Mathematics - 2024-05-02

EE18

00:40

Reading mathematical claims in an eng book after reading math books makes you want to cry

Xander Henderson

00:54

@EE18 Who's this "you", white man?

EE18

01:07

me myself and i

leslie townes

xander: oh, and the kids are going to get that reference. OK.

Xander Henderson

@leslietownes Okay, so we're both old.

Koro

02:03

can anyone please explain where the contradiction is?

Thorgott

god, what a convoluted proof

anyway, the contradiction comes from writing down the induced maps on cohomology rings

Koro

In past, I have done it using fibrations.

Thorgott

think about what $f$ does on the generator

Koro

But this time fibration sequences are not allowed [I didn't impose this condition].

anyways, I was thinking: f o h= g so passing onto cohomology: h* o f*= g* and on S^{2n+1}, g* being nullhomotopic so 0 so h*o f*=0 on S^{2n+1}.

this gives j* o pi* o f* =0 whence j* o (f o pi)*=0

But since f o j= id, we have j* o f*= id

this contradicts the earlier equality.

But there must be something wrong in it. :(

Thorgott

02:26

the equality $f\circ h=g$ doesn't imply anything meaningful at the level of homotopy invariants, because $D^{2n+2}$ is contractible

you can derive a contradiction from $f\circ j=\mathrm{id}$ and that alone

Koro

04:15

if pi: E-->M is a fiber bundle, can it be said that M is a deformation retract of E? If yes, then why?

copper.hat

hi @koro!

Koro

Hi @copper.hat!!

leslie townes

04:43

hi @copper.hat!

and @koro!

Koro

05:01

hi @leslietownes!!

robjohn

I think you're all hi!

copper.hat

05:53

hi @leslietownes @Koro, lo @robjohn

Jakobian

06:51

@koro About 2-dimensional problem of this, $\{x\in\mathbb{R}^2:\text{exactly one of }x_1,x_2\text{ is rational}\}$, then this is easily seen to be totally disconnected since its complement contains $(r_1, r_2)+(\pm x, \pm x)$ where $r_i$ are any rationals and $x\in\mathbb{R}$

But this argument doesn't quite work for $\{x\in\mathbb{R}^3 : \text{exactly one of }x_1, x_2, x_3\text{ is rational}\}$

Because such lines don't enclose a 2-dimensional shape

Soumik Mukherjee

It was Derivative who asked the question

Jakobian

07:14

@SoumikMukherjee I know that

But Koro asked about the 2d version

I don't have dementia

Swan

07:37

Is this solution correct? I've looked up the solutions to this problem and every solution uses more complex approaches so I'm unsure if there is some basic flaw in this solution that I'm missing out on.

leslie townes

you write b <= a, c <= a, d <= a in succession (where a = cos(x) and b = a^4 and c = a^2 and d = a), then you write just below that, "implies" b - c + d <= a. what is going on with the "implies"?

Jakobian

@Swan its not correct

leslie townes

note that if c <= a then -c >= -a, but this inequality "goes the wrong way" for it to be so easily added to the others

that's what jumped out to me. i haven't looked at the rest of it

Jakobian

But if you argue based on cos^4n <= cos^2n then it should be fine

Soumik Mukherjee

Oh I missed that koro asked for 2d, my bad

Swan

07:48

@leslietownes Yeah, I got it. Sometimes I do glance over these very fundamental errors and need assistance to spot the mistake. Thanks :)

@Jakobian Thanks :)

Koro

08:00

S^1----> S^2n+1 ----> CP^n is a fibration so writing its long exact sequence gives: pi_k S^2n+1= pi_k CP^n for all k>2

this isomorphism is induced by pi.

Hopf theorem: pi_k S^2n+1= Z when k= 2n+1.

But if pi were homotopic, then pi* would be the 0 map which can't result in isomorphism.

Koro

08:16

that's one way to look at it.

note that f*=0

6 hours ago, by Koro

can anyone please explain where the contradiction is?

so id =0 is the contradiction.

that's another way to look at the question I asked.

@Jakobian thanks. I'll look into it in a while.

Thorgott

11:34

@Koro in general, this question is not quite well-posed because $M$ is not a subset of $E$. the LES in homotopy implies that a necessary condition for $\pi$ to be a homotopy-equivalence is that the fiber $F$ is weakly contractible (hence contractible under mild technical assumptions). if the fiber $F$ is contractible, it turns out (under mild technical assumptions) that $\pi$ has a section which turns $M$ into a deformation retract of $E$.

(in particular, this is true for vector bundles, where the proof is of course much easier)

@Koro yeah, that's a nice argument

another argument that uses cohomology instead of homotopy, but is more convenient than the one you've screenshotted, is that if $\pi$ were null-homotopic, you would get homotopy-equivalent attaching spaces $\mathbb{CP}^{n+1}\simeq\mathbb{CP}^n\lor S^{2n+2}$, but these have non-isomorphic cohomology rings

Ryder Rude

13:13

"the way to study science/math is to be a frog in one field and a bird in the others"

do u agree with this

user 70432

no

Ryder Rude

why

Xander Henderson

@RyderRude I don't know what that is supposed to mean.

Ryder Rude

it means one should know the details of one field and an overview of others

Soumik Mukherjee

bird's view in the sense of seeing the big picture connecting everything?

Ryder Rude

13:20

yes

user 70432

books.google.ca/books/about/…

Ryder Rude

yeah...this is where i got it from

not the exact quote, but the thing about birds and frogs

user 70432

have you read it?

Ryder Rude

i read one paper of Dyson on it

user 70432

he writes like Dirac, very concisely

42

Q: Dirac notation for English

Dirac notation is very expressive: not only does it allow to write quantum states and operations more concisely, it can also be used to shorten written English words and phrases! Here are some examples: What is shown on the hidden photo?

YourLordJoyBoy

13:33

@Shaun Funeral for my grandma was yesterday, it went well but omgosh. I'm beat...

user 70432

> In 1941, as an undergraduate at Trinity College in Cambridge, Dyson studied physics with Paul Dirac and Arthur Eddington and found an intellectual role model in the famed English mathematician G.H. Hardy, who had previously mentored the mathematical prodigy, Srinivasa Ramanujan

ZaWarudo

I was thinking if it is possible that a sequence of functions $\{f_n\}$ converges uniformly to a continuous function $f$ without $f_n$ being continuous. The example I built is the following: define $f_n: \mathbb{R}\to\mathbb{R}$ by letting $f_n(x)=\begin{cases} 1/n, && x \ge 0 \\ -1/n, && x <0\end{cases}$. We have $f_n(x) \to 0$ uniformly on $\mathbb{R}$ with $0$ being continuous on $\mathbb{R}$ but $f_n$ is discontinuous in $x=0$. Is this correct?

Ryder Rude

@user70432 this would b great for a Sherlock story

user 70432

yup

Ryder Rude

theres some nice interviews of Dyson

user 70432

13:38

RIP

Ryder Rude

Freeman Dyson - Why is the Quantum so Mysterious?

►

there r a bunch on the Closer to Truth channel

@user70432 RIP

user 70432

he didn't believe in struggling with a problem for years to get a PhD

Ryder Rude

oh

he does not have a Phd

he contributed a lot anyway

EE18

13:55

Possible to get some help here? In particular, I don't follow their argument that $\bigcup A \neq Q$. Since $Q$ an upper bound on $A$ the $\bigcup A \subseteq Q$ is clear..

But how does it obtain that it must be a proper subset via their argument?

Thorgott

as they say, $\bigcup A\subseteq z$ for an upper bound $z$ of $A$

and $z$ is a proper subset of $\mathbb{Q}$ of definition

EE18

Just so I understand, we are here using the hypothesis that there exists an upper bound $z$ (in $R$) so that by definition of $R$ we have $z \subset Q$?

Thorgott

14:17

yes, that's what I'm assuming (I don't have the context, but this looks like a Dedekind construction)

EE18

it is, yup

thanks very much thorgott

(By hypothesis I meant from the statement of the theorem)

Jakobian

@XanderHenderson its for science ppl. Not mathematicians

Science is full of collaboration so it probably makes more sense

Across different fields of science that is

Sahaj

14:38

Hi jakobian

psie

19:26

I'm reading Folland, the section on Borel measures on $\mathbb R$, and at one point between the theorems he makes a couple of remarks. In particular, he says that for each increasing and right-continuous function we obtain a complete measure whose domain includes the Borel sigma algebra.

Then, when he wants to discuss some of the regularity properties, he fixes some complete measure $\mu$ with domain $\mathcal M_\mu$ and writes that for $E\in\mathcal M_\mu$ we have $$\mu(E) = \inf\left\{\sum_{1}^{\infty}\mu ((a_j,b_j]):E\subset \bigcup_{1}^{\infty}(a_j,b_j]\right\}.\tag1$$

I know every outer measure induces a complete measure via Caratheodory's theorem, but here he starts with a complete measure and so I wonder, if conversely, every complete measure is associated with an outer measure? For me the formula $(1)$ is a bit sudden and not motivated.

Page 35 in the book by the way.

Thorgott

he's not talking about any complete measure, he talks about a complete Lebesgue-Stieltjes measure

psie

true, but the formula for $\mu(E)$ reminds me of an outer measure. If we have a complete Lebesgue-Stieltjes measure, then is it the restriction of the Lebesgue-Stieltjes outer measure? He has never talked or mentioned the Lebesgue-Stieltjes outer measure as far as I remember, so I don't even know what it is

Thorgott

yes, cause it is induced by an outer measure, by definition

you start with $F$, construct a premeasure from that, then apply Caratheodory extension

so the resulting Lebesgue-Stieltjes measure is by definition the restriction of an outer measure to an appropriate $\sigma$-algebra

psie

19:43

ok, I'll have to ponder on this some more, thanks

Obliv

from FTC, $\int_a^b f(x)dx = F(b)-F(a)$ how to think about the parent function $F(x)$? It doesn't explicitly give the area of $f(x)$ since $x$ is a point and $F(x)$ is a vertical line i.e., it has no area, but $f(x)$ is the slope of this function evaluated at $x$

Thorgott

$F(x)=\int_{x_0}^xf(x)dx$, where $x_0$ is some fixed point of reference

different choices of $x_0$ lead to different $F$, so there is no such thing as "the" parent function

(the indeterminacy here is the "constant of integration" or whatever you wanna call it)

psie

@Thorgott in the steps you mentioned here, at what point do we construct the outer measure? Or does it always exist, we don't even need to construct it? Seems like the steps you mentioned don't even require an outer measure, since Caratheodory's extension is about extending the premeasure.

Caratheodory's extension probably involves the outer measure, so we need it there.

Obliv

@Thorgott what do you mean by no such thing as a parent function?

Suppose $F(x)$ is differentiable on an interval $[a,b]$, so that $f(x)$ is well-defined. Even if the parent function isn't finite, can't we construct a fourier series representation of $F(x)$

so it "exists" in that sense

psie

19:59

> The most common way to obtain an outer measure is to start with a family $\mathcal E$ of "elementary sets" on which a notion of measure is defined (such as rectangles in the plane).

I wonder though, what Folland means by "notion of measure". Seems a bit vague.

Xander Henderson

@Obliv No, you've missed the point.

There is no single primitive. $F(x) = \int_{c}^{x} f(t)\,\mathrm{d}t$ depends on a choice of $x_0$.

Every choice of $c$ gives a different $F$.

But, in general, once you fix some $c$, you can interpret $F(x)$ to be the area under the graph of $f$ between $c$ and $x$.

I have no idea what the relevance of a Fourier series representation is supposed to imply.

Thorgott

There's two ingredients. Caratheodory's theorem tells you that any outer measure restricts to a complete measure on an appropriate $\sigma$-algebra. The other ingredient is the construction that takes a premeasure and returns an outer measure. The former is Theorem 1.11, the latter is Proposition 1.13 and they are synthesized into Theorem 1.14.
In section 1.5, Proposition 1.15 tells you how an increasing right-continuous $F$ gives rise to a premeasure and then Theorem 1.16 constructing the corresponding Borel measure does it simply by applying Theorem 1.14 to this premeasure. In the followi…

@psie it's meant to be an intuitive description

psie

ok, thank you 🙏 I like the exposition, but I struggle with the synthesizing bit

the outer measure remains kind of mysterious to me though...

Thorgott

you have a construction that turns a premeasure into an outer measure (Proposition 1.13, using the formula you've already seen to define outer measures in Proposition 1.10) and a construction that turns an outer measure into a (complete) measure (Caratheodory's theorem), so if you combine them, you have a construction that turns a premeasure into a (complete) measure

psie

20:26

ah ok, I think I'm grokking it more now. Folland starts his exposition (in section 1.4) with outer measures and hence I've always thought this is what "comes first". Maybe there's no such thing as "comes first", but the whole starting point in the construction, as you outlined it, seems to be a premeasure, since it induces an outer measure, and by Caratheodory's theorem, we can restrict to a complete measure. Bingo!

Thorgott

20:40

yeah

John Zimmerman

23:44

Is there any mathematical validity to a construct that says A and B are equivalent in some sense but in our construct there is some obstruction to allowing one to equate A and B formally?

leslie townes

john one very low tech way of doing this is finding a function f from something containing A and B to something else and asserting that f(A) = f(B)

and if it not already clear to see that A and B are not equal, often one might show that by finding some function g from a similar place to something else and showing that g(A) != g(B)

John Zimmerman

For example say $A$ is isomorphic to $B$ but I don't want to actually use that isomorphism because it will destroy a certain property

for example

$S^2/\lbrace \mathrm{2~ points} \rbrace$. Clearly this is topologically equivalent to the once punctured plane

hmm maybe I'm thinking of metrics..

leslie townes

if two things happen to be isomorphic in some way, there is nothing forcing you to 'use' the isomorphism

{e} and {pi} are homeomorphic as subsets of the real numbers, we can ignore that and treat them as different

John Zimmerman

that's true I was somehow thinking i had to use the isomorphism. good point

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