Mathematics - 2020-05-04 (page 1 of 2)

Joe Shmo

00:24

if you'd like, we can formally restrict ourselves to $H$ compact for starters

Ted Shifrin

00:41

Hi Stan

Joe Shmo

Ted, did you enjoy being quoted?

Ted Shifrin

LOL, I did notice.

Joe Shmo

just making sure ;)

Ted Shifrin

Did that absolve you of your guilt?

Joe Shmo

it did, I passed on the karma to someone new now

Thorgott

00:45

man, grading homework online isn't all that fun

Ted Shifrin

How is the online homework presented?

Thorgott

some are nice enough to use $\LaTeX$, others scan their work

Ted Shifrin

So you have to print it out and mark it with pen?

Thorgott

no, I just look over it, enter the points into the system and write a comment in a comment box if needed

I miss being able to mark stuff with pen and write pedantic remarks in every nook and cranny

Ted Shifrin

Oh, so not that arduous.

Yeah, even when students sent me scanned homework, I typically printed it out and graded it as usual. But those were the old, old days.

Thorgott

00:48

yeah, it's not difficult work (which I'm glad for), but also not particularly fulfilling

also suboptimal for the students, cause the feedback isn't that precise anymore

at least for those students that appreciate detailed feedback

Joe Shmo

hot take: one day everyone will be submitting their homework in Agda 😈

Ted Shifrin

I agree. And, surprisingly, a number of my students over the year commented in letters afterwards about how much my careful grading helped them and improved them.

What's Agda?

Joe Shmo

a proof verifier

Thorgott

yeah, that's great

it's those students that make it worth

Balarka Sen

01:37

Quiet

I woke up too early. Have to decide what to do

Thorgott

draw some diagrams

Balarka Sen

03:14

Let $(K, d)$ be a cochain complex filtered by subcomplexes $(K_p, d)$, $K_{p+1} \subset K_p$. Then the exact sequence $0 \to \bigoplus_p K_p \to \bigoplus_p K_p \to \bigoplus K_p/K_{p+1} \to 0$, first map raising grade by $1$, gives rise to an exact couple $\bigoplus_{n, p} H_n(K_p) \to \bigoplus H_n(K_p) \to \bigoplus H_n(K_p, K_{p+1})$ which we iterate to get the spectral sequence. $A^1_{n, p} = H_n(K_p)$, $E^1_{n, p} = H_n(K_p, K_{p+1})$.

An element of $E^1_{n, p}$ has a representative $\alpha \in K_{n, p}$ such that $d\alpha \in K_{n+1, p}$, and $d_1 : E^1_{n, p} \to E^1_{n+1, p}$ is exactly $d_1([\alpha]) = [d\alpha]$. Iterating once more, an element of $E^2_{n, p}$ has a representative $\alpha \in K_{n, p}$ such that $d\alpha = d\beta + \gamma \in K_{n+1, p}$ where $\beta \in K_{n, p}$ and $\gamma \in K_{n+1, p+1}$. $d_2 : E^2_{n, p} \to E^2_{n+1, p+1}$ is exactly $d_2[\alpha] = [\gamma]$.

In general, $d_r : E^r_{n, p} \to E^r_{n+1, p+r-1}$ should be obtained as follows. Pick a representative $\alpha \in K_{n, p}$ of a class $[\alpha] \in E^r_{n, p}$. Then $d_{r-1}([\alpha]) = 0$ should translate to a staircase equation:

$d\alpha = d\beta_1+ \alpha_1 \in K_{n+1, p}$ where $\alpha_1 \in K_{n+1, p+1}$, $\alpha_1 = d\beta_2 + \alpha_2 \in K_{n+1, p+1}$ where $\alpha_2 \in K_{n+1, p+2}$, ..., $\alpha_{r-2} = d\beta_{r-1} + \alpha_{r-1} \in K_{n+1, p+r-2}$ where $\alpha_{r-1} \in K_{n+1, p+r-1}$.

And we set $d_r[\alpha] = [\alpha_{r-1}]$.

Should be essentially correct.

Balarka Sen

03:49

Ah, I suppose I switched the direction of the differentials. It's OK

Nah, this is fine.

Balarka Sen

05:46

Let's do this with a concrete example, homological indexing this time. $K_{n, p} = C_n(X_p)$, for some filtration $\cdots X_{p-1} \subset X_p \subset \cdots$ of spaces. $A^1_{n, p} = H_n(X_p)$, $E^1_{n, p} = H_n(X_p, X_{p-1})$, $$d_1 : H_n(X_p, X_{p-1}) \to H_{n-1}(X_{p-1}) \to H_{n-1}(X_{p-1}, X_{p-2})$$ the cellular boundary map, defined by $d_1[\alpha] = [\partial \alpha]$, where the absolute cycle $\partial \alpha \in C_{n-1}(X_{p-1})$ is thought as a relative cycle in $(X_{p-1}, X_{p-2})$.

If $[\alpha]_2 \in E^2_{n, p}$, $d_1[\alpha]_1 = 0$ translates to $\partial \alpha = \partial \beta_1 + \alpha_1 \in C_{n-1}(X_{p-1}, X_{p-2})$ for some $\beta_1 \in C_{n-2}(X_{p-1})$ and $\alpha_1 \in C_{n-1}(X_{p-2})$. Then $\partial \alpha_1 = 0$ so defines an absolute cycle.

We'd like to set $d_2[\alpha]_2 = [\alpha_1]_2$, should be easy to check if this is well-defined.

I meant $\beta_1 \in C_n(X_{p-1})$, whoops.

Say $\partial \alpha = \partial \beta_1 + \alpha_1 = \partial \beta_1' + \alpha_1'$ hence $\alpha_1 - \alpha_1' = \partial(\beta_1' - \beta_1)$, which means $\beta_1' - \beta_1 \in C_n(X_{p-1}, X_{p-2})$ is a relative cycle hence gives rise to an element in $H_n(X_{p-1}, X_{p-2})$ and $[\alpha_1 - \alpha_1']_1 = d_1[\beta_1' - \beta_1]$, so $[\alpha_1]_1 = [\alpha_1']_1$ and we're through

$d_1[\alpha_1]_1 = 0$ because $\partial\alpha_1 = 0$, therefore $[\alpha_1]_2$ is a well-defined class as well. OK

Leaky Nun

06:04

@BalarkaSen aha I caught you at the right moment

Balarka Sen

@LeakyNun Figured out the higher differentials, time to figure out Euler class

Haha yeah

Leaky Nun

tldr?

Balarka Sen

Take a filtration of spaces $\cdots \subset X_{p-1} \subset X_p \subset \cdots$ and run specseq on $C_n(X_p)$. Then $d_r : E^r_{n, p} \to E_{n-1, p-r+1}$ can be understood as follows: Take any class $[\alpha]_r \in E^r_{n, p}$ represented by $\alpha \in C_n(X_p)$.

Leaky Nun

why isn't that guy fxxxxxx banned?

Balarka Sen

Sorry I meant $d_r : E^r_{n, p} \to E^r_{n - 1, p - r}$. Got confused by wrong indices for a second

The equation $d_{r-1}[\alpha]_{r-1} = 0$ can be unraveled to be the following sequence of equations: $\partial \alpha = \partial \beta_1 + \alpha_1 \in C_{n-1}(X_{p-1}, X_{p-2})$ where $\beta_1 \in C_n(X_{p-1})$ and $\alpha_1 \in C_{n-1}(X_{p-2})$, $\alpha_1 = \partial \beta_2 + \alpha_2$ where $\beta_2 \in C_n(X_{p-2})$ and $\alpha_2 \in C_{n-1}(X_{p-3})$, so on and so forth until $\alpha_{r-1} = \partial \beta_{r-1} + \alpha_{r-1}$ where $\alpha_{r-1} \in C_{n-1}(X_{p-r})$

$\alpha_{r-1}$ can be checked to define a well-defined class in the $E^r$ page. $[\alpha_{r-1}]_r \in E^r_{n-1, p-r}$, and we set $d_r[\alpha]_r = [\alpha_{r-1}]_r$

Akiva Weinberger

06:17

Find all solutions of $y^2-xy-x^2=\pm1$ for integer $x,y$.

Balarka Sen

This is for $r > 1$. For $r = 1$ we all know what $d_1$ is, just the cellular boundary map, $d_1[\alpha]_1 = [\partial \alpha]_1$

I am pretty sure this will give the obstruction theory interpretation of the Euler class

Leaky Nun

@AkivaWeinberger $\Bbb Z[\phi]$ is UFD

Akiva Weinberger

And therefore? @LeakyNun

Balarka Sen

Therefore anyone can do it

Leaky Nun

I guess it doesn't matter

the unit group matters more

Balarka Sen

06:20

Unit group of $\Bbb Z[\sqrt{d}]$ for $d > 2$ is pretty simple

Leaky Nun

ibid. the unit group is generated by $-1$ and $\phi$

does this mean Fibonacci time

Akiva Weinberger

Oh, correction: for all positive integer $x,y$. Otherwise the solution isn't as nice

Leaky Nun

the conjugate of $\phi$ is $1-\phi$ I think

Balarka Sen

Hi @Ted

Ted Shifrin

Howdy.

Balarka Sen

06:23

Figuring out transgression homomorphisms

Leaky Nun

$(a+b\phi)\phi = a\phi+b(1+\phi) = b+(a+b)\phi$

Akiva Weinberger

I'm gonna sleep. Ping me with the answer lol

Leaky Nun

yeah so they're the Fibonacci numbers

let's try 5 and 8

$8^2-5 \times 8 - 5^2 = 64 - 40 - 25 = -1$

@AkivaWeinberger yeah so adjacent Fibonacci numbers

$y = F_{n+1}$, $x = F_n$

@BalarkaSen it's like a $\delta$-expansion

your favourite: Taylor series

Balarka Sen

That's exactly how people think of spectral sequences

Leaky Nun

@TedShifrin why isn't that guy fxxxxxx banned

@BalarkaSen can we work with an example for good measure

Akiva Weinberger

06:30

@LeakyNun Why are those the only solutions

Leaky Nun

@AkivaWeinberger because the equation is $N(y-x\phi) = 1$

so $y-x\phi$ is in the unit group

(or you can use Fermat descent by multiplying by $\phi$ or $\phi^{-1}$

i.e. if $(x,y)$ is a solution then so is $(y-x,x)$ etc

Balarka Sen

@LeakyNun Well, that's what our goal is: To understand the Gysin transgression in this context

Leaky Nun

right

Balarka Sen

Good example to figure out

I'm thinking

@LeakyNun rekt

i imagine this must be part of the appeal of number theory

to rekk people when they give you diophantine equations

Leaky Nun

precisely

Balarka Sen

06:39

So much information man, I'm getting lost in pinning down what's happening in the Gysin transgression

Akiva Weinberger

So for $n>0$, $n$ is a Fibonacci number iff $\exists x,(n^2-nx-x^2)^2=1$

where $x$ ranges over the integers (I think I don't need to specify positive)

EnjoysMath

0

Q: What is a subset of a ring that mult. absorbs the whole ring, and is closed for all differences except finitely many?

Let $M$ be a subset of the ring $\Bbb{Z}$ say, and suppose that $rM \subset M$ for all $r \in \Bbb{Z}$ and that there are only finitely many values $x - y$ that are not in $M$ over all the $x,y \in M$. Is such a thing a union of ideals? Looking at $2\Bbb{Z} \cup 3\Bbb{Z}$ we have that $1 = 3 - ...

Okay, got a good one for you all

Leaky Nun

@BalarkaSen rip

EnjoysMath

This is related to twin primes, though I held back stating it in the post

Leaky Nun

I don't need to see the post to know this

Balarka Sen

06:41

Lol

EnjoysMath

Suppose that you have a subset $M \subset \Bbb{Z}$

and $\Bbb{Z}M \subset M$

and $x,y \in M \implies x - y \in \Bbb{Z}\setminus M$ happens only for finitely many values $x - y$

It's almost an ideal

In fact a generalization of ideal since you take finitely many to mean $0$ and you get an ideal structure

I'm hoping it's something different than $\{ x \in \Bbb{Z} : |x| \geq n\}$

Balarka Sen

So the bundle is $S^{n-1} \to E \to B$ and $H_n(B; H_0(S^{n-1}))$ is the $E^{2=n}_{n, n}$-th term, so this would be represented by something in $C_n(\pi^{-1} B^{(n)})$ since we filtered by $X_p = \pi^{-1} B^{(p)}$

I take it's boundary, which goes inside $C_{n-1}(\pi^{-1} B^{(n)}, \pi^{-1} B^{(n-1)})$, and extract out the $C_{n-1}(\pi^{-1} B^{(n-1)})$ term.

Oh I see I am bad

I wrote the staircase equation wrong. Thought something was wrong.

Let's do one more iteration. $d_2[\alpha]_2 = 0$ implies $[\alpha_1]_2 = 0$, so $[\alpha_1]_1 = d_1[\gamma]_1 = [\partial \gamma]_1$

Oh this is fine. Then $\alpha_1 - \partial \gamma$ vanishes in $H_{n-1}(X_{p-2}, X_{p-3})$, so must be a relative boundary ie $\alpha_1 - \partial \gamma = \partial \eta + \delta$

$\delta \in C_{n-1}(X_{p-3})$. This is our $\alpha_2$

OK, everything is correct

Leaky Nun

07:12

@BalarkaSen imagine winning your own tournament

Leaky Nun

07:50

"Magnus Carlsen wins the Magnus Carlsen Invitational brought to you by Magnus Carlsen"

geocalc33

I'm quitting math for 2 days

CG.

Hi, I'm looking through old exams and have a few geometry questions (true/false type questions). The solution manual for exams prior to 2015 are not available, so I have a few questions. Are they on topic here?

In Neutral geometry: If there exists a triangle with angle sum = 180 degrees, then all triangles will have a circumscribed circle.

Manjoy Das

08:24

Ok i have a question.. in math.stackexchange i found that a polynomial can never have a negetive degree unless it is a zero polynomial.. so $ \dfrac 1x$ should not be a polynomial in $ x$. So why do we write $ f(x)=\dfrac 1x$?

Why do we say that $ \dfrac 1x$ is a polynomial in x?

Alessandro Codenotti

10:53

@ManjoyDas We don't, that's a rational function in $x$, or just a function

geocalc33

11:06

Possible to convolute the integral curves of two vector fields to create a new vector field?

1

Q: How do you show that the Lorentz metric is preserved for $\zeta^{3,1}?$

The following is a visualisation of a Lorentz transformation: https://www.desmos.com/calculator/u5qpd135uc. The lines of constant time in Minkowski space, $\Bbb R^{1,1},$ are hyperbolas: $$ f_T(x)=\frac{T}{x}$$ Where $T$ generates the set of curves. $T=\{t^2:t\in \Bbb R \cap(0,\infty)\}.$ Th...

special-relativity

Stupid Kid

12:17

@geocalc33 quitting math is good for mental health for 2 days lol

Rudi_Birnbaum

Hi all, I have posted here (a what I considered simple little question) math.stackexchange.com/questions/3658124/…, it got for whatever reasons bad ratings.

Stupid Kid

@Rudi_Birnbaum When is elementary number theory taught in undergraduate. My first and second yr course has not include it.

Rudi_Birnbaum

I have no idea, is it at all?

The term "prime number" is probably "elementary number theory" like gcd etc...

Stupid Kid

@Rudi_Birnbaum Is it included in math foundation.

Rudi_Birnbaum

What is "math foundation"? Sorry I went to school in Germany.

Stupid Kid

12:25

Well then I think I shall not continue my conversation since we might misunderstand each other.

Rudi_Birnbaum

Well this is usually the point when one should start the conversation, rather then stop it, isn't it?

Stupid Kid

I will try to understand the problem you posted. I got like 5 min break.

Balarka Sen

@LeakyNun I see what this map has to do with Euler class. The picture is vague but it's absolutely certain.

Rudi_Birnbaum

@StupidKid Thanks!

Perturbative

For Stiefel-Whitney classes, it's said that one of the immediate consequences of the axioms is that for isomorphic vector bundles $\xi$ and $\eta$, it follows that $w_i(\xi) = w_i(\eta)$, but isn't it the case that there's just some isomorphism $f^* : H^i(B(\xi), \mathbb{Z}/2) \to H^i(B(\eta), \mathbb{Z}/2)$ for which $f^*(w_i(\xi)) = w_i(\eta)$?

Like they're equal up to isomorphism but not literally equal

Balarka Sen

12:46

@Perturbative Surely the statement is for vector bundles over the same base, isomorphism being isomorphism which restricts to identity on the base

Perturbative

Ah I didn't realize that the statement was only for vector bundles over the same base

Stupid Kid

@Rudi_Birnbaum I think jam is right. You should write the series as double product to make it more clearer.

Perturbative

Oh well I guess I should've realized that

Balarka Sen

If you want to get a general statement I think you'd just look at the Stiefel-Whitney numbers; for any collection of indices $i_1, \cdots, i_k$ with $i_1 + \cdots + i_k = \dim(M)$, the Stiefel-Whitney number of the vector bundle $E/M$ with respect to those indices is $(w_{i_1} \smile w_{i_2} \smile \cdots \smile w_{i_k})[M]$

This gives an actual number rather than cohomology classes. If you have two vector bundles which are isomorphic (not fixing base now), the Stiefel-Whitney numbers are the same

This makes it base-free

Perturbative

Thanks for that, I'll take a bit of time to digest that though

vesii

12:58

Hi guys, why is the following limit is not corrent:
$\lim_{n\rightarrow\infty}\left(\frac{1^{2}+2^{2}+...+n^{2}}{n^{3}}\right)=\lim_{n\rightarrow \infty} \left(\frac{\frac{1^2}{n^2}+\frac{2^2}{n^2}+...+1}{n}\right) =\lim_{n\rightarrow \infty} \left(\frac{0+0+...+1}{\infty}\right)=0$

feynhat

The second equality is garbage.

vesii

Why? it looks correct to me. You divde each part by $n^2$

feynhat

That is fine. I am talking about the second equality.

Mike Miller

@BalarkaSen Alternatively, let Stiefel-Whitney classes lie by defn in the orbit set of H^*(M;Z/2) under the action of Homeo(M). :D

Balarka Sen

Hahah

@LeakyNun Let $S^{k-1} \to E \to M$ be an oriented sphere bundle over $n$-manifold $M$, and suppose you're trying to construct a section. Here's an approach: Choose a triangulation of $M$; over each $n$-simplex the bundle trivializes, so choose sections. But these need not patch up to anything meaningful; take sum of these simplices, considered as an $n$-chain in $E$ and take boundary. This gives an $(n-1)$-chain $E$.
Again, over the $(n-1)$-simplices in $M$, the bundle trivializes, so relative to the $(n-2)$-skeleton this chain should be homologically trivial. So this chain is bounded by s…

This is exactly the staircase diagram for $d_n$

The statement that this bundle has a section is the statement that $d_n = 0$, because you'd be able to match these sections after inducting downwards to $0$-skeleton

Thus, Euler class.

Rudi_Birnbaum

13:07

@StupidKid OK I will attempt an answer.

vesii

@feynhat sorry, I meant without the limit prefix. But I still get zero. And I think it should be $\frac{1}{3}$.

Balarka Sen

It's funny to see that at every stage you're making newer $(n-1)$-chains in $E$ which lie over the $(n-i)$-skeleton of $M$ and intersects the fibers in $(i-1)$-chains. So every stage you're "populating" the fiber $S^{k-1}$ by one dimension. This is precisely the geometric meaning of the index change that $d_i : E^i_{p, q} \to E^i_{p - i, q + i - 1}$ does.

Anyway, at the end you'd remain with a $(k-1)$-chain lying over the $0$-skeleton of $B$, populating $S^{k-1}$ in full dimension.

That's the guy in $H_n(B; H_0(S^{k-1})) \to H_0(B; H_{k-1}(S^{k-1})$

This is the pictorial meaning of spectral sequences

Needs a lot of polish before I come up with a precise story, but I'm quite satisfied for now.

I am curious to know if there are transgression interpretations of other obstruction classes. This is surprisingly clean.

Leaky Nun

@BalarkaSen I didn't understand your interpretation

Balarka Sen

13:23

Ok, I'll try to write it in greater detail and with more precision afterwards, maybe in garbology. Gotta finish a point set topology assignment lol

Leaky Nun

ok thanks

feynhat

@BalarkaSen How many of these assignments do you get? I see you doing one everyday.

@vesii $\infty$ is not a real number. You don't divide by it.

geocalc33

I'm quitting math because my mental health is questionable

Balarka Sen

That's because I keep piling them until the date of submission

geocalc33

just kidding my health is really good

Balarka Sen

13:30

So far I am just getting point set topology assignments. Very annoying garbage

feynhat

understandable

geocalc33

point set topology sounds annoying

Balarka Sen

Point set topology is quite fine. Routine crap is annoying, like verifying the Sorgenfrey plane is not normal for hundred pages where it's clear that it isn't

geocalc33

ohh

Balarka Sen

Proper point set topology is

Thorgott

13:34

Last week I had to verify that $\mathbb{R}\mathbb{P}^2$ (defined as quotient of $S^2$) is homeomorphic to $D^2$ with a copy of $S^1$ attached via $f\colon S^1\rightarrow S^1,z\mapsto z^2$. It was more annoying than the obviousness would suggest.

Balarka Sen

If $X, Y$ are compact metric spaces, $X \times \Bbb R \cong Y \times \Bbb R$ then $X \times S^1 \cong Y \times S^1$

geocalc33

does anyone want to hear a song?

feynhat

Quotients and product with compact spaces commute?

Leaky Nun

@BalarkaSen wow I didn't know that there are manifolds in Lean

Balarka Sen

There's no reason the homeomorphism X x R -> Y x R preserves the Z-action

geocalc33

13:36

darn I thought that was chess

feynhat

oh. right.

Leaky Nun

@geocalc33 am I that predictable?

geocalc33

@LeakyNun well I have been playing chess lately

I beat my brother

Mike Miller

@Thorgott I don't think it should be. You know already that the map $D^2 \to \Bbb{RP}^2$ from the upper hemisphere is biective on the interior of that cell, and a covering map to $\Bbb{RP}^1$ on the boundary --- that covering map given by $z^2$. That gives you a map $S^1 \cup_{z^2} D^2 \to \Bbb{RP}^2$ which is a bijective continuous map.

Stupid Kid

@geocalc33 Do you ever have this feeling. You complete real analysis course and when someone calls u to prove lim x to 3 f(x) and u can't do it lol.😂😂😂

geocalc33

13:42

@StupidKid Yeah, but I haven't taken real analysis

Stupid Kid

@geocalc33 So what u doin in math u should be in physics. Just jkin

geocalc33

I could have gone pro in chess if my parents started me earlier

Stupid Kid

I always loose in chess

geocalc33

@StupidKid haha that was a good joke

Stupid Kid

Too many possible moves

geocalc33

13:43

yeah there's something like n!cos(10) or somethin I forget

Stupid Kid

Today I nearly forgot definition of continouty

And did something stupid

geocalc33

Sorry I can't talk anymore

I have to release a song. Actually should only take a few minutes to upload

Stupid Kid

@geocalc33 have u take qft in physics?

geocalc33

I took quantum mechanics @Stupid

not quantum field theory though

Stupid Kid

@geocalc33 how was it?

was it painful?

geocalc33

13:48

it was good. the first exam I tried and basically got a C on it. but on the final I guessed on every question, finished in ten minutes and got an A. So I think I know the randomness of quantum principle

Stupid Kid

@geocalc33 XD.

feynhat

@BalarkaSen One point compactification?

Stupid Kid

I have bad memory and am very slow thinker. So should I go get a job for toilet cleaning service?

geocalc33

@StupidKid I am an extremely slow thinker

Edward Evans

Implicit shaming of the working class right there

geocalc33

13:51

lol

Stupid Kid

@geocalc33 Then both of us should get job at Bathroom Cleaning service lol. 😂😂😂 And doing some experiment about fluid dynamics there.

Edward Evans

-.-

geocalc33

@StupidKid @StupidKid ...

Stupid Kid

Well My love with physics began from bathroom actually when I saw the love shape in bucket

geocalc33

I have a question for you @stupid. Can a series of non-logical statements computed by some brain $B$ ever converge to a completely true piece of information.

and by non-logical I should probably say, partially true.

Thorgott

13:57

@MikeMiller it's a second week in a first course on topology exercise. no covering maps, no "continuous bijections from compact spaces are automatically homeomorphisms", only definitions.

Stupid Kid

@geocalc33 Well I have to think about that

Thorgott

the geometric picture is the one you describe, of course, it's just working out the details that's kind of annoying

Stupid Kid

btw the shape was like this r=1-sin(\theta)

Mike Miller

I mean, the phrase "covering map" there was inessential

You just need to know that the map on that boundary circle is your $z^2$

feynhat

No. That doesn't work. I was thinking one-point compactification of product of spaces is product of one-point compactification. This is obviously false. Take $\Bbb R \times \Bbb R$.

Mike Miller

14:00

@feynhat Write $\hat X$ for the one-point compactification. It is a pointed space. If $X$ and $Y$ are both locally compact Hausdorff (maybe some point-set requirements? didn't check before writing this) then $\widehat{X \times Y} \cong \hat X \wedge \hat Y$, where that last thing is the smash product. The smash product of pointed spaces $X, Y$ is $X \times Y / X \vee Y$.

For instance, if $X = \Bbb R^n$ and $Y = \Bbb R^m$, this gives the formula $S^n \wedge S^m \cong S^{m+n}$.

Stupid Kid

@geocalc33 Yes . Yes it can.

Anyone ? Ever wondered why antman can still breathe when he becomes smaller than oxygen atom ?

2

geocalc33

no

feynhat

@MikeMiller Oh. Thanks. I remember seeing smash product where they were trying to generalize suspension.

geocalc33

Can you compactify $\Bbb R^n$ to be dense in an n-cube

Stupid Kid

May be this is wrong place to put ur jokes.

Thorgott

14:10

$\mathbb{R}^n$ is homeomorphic to the interior of an $n$-cube, so yes

Stupid Kid

ok I need to go study instead wasting moui time joking see ya guys.

see ya at 3.14 pm

feynhat

How do I see that $S^1 \wedge S^1$ is $S^2$? I want to collapse the two circles. Collapsing one, I get the sphere with its poles identified. Collapsing the other...

geocalc33

it would be better to have a diffeomorphic condition?

so I guess I have to figure out if $\Bbb R^n$ and the $n-$cube are diffeomorphic

I know that the unit ball in $\Bbb R^2$ is diffeomorphic to $\Bbb R^2$

so I'm tempted to say that yes, the $\Bbb R^n$ and the $n-$cube are diffeomorphic.

as long as you don't include the boundary of the square

because that has corners and would make it just homeomorphic

Mike Miller

14:25

@feynhat Actually, think about it like I said above.

If S^n is in some sense R^n union a point at infinity, write out their product in these terms, then their wedge

(Can be made more formal than this)

Alessandro Codenotti

@MikeMiller (I think no extra point set requirements are needed here apart from locally compact + completely regular, which follows from locally compact + Hausdorff)

Thorgott

$f\colon(-1,1)^n\rightarrow\mathbb{R}^n,\,(x_1,...,x_n)\mapsto(\tan(\frac{\pi}{2}x_1),...,\tan(\frac{\pi}{2}x_n))$

geocalc33

14:54

It's clear that the open $n-$cube is diffeomorphic to $\Bbb R^n$ via $f\colon(-1,1)^n\rightarrow\mathbb{R}^n,\,(x_1,...,x_n)\mapsto(\tan(\frac{\pi}{2}x_1),...,\tan(\frac{\pi}{2}x_n)).$

Another set of piecewise maps seems to work as well:

$\rho:\Bbb R^n\to(0,1)^n, (x_1,...,x_n)\mapsto(\exp(x_1),...,\exp(x_n))$

But, for $\Bbb R^1$ I think we need $2^1$ maps and for $\Bbb R^n$ I think we need $2^n$ maps to achieve desired symmetry inside the cube.

For example with $\Bbb R^1$ we can do $g:\Bbb R^1\to(0,1), (x_1)\mapsto(\exp(x_1)).$ But another map is needed for symmetry in the unit interval…

so your map @Thorgott is much more succinct. I'm just wondering if there's a problem with my construction...

Thorgott

an immediate problem would be that the range of $\exp$ is not $(0,1)$

geocalc33

but $\exp(x_1)$ takes all negative valued $x_1$ to the unit interval

and maps 0 to 1

I'm ignoring the y coordinate

it just stays 0

basically you can just use $f:\Bbb R^2 \to \Bbb R^2$ with $(x_1,x_2) \mapsto (\exp(x_1),\exp(x_2))$ and then use symmetry to flip things around and make copies of the structures

feynhat

15:12

$\hat{\Bbb R} \wedge \hat{\Bbb R} = \cfrac{ (\Bbb R \cup \infty) \times (\Bbb R \cup \infty ) }{(\Bbb R \cup \infty) \vee (\Bbb R \cup \infty) } \cong \cfrac{ (\Bbb R \times \Bbb R) \cup (\Bbb R \times \infty) \cup (\infty \times \Bbb R) \cup (\infty, \infty) }{ (\Bbb R \times \infty) \cup (\infty \times \Bbb R) } \cong \Bbb R \times \Bbb R \cup (\infty, \infty)$.

Is this what you meant @MikeMiller?

Edward Evans

ew

lol

feynhat

where the last homeomorphism comes from the fact that $(A \cup B)/B \cong A $ (where B is collapsed to some $a_0 \in A$). In our case, $(\Bbb R \times \infty) \cup (\infty \times \Bbb R)$ is collapsed to $(\infty, \infty)$ which sits in the space upstairs.

Mike Miller

Yeah, @feynhat, though one has to be very careful with the "union" notation.

The formal way one would write it is this. You have a map $\hat X \times \hat Y$ to $\widehat{X \times Y}$ as follows. For points of the form $(x,y)$, we send that to the point $(x,y)$ in $\widehat{X \times Y}$. Any point of the form $(\infty, y)$ or $(x, \infty)$ is sent to $\infty$ in the codomain. Continuity follows from the definition of product topology and so on.

Then it's clear that the map above is a continuous surjection which collapses the $X \vee Y$ at infinity to a point. So it descends to a homeomorphism $\hat X \wedge \hat Y \to \widehat{X \times Y}$.

feynhat

15:33

yes. what i wrote above doesn't make much sense if i don't describe the maps.

Thanks.

But I still don't see how to collapse two circles on torus to get a sphere.

Mike Miller

If you pinch the meridian on the torus what you get looks like a half-moon with the vertices touching.

Now imagine slowly shrinking the meridian (a loop that passes through the vertex of the half-moon) and see what happens

feynhat

16:14

Yes. What I don't see is why there won't be a problem at the base point when I shrink the second circle. I can see homotopy, because the half-moon thing that you talked about is just $S^2 \vee S^1$ (in homotopy), then we shrink that $S^1$.

oh

Never mind.

Mike Miller

I'm suggesting not passing through the homotopy

I was suggesting imagining it while the half-moon was still a proper half-moon

Instead of elongating it to there being a piece of string between the ends of the moon

Alessandro Codenotti

@Balarka I don't know if you remember but a while ago we discussed conditions on the action of a group $G$ on a metric space $X$ such that $X/G$ is metrizable. You said compact orbits is probably enough. I just realized that it is indeed, because then $X/G$ is metrized by the Hausdorff metric on orbits

Anonymous

@MikeMiller I understand that each 2-cycle in $S_5$ would give us a legitimate section $s$. Say, we get the section $s_1$ corresponding to the 2-cycle $(12)$ and the section $s_2$ corresponding to the 2-cycle $(23)$. Would both $s_1$ and $s_2$ give us the same semi-direct product $A_5 \rtimes C_2$ as $(12)$ and $(23)$ are conjugate elements in $S_5$? (Sorry for annoying again; I was going through the comments you made yesterday.)

user193319

Let $G$ be a locally compact group and $Q$ a compact subset such that $G = \langle Q \rangle$. If $K$ is another compact subset, does there exist $n \in \Bbb{N}$ such that $K \subseteq (Q \cup Q^{-1})^n$?

Mike Miller

16:30

Yeah, exactly

infinite-blank-

0

Q: Finding plane which cuts prism in a section which forms an equilateral triangle

We are given a rectangular prism having $x+y=0,x-y=0$ and $x=1$ as its faces. We have to find the normal vector to the plane which cuts this prism in a section that forms an equilateral triangle. Now I couldn't visualize it properly so I used a 3-D calculator: https://www.geogebra.org/3d/trapevz...

vectors analytic-geometry 3d

This is a question I posted yesterday but didn't get any attention so I thought of posting it here. Please provide your views.

Anonymous

@MikeMiller Were you replying to me or user193319? :P Btw do you have any quick proof of the fact that choosing conjugate elements for a section results in identical semi-direct products?

Anonymous

I mean, I can't immediately see a way to conclude that choosing either (12) or (23) will result in the same semi-direct product

Mike Miller

16:47

that's the whole discussion about the semidirect product only depending on $\varphi$ up to conjugacy

Anonymous

@MikeMiller Is there a thread on Math SE discussing this?

Anonymous

Or could you give me some reference?

Tobias Kildetoft

@MikeMiller Well, except that these are taken inside $S_5$, so there is no way they could not give the same result.

Mike Miller

"if there is a homomorphism $f: H \to N$ so that $\varphi, \psi: H \to \text{Aut}(N)$ satisfy $\varphi(h)(n) = f(h) \psi(h)(n) f(h)^{-1}$, then $N \rtimes_\varphi H \cong N \rtimes_\psi H$"

@TobiasKildetoft fair enough

Tobias Kildetoft

But for external ones, it is a good exercise to go through

Mike Miller

16:49

I don't know what's on Math.SE, if I need to know a fact I don't know the reference for I just google various keywords until I do find it

I don't know a reference for the above fact but that's because it gives you enough information to just write down the isomorphism

Tobias Kildetoft

I think I asked about something long these lines a long time ago. Let me look for it

Anonymous

Okay, np. Thanks for stating the theorem in full :)

Tobias Kildetoft

7

Q: How to determine if two semidirect products are isomorphic?

In answering Classify all groups of order 182 I realized that I don't actually know any good general results that will allow me to determine whether or not two given semidirect products are isomorphic without resorting to various ad-hoc methods (like in that question where I counted elements of o...

group-theory

There is a link in a comment to some notes on the topic

Mike Miller

the converse should be true too, except you should say "...by an isomorphism fixing $N$ and whose preserving the map to $H$" (aka, if they're isomorphic as extensions). i only say should because i didn't check, not because it's plausibly false

Arbuja

17:06

-2

Q: Coming up with a rigorous definition for a Riemman-like sum which is easier to compute?

Continuing from my last question, I understand my last definition was unclear, so my tutor modified it. Forgive him if it's still unclear. He's the best I have. Definition Consider $f:A\to[0,1]$ where $A\subseteq[0,1]$. Finite-partition $[0,1]$ into $n$ non-empty subintervals such that each...

integration measure-theory summation definition partitions-for-integration

I need help improving my question. Anyone willing to help.

Ted Shifrin

17:17

We have three almost identical lavender avatars now.

Tobias Kildetoft

I guess at some point I ought to get a custom one, but ehh, effort.

It's not like I have been using chat for a long time or something like that :)

Ted Shifrin

Define long time.

Tobias Kildetoft

I mean on a geological scale of course

Ted Shifrin

Oh, well, I remember you from the mesozoic era.

Alessandro Codenotti

@TedShifrin Lukas's one is also kinda similar

Ted Shifrin

17:23

At least I can claim I'm unique :P

Anonymous

@TobiasKildetoft Thanks, those notes look helpful. By any chance, could you answer this? I'm trying to do the proof myself and this seems like an essential component to figure out first

Astyx

Hi lavender comrades

Alessandro Codenotti

lavender comrades sounds like you put communism in the washing machine and it got a bit discoloured

Ted Shifrin

washed out, you mean?

Alessandro Codenotti

I think so

I can't English

Ted Shifrin

17:35

Oh, your English were fine.

Astyx

Is it just me or are there more pinkish automatically generated profile pictures than other colors ?

Ted Shifrin

That was what I was wondering, I guess.

Astyx

It might just be cognitive bias

Or just a strange correlation between coming to this chat are having a pink auto generated profile picture

user777

17:54

Hi! I have a small question: From discrete math, we have the following question:

*Show that the set of functions from the positive integers to the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} is uncountable. [Hint: First set up a one-to-one correspondence between
the set of real numbers between 0 and 1 and a subset of
these functions. Do this by associating to the real number
0.d1d2 … dn … the function f with f(n) = dn.]
the answer is clear, but I wonder to solve the question by not showing a bijective function with positive integer number. How would you to show that this this way?

I mean how to prove it that it doesn't have a bijective function with the set of functions f, where f: N --> [0-9]

Thorgott

18:08

but it does

actually, nevermind, I think I'm not entirely sure what you're asking

user777

@Thorgott Okay, my question is: Is it possible to show that " set of functions from the positive integers to the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}" is uncountable by showing a not bijective function with positive integer? Or Is it impossible since there are infinite number of functions and you need to test each one of them to see whether if they fail to have a bijective function?

Manjoy Das

can i say $ A=\{x\in S:x\in T\}\implies A= S\cap T$?

Leaky Nun

sure

Manjoy Das

then where's the difference between $ S \subset T$ and $ S \cap T$?

$ S\cap T \subset S \subset T$?

Thorgott

"by showing a not bijective function with positive integer"
I'm not sure what exactly this is supposed to mean. Are you talking about a positive integer or the set of positive integers? Are you asking whether it is enough to exhibit one such non-bijective function or whether it is enough to show that no such function can be bijective?
"Or Is it impossible since there are infinite number of functions and you need to test each one of them to see whether if they fail to have a bijective function"
What does it mean for a function to have a bijective function?

Manjoy Das

18:25

Actually i was proving 2nd ring isomorphism theorem. so i needed to prove that $ \theta:S\rightarrow(S+I)/I$ is an isomor. and ker{$ \theta \}$= $ S\cap I$. So how do i prove that ker{$ \theta \}$= $ S\cap I$?

Thorgott

In fact, $\{x\in S\colon x\in T\}$ (or, equivalently, $\{x\in T\colon x\in S\})$) is how one may define $S\cap T$

"$S\subset T$" is a statement, "$S\cap T$" is a set

Manjoy Das

what i did is: let $ i \in ker{\theta\}\subset S$. now $ \theta (i)=I\implies i+I=I \implies i\in I$ thus ker$ \theta = \{i \in S:i\in I\}\implies \text{ker} \theta= S\cap I$.. is it ok?

user777

@Thorgott "by showing a not bijective function with positive integer" I mean: the set of positive integer. So, again my question: to show that this set:" set of functions from the positive integers to the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}" and this set: "set of positive integer" have non-bijective function? if we can, how we do it?

I'm really sorry if my question isn't clear! see here a solution to the question above: slader.com/discussion/question/…

in the answer, they use a set of real number to show a bijective, and therefore this set: " set of functions from the positive integers to the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}" is uncountable

Thorgott

It is not enough to show that there is a non-bijective function. You need to show to show that any such function is not injective.

Manjoy Das

and with this, are the elements of$ (S+I)/I$ equal to the elements of $ S/I$?

Thorgott

18:39

what would $S/I$ be?

Manjoy Das

$ S/I= \{s+I:s \in S\}$

user777

@Thorgott So in this case If I have a set A and I don't know if it is infinitely countable or uncountable, then I test with real number, if there is a bijective between real numbers and set A, then this set A is uncountable. Otherwise, we test it with positive integer number and if this set A is bijective with positive integer number, then we say set A is infinity countable.

Manjoy Das

and $ (S+I)/I=\{(s+i)+I:s+i \in S+I\}=\{(s+(i+I)):s+i \in S+I\}=\{s+I:s \in S\}=S/I$ because $ i+I \in I$

Thorgott

@ManjoyDas sure, these sets are the same, but the notation $S/I$ is problematic, because it suggests this is a quotient ring, which it is not (this is why we talk about $(S+I)/I$ in the first place)

@user777 I don't know what "infinitely countable" means. What's true is that if a set is in bijection with the real numbers, it is uncountable, and if it is in bijection with the natural numbers (the set of positive integers), it is countable. There's a lot more that can potentially happen, though, and other ways of determining the cardinality of sets.

Manjoy Das

is $ (S+I)/I$ not a quotient ring?

Thorgott

18:46

@ManjoyDas it is

user777

I wonder if the set A doesn't have a bijective function from the real number to set A nor natural number to set A!

Manjoy Das

@Thorgott enumerable sets are also called countably infinte sets.

Thorgott

that can happen

user777

@Thorgott what do we call this set? Is there a name to look for

Countable Infinitely means if the set contains an unlimited number of elements and there is a bijective function with the positive integer

Thorgott

I now what "countably infinite" means, my point was that that's not the same as "infinitely countable", which doesn't make sense.

Manjoy Das

18:50

@Thorgott i didn't understand t. why $ S/I$ not a quotient ring but $ (S+I)/I$ is?

Thorgott

If you're interested in which sets are in bijection with one another, the keyword to look up is "cardinality"

@ManjoyDas $I$ is not necessarily a subset of $S$

user777

@Thorgott Thank you! I really enjoying chatting with you! I really understand something here.

Manjoy Das

oh yeahhh

ok.. but i can write the elements of $(S+I)/I$ as $s+I, s \in S$ right?

because $I$ is an ideal of $S+I$

Thorgott

you can, that's what your above argument shows

it's just that thinking of it as $(S+I)/I$ tells us what the ring structure on it looks like

Manjoy Das

what about the kernel that i showed above?

Kay Bei