n$, $n^n$

My current ranking is as follows:

$(log(n))^3$ $<$ $10\sqrt(n)$ $<$ $nlog(n)$ $<$ $n\sqrt(n)$ $<$ $n^4 + n^3$ $<$ $(2.1)^n . n^2$ $<$ $2^n . n^3$ $<$ $3^n$ $<$ $n! + n$ $<$ $n^n$

Is my ranking of the functions in increasing order of complexity correct?

Dirichlet

08:52

@user714630 the book uses original defn of f'(x) i.e $lt_ {h->0}\frac{( |cos2(\pi/4+h)|-|cos(2\pi/4)|)}{h}= lt_ {h->0} \ sin(2h)/h=2$ ,hence differentiable.

Edward Evans

10:48

@Alessandro sanity check: if I wanna show a map is continuous, is it enough to check it's continuous on a basis for the topology?

Even more, is it enough to check it's continuous on, say, a neighbourhood of 1?

also, yo

user2103480

Neighborhood of 1? Is this a group?

Edward Evans

yeah sorry

that's all I'm thinking about atm

and have been spamming about it in here hahaha

user2103480

But yes the basis suffices

Edward Evans

thought so hehe

ty

user2103480

Every open set is a union of base elements and the inverse imagd of a union is the union of inverse images

Edward Evans

11:02

indeed

user2103480

@EdwardEvans wouldnt we need some property like linearity for that?

Do you mean for a homomorphism?

Edward Evans

well I can get all other neighbourhoods by translation so I guess a homomorphism yet

ye*

but yeah that makes sense, if I have a neighbourhood basis of 1 then I can just check a homomorphism is continuous on 1

or

on a neighbourhood of 1

*

porridgemathematics

is it true that if X is a topological space, and Y,Z are (non-empty) sets, that (X^Y)^Z is homeomorphic to X^(YxZ) (both with the product topology)?

maths researcher

12:18

an eulerian graph G is disconnected if and only it has isolated vertices

right?

Alessandro Codenotti

@EdwardEvans yes

Usually you only need to check stuff around 1 in a topological group, because the rest of the points have nbhds basis obtained by translating around a nbhds basis at 1

Edward Evans

yeah this is what I thought

tyty

Alessandro Codenotti

12:36

There are a bunch of results that are usually stated in terms of nbhds of 1 for that reason (an Hausdorff $G$ is metrizable iff it has a countable nbhd basis at 1 for example)

Edward Evans

aight because it seems that I can give a locally profinite group just by specifying a neighbourhood basis of 1 consisting only of compact open subgroups

which is cool

shrugs arithmetically

MathStudent

12:51

$X_n = o_{P_{\theta_0}}$ means for all $\epsilon > 0$ it holds that $\lim_{n \to \infty} P_{\theta_0}(|X_n| > \epsilon) = 0$. That means that $o_{P_{\theta_0}} (1) = o_{P_{\theta_0 + h n^{-1/2}}(1)$, right?

Sayan Chattopadhyay

13:03

@BalarkaSen you maybe interested in this

home.icts.res.in/~anirban/hdp-2021

Balarka Sen

13:47

Thanks! This is cool

MathStudent

Actually, I think I can't write it like that since it may hold that $X_n \neq Y_n$ but both $= o_{P_{\theta_0}}(1)$. So, I think what rather is true is: $X_n = o_{P_{\theta_0}}(1)$ holds iff $X_n = o_{P_{\theta_0 + h n^{-1/2}}(1)$.

skullpatrol

in The h Bar, 3 hours ago, by skullpatrol

JEE-Advanced result: Chirag Falor to skip studying at IITs, will head to MIT

MathStudent

14:03

I think I found a solution to Problem 13.12 in Lehmann and Romano's "Testing Statistical Hypotheses" but only when using a stronger assumption (freely available here: sites.stat.washington.edu/jaw/COURSES/580s/582/HO/… ). If anyone could check whether it's correct, that would be great (or even better if you can show it without using the stronger assumption). I've posted it in the EDIT here:

0

Q: Show uniform convergence in probability given pointwise convergence in probability and an upper bound

Suppose $X_1,\dots,X_n$ are i.i.d. according to a family $\{P_{\theta}, \theta \in \mathbb{R}^p\}$ with a continuously differentiable density function $p_{\theta}(x) = dP_{\theta}(x)/d\mu(x)$. Let $I(\theta) \in \mathbb{R}^{p \times p}$ be the Fisher information matrix defined by $I_{m,l}(\theta)...

Tobias

16:17

Hello all - I'm new to this community. I have a question about how to parse a certain relation in my book, namely: $\overline{\lim_n} f_n = \inf_k \sup_{n \geq k} f_n$. My problem with this is that the supremum is one element (= real number), so the prepended infimum doesn't add anything: it's just the same element again?

Ted Shifrin

@Tobias First, these are functions, not numbers. Second, where do you see one element?

Tobias

@TedShifrin Here's how my book defines it: $(\overline{\lim_n f_n)(x) = \overline{\lim_{n\to\infty}} f_n(x)$

Ted Shifrin

Yes, you typed that already.

Tobias

So, given $x$, what is the supremum of the set $\{ f_n(x) : n \in N \}$

Ted Shifrin

No, you do $n\ge k$ and the answer depends on $k$.

So you get a non-increasing sequence as $k$ varies.

Tobias

16:30

@TedShifrin Can you rephrase it please? The entire thing, because I'm having problems parsing it

This: $\overline{\lim_n} f_n = \inf_k \sup_{n \geq k} f_n$

Ted Shifrin

You are ignoring the subscript of sup.

Tobias

The least $k$ such that what?

Sayan Chattopadhyay

@Tobias that supremum is over all $f_n$ where $n$ starts from $k$, and not $1$

Ted Shifrin

No. Not the least $k$. For each $k$ you get $a_k(x)$. Now you take the inf of that set as $k$ varies to infinity.

Tobias

@SayanChattopadhyay over all $x$?

and*

Ted Shifrin

16:34

Do an example with numbers first, not functions.

Sayan Chattopadhyay

Yes, wherever your $f_n$'s are defined

Tobias

@SayanChattopadhyay the supremum of a set is one element, right

Ted Shifrin

Take the sequence $b_n = (-1)^n + 1/n$.

Sayan Chattopadhyay

@Tobias look at Ted's example

Tobias

the sup of b_n is 2

the inf is -1

Ted Shifrin

16:38

You are not listening to what I said.

I want $a_k$, as above.

Sayan said it too. You have to listen when we say things.

Text is more difficult than talking with blackboard.

Tobias

ok that makes sense

porridgemathematics

@Tobias you may want to google 'limsup' because $\overline{\lim_n} f_n$ is denoting $limsup_n f_n$, which if you understand for sequences, is the same exact thing for functions

Ted Shifrin

So what is $a_k$ in my example?

Tobias

you didn't provide a_k

Ted Shifrin

I defined it way above.

Tobias

16:47

Do you mean b_n?

Ted Shifrin

Scroll up, man. I mean the sup for $n\ge k$. You need to pay attention

Tobias

thanks for the help

Joe Shmo

Hello chat

Sophie

Hello

Joe Shmo

how's everyone doing

Sophie

17:00

If I have a group generated by $a,b$ and $a^2=b^3=(ab)^2=e$, is there an easy way of proving that it has order 6?

(this is $S_3$, just for the sake of example)

@JoeShmo Fine, thanks. And you?

Joe Shmo

I'm doing well, trying to parse my European prof's cryptic problem statements. Great guy, otherwise

and yourself?

Sophie

Studying group theory, currently confused about generators as you can probably tell

athos

ab = ba. then $ab^{-1}$ can be the generator

Ted Shifrin

@athos Where did you get that?

@Sophie: I would try to give the isomorphism to $S_3$. In general, things with generators and relations are nontrivial.

athos

get what?

Ted Shifrin

17:15

What you typed earlier.

Astyx

hi chat

Ted Shifrin

Hi Astyx.

Joe Shmo

Hi Astyx, Ted

Ted Shifrin

Hi Shmo.

athos

ah sorry it's not necessary $ab=ba$. pls ignore me.

porridgemathematics

17:18

you could show that any product of a....ab....b is e,a,b,b^2,ab,ab^2 and then show that these are all distinct, and form a group, which would do it

not sure if that is 'easy' though, but it works

*is one of , is what I mean

forgive me if this is wrong, but by 'is the generator' are you not implying that S_3 is cyclic saying that?

Sophie

No because it's generated by 2 elements, if it was cyclic it would be only one

porridgemathematics

yes I was referring to what @athos said

Ted Shifrin

What athos said was just wrong.

porridgemathematics

anyway, one you've shown a....ab...b is one of the things in {e,a,b,b^2,ab,ab^2} and that that list is a set, it only remains to show that {e,a,b,b^2,ba,b^2a} = {e,a,b,b^2,ab,ab^2} , and then you would be done

i dont think this is too hard, its mostly tedious application of the rules you've been given about how a and b interact

right its certainly wrong, but the red flag may as well be that one generator implies the thing is cyclic, which it can't be if its iso to S_3

is what I was trying to say

Sophie

I mean I don't actually need a proof, I'm more like trying to understand what happens if you pick different sets of generators, for example for $S_3$ you may pick $(1 2)$ and $(2 3)$ or $(1 2)$ and $(1 2 3)$

porridgemathematics

17:24

i thought you were interested in an easy way to prove it gives a group of order 6

Sophie

I mean I asked myself that question to see if I really understood what was happening

Astyx

If I'm looking at the sheaf of continuous functions on the real space, is the stalk at a given $x\in \Bbb R$ just $\Bbb R$ or am I missing something ?

porridgemathematics

you can pretty much brute force these properties of S3, there are probably high level more sophisticated ways of figuring out it only has two possible generating sets of order 2, but for groups of such a small order it doesn't even take long to just brute force

Ted Shifrin

Astyx. Way huger than that. A germ of a continuous function is not determined by the value. Here's a hint. This is not a Hausdorff space.

Astyx

I'm having trouble understanding what a germ is

Ted Shifrin

17:33

It's an equivalence class of functions on neighborhoods of the point.

Astyx

Oooh

Ok that makes a lot of sense

Ted Shifrin

Can you give me two functions whose germs at $0$ do not have disjoint nbhds?

Astyx

What do you mean by "do not have disjoint neighborhoods" ?

Ted Shifrin

Oh, maybe you don't know this sheaf is a topological space.

Projection to the domain is a local homeomorphism. Very strange.

Sections are local homeomorphisms.

Astyx

If I take holomorphic functions instead, then the stalk at $0$ is $\Bbb C^{\Bbb N}$ because locally functions are given by each of their derivatives ?

And I guess it's much bigger for $C^\infty$ real functions, and even larger for just continuous functions, because we don't have an easy way of describing how a function behaves locally

Ted Shifrin

17:40

Well, a subset of that. You need a convergent power series on some nbhd.

Astyx

Oh yeah, that's true

anakhro

Just asking for a double check: universal coefficients theorem easily deduces that $H^1(X;\mathbb Z/2\mathbb Z)$ is trivial where $X$ is the closed $n$-ball, $n>3$? $X$ is path connected so the Ext term is trivial, and $X$ is contractible so the Hom term is trivial.

Ted Shifrin

Why $n>3$?

anakhro

I meant $n\geq 3$, but I just wanted those cases for my use. :P

I didn't want to think about any lower n.

Ted Shifrin

OK

anakhro

17:43

2 works the same, and 1 does, too, I guess.

They still satisfy the path connected and contractible

Ted Shifrin

Yup. Contractible means $0$ cohomology for any coeffs, I think.

Astyx

And the inductive limit to define the stalk of a general sheaf is given by some universal property that lets you factor functions that survive restriction (?)

anakhro

I think you are right about that, Ted.

Astyx

contractible means you have the same cohomology as a point, so 0 everywhere but at order 0

Ted Shifrin

Not sure I know what “survive restriction” means.

Astyx

17:49

I mean $\phi_U:F(U) \to Y$ for each open set $U$ such that if $U\subset V$ and $\rho:F(V)\to F(U)$ is the restriction, $\phi_V = \phi_U \circ \rho$

I typed that wrong

Should be fixed

I'm gone for now

Bye

Ted Shifrin

Bye

anakhro

Ted, did you see that documentary about Maryam Mirzakhani?

"Secrets of the surface" I think it was titled.

Mike Miller

18:09

Don't use universal coefficients... cohomology with any coefficients is a homotopy invariant and you're assuming X is htpc to a point ...

anakhro

This is a good point. Is there any extra considerations to make with arbitrary coefficients when trying to prove homotopy invariance?

Hawk

can someone explain how shown M_1 is torsion free implies M_1 is free here?

Ted Shifrin

No, although I think I knew they'd made it.

Hawk

M is torsion free and finitely generated here*

anakhro

@TedShifrin it was pretty good I think, as far as documentaries go on mathematicians. The explanations for the layfolk were pretty good, though I really think they should have given more insights into her work since the lay-explanations were IMO the best part.

Ted Shifrin

18:16

Howdy @MikeM. How's the kidlet doing with diff geo?

@Hawk: It's free on the generator $m_1$, no?

Hawk

@TedShifrin i was thinking about showing independence on the entire subset of {m_1, \dots, m_i \} because they introduced the generators but only used m_1, but then I realize what they were trying to say is that the map 1 \to m_1 gives the isomorphism R/ann(m_1) = R/(0) = M_1

Ted Shifrin

There is just one generator here.

Hawk

But they say it is generated by {m_1, \dots, m_i} (extracted from M)

why did they introduce this set when they only used m_1

Ted Shifrin

But $i=1$ here.

They introduced the set for later parts of the problem, I have to assume.

Hawk

oh ok

Mike Miller

18:28

@anakhro The proof is completely unchanged.

@TedShifrin He's still exploring what he wants to do. We'll only meet some amount this semester and really get into gear next sem.

Ted Shifrin

Ah, nice to have a leisurely approach.

Mike Miller

Yeah, he's not as concerned with his CV and just wants to have fun learning something.

Ted Shifrin

I'll try not to butt in further :)

Mike Miller

Feel free to butt in.

Balarka Sen

18:50

@anakhro That documentary was sort of meh

Ted Shifrin

Howdy, a @Balarka!

Balarka Sen

Hi Ted

Astyx

I'm back

Ted Shifrin

@Astyx: You left as I was about to ask what that $F(U)\to Y$ nonsense was, anyhow.

Astyx

I was talking about the inductive limit

Trying to put this diagram in words

Limite inductive

En mathématiques, et plus particulièrement en théorie des catégories et en algèbre universelle, la notion de limite inductive généralise à des structures la notion classique de limite issue de l'analyse[réf. nécessaire]. == Avant-propos == En analyse, l'efficacité du concept de limite n'est plus à démontrer. L'idée de la limite inductive (et de sa duale, la limite projective) cherche à généraliser à l'algèbre ce qui fonctionne si bien en analyse [réf. nécessaire]. Un premier point clef est la notion de passage à la limite. Une limite de nombres réels positifs est positive. La limite de la racine…

Ted Shifrin

18:56

I know what this is. Yes, the restriction map is $F(V)\to F(U)$ when $U\subset V$. I have no idea who your $Y$ is. The maps to whatever are built into $F$.

@Balarka: I'm not missing something, am I?

Astyx

Y is meant to be the same Y as in the diagram

Ted Shifrin

Yeah, but it has no context in our discussion. $F(U)$ is whatever-valued functions on $U$.

Astyx

I was talking about general sheaves, not necessarily functions

Ted Shifrin

Well, you'd have to give me an example to convince me. But whatever.

Astyx

That's how the stalk is defined in my course, as the inductive limit of the sheaf at a given point

However the inductive limit is itself not defined

Ted Shifrin

19:06

Yes, the stalk is the inductive limit. I guess you can do this with sheaves of groups, rings, whatever, and have the categorical terminal element. Ignore me.

Mike Miller

Is there a simple question I can answer here

Vishesh Mangla

how to find number of subgraphs in a graph?

ibb.co/fCzKFr4

here there are 2

Balarka Sen

@TedShifrin No I think you're right

Ted Shifrin

Thanks, @Balarka :)

Vishesh Mangla

it's basically a chemical engineering problem design

Ted Shifrin

19:15

@MikeM, if that was to me, it was very elementary double complex thingy in Bott-Tu.

Their proofs are sometimes elliptical.

Someone just gave me a negative vote on an answer from 2013. And I was right. Who cares.

Oh, but I suppose I didn't answer the question completely. Meh.

Mike Miller

Oh, I just figured Astyx had a question that was unresolved that I could maybe answer.

Vishesh Mangla

oops I said the word wrong , it isn't called subgraphs, I just revised the definition. I don't know what it is called but well independent components. Like the graph can be split into two independent graphs

Mike Miller

That's one benefit of not having an MSE account anymore, @TedShifrin, though I always found few upvotes more irritating than downvotes.

Ted Shifrin

LOL, I am not that annoyed, more amused. I usually put a piercing comment or question before I downvote, and only downvote when the OP or answerer ignores me. .... Oh, Astyx is trying to understand, e.g., the sheaf of germs of various sorts of functions. But he has a categorical definition of inductive limit.

Mike Miller

Junk!

He should learn about germs concretely and then see the general case as being about germs. :)

Ted Shifrin

19:28

I agree, of course. Except occasionally for tensor products, I rarely think in terms of universal properties.

It still is mind-blowing that sections of a sheaf (as a topological space) are local homeomorphisms. My students always had trouble getting over that hurdle.

Mike Miller

I'm trying to emphasize those in topology without the categorical phrasing, because that or tensor products are the first place it matters.

The subspace topology is important because a cts map to X factoring through S set-theoretically does so topologically... the product topology is important because a bunch of continuous maps to X_i factor through the product... the quotient topology is important because a map factoring through the quotient does so continuously... etc.

Obviously I'm being far terser here.

Is terser a word? More terse.

Ted Shifrin

Yes, of course we want to understand the product topology as the coarsest (?) one that makes the mapping continuous if the component functions are. Even I believe in that much categoricalness.

Mike Miller

I've never introduced the notion of coarser or finer topologies and I never will. :)

I don't like that language. Never found it useful.

Ted Shifrin

Well, comparing topologies on a fixed set is quite reasonable, so I do it — also because if I'm following Munkres, he uses the terminology not just once.

Vishesh Mangla

hey, can otherwise if you guys can tell me what a manifold is?

Mike Miller

19:33

I'm suggesting readings from Munkres but not following it. I don't like that book.

I use Hatcher's brief notes and produce notes of my own, whichever people find more useful.

Vishesh Mangla

I looked at various definitions, read a few books but couldn't get it

Mike Miller

I dunno what you're looking for.

Ted Shifrin

I actually choose books for their exercises, and I think Munkres is excellent. I add a few of my own. But I don't love it.

Mike Miller

Oh, I produce my own exercises too. I agree that Munkres has good exercises.

Ted Shifrin

@Vishesh: You probably should start with manifolds that live in $\Bbb R^n$ and not try to read the most general thing.

Mike Miller

19:34

A topological manifold is a kind of shape where if you zoom in near any point, it looks like Euclidean space.
A smooth manifold is like that but better: because it looks like Euclidean space near any point, and you know how to do calculus in Euclidean space, a smooth manifold is a place you can do calculus.

Figuring out what "looks like" means is hard.

Ted Shifrin

Do you know the implicit function theorem, @Vishesh?

Vishesh Mangla

consider a hyperboloid, if I looks very closely at a point at it, well around that point I can think the space around in a small neighbourhood as flat

is that what it means?

@TedShifrin sorry nope, but it wiki tells it has something to do with graph of a function

I have a course of functional analysis but it's been quite a bit of time

Mike Miller

To learn about manifolds you gotta start with a firm grounding in multivariable calculus, which is what the IFT is part of.

@VisheshMangla Depending on what you mean by flat, yes.

Vishesh Mangla

well Euclid defined everything as planar

Mike Miller

I'd say: pick a point on the hyperboloid. You can flatten it out by projecting it to the plane. That's a continuous (smooth actually) operation. You can also unflatten the plane (curve it a bit into the hyperboloid).

That's also continuous (smooth).

Something is a topological manifold if you can do this sort of continuous flattening near every point (where it's still continuous when you 'undo' the flattening). A smooth manifold has both the flattening and the unflattening smooth.

Vishesh Mangla

19:41

if I consider a circle on the hyperboloid, it's on a surface and it's definately not planar

now if I decrease the boundary of the circle , well as the area will decrease the world inside the boundary of the circle will become more and more planar

is this manifold and what you were saying?

@MikeMiller what is the plane you are referring to here? I can visualize the hyperboloid but I can't find where is the plane on which you are projecting the point to

Mike Miller

You should have this conversation with someone in person (or on Zoom) where you can draw pictures. I don't know what you're saying there. Someone with more patience might try, but I'm out.

Astyx

Can we say anything about the intersection of prime ideals of a (commutative) ring in general ?

Vishesh Mangla

@MikeMiller sorry for the obfuscation , but thanks, I can try to understand it

Mike Miller

Nothing to apologize for, I'm just an impatient man and I've got stuff I have to get to. Seems like this would be a long convo.

Ted Shifrin

@Vishesh: He's projecting on the tangent plane to the hyperboloid at the center of your circle. Maybe start with a paraboloid instead of a hyperboloid. Easier to visualize.

Vishesh Mangla

19:47

oh ok

Ted Shifrin

But you can watch some of my YouTube videos on the basics of manifolds if you're willing to have multivariable calculus in there. The motivation is that in the differentiable setting linear algebra does all your work.

Vishesh Mangla

channel?

Are you Theodore Shifrin?

Astyx

he is

Vishesh Mangla

oh, so you are a professor.

thanks for the lectures

Ted Shifrin

Was. :)

There are two in the first set and two in the second set, but for your purposes probably just the first set will be enough for intuition.

MathStudent

19:53

Hello. I can't seem to find a citable source for a definition of equicontinuity for sequences of general functions (not restricted to functions that are in C(T) with T a metric compact space or functions from R^k to R^k). Does anyone know of a source off hand or have a guess on a source where I could find it?

Ted Shifrin

How can discontinuous functions be in an equicontinuous family?

Oh, you want the domain more general.

You only need the domain compact and the range metric, I think.

MathStudent

Yeah

Ted Shifrin

This is certainly in Munkres's topology book.

You might look at Simmons's book on Topology and Analysis, too. I no longer have it, so I can't check how general he is.

MathStudent

@TedShifrin Hm not sure if I have the range metric. This paper I'm looking at doesn't seem to assume it in a theorem where it uses equicontinuity.

@TedShifrin Okay, thanks for the sources! I'll see if I can look into them.

Unfortunately, my uni doesn't give me online access to either of them. :( But I guess since both books you stated are topology I should look for a topology book that they do give me online access to

porridgemathematics

20:12

@MathStudent for a sequence of functions between topological spaces X -> Y, if Y is metrizable, then the definition is the sequence is equicontinuous at a point x \in X iff for all eps > 0, there is a nbhood U_x of x for which if a \in U_x then d_Y(f(a),f(x)) < eps for all f in the sequence, if we drop the metrizabilitiy of Y, then you can say its equicontinuous at x \in X then the definition looks a little uglier

*but the

but im not sure how you would define uniform equicontinuity without a metric on X, since we would need some way to measure the diameter of the neighbourhoods

MathStudent

Thanks! I think I can probably actually just assume it is a metric space since the theorem should be true then too even if the paper doesn't assume it is. And for my purposes stating it for a metric space should be enough I think.

Alessandro Codenotti

21:41

@porridgemathematics you need to introduce a uniformity compatible with the topology

If your spaces are compact there is a unique uniformity, if they are not it gets messy

Nike Dattani

Of interest to anyone working with differential equations (existence / uniqueness): mattermodeling.stackexchange.com/q/2424/5

3

Q: Does the Schrödinger equation yield a unique wave function and density?

I am learning DFT and the Hohenberg Kohn Theorem of Existence. It says that there is a one-to-one correspondence between the external potential and the density. However the proofs that I have seen only show that potential gives a unique density. How do we know that a density gives a unique potent...

density-functional-theory quantum-chemistry quantum-mechanics from-phys-se wavefunctions

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