Mathematics - 2018-01-08 (page 4 of 5)

@LeakyNun That's why I said it's non obvious!

You have to compute $H_2(S^2)$ at least.

@BalarkaSen do you have a proof-sketch?

@LeakyNun homology, $\pi_2$

@MatheinBoulomenos well

do you have a more elementary proof?

so being simply connected doesnt mean contractibillity

8:01 AM

@LeakyNun Fine, I do. Come up with a closed 2-form on $\Bbb R^3 \setminus 0$ which integrates to something nonzero on the unit sphere.

im amazed :p

@ManolisLyviakis right

If $S^2$ was contractible, you would get two linearly independent vector fields that vanish nowhere

That gives an example of a closed 2-form on $\Bbb R^3 \setminus 0$ which is not exact.

That can't happen if $\Bbb R^3 \setminus 0$ was contractible (prove)

So you're still doing homology @Balarka

8:02 AM

Yes, but the elementary version.

@BalarkaSen isn't $\pi_1(W)=\Bbb Z$? It should be weak homotopy equivalent to $S^1$

Cohomology actually

@Alessandro The Warsaw circle is simply connected

DR cohomology for the boys

hmm so S^3 is contractible? @MatheinBoulomenos since you can get a vector field that vanishes nowhere?

There is no loop that goes around the whole Warsaw circle, because it you have to "pass through" the bad point on the topologists's sine curve

8:03 AM

@ManolisLyviakis it isn't

Yeah what Mathei just said is bollocks.

no, this argument doesn't generalize @Manolis

damn :p

Lots of parallelizable but noncontractible manifolds

@BalarkaSen oh, we're thinking about different circles, the pseudocircle is the one with 4 points

8:05 AM

@Alessandro Oh, is that it? I didn't know that had a name.

So contractible i need identity homotopic to a point

@Manolis have you heard of Sard's theorem?

can you tell me how you intuitions works?

Then you're right.

$S^1$ and that thing are weakly homotopy eq

Pseudocircle

The pseudocircle is the finite topological space X consisting of four distinct points {a,b,c,d} with the following non-Hausdorff topology: { { a , b , c , d } , { a , b , c } , { a , b , d } , { a , b } , { a } ...

what the hell did I just read

8:06 AM

@BalarkaSen how does that make my argument bollocks? I only need the implication contractible => parallelizable

@Daminark nope :/

@Leaky Separate $S^1$ as union of 4 closed arcs. Now quotient the interiors of the arcs

@Mathei Oh I see. Alright.

This Warsaw circle is a weird space though, didn't know about it

@BalarkaSen hmm

It's a very nonHausdorff space but still has the homotopy groups of $S^1$, like Alessandro said

8:08 AM

Contractible implies paralellizable for a manifold $M$ because isomorphisms of vector bundles are in bijection with homotopy classes of maps $M \to G$ where $G$ is something.

Grass(infty, n) if you want

@BalarkaSen are you saying that the pi_1 of the pseudocircle is Z?

@LeakyNun Quite right

I can't believe it

@Balarka Which is why I used the letter G

8:08 AM

In fact the quotient map is exactly the generator

hmm 2 spaces are homotopic if there exists a deformation retract

so if i wanted S^2 to be contractible to a point it should be homotopic to a point meaning there exists a deformation retract

and thats how i can get the intuition of it

i cant expand a point to a sphere

:p

group rep theory and ANT and Galois theory today :O

group rep on 0900 which means that I need to go out now

bye everyone

bye @Leaky

bye @LeakyNun

@MatheinBoulomenos do you have a proof sketch of "non-zero analytic function's zero set cannot have limit point"?

8:15 AM

identity theorem

@Mathein I think he's kinda asking for a proof of the identity theorem sorta

You'd want to Taylor expand the function at the limit point and see what happens

does analytic mean analytic everywhere?

@Leaky the idea is that if you have a sequence of zeroes converging to a point, you expand around that point

Assume it vanishes with a finite order of vanishing, get contradiction by looking at all the cluttering zeroes

8:17 AM

@Leaky that's called entire

@LeakyNun In this case your function is analytic on a connected domain containing the sequence and it's limit point, I believe

For this sort of thing you really just need connectedness

Sniped

welp

lets say C

But the idea is this

8:19 AM

If $f$ vanishes with finite order at the limit point $a$ of the sequence $a_n$, $f(z) = (z - a)^r g(z)$ for some $g \neq 0$ on a ball around $a$.

You say alright, $z_n \to z_0$ and $f(z_n) = 0$ for all $n$

Oops. you go on

Dammit

I'll just go back to listening to Dragostea Din Tei vaporwave cover instead of sniping people

(I wanna die)

But yeah so by continuity $f(z_0) = 0$, so you take the smallest $n_0$ such that $f^{(n_0)}(z_0) \ne 0$. Then $f(z) = (z-z_0)^{n_0}g(z)$. But then you know $f(z_n) = (z_n-z_0)g(z_n) = 0$

But that forces $g(z_n) = 0$

But $g(z_n) = f^{(n_0)}(z_0)$ so rip

Now, all the derivatives of $f$ at $z_0$ are $0$

But then when you write the power series for $f(z)$ where $z\in D$, all the coefficients are $0$, so that $f(z) = 0$ for all $z\in D$

8:27 AM

Oh god

ナマー・ナマーｎｕｍａｎｕｍａ

►

There is actually such a thing

Hi, if we have $|f_k (x)-1| \leq \frac 1 k \sum \frac {x^{2n}} {n!}$. From here how can we explain that $\lim_{k\to+\infty}f_k(x)=1?$

Well what happens to the RHS?

lmao. This is a page about the Tits group: "We apologise to the eminent mathematician whose name is usually attached to this group for removing his name from this page and those linked to or from it. The reason is that certain webcrawlers which have been scanning these pages have misinterpreted the occurrence of this name as an indication of quite a different content on these pages from that which actually pertains." web.mat.bham.ac.uk/atlas/v2.0/exc/TF42

Hahahaha

RIP Jock

really unfortunate name though

i didn't know how to spell a Tits complex when I first saw it

8:33 AM

It's pronounced teeyts approx

Yea I know

but I got to know about that much later

Which I didn't know when I first mentioned I was looking at it to my adviser

Well the "s" is silent really

@Narcissusjewel l m a o

Worse I said 'jack'ee's tits'

I know that the series on the RHS is convergent, but so then what happens? @Narcissusjewel

8:34 AM

A lot of people say tits.

That's how I first read it

the Bra-tits building.

@LeylaAlkan So as $k\to\infty$ the RHS goes to 0, so |f_k(x)-1| which is positive via the abs values goes to 0, so f_k(x)-1 must have absolute value going to zero, so f_k(x) goes to 1?

Okay thanks

No problem

8:54 AM

IF there exist a deformation retract then the 2 Spaces are homotopic. IS the reverse true?

Depends on what you even mean by "reverse". I think it is true that two CW complexes are homotopy equivalent if they are deformation retracts of a common CW complex.

But that is very much not a reverse

I think reverse was meant to be converse?

Yes, I am asking him to carefully state the converse

(The proof of what I said is just the mapping cylinder construction, I think)

[I'm highly non-sober at this point, I should probably sign off and finish this marking] - back later

i mean If they are homotopic then there exists a deformation retract

8:59 AM

@ManolisLyviakis Okay, that statement is nonsense. If $X$ is homotopy equivalent to $Y$, it does not make sense to talk about deformation retracts in general because $X$ doesn't need to be a subspace of $Y$ (vice versa)

if Y is a subspace

*

In that case it's false. Consider the Hawaiian earring to be $X$ and a subspace $Y$ to be given by deleting finitely many circles.

$X$ and $Y$ are homeomorphic (which is more than homotopy equivalent), but $X$ does not deformation retract to $Y$

I need to prove that If Adeformation retract of X and B deformation retract of A prove that B is deformation retract of X

That is true.

so if they are homotopic doesnt mean there exists a deformation retract

dayumn you come up with crazy examples haha

9:02 AM

No. (Also you mean homotopy equivalent.)

yes

It doesn't need to be the Hawaiian earring, you could take $X$ to be the infinite wedge of circles.

And do the same construction

The earring is just easier to see

what i need to prove essentialy is the transitivity of homotopy equivalnce because if there exists a deformation then they are homotopy equivalent

Lozansky

If I define $\langle\mathbf{z}, \mathbf{w} \rangle = z_1 \overline{w}_1+...+z_n \overline{w}_n$ as the inner product for $\mathbf{z},\mathbf{w} \in \mathbb{C}^n$, then how do I orthogonalize $(1,2,3), (3,1,4), (2,1,1)$ using Gram-Schmidt?

lol it’s next week

9:07 AM

@BalarkaSen Two topological spaces$ X$ and $Y$ are homotopy equivalent if there exist continuous maps$f:X->Y$ and $g:Y->X$, such that the composition $ f \circ g$ is homotopic to the identity $id_Y$ on $ Y$, and such that $g \circ f $is homotopic to $id_X$ just what i mean homotopic spaces

@Daminark nice

I didn't ask for the definition of a homotopy equivalence... but that is correct

So homotopy equivalnce doesnt mean existance of deformation retract when $Y$ is subspace

No.

ok

thanks

9:09 AM

Yeah complex analysis is dank stuff

The vaporwave of math

Truly

I would get so much flack for that description

9:27 AM

planetmath.org/sites/default/files/texpdf/37683.pdf

can some explain me how $G$ with the cap is defined?

imnot used with these products inside functions

what they mean

$iG(id_I \times r) $

wtf is this?

$\tilde{G} =iG(id_I \times r) $ where $i$ is an inclusion map

id is the identity on [0,1]

$r$ is retract

does it mean $i \circ G(id_I \times r) $

Vrouvrou

11:19 AM

hello somwone here speak portugais ?

11:48 AM

in The h Bar, 22 mins ago, by Slereah

https://web.math.rochester.edu/people/faculty/doug/otherpapers/blumberg-burnside‌.pdf

ack, the Incomprehensibles!

philmcole

12:49 PM

Hi guys. In the ratio test for series one often uses the normal $\lim$ instead of $\limsup$ and $\liminf$ like in the root test. Why is that sufficient?

Is it because $\liminf \le \lim \le \limsup$ and therefore if $\limsup \lt 1$, so is $\lim \lt \limsup \lt 1$ and if $\liminf \gt 1$, so is $\lim \gt \liminf \gt 1$?

If the limit exists, then the limits superior and inferior both exist and coincide with the limit.

So for the limsup/liminf version to work it is sufficient (but not necessary) for the limit version to work

philmcole

@Semiclassical Ok thanks!

Np

1:17 PM

Please give an example of uncountable open subset of $\Bbb R$ that does not contain a perfect subset. Inspired from here

There isn't any

oh, so why Brian in that answer mentioned closed ?

Note that “not closed” isn’t equivalent to “open”

I don’t know if that resolves it, but the two conditions don’t work that way

Oh! Thanks for reminding. Then, eg, set of all irrationals is an uncountable subset of R that does not have a perfect subset. Thanks!@Semiclassical

Am i right?

11 hours ago, by Silent

How do we get $\limsup \max (a_n, b_n) \leq \max(\limsup a_n, \limsup b_n)$ here?

Please someone have a look at this, too!

@Silent The irrationals contain a perfect subset. (Exercise: find it.)

Hm, it seems that you need the axiom of choice to prove that there exist uncountable sets with no perfect subset

1:34 PM

@AkivaWeinberger, My god! you are so true! I had done this in Rudin's exercises, a set that is similar to Cantor set and has all the irrationals only. If $C$ is a set that contains decimals $0.a_1a_2\dots$ where $a_1$ , $a_2$ etc can take only 4 or 7, and $a=0.1010010001...$ then $a+C$ is desired perfect set.

meaning that it's consistent with ZF that every uncountable set has a perfect subset

That's weird

@Silent In particular it does contain a perfect subset

@AkivaWeinberger What's weird?

$a+C$ answer?

No, the fact that you need the axiom of choice to construct an uncountable set with no perfect subset

oh :)

1:37 PM

Is it due to Baire category stuff?

That's one way to show that the irrationals contain a copy of the Cantor set, but there are others

What are perfect subsets use for, why is it important to know about a set that contains only limit points?

the retract of a space X say A is homotopic equivalent with X right? cause of inclusion and the retract function i get $r \circ i=id_A $ and $i \circ r=id_X$

@Secret It's probably important in descriptive topology, which I know nothing about

The terminology SE uses for duplicate questions annoys me at times

1:50 PM

Wikipedia says Cantor proved that every closed set is the union of a countable set and a perfect set

(Closed)

Insofar as “a duplicate of a question with an answer” seems to exclude people who repost their questions when they dislike the response they got

well, uh, I don't see how you can produce an infinite dedekind finite set from that union (since no infintie dedekind finite sets can contain a countable subset). That said, I have not investigated what it means for an infinite dedekind finite set to be closed

@ManolisLyviakis don't you need a deformation retract here?

@Secret Closed subsets of the real line

But that seems a duplication as well

1:53 PM

Ah I see, then we are good

Actually - can you have a Dedekind finite subset of the real line? I don't know. I doubt it

i can prove homotopy equivalance either with deformation retract or finding 2 functions that give me identities @AlessandroCodenotti

Q: Can $\mathbb{R}$ be partitioned into dedekind-finite sets?

9

Assuming $ZF$ itself is consistent, it is consistent that there are sets $D$ which are infinite but cannot be placed in bijection with any of their proper subsets; such sets are called "strictly Dedekind-finite." Consistently, there is even a Dedekind-finite set of reals. My question is, is i...

As for the usual reals, I am not sure

@Secret WAAA

@Secret ? That is the usual reals

WRYYYYY

1:56 PM

Ah, I forgot to read this line:

> Now use a bijection of R×A with R to transport this partition to a partition of RR without changing the cardinality of its parts.

thus it is indeed the usual reals

lol, we have infinite ONIONS in the reals!

Onions?

infintie dedekind finite sets looks like onions to me because of the property that any subset of an infinite dedekind finite set is infinite dedekind finite

and also, that its cardinality can decreases indefinitely as you remove elements one by one from it

they can support pretty rich structures such as linear ordering and there a known borel set that is infintie dedekind finite. However that set is produced from borel sets that are $F_{\sigma \delta}$ and I had great trouble visiualising these sets

@AlessandroCodenotti im wrong since $i \circ r$ is not the identity of X ( where $r$ is retract and $i$ is inclusion

Correction: any subset of an infinite dedekind finite set is dedekind finite

(always forgot that finite sets are dedekind finite)

Just a quick question we know $\sum_{n=0}^\infty \frac {x^{2n}} {n!}$ converges to $e^{x^2}$ . So can we say that then $\sum_{n=1}^\infty \frac {x^{2n}} {n!}$ converges to $e^{x^2}-1 $

2:16 PM

@LeylaAlkan Yes

[Random]

I am trying to think about good ways to check whether a limit of a sum is "separable" i.e.:

$\lim_{x \to a}\sum_{n=1}^{m} f_n(x) = \sum_{n=1}^{m}\lim_{x \to a} f_n(x)$

The simplest such case will be the criteria for $f$ and $g$ in order for the following to hold:

$\lim_{x\to a} f(x)+g(x) = \lim_{x\to a}f(x) + \lim_{x\to a}g(x)$

It is known a necessary criteria is that both limits exists and bounded (no shooting to infinity), but is it sufficient?

2:50 PM

$(f \times f' ) \circ (g \times g')=(f \circ g ) \times (f' \circ g') $

is that right?

where f,g f',g' are functions

i cant find properties of things like this anywhere

can some1 point me where to read about properties of things like that

or Composition of functions and products inside

like $g \circ f(X \times Y) $

@AlessandroCodenotti

3:19 PM

@ManolisLyviakis Should be true

Try shoving inputs into them, like $(x,y)$

You should get $\big(f(g(x)),f'(g'(x))\big)$ for both

ye thats what i did

4:04 PM

Hi all, I want to ask a question, but it's about comprehending a couple of sentences in a proof

Let $f_k(x) = 1 - \frac{x^2}k+\frac{x^4}{2! k(k+1)}-\frac{x^6}{3! k(k+1)(k+2)} + \cdots \qquad (k\notin\{0,-1,-2,\ldots\})$

Claim : If $x\neq 0$ and if $x^2$ is rational, then $(\forall k\in\mathbb{Q}\setminus\{0,-1,-2,\ldots\}):f_k(x)\neq0\text{ and }\frac{f_{k+1}(x)}{f_k(x)}\notin\mathbb{Q}.$

When he proves this claim he starts with..

Otherwise, there would be a number $y\neq 0$ and integers $b$ such that $f_{k}(x)=ay$ and $f_{k+1}(x)=by$. In order to see why, take $y=f_{k+1}(x)$, $a=0$ and $b=1$ if $f_{k}(x)=0$

first of all does he assume here that $f_k(x)=0\text{ and }\frac{f_{k+1}(x)}{f_k(x)}\in\mathbb{Q}?$

I know he writes this "there would be a number $y\neq 0$ and integers $b$ such that $f_{k}(x)=ay$ and $f_{k+1}(x)=by$ "because he assumes what if
$\frac{f_{k+1}(x)}{f_k(x)}\in\mathbb{Q}.$ but does he also assume that what if $f_k(x)=0\text ?$

4:23 PM

I mean what he is doing here is the use of contrapositive of the statement?

I just want to know how is the process of thinking made here

cant find the homotopy no clue where to look for

4:41 PM

If we say "suppose $x$ is a rational number, then $x={a \over b}$ ,where $a\in\mathbb Z , b\in\mathbb Z -\{0\}$ .Otherwise, $x\neq {a \over b}$ I want to see something like this. Against which sentence this otherwise statement is made?

@ManolisLyviakis can you visualize it?

which part of it?

i have a cylinder

and a circle

How could you continuously deform a cylinder into a circle?

something like that

with lines

down to a circle

Exactly, you just want to crush the cylinder onto its base

4:49 PM

visualizing its one thing

:p

writing it down is another

well, could you do this for a line sitting above a point?

line segment*

hm...

Why don't you first try to deform $I=[0,1]$ into $\{0\}$.

ok

deform meaning

make a homotopy equivalence

4:52 PM

that its homotopy and a retract

i have 2 definitions

for homotopy equivalance

one is find 2 functions

and the other find a deformation retract

i was trying to use the first one fro the exercise

but i didnt get the identity of the second half

but if i find a homotopy to the identity im fine

ok ill try prove $I$ is homoty eq to ${0}$

If you can do that, then the result for your cylinder is almost immediate.

I'll have this screen minimized so @ me if you need me

Sha Vuklia

Guys, Spivak says the following: say we have $f\colon V\to W$, a linear transformation, then we can define $f^*\colon \mathcal{J}^k(W)\to \mathcal {J}^k(V)$ by $f^* T(v_1,\dots,v_k)=T(f(v_1),\dots,f(v_k))$, where $T\in\mathcal J^k(W)$ and $v_i\in V$. Here $\mathcal J^k(V)$ is the set of all $k$-tensors on $V$. To me it seems like we have $f^*\colon \mathcal J^k(V)\to\mathcal J^k(W)$, instead of the other way around. Is this a typo of the book?

I can't render LaTeX in chat, so I can't help you –– sorry.

@Antonios-AlexandrosRobotis $f:I \rightarrow x_0 $ $f(x)=x_0$ and $g:x_0 \rightarrow I$ $g(x)=0$

f constant function

g constant to zero

Why is $f$a constant function?

5:01 PM

f composite g is =0

which is homotopic to the identity map in I with the straight line homotopy

Just give me a second to write something down

f composite g = $x_0$ which is the identity map in $x_0$ and g composite f is zero which is homotopic to the identity map with the straight line homotopy

so I homotopy equivalent to a point

My only question is why are you using $0$ and $x_0$?

i dont write in latex since you cant render it

I might just be misreading

I can read short strings of LaTeX fine, just not a whole paragraph

5:05 PM

oh ok

on that note: haaaaate you latex

(ignore me)

to prove homotopy equivalne i need 2 maps

Good news @ManolisLyviakis I fixed the problem by using firefox

I can officially render haha

let me reread now

such that $f \circ g=id_Y$ and $g \circ f=id_X $

More generally, you need $f\circ g\simeq id_Y$ and $g\circ f\simeq id_X$.

5:07 PM

$g :Y \right arrow X $ $f: X \rightarrow Y $

yeap

the 2 maps i told u

one composition goes to the identity of $x_0$

the other to zero but with straight line homotopy im ok

Yep. Sorry, I was being daft.

so i get $I$ equiv to $xo$

Your solution is great.

Indeed. So, $I\simeq \{x_0\}$.

(We might as well just take $x_0=0$, but in any case)

Now, how do you think we can use this to solve your cylinder problem?

i know that $S^{1} $ eq with himself

and I with the point

Right, so you want to do this homotopy coordinate-wise.

5:12 PM

i can take

the product

$S^{1} \times I \simeq S^{1} \times \{x_0\} $

That was a poor way to explain it.

Exactly what you said.

Try to think about working coordinate by coordinate and you'll get the answer.

and the functions will be

the products

now i got 2 questions

why $S^{1} \times \{x_0\} \simeq S^{1} $

@ShaVuklia Sorry for the delay. Spivak is correct. $f^*$ takes the $k-$tensor $T$ which acts on elements of $W$ and teaches it how to act on $k-$tuples of vectors from $V$. So, $f^*: \mathcal{J}^k(W)\to \mathcal{J}^k(V)$. Note that an upper star $(f^*)$ usually denotes a contravariant relationship.

@ManolisLyviakis Actually, $S^1\times \{x_0\}\cong S^1$.

ohh right

ok !!

and the other one is

i had the functions

before if you saw

I'll go look real quick

5:17 PM

if you see the photo

yep I do

but one composition doesnt lead to identity so i need a homotopy

to the identity

of the (s,0)

although i can prove that the homotopy equivalance holds in products so i dont even have to do that but for practice

i was finished when i proved what you said

Think about it. I'll write it this way: $\pi: S^1\times I\to S^1$ and $\iota: S^1\to S^1\times I$.

Now, it's just what we did.

As you correctly wrote, $\iota\circ \pi =id_{S^1}$.

The other way gives us $\pi\circ \iota$ which basically just flattens the cylinder, right?

We want to show that $\pi\circ \iota \simeq id_{S^1\times I}$.

yeap

Try this (this is actually what I originally intended lol). Define a homotopy
$H: I\times S^1\times I\to S^1\times I$ by
$H(t,(x,y))=(x,y(1-t))$.

5:23 PM

ohh

thats the one

i was asking for

:)

If I didn't do something silly here, what we see is that $H(0,(x,y))=(x,y)=id_{S^1\times I}$ and $H(1,(x,y))=(x,y(0))=(x,0)=\pi\circ \iota$.

yeap yeap

thats the one

ok ill have in mind those 1-t tricks

Now, the reason why this is conceptually the same, is that all we did here was apply the straight-line homotopy you already understand on each of the "lines" sitting above points on $S^1$.

yess

so whenever im looking for straight line homotopy

ill look for 1-t things

Lol basically.

5:26 PM

xDD got it i think

The idea is that a convex combination of points $p$ and $q$ in a vector space is given by $p(1-t)+tq$.

thanks man

That parameterizes a line segment from $p$ to $q$.

ohhh

ye

but we didnt need the tq

part

Sha Vuklia

@Antonios-AlexandrosRobotis yea thanks! I figured it out later. Their notation was a bit confusing, they could have written $f^*(T)(v_1,\dots,v_k)$ to make this clear, but it's ok now!

5:27 PM

the $tq$ is surpressed because $q=0$ @ManolisLyviakis

ohhh

No worries @ShaVuklia.

Keep practicing @ManolisLyviakis ;)

what lessons do you have this semester ?

CryinShame

5:44 PM

I found this theorem in Hubbard & Hubbard's Vector Calculus book and was wondering if it's wrong:
(Mean Value Theorem for functions of Several Variables) If $U \subset \mathbb{R}^n$ is open and $f:U \rightarrow \mathbb{R}$ is differentiable, and the segment $[\vec{a}, \vec{b}]$ joining $\vec{a}$ to $\vec{b}$ is contained in $U$, then there exists $\vec{c} \in U$ such that $f(\vec{b})-f(\vec{a}) = [Df(\vec{c})](\vec{b}-\vec{a})$

The counter-example is the typical $f(x,y) = (e^x \cos(y), e^x\sin(y))$. We can show $det\: f'(x,y) \neq 0 $ for all $(x,y)$ with $f$ not being one-to-one. This means we can find $\vec{b} \neq \vec{a}$ such that $f(\vec{b})-f(\vec{a}) = (0,0)$. Hubbard & Hubbard's theorem would imply that $det \:f'$ would have to be zero somewhere wouldn't it?