Mathematics - 2025-01-03

leslie townes

01:55

psie i have not thought about it in detail but the linchpin of that type of argument would only be that sets are assigned the same measure, not that the sets are literally equal to one another

leslie townes

04:18

as a fix for the thing you were guessing up above something like f^{-1}(x + A) = f^{-1}(x) + f^{-1}(A) might be more likely to be true than the identity conjectured above. as always, much of the relevant content is present in the d = 1 case of R^d :)

depending on how you set things up (and account for things like 'signed' volume) you could potentially use something like the above theorem to define the determinant

ModularMindset

05:21

-1

Q: What is $CX_V$ homeomorphic to?

Foliations: Take $X=(0,1)^3.$ Fix points $p,q$ s.t. $\text{dist}_3(p,q)=\sqrt{3}.$ Construct a smooth regular foliation of $X$ with $(3-1)-$dim. leaves which are topologically $(0,\sqrt{3})\times S^{3-2} $ accumulating to $p,q.$ This is equivalent to existence of a smooth foliation of $\Bbb R^3$ ...

ignoring the ignorant downvoter - am I on the right track?

and close voter

leslie townes

05:33

how does what you have put there define what "CX_V" is, as a set? let alone as a topological space (which would additionally specify a topology on that set). "CX_V" seems to depend on unspecified choices. we're broadly invited to consider assigning things to other things, but there is no guidance given as to how to do that, why it is even possible to do that, or even whether you claim that the resulting space CX_V does not depend on any choices that one makes when doing that

"am i on the right track" is too unfocused of a question for the site

i voted to close as "needing details or clarity" it is gibberish now

at a minimum i would suggest adding context identifying what CX_V actually is, as a set of points, and how you propose to topologize it

this is too long for a comment or i would have commented on the question

ModularMindset

05:52

@leslietownes I'm gonna add more details

handan_toddler

please

leslie townes

if the thing is embedded in R^3 then probably it is locally a 3-manifold, at least away from these points you are considering

for what it's worth, i didn't downvote. i conceive of downvotes as mildly punitive and on a different plane than votes to close

ModularMindset

almost done the editing process

handan_toddler

👍 votes to close are on another plane

🙈🙉🙊 >_>

leslie townes

06:12

absent an indication that a space exists within some collection that has some well known classification associated with it, "what is X homeomorphic to" is generally kind of an underspecified question. it's homeomorphic to itself, for example. why isn't that an answer? what's an answer?

ModularMindset

06:25

hmm

Daniel Donnelly

Anyone heard of mixed homology theory?

It plays well with Snakes (lemmas)

If the middle column of the snake diagram has reverse homology.

Snakes... I hate snakes (Indian Jones)

🐍 lemmas

It's not snake oil. The Snake lemma proof just goes through.

ModularMindset

06:58

1

Q: What is $CX_V$ homeomorphic to?

Edit: Added more details at the end (in order to share the thought process behind this question). Foliations: Take $X=(0,1)^3.$ Fix points $p,q$ s.t. $\text{dist}_3(p,q)=\sqrt{3}.$ Construct a smooth regular foliation of $X$ with $(3-1)-$dim. leaves which are topologically $(0,\sqrt{3})\times S^{...

general-topology continuity differential-topology foliations

finished

Daniel Donnelly

07:49

Well this room ☠

@ModularMindset what is that algebraic topology?

What could you do with reversible homology measurements?

Daniel Donnelly

08:52

@ModularMindset I upvoted

But I can't crack it

Pizza

09:03

Hi

Pizza

09:24

@SineoftheTime but could the exercise from the other time also be done like this?

$f(x)=(x^2−10x+25)e^{−|x+5|} =(x−5)^2e^{−|x+5|}$
$g(x) = x^2 e^{-|x|} \Rightarrow g(x-5) = f(x)$

$F(g(x))(\omega) = F(x^2 e^{-|x|}) = \frac{1}{(-i)^2} \frac{d^2}{d\omega^2} \hat{f}$ where $\hat{f} = F(e^{-|x|})$

$\frac{d}{d\omega} \hat{f} = -\frac{4\omega}{(1+\omega^2)^2} \quad \frac{d^2}{d\omega^2} \hat{f} = -\frac{4(1+\omega^2)^2 - 8\omega(1+\omega^2)\cdot 2\omega}{(1+\omega^2)^4} \Rightarrow F(x^2 e^{-|x|}) = \frac{4+4\omega^2-16\omega^2}{(1+\omega)^3} = \frac{4-12\omega^2}{(1+\omega)^3}$

$F(f(x))(\omega) = F(g(x-5)) = e^{-i5\omega} \cdot \frac{4-12\omega^2}{(1+\omega)^3}$

Dec 22, 2024 at 18:00, by Pizza

Calculate the Fourier transform of:
$f(x) = (x^2 - 10x + 25)e^{-|x-5|}$

psie

10:05

I'm reading about the regularity of a finite measure on a metric space $E$ equipped with the Borel $\sigma$-algebra $\mathcal B(E)$. The proof goes like this. Let $\mathcal O$ be the collection of all open subsets of $E$ and let $\mathcal C$ be the collection of all sets $A\in\mathcal B(E)$ for which the theorem is true, i.e. for which \begin{align*}\mu(A)&=\inf\{\mu(U):U\text{ open set},A\subset U\} \\&=\sup\{\mu(F): F\text{ closed set},F\subset A\}.\end{align*}

Since $\mathcal B(E)$ is generated by $\mathcal O$, it suffices to show that $\mathcal O\subset\mathcal C$ and that $\mathcal C$ is a $\sigma$-algebra. And then they claim, if $A\in \mathcal O$, the first equality above is trivial. If we denote the infimum by $m$, then obviously $\mu(A)\leq m$, but how do I get the reverse inequality?

Jakobian

10:18

this is not always what people mean by regularity

there is, sometimes called tightness, where every set is a supremum of measures of compact sets

back to your question, $\mu(A)\leq \mu(U)$ for all open $U\supseteq A$

so $\mu(A)$ is the minimum of the set $\{\mu(U) : U\text{ open }U\supseteq A\}$

psie

@Jakobian ah yeah, $A$ contains itself, of course. Thank you :)

Jakobian

11:22

@AlessandroCodenotti is there some way to embedd Cantor cube in itself as a nowhere dense set?

handan_toddler

@leslietownes did you watch the university challenge clip I posted about 5 + √4

yesterday, by think_meaning_buildß

in this clip from University Challenge, he asks "What are the two possible answers to the calculation 5 plus the square root of 4?"

> surprising how much life boils down to things like completing the square and squares being nonnegative

Perhaps, if it was presented as x²≥0,
rather than x²=b there would be less misunderstanding.

xⁿ=b still looms on the horizon.

Jakobian

ah yeah. If you take an infinite set $D\subseteq \kappa$ with $|\kappa\setminus D| = \kappa$ then $A = \{x\in 2^\kappa : x_d = 0\forall_{d\in D}\}$ is a nowhere dense set homeomorphic to $2^\kappa$

Soumik Mukherjee

13:03

I don't understand what canonical map they are talking about

Sine of the Time

@Pizza if the formula for $\frac{d^2}{dx^2} \hat f$ is correct then you can use that approach

In the convention of FT I use: $D\hat f=-i\widehat{xf}$; denoted by $g=xf$ then $D^2 \hat f=D(-i\hat g)=-iD\hat g=-i (-i)\widehat{xg}=-\widehat {x^2f}$

Dec 22, 2024 at 18:10, by Sine of the Time

I'm in a hurry right now, but it seems to me that after one integration by part you can apply the property $D \hat f =-i\widehat{xf}$

I was familiar with the first derivative of FT but I don't recall ever using the second derivative

Thorgott

13:23

@SoumikMukherjee who? where?

Sine of the Time

I've added here a while ago a method to compute the FT of $(1+x^2)^{-2}$ using $D\hat f$ and $\widehat{Df}$; if you're interested see here

Soumik Mukherjee

@Thorgott I am unable to upload the picture from my mobile.

Thorgott

13:37

ah, I never figured out how to do that either

perhaps just try uploading on imgur directly or something

Soumik Mukherjee

I see, but nvm, I figured it out, I was being an idiot, the map was just inclusion

Jakobian

@SoumikMukherjee @Thorgott switch to PC mode (you can do that either in your browser options, or by clicking a button in the menu on the site). Then click upload button

Soumik Mukherjee

Pc mode means desktop mode?

Jakobian

sure, however its called

Soumik Mukherjee

13:55

Oh, I am already at that mode

Maybe it's just network issues

ModularMindset

14:18

@DanielDonnelly Yes

ModularMindset

14:36

In particular $\mathcal I_3$ plays nicely with Dehn twists

hope you are still doing good

Jakobian

15:21

@leslietownes @VladimirLysikov I got an answer under a question of mine. It turns out that $E(2^\mathfrak{c})$ can be embedded in any $E(2^\lambda)$ for $\lambda \geq \omega_0$, so that in my case of $X = (\omega_2+1)_\delta$, if $\mathfrak{c}\leq \omega_2$ then $X$ can be $C^\ast$-embedded into $E(2^{\omega_2})$ directly, and if $\omega_2\leq \mathfrak{c}$, then we can $C^\ast$-embed $X$ in $E(2^\mathfrak{c})$ which then gets embedded into $E(2^{\omega_2})$

so it turns out that it's not as straightforward as Dow was claiming it to be in his article, but is actually true

Or, em, $E(2^\kappa)$ can be embedded into any $E(2^\lambda)$ for $\lambda \geq \omega_0$ as long as $\kappa \leq \mathfrak{c}$

psie

15:46

Let $E$ be a metric space equipped with the Borel $\sigma$-algebra $\mathcal B(E)$, and $\mu$ a finite measure. Then \begin{align*}\mu(A)&=\inf\{\mu(U):U\text{ open set},A\subset U\} \\&=\sup\{\mu(F): F\text{ closed set},F\subset A\}.\end{align*}

The proof goes like this. Let $\mathcal O$ be the collection of all open subsets of $E$ and let $\mathcal C$ be the collection of all sets $A\in\mathcal B(E)$ for which the theorem is true. Since $\mathcal B(E)$ is generated by $\mathcal O$, it suffices to show that $\mathcal O\subset\mathcal C$ and that $\mathcal C$ is a $\sigma$-algebra.

To show the second equality for $A\in\mathcal O$, they put $F_n=\left\{x\in E:d\left(x{,}A^c\right)\ge \frac{1}{n}\right\}$. The $F_n$'s are contained in $A$ and are closed. Moreover, $A$ is the increasing unions of the $F_n$'s. So $$\mu(A)=\lim\uparrow\mu(F_n)\tag1$$which proves the second equality. I don't understand how $(1)$ shows the second equality. Is this clear to someone?

Of course, $\mu(A)\geq\mu(F)$ for any $F\subset A$ closed. Does the limit in $(1)$ give us the reverse inequality?

Vladimir Lysikov

The limit of an nondecreasing sequence is the same as supremum

so $\sup \{\mu(F)\colon F \subset A, F \text{ closed}\} \geq \sup \{\mu(F_n)\} = \lim \mu(F_n) = \mu(A)$

psie

ah yes! nice observation :) makes sense, thanks!

Soham Saha

17:05

A cool thing happened today. I was trying to do a [Fermi estimate](https://en.wikipedia.org/wiki/Fermi_problem) about the number of comics on [XKCD](https://xkcd.com/).

I saw a comic again today, which I had sent to a friend on 9th September last year, and hadn't seen again before today. That's about 100 days ago. I see about 30 comics a day, clicking on the "Random" option. So after 3000 views, I found the same comic again.

Let there be $N$ comics in total. Let $p$ be the number of (random) views needed to see a specific comic after having selected one.

ILikeMathematics

Do you pronounce $\tilde a$ as "tilde $a$" or "$a$ tilde"?

Soumik Mukherjee

a tilde

leslie townes

the latter sounds more natural to me, although i would go to considerable length not to have to verbalize something like that

mo-_-

17:21

hi

Jakobian

17:45

my understanding of everything increased and decreased at the same time

Thorgott

a tilde

ILikeMathematics

@SoumikMukherjee Thanks

@leslietownes Well, if you write something down and then talk about it in detail to someone, then you don't want to always describe what it is, I guess

leslie townes

when i had lecturing as a part of my job i would absolutely go out of my way to avoid the use of 'decorated' versions of letters i was using if it was at all possible to avoid it, just because it felt silly to be reading typesetting aloud

on the same principle i'd use a' ("a prime") before a tilde just because that's one less syllable, unless there's some reason why that wouldn't work

ILikeMathematics

Ah, alright

leslie townes

but if i had to read \tilde a it's "a tilde" and not "tilde a"

ILikeMathematics

17:55

The universal property of tensor products says that if we have a multilinear $\varphi: V_1 \times \dots \times V_n \to W$ then there exists a $\tilde \varphi: V_1 \otimes \dots \otimes V_n \to W$ with $\varphi(v_1, \dots, v_n) = \tilde \varphi(v_1 \otimes \dots \otimes v_n)$.

Doesn't this kinda feel backwards? Don't we want the other direction; simplifying the tensor products

Like this, it feels like it's just making it more complicated for no reason

Why do we ever want to use this?

Thorgott

you've turned a multilinear map into a linear map, that's simpler

ILikeMathematics

Oh

Thorgott

also you forgot "unique" after "there exists a"

ILikeMathematics

True, thanks

psie

18:52

> A real function $h$ defined on $[a,b]$ is called a step function if there exists a subdivision $a=x_0<x_1<\cdots<x_N=b$ of the interval $[a,b]$ and reals $y_1,\ldots,y_N$ such that, for every $i\in\{1,\ldots,N\}$, we have $h(x)=y_i$ for every $x\in (y_{i-1},y_i)$.

I'm reading about the Riemann integral in connection with studying the Lebesgue integral, and the above is an extract from the book. Is it a typo that $x\in(y_{i-1},y_i)$? Shouldn't it be $x\in (x_{i-1},x_i)$? Moreover, is a step function Borel measurable? I don't quite see the difference between simple and step function...

Vladimir Lysikov

@ILikeMathematics I like to think about this as follows: the definition of the tensor product tells us two things: how to construct tensors ($\otimes$ map constructs an elementary tensor from vectors, and vector space structure gives also linear combinations of elementary tensors) and how to use them (take a multilinear map and apply it as a linear map to a tensor)

Jakobian

@psie yes. Yes. Yes. A step function is a simple function but not conversely

the difference is that here the sets we take are not arbitrary measurable sets, but intervals

psie

ok, cool beans! :) so all the properties of simple functions, like independence of representation when it comes to the integral or monotinicity, carry over to step functions

Jakobian

not sure what you mean by monotonicity

psie

well, that if $h\leq h'$ then $I(h)\leq I(h')$ where $$I(h)=\sum_1^N y_i(x_i-x_{i-1}).$$

Jakobian

19:04

yes, if you define $I$ this way then all those properties hold

psie

great 👍

ILikeMathematics

20:20

@VladimirLysikov Oh, yeah that works. My lecture notes define them differently though, and then prove the universal property

They first define $\mu_{v_1, \dots, v_n} \in \mathcal L(V_1, \dots, V_n; K)^*$ as $\mu_{v_1, \dots, v_n}(f) = f(v_1, \dots, v_n)$ and then $$V_1 \otimes \dots \otimes V_n = \left \langle \mu_{v_1, \dots, v_n} \mid v_i \in V_i \text{ for $1 \leq i \leq n$}\right \rangle$$ and $v_1 \otimes \dots \otimes v_n = \mu_{v_1, \dots, v_n}$

Let $v_1 = (1, 0, 0)^T$, $v_2 = (-3, 2, 0)^T$ and $v_3 = (3, -6, 3)^T$ be a basis of $\mathbb R^3$. Determine the dual basis $\{\lambda_1, \lambda_2, \lambda_3\}$ to it; thus coefficients with $$\lambda_j(x_1, x_2, x_3)^T = a_{1j}x_1 + a_{2j}x_2 + a_{3j}x_3$$ and spot a general pattern.

Vladimir Lysikov

@ILikeMathematics Yes, people often do similar thing. I personally don't like it because it is subtle, it needs thinking similar to the double dual and because of this works only for finite-dimensional spaces

I first introduce the explicit construction in bases, and then explain that if you don't choose the bases we can use the quotient construction

ILikeMathematics

To obtain the $\lambda_i$, just set $\lambda_i(v_i) = 1$ and $\lambda_i(v_j) = 0$ for $i \neq j$. If we write the coefficients as above, we get $$A = \begin{pmatrix} 1 & 0 & 0 \\ 3/2 & 1/2 & 0 \\ 2 & 1 & 1/3 \end{pmatrix}.$$ One can notice that this is the inverse of the matrix with $v_1, v_2, v_3$ as rows. I'm not sure if that's what they mean by 'general pattern' though.

Vladimir Lysikov

Yes, this is probably what they want you to say here

ILikeMathematics

I.e. if $n$ linearly independant vectors $v_1, \dots, v_n \in K^n$ are given, they want some general pattern to determine $(K^n)^*$. Is determining the inverse as I described the best way to do this or is there some 'better' pattern?

@VladimirLysikov Alright, thanks

psie

20:43

I'm reading about the fact that the Riemann integral on $[a,b]$ for a bounded function $f$ coincides with the Lebesgue integral, and $f$ is Lebesgue measurable. In the proof we have an increasing and decreasing sequence of step functions $(h_n)$ and $(\tilde{h}_n)$ respectively that are bounded above and below by $f$, i.e. $$h_\infty=\lim\uparrow h_n\leq f,\quad \tilde h_\infty=\lim\downarrow \tilde h_n\geq f.$$

Then they claim $h_\infty,\tilde h_\infty$ are bounded. I have hard time accepting this; a decreasing sequence converges to its infimum, which is bounded below by $f(x)$, a finite value. Does this show boundedness of $\tilde h_\infty$?

Thorgott

$h_1\le h_{\infty}\le f\le\tilde{h}_{\infty}\le\tilde{h}_1$

step functions are bounded

a function bounded between two bounded functions is bounded

psie

ah nice, ok, thanks Thor! 👍

ILikeMathematics

22:03

How does an example of a map from $V^*$ to $V^{**}$ (dual space to bidual space) look like?

Thorgott

what kind of answer are you expecting?

ILikeMathematics

@Thorgott Does $\psi: V^* \to V^{\text{two stars}}$ with $\lambda \to \varphi_{\lambda}$ with $\varphi_{\lambda} \in V^{**}$ with $\varphi_{\lambda}(v) = \lambda(v)$ work?

Ok no that's not in $V^{**}$

Can we fix this?

Hm, actually maybe that works

Jakobian

22:28

@ILikeMathematics yeah, the definition is not okay

Thorgott

that definition does not make sense to me

ILikeMathematics

@Jakobian How to fix it?

leslie townes

people usually find themselves resorting to structure other than that provided by 'abstract vector space' to define a map of a vector space into its dual without making weird and arbitrary choices

Jakobian

@ILikeMathematics I don't know what you're trying to do

ILikeMathematics

@Jakobian I just want to see how some arbitrary map from $V^*$ to $V^{**}$ looks like

leslie townes

22:34

if you allow yourself to do things like choose bases for abstract vector spaces this question has no content

and if you don't, you're maybe where you are now

Jakobian

@ILikeMathematics a (linear) map between (finite-dimensional) vector spaces looks like a matrix

there is not really much more to say, it's not that $V^\ast$ and $V^{\ast\ast}$ are much more related

ILikeMathematics

@Jakobian I meant an explicit way to map elements from $V^*$ to $V^{\text{double star}}$. Let me try correcting the above. Define $q_{\varphi}: V^* \to K$ with $\lambda \mapsto \lambda(v) \cdot \varphi(v)$. Now define $\psi: V^* \to V^{**}$ with $\lambda \mapsto q_{\lambda}$. Does this look fine?

As I said I just wanted some random map

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$Mathematics$ Mathematics