Mathematics - 2021-02-18 (page 2 of 3)

Astyx

16:00

Ok, so do this: let $x_i$ be the salary of each individual, with the first 50 being the mean (index 1 to 50) and the last 40 women (index 51 to 90)

Rover

Yes ok

Astyx

Write out the means $m_{tot}, m_{men}, m_{women}$ in terms of the $x_i$ and SD $\sigma^2_{tot}, \sigma^2_{men}, \sigma^2_{women}$ in terms of the $x_i$ and $m_i$

@BalarkaSen I'll try watching it if I find the courage lol

Rover

Yes done

Astyx

I mean write them here in the chat

So I can check we're talking about the same things

Balarka Sen

wait do you know who killed laura palmer yet

Rover

16:05

m(total)=590 , m(men)=630 and m(women)=540

Astyx

Huh, I think I remember it's the guy with the truck but I'm not sure

@Rover that's not what I asked

Rover

In terms of x , its difficult to write in chat , can I send a pic?

Balarka Sen

ah ok so its not revealed yet. you should proceed with S2 until that is revealed

(it's not the truck guy)

Astyx

@Rover whatever makes me understand

Oh yeah there's the mysterious guy that shoots the agent at the end of season 1

Balarka Sen

right

name of the giant is the fireman, btw

dont ask me why but its all subliminal

Paul White

16:08

@eryceriousmatherfunker Can I help you with something?

4

Astyx

That's one of the reasons I didn't continue, it started taking a weird alien/ghost/something turn

Balarka Sen

it'll get very surreal by season 3 but thats part of why i enjoy it :)

Ted Shifrin

Howdy, a @Balarka, @Astyx.

Astyx

hi Ted!

Balarka Sen

Hi @Ted.

Rover

16:13

Astyx

Nice handwriting. can you write those with sigmas? It might make things easier

Rover

Yes ok

Astyx

You've said you knew how to find m_tot with m_m and m_w right? So we just need to find s^2_tot in terms of everything else

Rover

Yes

@Astyx done

Astyx

I'm claiming that the hard part about s^2 is the x_i^2 you get when developping the squares, but luckily we can find a way to cancel them

Rover

16:17

Yes..How?

Astyx

We you have them in s^2_women and s^2_men

Rover

@Astyx Yes

Astyx

So I claim you can compute 90s^2_tot - 50s^2_men - 40s^2_women in terms of everything else

Rover

Yes

Astyx

So you're done? :)

Rover

16:24

?

Astyx

What's bothering you?

Rover

@Astyx Why are we doing it ?

Astyx

BEcause then we get s^2_tot = (50s^2_men + 40 s^2_women + result)/90

Rover

ok I will try out as you said

Akiva Weinberger

What is the degree of an order-n cyclic branched cover of a trefoil complement?

Jeff

16:42

How do you show the convergence of $\int_1^{\infty} \cos(x)/x^2 dx$ using either direct or limit comparison tests? Can't use direct because $\cos(x)/x^2$ is not nonnegative and can't use limit because limit of $\frac{\cos(x)/x^2}{1/x^2}$ does not converge.

123

Hi All.

Hello @copper.hat @Thorgott

1drv.ms/u/s!AozWlUoG8z4tihC0oAzv-sVu3WJj

Pls see the link. Where I tick I think there is some mistake in solution. Pls check and clear me.

If differiation is there why they did not add power to sqrt(sec^2 - 1)

Balarka Sen

@AkivaWeinberger What is the "order-$n$" cyclic branched cover? The branching index at the circle is $n$?

Chaos

Hi guys, good afternoon. Hope you can give me an answer for this, in general (see Wikipedia for instance) an "homeomorphism" is defined as a "topological isomorphism", and this of course implies that it's a bijection. In a book about Quantum Mechanics it's stated that a -homemophism (between C-algebras ) is a called *-isomorphism if it's also a bijection.

there are some "*" missing... but I was wondering isn't a homeomorphism a bijection by definition? what's the point of saying "if it's a bijection then it's a *-isomorphism"?

Akiva Weinberger

@BalarkaSen I think yes?

Chaos

is this some physicist convention?

Balarka Sen

16:56

@Akiva So if you take a little meridian around the knot, the covering space above is $S^1 \to S^1$, $z \mapsto z^n$?

Or do I have this wrong

Akiva Weinberger

@Chaos You're confusing "homeomorphism" and "homomorphism", probably

@BalarkaSen That sounds right

Balarka Sen

So why is it not degree $n$

Chaos

@AkivaWeinberger oh shit you are right! ahah thanks!

Akiva Weinberger

@Chaos hom-eo-morphism is a topological isomorphism; hom-o-morphism is any function between groups that preserves the group structure (a group-theoretic morphism)

@BalarkaSen Remember this thingy?

Trefoil knot in KnotPortal

►

Balarka Sen

yeah vaguely

Akiva Weinberger

16:58

That's order-2 because going around any bit near the edge twice gets you back to where you started, but it's degree 6

Thorgott

note, however, that a C*-isomorphism between C*-algebras is automatically a homeomorphism

Balarka Sen

ah yes, so something strange is happening

Akiva Weinberger

Balarka Sen

yeah i see it

Akiva Weinberger

(I think that diagram is missing two more lines through the center)

Balarka Sen

16:59

there are three worlds inside three of those loopies and going around each loopy 1/2 gets you to a different world

so 2*3 = 6

Akiva Weinberger

Order-3 is 24 worlds apparently

Balarka Sen

so let me think

Chaos

@Thorgott that makes sense, thanks! I am getting a bit lost with all this considerations, I work in Stochatic Analysis and I decided to follow a course in Quantum field theory. Bad idea ahaha

Balarka Sen

17:41

@AkivaWeinberger Is it obvious that there is a unique such cover? It seems to me that you're doing the following: Say $T$ is the trefoil, take a Siefert surface $S$ of $T$ in $S^3$. Cut the manifold along $S$ to get two copies of $S$ as boundary now (with corners in $K$), let's say $S_+$ and $S_-$.

Then you take two copies $S^3 \setminus T_1$, $S^3 \setminus T_2$, cut both along their respective Siefert surfaces $S_1, S_2$ and then glue $S_{1, +}$ with $S_{2, -}$ by a $1/3$ rotation (The Siefert surface has a natural action of $\Bbb Z_3$), and $S_{2, +}$ with $S_{1, -}$ by a $1/3$-rotation as well.

Akiva Weinberger

I think the cyclic branched cover is defined to be the "largest" ("universal"?) one

Balarka Sen

Aha, so that makes sense.

Is my description correct? It seems you get $6$ worlds this way.

Maybe don't do $1/3$-twist when you glue $S_{2, +}$ with $S_{1, -}$. Just do it for the first pair.

Akiva Weinberger

I don't think I have the visualization skills for this

Balarka Sen

No way I'm just not communicating properly I think

I'll try again after dinner

Akiva Weinberger

1

Q: What is the degree of an n-fold branched cover over a trefoil?

The order-2 cyclic branched cover over a trefoil has degree 6, meaning the preimage of any point off the trefoil has cardinality six. (You can find a wonderful video of this here, made by Moritz Sümmermann.) The order-3 cyclic branched cover over a trefoil has degree 24. What is the formula for t...

I asked on MSE

Snapdragon-X

18:03

22 hours ago, by robjohn

@Snapdragon-X The contrapositive of $A\ne0\implies\left(\exists x\in\mathbb{R}^n:Ax\ne0\right)$ is $\lnot\left(\exists x\in\mathbb{R}^n:Ax\ne0\right)\implies\lnot(A\ne0)$

Balarka Sen

@AkivaWeinberger Back. Want to understand my construction?

Snapdragon-X

@robjohn my question is why can I not say so...I mean it sounds good that

there exists no A such that A not equal to 0....doesnt this imply that A has to be zero?....I can't get a good grip over here

Akiva Weinberger

No I have class @BalarkaSen

Sorry

Balarka Sen

k

Snapdragon-X

@Snapdragon-X When I asked a TA, they said I cant make claims on existence....but isnt saying A=0 also a claim of existence that A will exist and be 0?

Hi Ted

You asked me not to use not too much because not not yes

but I don't get the logic here...:)

Ted Shifrin

18:08

Huh?

Snapdragon-X

22 hours ago, by Ted Shifrin

@Snapdragon-X: Better form is not to put so many "not"s. Not for every x, P(x) is not useful. Better to say there exists an x so that P(x) fails.

Ted Shifrin

Yes, I remember what I said. But your recent sentence is crazy.

Snapdragon-X

notnotnotnot yes

robjohn

@Snapdragon-X You are not looking for the existence of $A$. $A$ is a given operator. Saying $\lnot\exists A$ is not meaningful

Ted Shifrin

It I want to negate the sentence "Every person in this chatroom is from Croatia" I am saying that saying "Not every person in this chatroom is from Croatia" isn't useful.

Thorgott

18:10

can you prove that not every person in this chatroom is from Croatia?

Ted Shifrin

It is useful to say "There is some (at least one) person who is from India." Or "There is at least one person who is from Germany."

Yes @Thor

Snapdragon-X

I see, now I somewhat get it....so by saying "whatever i said" I may also mean that A doesnt exist? right?...

@robjohn Oh.. Okayy

Ted Shifrin

The contrapositive should be "If $Ax=0$ for all $x$, then $A=0$."

robjohn

22 hours ago, by robjohn

So the contrapositive is equivalent to $\left(\forall x\in\mathbb{R}^n,Ax=0\right)\implies A=0$

@TedShifrin indeed

Snapdragon-X

Yeah I get it...Thank you @rob and @Ted !

Ted Shifrin

18:13

Shocking. @robjohn and I agree for once. And @Thor and I have agreed several times. I am redundant.

robjohn

@TedShifrin come back tomorrow, maybe we'll need you then ;-)

Ted Shifrin

Maybe.

Snapdragon-X

May there be be.

robjohn

To be be or not to be be...

Do be be, do be be, do

Ted Shifrin

moans because it is too early in the day for a martini

Astyx

18:16

You're orange

Snapdragon-X

$\forall \text{to} \in \text{be} \exists \text{be}$

Thorgott

revolting!

Ted Shifrin

@Astyx Who?

robjohn

@TedShifrin actually, we don't agree totally: the contrapositive is equivalent to what I said, but...

22 hours ago, by robjohn

@Snapdragon-X The contrapositive of $A\ne0\implies\left(\exists x\in\mathbb{R}^n:Ax\ne0\right)$ is $\lnot\left(\exists x\in\mathbb{R}^n:Ax\ne0\right)\implies\lnot(A\ne0)$

Astyx

robjohn

Ted Shifrin

18:19

I'm talking about useful negations, a concept I had to convey every time I taught "Intro to Higher Math."

Astyx

the song goes "i'm blue, da be dee da be da", not "I'm orange da be dee da be da"

Ted Shifrin

hides

Balarka Sen

@Thorgott Easy by pigeonhole principle, since Croatia has like a population of 10.

Ted Shifrin

a @Balarka: Not true. I was there a few years ago and I can prove that's false.

Thorgott

LOL

robjohn

18:20

@TedShifrin it was false...

Ted Shifrin

LOL, point taken.

Balarka Sen

thanks for backing me up robjohn

robjohn

. o O ( gotta stop being so precise )

Snapdragon-X

@robjohn but you don't know if Ted's in Croatia

Astyx

Is everyone from Croatia in this chatroom?

robjohn

18:22

@Astyx how did you guess?

Balarka Sen

@Astyx Asking the right question

Ted Shifrin

So I wonder if those same people have posted more differentiable manifolds/geometry takehome exam questions.

robjohn

@TedShifrin where cheating is legal? (I call them Trump exams)

Ted Shifrin

I think that, sadly, there are far more cheaters than just those in his sphere.

robjohn

True

Astyx

18:24

Did Trump ever take an exam?

Thorgott

Is this chatroom in croatia?

robjohn

but the wounds have not healed yet.

@Astyx he probably had one when he had Covid

Balarka Sen

@Thorgott Hold up you're changing the category of the embeddability problem

Astyx

Man woman person camera TV

Thorgott

dual problem

geocalc33

18:27

I've never been to Croatia but I'm from Croatia

Balarka Sen

That is logically impossible, by reversal of the direction of time

robjohn

@TedShifrin yes, but do they consider the cheating they do to be legal?

Ted Shifrin

Clearly, these students seem to.

robjohn

@TedShifrin That is why... never mind.

Ted Shifrin

Is there any actual mathematics in this room? :D

robjohn

18:30

What?! you expect math in this chatroom?

Why?

Balarka Sen

I gave a description of a branched cover of $S^3$ branched over the trefoil knot a few messages up

At least, I'd like to think I did

Ted Shifrin

If it required too much visualization skill for DogAteMy, it's well beyond my pay grade.

Thorgott

I did some math in here like yesterday

Ted Shifrin

How was it like yesterday?

Thorgott

it was a different time back then

those were the days

Balarka Sen

18:34

Here's some simple math then: If you consider the family $f_t : \Bbb R^2 \to\Bbb R$, $f_t(x, y) = x^3 - tx + y^2$ then $F : \Bbb R^3 \to \Bbb R$, $F(x, y, t) = f_t(x, y)$ has a gradient flowline which connects through all the critical points of $f_t$ on the slices $\Bbb R^2 \times \{t\}$.

Ted Shifrin

Too hard for Ted.

geocalc33

should we have a simple math day where we just do very simple math and everyone understands and is happy

Balarka Sen

I disagree, more likely I did not communicate properly

@geocalc33 everyone should only do simple math

Astyx

The hard part is understanding why the simple math is simple

Thorgott

I'm too bad at simple math

Ted Shifrin

18:37

@Thor: Sounds like Nana Mouskouri. I was trying to find a YouTube of her recording but haven't succeeded.

robjohn

@TedShifrin NSFT

Ted Shifrin

A curve passing through the point?

@robjohn Also too hard for Ted. ??

Astyx

I read that wrong

robjohn

@TedShifrin I just extended it to "Not Safe For Ted"

Astyx

Why does it connect the critical points?

Balarka Sen

18:41

@Astyx I mean to say, the critical points of $f_t$ for $t > 0$ are $(\pm t^{1/2}/3, 0)$, which merge together at time $t = 0$ and then for $t < 0$ there are no critical points

Then I think that $(\pm t^{1/2}/3, 0, t)$ is a flowline of $F$

Astyx

18:51

Hmm that's not what I get

I'll check after dinner

Ted Shifrin

Bon appétit!

Balarka Sen

@Astyx It's because I messed up, the critical points are at $\pm (t/3)^{1/2}$

Enjoy!

Astyx

19:19

I think I'm missing something

The gradient in $\Bbb R^3$ at a critical point of $f_t$ in $P_t = \Bbb R^2\times\{t\}$ is always going to be orthogonal to $P_t$

So if there is a gradient flowline connecting the critical points, they're going to form a line perpendicular to the xy plane

@BalarkaSen

In fact that's a characterization of critical points

schn

@robjohn Regarding my problem from yesterday, $\int_{\mathcal{C}} \frac{e^{-izt}}{z+iA+B} dz$, where $\mathcal{C}$ is the D lower contour and we are integrating clockwise. Consider the attached screenshot again. You were saying clockwise integration inserts a minus sign when using the Residue theorem. Is then equation (8) not correct? What is meant by the minus compensates?

We would get $$\int_{\mathcal{C}} \frac{e^{-izt}}{z+iA+B} dz=\int_{-R}^R \frac{e^{-ixt}}{x+iA+B} dx+\int_{\mathcal{C_R}} \frac{-iRe^{-iRe^{-i\phi}t}}{Re^{-i\phi}+iA+B} d\phi$$ using clockwise integration, where $\mathcal{C_R}$ is the lower half circle of radius $R$.

The second integral is bounded by Jordan's lemma and presumably goes to 0 for R to infinity.

robjohn

@schn I think that they are wrong. The residue at $z=-B-iA$ is $e^{(iB-A)t}$, so the integral should be $-2\pi ie^{(iB-A)t}$

schn

@robjohn Thanks for the reply. So presumably the "minus sign compensating" does not make a lot of sense?

robjohn

19:35

@schn Yeah, I even double checked with Mathematica and the residue is what I said.

Ted Shifrin

Integrating clockwise?

robjohn

@TedShifrin yep

downward bulging D

Ted Shifrin

Still don't know why you would integrate clockwise.

robjohn

@TedShifrin to make the exponential vanish, we want to go into the lower half plane with the circular contour

Ted Shifrin

Oh, but the integral along the real axis is still left-to-right. OK.

robjohn

19:38

yeah

Ted Shifrin

I haven't been paying attention, of course.

user2103480

@TedShifrin you have nothing to lose but your chains

Ted Shifrin

Is that a literary allusion, @user2103480?

geocalc33

do two distinct real analytic space curves $s_1,s_2 \in \Bbb R^3$ when analytically continued, lie in the same complex plane?

robjohn

Workers of the world, unite!

The political slogan "Workers of the world, unite!" is one of the most famous rallying cries from The Communist Manifesto (1848) by Karl Marx and Friedrich Engels (German: Proletarier aller Länder vereinigt Euch!, literally "Proletarians of all countries, unite!", but soon popularised in English as "Workers of the world, unite! You have nothing to lose but your chains!"). A variation of this phrase ("Workers of all lands, unite") is also inscribed on Marx's tombstone. The essence of the slogan is that members of the working classes throughout the world should cooperate to defeat capitalism and...

user2103480

19:40

@TedShifrin if you call the communist manifesto prose

then sure

Ted Shifrin

Oh. Why is this relevant?

I was not complaining of political oppression. Just complaining about horrid humo(u)r in here.

user 726941

Meanwhile, in the real world:

U.S. life expectancy plummeted in 2020 and the drop was particularly steep for Black Americans, with life spans declining by 2.7 years. And Black and Latino Americans are not getting equal access to vaccines.

2

user2103480

@TedShifrin it was a joke about not drinking early being a form of oppression

Ted Shifrin

Yes, access to vaccines is still a crapshoot at best.

OK, OK, @user2103480.

robjohn

The Griffith Observatory is going to be covering the Perseverance landing starting in 30 minutes. I'll be watching that and I might be very slow to respond during.

@TedShifrin Here, people over 65 haven't had too much trouble getting the vaccine. Cal State Northridge and Dodger Stadium have been the two major mass sites.

I don't know when I will be able to get mine (not over 65 yet).

Ted Shifrin

19:47

Well, my duly designated #2 was delayed 8 days because of non-delivery of vaccine. I lucked out and got into CVS before they were overrun. But for POC and for elderly without computers and/or transportation, this is grim. And then Texas ...

robjohn

@TedShifrin Yeah, the need for a computer to sign up is bad for people without

Ted Shifrin

Yeah, I should watch that. Somehow I'm not as excited as I was with the moon landing in 1969.

robjohn

@TedShifrin I am of the mind that they are rushing human exploration of Mars too soon. I think we may have a number of corpses on Mars, if the corpses even make it that far.

Ted Shifrin

Well, might be more exciting than dying of COVID-19.

robjohn

possibly

geocalc33

19:51

anyone have any feedback on my question?

robjohn

@geocalc33 about easy math day?

12 mins ago, by geocalc33

do two distinct real analytic space curves $s_1,s_2 \in \Bbb R^3$ when analytically continued, lie in the same complex plane?

oh

do the curves in $\mathbb{R}^3$ lie in the same plane there?

geocalc33

no

Ted Shifrin

What do you mean by "complex plane"?

robjohn

Then, why would they lie in the same complex plane?

Ted Shifrin

Even if they were in a real plane, the complexification would live in a $\Bbb C^2$, not a $\Bbb C$.

robjohn

19:55

@TedShifrin I was going to ask that next ;-)

geocalc33

really?

Ted Shifrin

Try any example other than a line.

geocalc33

okay I get it

Balarka Sen

@Astyx I didn't understand what you said but let me try to compute. The $y$ part is a filler, so just look at $f_t(x) = x^3 + tx$. For $t = -1$ we have two critical points which are destroyed at $t = 0$ in pairs and by $t = 1$ we have no critical points. So I guess what do the flowlines of $F : \Bbb R^2 \to \Bbb R$, $F(x, y) = x^3 + xy$ look like? They're solutions to $x'(s) = 3x(s)^2 + e^s$ and $y(s) = e^s$. That looks like an awful ODE

Astyx

Isn't the flow line defined such that the tangent at every point is colinear to the gradient?

Balarka Sen

20:05

Yeah

Which is the equation I wrote down

Astyx

Do you agree that the gradient at a critical point is orthogonal to the xy plane?

Balarka Sen

I don't understand what that means. There is a family of functions here, then the movie function F, then critical points of the family of functions, and the xy plane is the domain of F

geocalc33

so @TedShifrin and/or @robjohn Do either of you happen to know whether you can "glue" together these complexifications?

Astyx

I mean that if you compute the gradient of F at a critical point $(x_t, y_t, t)$, it's going to be along the t direction

Balarka Sen

I am having a very hard time understanding notation but I also see your point. Sorry for being a dumbass, but let me try to show you a picture.

Thorgott

20:16

I like that you apologize in advance for showing a picture

Balarka Sen

grapher.mathpix.com/?latexList=%5B%22z%3Dx%5E3%2Bxy%22%5D

Astyx

My notation sucks dw

Balarka Sen

This is the graph of the movie of the homotopy $f_t(x) = x^3 - tx$.

There is an obvious gradient flowline on the graph aka a path of least steep descent: the part of the graph over the parabolic trajectory

Am I wrong?

This seems to connect through all the critical points of the slices of the graph

user2103480

@TedShifrin i find it cute how mathematicians talk about the objects they work with

"this guy lives in"

"the model believes that"

Astyx

Which parabolic trajectory?

user2103480

20:21

heckin platonists

Balarka Sen

I see, this graph is perhaps not ideal. I really should do path of steepest decent, but for that the graph should be a little more spread out

Thorgott

@user2103480 meanwhile algebraists: "this map kills the kernel"

user2103480

@Thorgott brutal

Balarka Sen

@Astyx grapher.mathpix.com/…

geocalc33

Take a real analytic surface embedded in $\Bbb R^3$ ,homeomorphic to a sphere, and analytically continue each real analytic function (the union of which equals the surface) to the complex space...When if ever does the union of extended analytic functions form a complex manifold?

Astyx

20:24

algebraists like to kill ideals as well

Thorgott

one of the profs uses that phrase regular

geocalc33

so each analytically continued function lives in $\Bbb C^2$

Thorgott

he would literally use the word "kill" even tho the lecture was in german, which makes it even better

user2103480

@Astyx obligatory "radicals" and "blowing up points in a plane" remark

Me in scholarship application:
"The interdisciplinary cooperation and communication in the courses that I took outside of my subject was a challenge at first."

Me in reality:
"$\varphi_{k}(t):=e^{i k \cdot x(t)}=\int e^{i k y} p(t, y) d y$"? Wtf thats not how this works stop it

Astyx

@BalarkaSen I don't get what you're trying to say

Balarka Sen

20:26

@geocalc33 Very rarely it'll be a manifold. You seem to be describing, for example, real links of complex algebraic varieties.

geocalc33

@BalarkaSen cool. What do you mean by "links?" As in linking two things together?

Balarka Sen

@Astyx I don't know how else to explain it. Look at the curve $(t, -3t^2, -2t^3)$ on the graph of $z = x^3 + xy$. I am saying this is a gradient flowline, i.e., direction of steepest descent. This is not quite true, but I think an obvious modification works.

Astyx

My point is that if there is a gradient flowline connecting critical points, then it's a straight line along the t coordinate

Balarka Sen

There's some problem with the formalism I did, then, because that is nonsense. There are plenty of gradient flowlines between critical points typically, for a function on a manifold

OK, final picture:

grapher.mathpix.com/…

That curve is always in the direction of steepest descent, right?

Astyx

Oh wait

You don't mean that the flowline consists only of critical points?

Balarka Sen

20:40

But my curve in the picture above does, right? $(t, -3t^2)$ are all critical points of $f_y(x) = x^3 + yx$ where $y = 3t^2$. So that's why I am confused.

Because differentiate: $f_y'(x) = 3x^2 + y = 0$ if $x = t$, $y = -3t^2$.

anakhro

Anyone good with elliptic functions present?

user2103480

@MikeMiller Exam advice time again. What are the topological properties that can be relatively straightforwardly computed via homology? E.g. I've seen these things:
1. Jordan curve & generalizations via computing reduced 0th homology
2. Showing an embedding of S^n into S^n is an iso via showing $\tilde{H}_{-1}(S^n\setminus f(S^n)) = \Bbb Z$
3. Brouwer via functoriality
4. Orientability via n-th homology
5. Uhh does one ever use that first homology is the abelisation of the fundamental group?

Brouwer is arguably not a topological property but I hope you get what I mean. Types of problems.

Mike Miller

Yes one uses (5). Invariance of domain as well, but that's a variant of the ideas in the JCT stuff

(2) seems ridiculous to me

The idea is invariance of domain, injections of manifolds of the same dimension are open maps

user2103480

@MikeMiller It is, but if there is a given formula, that can be used

Thorgott

my mans computing -1st homology

user2103480

20:54

@MikeMiller Ah!

Mike Miller

I mean given that your statement is literally the same as $f(S^n) = S^n$ I am not sure how you get there

Must be (1) somehow I guess

Balarka Sen

What is $H_{-1}$

Mike Miller

Seems nuts, no content

Balarka Sen

What the hell

Thorgott

-1st homology, duh

anakhro

20:54

Means n-1?

user2103480

We used the previously proven formula

Mike Miller

No it's not -1st homology

user2103480

reduced homology bruhs

Mike Miller

That's always zero

But reduced homology sticks a guy in degree -1

Astyx

Mike Miller

20:55

The statement is that $\widetilde H_{-1}(X;\Bbb Z) = \Bbb Z$ iff $X = \varnothing$

It's the kind of shit a German would write

user2103480

$\tilde H_{n-r-1}(S^n \setminus f(S^r)) = \Bbb Z$ is what we used, for an embedding of $S^r$

Mike Miller

You mean $\Bbb Z$ in degree $n-r-1$ lol

I dunno how you prove that I never remember the JCT stuff

user2103480

fucc yes

Mike Miller

Alexander duality is the right way to think about it

@user2103480 Good notation is $\Bbb Z[n-r-1]$

Thorgott

:deepfriedjoy:

Balarka Sen

20:57

Z^(n-r-1)

Wtf

H_-1(S^n - S^n) = Z^(n-r-1)

user2103480

homolgye

Astyx

@BalarkaSen blue/orange gradient is the density plot of the function, arrows is gradient vector field, blue line is critical points

As I claimed, the gradient vector field on the blue line is along the t-axis (arrows are vertical)

user2103480

@MikeMiller to come back to this

Balarka Sen

@Astyx I agree, but how do you explain that 3D plot then?

I'm very confused

Astyx

What do you mean?

This is the same thing as the 3D plot

Balarka Sen

20:59

grapher.mathpix.com/…

I'm saying here the curve I drew is obviously a curve along which you go downhill the most

Right?

I mean

user2103480

Am I understanding correctly that statements such as "homeomorphic subsets of $\Bbb R^n$ are both open iff one of them is open" and the same for $S^n$ are special cases of this?

Mike Miller

Yes

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$Mathematics$ Mathematics