Mathematics - 2024-10-29

Thorgott

00:01

"a bit" as in, like, it's sort of a converse

Sine of the Time

00:39

math.stackexchange.com/q/4991102/1118406 @leslie

sku

for the question that there is a unique sup for a non empty set of real numbers, here is my proof (different from textbook). Let a and b be the two different sups. WLOG, a <= b. If a < b, then b is not a sup. If not they are equal. Is this correct?

Jakobian

The only property needed to have unique sups is that $x\leq y$ and $y\leq x$ implies $x = y$

sku

00:59

is my argument incorrect in some way?

IThinkHighlyOfEiligh

Aloha algebros and topologists

What’s cracka lackin

Xander Henderson

No.

IThinkHighlyOfEiligh

Depressive

nickbros123

09:25

Is there a metric $\tilde{d}$ such that $[0,1),|\cdot |$ be homeomorphic to $[0,1),\tilde{d}$ where $\tilde{d}$ is such a metric that $[0,1),\tilde{d}$ is a complete metric space?

I can see that if $x_n$ is a cauchy sequence in $\{[0,1),||\}$ then $f(x_n)$ must diverge, but cannot conclude anything more

part of me thinks this is possible

but another part thinks its not possible

Jakobian

09:44

@nickbros123 its possible

Do you want me to write you a solution

More generally this is equivalent to being a $G_\delta$ subset of $\mathbb{R}$

You might be surprised to know that irrational numbers can be made to be complete for example

Ryder Rude

hi

nickbros123

@Jakobian i just wanna know where i should start thinking

Jakobian

@nickbros123 what is $[0, 1)$ homeomorphic to?

nickbros123

[0,1) is Homeomorphic to $R$ among other things?

oh wait its not

Ryder Rude

i think (0,1) is homeomorphic to R

to push in an extra point, one needs a discontinuous function @nickbros123

nickbros123

09:56

yeah just noticed

Jakobian

@RyderRude huh?

Ryder Rude

@Jakobian i mean that the only bijections between [0,1) and (0,1) are discontinuous functions

Jakobian

Random

nickbros123

12:42

@Jakobian can I get another hint?

Sine of the Time

12:53

@nickbros123 hint

nickbros123

thanks for the hint

Sine of the Time

np

Jakobian

13:08

@nickbros123 $[0, 1)$ is homeomorphic to certain closed subset of $\mathbb{R}$

nickbros123

I think I have a function for [0,infty)

$2/\pi \arctan(x)$ works?

Jakobian

Sure

Now translate the metric structure from $[0, \infty)$ to $[0, 1)$

psie

15:43

I'm skimming through Spivak's Calculus on Manifolds. In the inverse function theorem, he states that $$(f^{-1})'(y)=[f'(f^{-1}(y))]^{-1}.$$I'm a bit rusty with his material. He uses $f'(a)$ to denote the Jacobian of the associated linear map $Df(a)$. So in this case, would it be incorrect to write $$Df^{-1}(y)=[Df(f^{-1}(y))]^{-1}?$$

Maybe my second equation mixes up the notation in a bad way...

But what I'm asking is if we can write the conclusion of the inverse function theorem in terms of the linear map instead of the matrix, and if yes, what would be the correct way.

Or maybe it is correct as written, both of them.

Thorgott

15:59

a matrix just records a linear map w.r.t. some bases, any statement about linear maps can be translated into matrices and vice versa

psie

ok 👍

nickbros123

16:13

@Jakobian you mean like $\delta(x,y)=|tan(\pi x/2)-tan(\pi y/2)|$ for $x,y\in [0,1)$ right?

I was able to show, under the map $\gamma(x)=x$ from $[0,1),|.|$ to $[0,1),\delta$ would be continuous and inverse also continuous

Jakobian

If $f:[0, 1)\to [0, \infty)$ is any given homeomorphism, I mean metric $(x, y)\mapsto |f(x)-f(y)|$ for $x, y\in [0, 1)$

so this is the metric on $[0, 1)$ induced by the one on $[0, \infty)$ via the map $f$

nickbros123

oh so you want it to come from arc tan itself

Jakobian

I don't want arc tan, I don't care which map you use

whats important is that then $[0, 1)$ with this metric becomes isometric to $[0, \infty)$ with Euclidean metric, and so properties like completeness are the same

nickbros123

@Jakobian ok the tan one satisfies this

@Jakobian I see..

Jakobian

If instead of $[0, 1)$ you had any set $X$ whatsoever and $f:X\to [0, \infty)$ was a bijection then you could do the same exact thing to put a metric structure on $X$ isometric with $[0, \infty)$

You can translate structures from one object to another like this

very simple concept that people don't really understand perhaps

you can do this with other types of structures, if $f:X\to\mathbb{R}$ is a bijection, you can make $X$ into a group isomorphic to $\mathbb{R}$ for example

well whats important is that here this bijection $f$ is also a homeomorphism so the topology on $[0, 1)$ with Euclidean and the metric induced by $f$ is the same

nothing complicated

this should be automatic, trivial, and obvious

Contemplate on this concept

If you never seen it before

Otherwise don't dwell on it

nickbros123

16:39

this is the first time im seeing this. I kinda get it now. id never thought about this before in this fashion- but the isometry falls rather neatly.

we have $f:\{[0,1),|.| \} \to \{[0,\infty),|.|\}$ a homeomorphism (one example is tan), then $[0,1), \delta(x,y)=|f(x)-f(y)|$ would be isometric to $[0,\infty),|.|$ trivially since $f$ itself would act as that isometry, $|f(x)-f(y)|=\delta(x,y)$ from simply the definition

Jakobian

17:12

I think across a mathematicians education it should be mentioned at least one, its an important concept, despite how trivial it is

nickbros123

i think I understood this thing now clearly. If $(X,d_X)$ is homeomorphic to $(Y,d_Y)$ under the function $f$ then there is another metric $\delta$ on $X$ given by $\delta(x,y)=d_Y(f(x),f(y))$ which would be isometric to $Y,d_Y$. I can just now see how trivial this thing is

and isometries are nicer than homeomorphism

Pizza

17:45

Hi

Pizza

17:57

@SineoftheTime Have you studied the application of the Laplace transform applied to differential equations?

Sine of the Time

@Pizza I took a look at LT on my own a couple of months ago

why?

Pizza

I was just checking it out

Sine of the Time

Is it hard? For computation purpouses, I don't think it's that difficult

Pizza

18:13

Now I'll try to see

Michael

Anyone interested in aeronautical engineering? I'm looking for a partner for my nonprofit

Building a human powered aircraft

Sine of the Time

19:13

$\cancel{x}$

why \cancel is not working?

leslie townes

sine: what is \cancel? i think that mathjax only supports only parts of "normal" latex (and maybe other stuff), but if that comes from some special package the odds of it working in chat probably go down

Sine of the Time

Maybe I figured it out

@leslietownes It's used to cross out thing see here

but even writing \require{cancel} does not solve the problem

I'm aware that in chat may not work, but even on main it's not working

with \begin{align} now it works

\begin{align}
\require{cancel}
\cancel{x}
\end{align}

Apparently, you have to write \require{cancel} and then skip a line

Sorry for the useless messages :)

leslie townes

oh, my highly subjective and personal take re \cancel would be just to never write that, it looks horrible

although it's mildly interesting/cool that chat lets you load specific packages

Ben Steffan

19:39

\require{tikzcd} when :(

Xander Henderson

@BenSteffan I learned pstricks way back. I've never bothered with tikz or anything related to it.

Ben Steffan

you also don't work in a diagram-heavy field I gather

Xander Henderson

@BenSteffan Heck, no!

Commutative diagrams are for LOSERS! :P

Ben Steffan

anyways, MathJax only supports amscd and amscd sucks

and also it's broken in the MathJax v3 beta right now

Xander Henderson

There are three small commutative diagrams in my thesis. I believe I used xymatrix?

Is that a thing?

Ben Steffan

19:44

uhh

apparently, but I had to google it

never heard of it before

Sine of the Time

@XanderHenderson real mathematicians study real analysis

Xander Henderson

Yeah, much lighter overhead than tikz.

Ben Steffan

it can do more than amscd, so that's good (although the barrier is rather low)

if all you do is commutative diagrams the tikz overhead is usually manageable, at least if your document doesn't contain upwards of, say, ~100 diagrams

but that's nowadays :)

but yeah the performance of tikz is a pain

I had to externalize all my tikz images in at least one set of lecture notes because of performance reasons

Xander Henderson

@BenSteffan My thesis was pretty long by the standards of mathematics (around 120 pages). It took a while to compile, and my general preference was to keep it as light as possible. Why load tikz when I am only using it for three diagrams?

F that noise.

Ben Steffan

@XanderHenderson yeah, especially if the diagrams are simple

fun fact: putting your hand on a hot plate is more fun than setting up externalization to play nice with latexmk

Xander Henderson

19:51

Ben Steffan

and also externalization does not work out of the box with tikzcd, and I never managed to set it up

Xander Henderson

That is the most complicated of the three diagrams.

Ben Steffan

the artifacts on arrows that aren't perfectly horizontal or vertical is something that tikz does a better job at avoiding :^)

but yeah that's fine

Xander Henderson

@BenSteffan That is an effect of the rasterization on the screen.

It looks better if you have higher resolution / zoom in.

leslie townes

well, maybe those effects vanish, but trust me, it doesn't look any better

Xander Henderson

19:54

@leslietownes I mean, it is a commutative diagram. Only so much you can do.

Ben Steffan

@XanderHenderson it is, but I believe the pdf has some say in how the rendering goes

Xander Henderson

@BenSteffan No idea. What I know is that the dead tree copies I have look fine.

Ben Steffan

at least I see these artifacts consistently in texts that don't use tikz as opposed to those that don't

yeah, it's digital only

leslie townes

xander it's funny you mention compile time as a consideration, i certainly never designed my documents around that, but i very much remember noticing compile time and sometimes not compiling as often as i would have liked just to avoid the disruption. you should try compiling your thesis today, on your phone, and see if you even notice the compile time

Xander Henderson

The whole point of LaTeX is to produce a print copy. Who cares how it looks on screen? :P

@leslietownes Heh.

Ben Steffan

19:56

@XanderHenderson the screen is your printer :)

who can afford paper, who can afford ink, in this economy

Xander Henderson

@BenSteffan My screen is not made of dead trees.

@BenSteffan I just spent \$80 on a pen. Ink is cheap.

(And that that pen is not the most expensive pen I own.)

(It's pretty average among my pens.)

Ben Steffan

@leslietownes github.com/bensteffan/algebraic-topology-1-notes-bonn-ws23 pull this, run latexmk and tell me about compile times again :^)

@XanderHenderson If I printed everything I wanted to read I would exceed $80 in ink costs rather quickly

$80 is not that much for a pen, in the grand scheme of things

I'm not in a position to spend that much money on one but if I was I might

Xander Henderson

@BenSteffan I work at a college, and no one checks how much toner I am using. However, I have a laserjet printer at home which is still on its original toner cartridge and has printed around 7000 pages.

Ben Steffan

I am conveniently omitting that my uni let's us print for free :)

but even then I think they'd be mad if I walked in to print a copy of higher algebra

Xander Henderson

A \$100 toner cartridge will, generally speaking, get you somewhere between 5 and 10k pages.

Ben Steffan

20:03

Printing both bibles of homotopy theory would eat through 2.5k pages alone, so I think my estimate is not unsound

Xander Henderson

@BenSteffan I don't read math on a screen---it just doesn't work for me. So I either print out or buy anything I want to read.

Ben Steffan

that's fine

ModularMindset

Consider a co-dimension $1$ surface of revolution $L^{n-1}$ and an embedding $e:L^{n-1} \hookrightarrow X^n$ for $X^n=[0,1]^n$ with points $p,q$ elements of $\partial X^n$ where $\partial X^n=X^n-(0,1)^n$ for $\mathrm {sup}~ \mathrm{dist}(p,q)=\sqrt{n}$. Assume $L^{n-1}$ has constant positive sectional curvature. Next replicate this symmetrical manifold, $L^{n-1}$, $2^{n-2}$ additional times. In each replication, the cone points on $L^{n-1}$ align with unique pairs of vertices on the boundary of the $n$-cube. Take the union of all these surfaces - there $2^{n-1}$ total. We'll call this unio…

Ben Steffan

I know one or two people of my generation who are like this, still

but in my field it is a bit impractical

leslie townes

@BenSteffan i feel like every warning i have ever been given about downloading strange things from the internet and running them at home was in preparation for this moment of me not downloading and compiling your algebraic topology notes

Xander Henderson

20:06

@BenSteffan I think that you are calling me old?

@leslietownes is old...

Ben Steffan

@leslietownes :)))

you should read the latexmkrc to improve your confidence

ModularMindset

So...is it path connected?

Ben Steffan

@XanderHenderson I'm pretty sure that you're older than, say, 25

ModularMindset

heh.

Ben Steffan

no offense but, :)

Xander Henderson

20:07

@BenSteffan Is that considered "old" now?

Ben Steffan

have I said "old"? No, I have said that you're not of my generation

leslie townes

i'm old, but also kind of the opposite of xander in this respect. in the 1990s i would sometimes use "did the latex compile" as a substitute for loading a document in a GUI and checking it visually prior to printing

Derivative

if $S$ is a CW complex and $X/S$ is a CW complex then is it true that $X$ is a CW complex?

ModularMindset

When I said "heh." i was not responding to anyone I was simply saying Heh.

for the record

Ben Steffan

@Derivative no

consider the warsaw circle

Derivative

20:09

why not?

Ben Steffan

wrong question

why should this be true?

I think it's a pretty extraordinary claim, so the burden of argument is on you :)

Derivative

I'm trying to prove that the pushout of a cellular map and a subcomplex inclusion is a CW complex

ModularMindset

that seems way harder than my question

Derivative

then someone told me that one of the steps of the proof is giving a CW complex structure to a quotient

Ben Steffan

that's something much more concrete than your conjecture

Derivative

20:11

but then idk where do I go from there

Ben Steffan

define an explicit CW-structure.

Derivative

?

ModularMindset

https://chat.stackexchange.com/transcript/message/66531492#66531492

Any thoughts?

Ben Steffan

define an explicit CW-structure. :)

In terms of the CW-structures on the constituent spaces of the pushout

Derivative

@BenSteffan I have CW structures on $X$ and on $Y$ but so far no guarantee that I can glue them together

ModularMindset

20:14

I think this is called the connected P conjecture

Michael

Hello everyone

ModularMindset

Hello

Derivative

hello

Ben Steffan

@Derivative that's sort of what you're supposed to figure out I suppose (the gluing)

and how "cellular map" and "subcomplex inclusion" help you do this

Michael

20:33

I'm looking for a partner and mentor for a human powered aircraft project. I was formerly a software engineer, now I'm studying aeronautics and biomechanics

Xander Henderson

@Michael So you've already said. Please don't spam the room.

ModularMindset

Are some mathematicians interested in objects that are topologically equivalent to a 2-torus, and smoothly non-equivalent to a 2-torus?

I'm trying to wrap my head around that

for example came up with a trivial example - a polygonal torus

So lettuce beef the question up and look at a more a-bun-dance set of examples

Ben Steffan

20:50

@ModularMindset Nobody is interested in this because no such objects exist :)

the first case of the disagreement between isomorphism in the smooth and topological categories happens in dimension 4

psie

I have read the inverse function theorem in Spivak's Calculus on Manifolds and I have hard time reconciling it with what Folland states in his book on the chapter on the $n$-dimensional Lebesgue measure (see second picture). In particular, why does Folland stipulate that $G$ is injective? Isn't this a consequence of the inverse function theorem (see first picture)? Is Folland using some other inverse function theorem that I am not aware of yet?

Ben Steffan

@psie If $G$ is not injective it obviously cannot have a global inverse

Folland uses the inverse function theorem that the inverse of $G$, which a priori might not even be continuous, is in fact a $C^1$-diffeo

he is using exactly the statement of the inverse function theorem from the first picture

psie

@BenSteffan really? That's a relief :)

ModularMindset

@BenSteffan The context is a surface that is topologically a 2-torus, but it's not diffeomorphic to a 2-torus because it intersects itself and therefore isn't differentiable at places. Are you claiming this doesn't exist?

Ben Steffan

I don't follow. Something that intersects itself isn't even manifold.

I am saying that any 2-manifold that is homeomorphic to the torus is diffeomorphic to the torus.

In fact, this is still true if you replace "homeomorphic" with "homotopy equivalent"

and I suppose also "diffeomorphic" with "diffeotopy equivalent," if you'd like

ModularMindset

21:13

Okay I see

psie

@BenSteffan the inverse function theorem in the picture says that the inverse may only be a $C^1$-diffeomorphism locally. What I still don't quite understand is that we have $G:\Omega\to\mathbb R^n$ and then he concludes that $G^{-1}$ will be a $C^1$ diffeomorphism on the entire $G(\Omega)$. Is that really what the inverse function theorem says?

Ben Steffan

@psie It is a consequence of the inverse function theorem, yes.

Hint: Being of class $C^1$ is a local property

...I guess that's less of a hint and more of a "giving away the punchline" :)

psie

Ok, hmm.

indeed, differentiable and continuous are local properties, and they imply some sort of global property in this case, is that what you are saying?

Ben Steffan

I am saying they are local properties, in the sense that a function has property $P$ if it has property $P$ at each point of its domain

psie

ah ok

The Empty String Photographer

21:23

$10^\infty=0$. Convince me otherwise.

Ben Steffan

@TheEmptyStringPhotographer $10^8 = 100000000$. Be a bit more careful with your 8s in the future :)

psie

21:37

@BenSteffan so Folland is applying the inverse function theorem to each point of the domain of $G$ and then the inverse function obtained at each point is kind of "glued together" to a single function $G^{-1}:G(\Omega)\to\Omega$. Maybe gluing is not the appropriate word here.

Ben Steffan

yeah, kinda

gluing is indeed not the right word but that's the idea

you already have an inverse function

psie

right

Ben Steffan

the inverse function theorem then tells you that around each point, you find a neighborhood such that the function is $C^1$ there

since this works for every point, and being $C^1$ is a local property, you conclude the inverse as a whole must be of class $C^1$

psie

ah, this makes a lot more sense now

ModularMindset

I just realized today that I am working with a stratified space.

My definitions line up with the actual definitions! :)

Thorgott

22:12

@Derivative if $(B,A)$ is a CW-pair and $f\colon A\rightarrow C$ a cellular map, define a filtration on the pushout $C\cup_fB$ via $(C\cup_fB)_k=C_k\cup_{f_k}B_k$ and show that this gives it a CW-structure

more precisely, the cells of $C\cup_fB$ will be the (canonical images of) the cells of $C$ and the cells of $B$ not in $A$

Claudio

23:14

guys I'm trying to solve this exercise: given a matrix $A \in \mathbb{R}^{m \times n} $ and $\mathbf{b} \in \mathbb{R}^m$ prove that $$g(\mathbf{x}) = \lVert A\mathbf{x}-\mathbf{b} \rVert $$ has an absolute minimum in $\mathbb{R}^n$

Jakobian

@Claudio what is an absolute minimum? A global one?

Claudio

yea

I've showed that $g$ is convex

Jakobian

Hilbert projection theorem

In mathematics, the Hilbert projection theorem is a famous result of convex analysis that says that for every vector x {\displaystyle x} in a Hilbert space H {\displaystyle H} and every nonempty closed convex C ⊆ H , {\displaystyle C\subseteq H,} there exists a unique vector m ∈ C {\displaystyle m\in C} for which ‖ c − x ‖...

should follow from this

Claudio

ok wait

can't use this

I need help in finding a stationary point for $g$

Jakobian

why not

Claudio

23:17

@Jakobian can't use something which isn't covered in the course hahaha

I mean I could

but I know how to solve this, I just need to compute the gradient of $g$, but I can't seem to see how as of now

Jakobian

You can try considering $x\mapsto \|Ax-b\|^2 = (Ax-b)\cdot (Ax-b)$ instead

your function is not always differentiable

but its square is

Claudio

i need to pay attention to matrix products tho

could you expand on these last .two messages

Jakobian

concretely what do you want from me

Claudio

where do we lose differentiability? :p

Jakobian

where is $x\mapsto \sqrt{x}$ not differentiable

or maybe $x\mapsto |x|$

at $0$ of course

Claudio

23:21

oh I see so $Ax = b$

Jakobian

that's where you lose differentiability, something that gets fixed with taking squares

you also can use formulas for differential of dot product

If $f(x) = f_1(x)\cdot f_2(x)$ then $f'(x) = f'_1(x)\cdot f_2(x) + f_1(x)\cdot f_2'(x)$

Claudio

Oh I wanted to expand the dot products instead

thanks for the hints

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$Mathematics$ Mathematics