Mathematics - 2025-01-05

ModularMindset

03:56

Hi

Bob

good evening

Daniel Donnelly

Evening

ModularMindset

04:11

Evening good

Bob

If I post a question on math stack exchange and it is voted down and I do not understand why, is it okay to ask about it on Math Stack Exchange Meta

ModularMindset

06:03

Hey

ModularMindset

06:41

3

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 $\Bbb R^3$ ...

any ideas?

@leslietownes would you consider undoing your close vote?

Vladimir Lysikov

07:01

@ILikeMathematics I assume you mean $u_i$ instead of $v_j$ in the first factor. What are $s_i$ and $\ell_j$?

@ILikeMathematics This is still not an arbitrary tensor. You need to take an arbitrary $T \in U \otimes V$ and prove that $\varphi(T) = 0 \Rightarrow T = 0$. Arbitrary means not just $u \otimes v$, but also linear combinations of these

A useful lemma is that if $\{u_i\}$ is a basis of $U$, then every tensor $T \in U \otimes V$ can be represented as $T = \sum u_i \otimes x_i$ for some $x_i \in V$

Soumik Mukherjee

07:36

Yesterday there was only one beehive, today there are two.

It's amusing how fast they make a new beehive.

ModularMindset

07:53

Hi

anyone know anything about 2-tori?

Vladimir Lysikov

Aren't 2-tori just the usual tori?

ModularMindset

@VladimirLysikov Yes I'm examining a family $\{\mathcal I_{t}\}_{t>0} \simeq T^2$ for all $t$.

Vladimir Lysikov

I imagine almost everyone here knows *something* about tori
You should be more specific

ModularMindset

these are topological tori for all t, but there is some anisotropic radial expansion ocurring

trying to determine how to stitch these slices together into a manifold or stratified space

Actually this is almost impossible to explain without at least 3 days one on one

The link that I linked above there gives some good info about this

there's just a lot of background material that most people don't have

I need a world leading expert in this stuff

Jakobian

08:17

@ZacU. additionally to what Thorgott said (which might be true but perhaps not always), there is also issues with occassional incomplete and/or half-assed resources. This also varies from topic to topic. Other reason (that I can think of) might be that you aren't mathematically mature

To split it down into three categories, this might be because 1) your skills and abilities, 2) the resources you are using, 3) because it just is like that for everyone no matter the skill or quality of resources

or some combination of the three

ModularMindset

learning math is no cakewalk I agree

Jakobian

I feel like issue number 2 on my list is especially more and more present the further down the line you go

the resources become more and more obscure, and your skills have to be honed sharp to be able to come up with arguments on the spot

people who begin to learn the topics don't know the quality of life that they get from learning a topic that people had the ability to refine the teaching of to the point that it's accessible to everyone

while 3 might be the case here, I think that in Thorgott's case, the topics he learns, this is especially the case, so take that with a grain of salt

psie

10:31

Consider a continuous, real-valued compact support function. According to the answer following the link below, $$f(X)\subseteq\{0\}\cup f(\mathrm{supp}(f)).$$There is a comment there that confuses me.

I take my above comment back, since $\{x\in X: f(x) =0\}$ need not be non-empty. Nevertheless, if $\{x\in X: f(x) =0\}\ne \emptyset$, then $f(X) = \{0\} \cup f(\text{supp}(f))$. On the other hand, if $\{x\in X: f(x) =0\}= \emptyset$, then $f(X) = f(\text{supp}(f))$. In either case, $f(X)$ is compact. — Satana Mar 24, 2019 at 1:32

See, Satana claims that if $\{x\in X: f(x) =0\}=\varnothing$, we have proper inclusion, but can $\{x\in X: f(x) =0\}=\varnothing$ occur for a compact support function? I thought the idea was that compact support functions vanish outside a compact set, i.e. equal $0$ there.

Jakobian

10:47

@psie $f:[0, 1]\to \mathbb{R}$, $f(x) = x+1$ for example

psie

@Jakobian ok, is $[0,1]$ compact in the subspace topology of $[0,1]$?

Soumik Mukherjee

11:44

I am reading lectures on formal and rigid geometry by Bosch and I am stuck on a proof on page 14

Jakobian

@SoumikMukherjee which part of it?

Soumik Mukherjee

The converse part

Where it's written f is of type 1-g

Here $K$ is a complete non-Archimedean field. $T_n$ is a Tate Algebra. $R$ is the valuation ring $\{a\in K : \lvert a\rvert \leq 1\}$ with maximal ideal $\{a\in K : \lvert a\rvert < 1\}$. $k$ is the residue field.

Jakobian

@SoumikMukherjee I think it has to somehow follow from $\tilde{f}$ being invertible

oh those are polynomials

for $\tilde{f}$ to be in $k^\ast$ it means that there is $h\in R\setminus \mathcal{m}$ such that $f = h+h_1$ where $h_1\in\mathcal{m}\langle \xi_1, ..., \xi_n\rangle$

right, I think that makes sense

because if you write $f$ as $\sum_{\nu\in\mathbb{N}^n} c_\nu \xi^\nu$ then $\tilde{f} = \sum_{\nu\in\mathbb{N}^n} \tilde{c_\nu} \xi^\nu \in k^\ast$

and the latter means that $\tilde{c_\nu} = 0$ for all $\nu\neq 0$

which means that $c_\nu\in\mathcal{m}$ for $\nu\neq 0$

now this means $|c_\nu| < 1$ for all $\nu\neq 0$, that is $|\sum_{\nu\neq 0} c_\nu \xi^\nu| = \max_\nu |c_\nu| < 1$

@SoumikMukherjee

Soumik Mukherjee

12:15

yes

I am following

Jakobian

yeah so here what you did is assume $f(0) = 1$ which is to say that $h = 1$

so $g = \sum_{\nu\neq 0} c_\nu \xi^\nu$ is your element

Soumik Mukherjee

$-g$ as they wrote

THANKS!!

Jakobian

ah, yeah, $-g$

no problem. It's actually something new to me

Soumik Mukherjee

$T_n$ is a Banach $K$-algebra. So maybe you'd be interested in this.

Jakobian

I don't think I'd like it. It does look nice how everything gets simplified though

Soumik Mukherjee

12:23

ooh

ModularMindset

12:59

cells interlinked

cells within cells. interlinked

ModularMindset

13:50

I don't understand the upvotes. oh well. interlinked

ModularMindset

14:45

In my research, I explore stratified systems of foliations on open manifolds with regular polytope boundaries. Specifically, the 1-dimensional strata of the foliated system forms a family of finite, regular, connected, undirected topological multigraphs. Each shell in this foliated system is topologically homeomorphic to a 2-torus. The foliated system is constructed such that these shells have vanishing Euler characteristic and compactness, ensuring their toroidal structure.

A key advantage of this framework is the ability to endow each 2-dimensional stratum with metrics of constant curvat…

Does that sound cranky?

interlinked.

Does that sound like I am a quackjob?

(that's bonafide kosher research so..)

ModularMindset

15:37

I think that my research is mostly differential topology

*most closely resembles

by the way "cells interlinked" is from bladerunner

Blade Runner 2049 Cells Interlinked Scene 1080p

►

ModularMindset

16:19

1

Q: Find resources/topics related to my interests

I am looking for peer-reviewed papers and topics that relate to what I'm interested (below) as much as possible. I think that my interests would mostly fall into differential topology. Appreciate the resources and exploring something new. In my research, I explore stratified systems of foliation...

general-topology differential-geometry reference-request differential-topology geometric-topology

made a little question out of it.

psie

16:39

I have a basic doubt. I sometimes see convergence in $L^p$, $p\in[1,\infty)$ stated as $$\lim_{n\to\infty}\int|f_n-f|^p\,\mathrm{d}\mu=0.$$This means $\|f_n-f\|_p^p\to0$ as $n\to\infty$. Now, this is equivalent to $\|f_n-f\|_p\to0$ as $n\to\infty$ by continuity of $g(x)=x^p$, right?

At least in a proof I'm currently reading, on several occasions the author concludes $\lim_{n\to\infty}\int|f_n-f|^p\,\mathrm{d}\mu=0$ and consequently says we can find an $n$ large enough such that $\|f_n-f\|_p<\epsilon$ or says the integral expression tending to $0$ means $\|f_n-f\|_p\to0$ as $n\to\infty$.

leslie townes

yes psie the fundamental thing i would focus on is the idea of convergence in a norm (however you define L^p, || ||_p is a norm on it). people will frequently work with pth powers of that norm because the formulas are easier to handle and it often doesn't affect the analysis (e.g. the ^{1/p} is needed to make the p-norm a norm, but as you point out/suggest ||f_n - f||_p^p goes to 0 if and only if ||f_n - f||_p goes to 0

the fundamental thing here is convergence in a norm, i would think of all of this ^p or ^{1/p} stuff as just calculationally specific stuff to make nicer formulas in the case of L^p spaces

psie

great :)

Jakobian

17:15

maybe not as much continuity of $x\mapsto x^p$, as it being a self-homeomorphism of $[0, \infty)$

leslie townes

17:30

yeah, continuity of the inverse map is what you use for the more interesting direction of that equivalence

psie

right, for $\|f_n-f\|_p\to0\implies \|f_n-f\|_p^p\to0$ we require the continuity of the inverse too. My bad!

Jakobian

17:56

uh, no, that's where you apply continuity of $x\mapsto x^p$

continuity of the inverse is in the opposite implication

psie

you're right :) my bad again!

leslie townes

for general p it's all kind of the same psie but even in the most interesting case of p = 2 you see a contrast between the squared norm relating directly to an inner product and being symbolically easier to analyze than anything involving square roots (which can, for example, interfere with differentiability of maps involving the norm)

Bml

18:58

Hi everyone. Where can I find an explicit proof that any continuous function defined on $\mathbb{Q}$ extends to at most one continuous function on $\mathbb{R}$ (i.e., it is unique)?

Jakobian

@Bml I can write you a proof

leslie townes

19:11

1

Q: Extension of a continuous function $f: \mathbb{Q}\to\mathbb{R}$ to a continuous function $g:\mathbb{R}\to\mathbb{R}$.

Prove or disprove that for all continuous functions $f:\mathbb{Q}\to\mathbb{R}$ there is a unique continuous function $g:\mathbb{R}\to\mathbb{R}$ with $g(x)=f(x),\ \forall x\in\mathbb{Q}$. Now, this is a task consisting of 2 subtask and this is the latter. In the first one (I think) I have proved...

analysis continuity

4

Q: Prove $f(x) = 0$ for all rational numbers implies $f(x)=0$ for all reals.

Let $F:\mathbb{R}\to\mathbb{R} $ be a continuous function such that $F(x)=0$, $\forall x\in \mathbb{Q}$. Show that $F(x)=0$, $\forall x\in \mathbb{R}$. I said: Let $k\in \mathbb{Q}$, $\varepsilon,\delta\in\mathbb{R}\setminus \mathbb{Q}$ such that $k-\varepsilon\le k \le k+\delta$ Since $F(k)=0$...

real-analysis continuity

Jakobian

20:02

If $f_1, f_2$ are two continuous functions with the same domain and codomain, then you want to look at the set $A = \{x : f_1(x) = f_2(x)\}$. If $\Delta$ denotes the diagonal of the codomain, $g(x) = (f_1(x), f_2(x))$ then $A = g^{-1}(\Delta)$ is a preimage of closed set by continuous function, and it contains a dense set, so is whole space

That is if $Y$ denotes the codomain then you want $\Delta = \{(y, y) : y\in Y\}$ to be closed

You can check this is true when $Y$ is a metric space

psie

20:51

> Theorem Let $E$ be a separable locally compact metric space. Then there exists an increasing sequence $(L_n)$ of compact subsets of $E$ such that, for every $n\in\mathbb N$, $L_n\subset L_{n+1}^\circ$ and $$E=\bigcup_{n\geq 1}L_n=\bigcup_{n\geq 1}L^\circ_n.$$

The proof starts off by showing that $E$ is the union of an increasing sequence of compact sets $(K_n)$. I understand this part.

However, I don't understand how they construct the $L_n$'s. First, take $L_1=K_1$ and if $L_n$ has been constructed, we find a cover of the compact set $K_{n+1}\cup L_n$ by a finite union $V_1\cup V_2\cup\cdots\cup V_p$ of open balls with compact closure whose centers belong to $K_{n+1}\cup L_n$, and we take $L_{n+1}=\bar{V_1}\cup\bar{V_2}\cup\cdots\cup\bar{V_p}$.

What confuses me is
(a) why such open balls exist?
(b) previously, the author has used $\bar{B}(x,r)$ to denote the closed ball, so I'm not sure what $\bar{V_i}$ denotes, if it's the closure or a closed ball?

leslie townes

presumably it's just the closure (in the metric topology) of V_i

Thorgott

the space is locally compact, so every point has a neighborhood basis of open balls with compact closure

in fact, that's the same thing as being locally compact in this context

leslie townes

separately from this (pun intended), you haven't asked this directly, but the closure of an open ball of radius r and center x in a metric space is not always the closed ball of radius r and center x

Thorgott

yeah, I think $\overline{B}(x,r)$ is bad notation for this reason

leslie townes

so "the closure of V_1" might indeed be the simplest way of describing that set in an arbitrary metric space

psie

20:57

ok, thanks

@Thorgott the definition of locally compact used is that every point has a compact neighborhood. How do we construct such a neighborhood basis of open balls with compact closure?

What is a neighborhood basis? :) I'll have to google this.

Thorgott

not necessary to know

point is, the point has a compact neighborhood

its a neighborhood, so it contains small enough balls

then they contain their closures (because compact implies closed in your setting) and a closed subset of a compact space is compact

psie

that makes more sense now, thanks

Transcript for

$Mathematics$ Mathematics