Mathematics - 2017-11-10 (page 3 of 4)

Balarka Sen

05:01

@Kevin There's a generalization of Poincare conjecture, but let's not get into that. What's important is Milnor found the generalized Poincare conjecture is false in dimension, uh, 7

And in fact, something more shocking came out

Kevin Driscoll

@Semiclassical Holy shit winning 2v1? Bodied dude. Roasted. Brutal. Savage. Wrekt. Like IM not sure I could compete again after that

Balarka Sen

He found a 7-manifold homeomorphic to S^7, but not diffeomorphic to it in the process, which was completely unintentional and infinitely confusing

Semiclassical

yuup

Daminark

@BalarkaSen I'm shocked!

Balarka Sen

@Daminark Every Poincare conjecture you knew, is false? I'm shocked!

Kevin Driscoll

05:02

@BalarkaSen And theres what like.... 18 weird $S^7$s or something like that

Semiclassical

though the swap mode matches are more interesting than this format

Balarka Sen

@KevinDriscoll 28

But yes

Gah ur no fun, you know the story

go away

shoo

Kevin Driscoll

Haha sorry

Etnyre spoiled me already

I know about 'fake' $\mathbb{R}^4$s too

Balarka Sen

Not a bad thing! He's a great mathematician

@KevinDriscoll Do you know about the small ones?

There are open subsets of the standard R^4 which are homeomorphic but not diffeomorphic to the standard R^4

Kevin Driscoll

But ya other than this kind of "Which manifolds are diffeomorphic to which other ones?" kind of thing I dont know what other question a contemporary topologist might ask

@BalarkaSen Wait whaaaaaaaaaaat.

Daminark

05:06

@KevinDriscoll that's an example of a question a contemporary topologist might ask

4

Kevin Driscoll

Is that it?

Balarka Sen

@KevinDriscoll More mindfuck: It's a theorem of Stallings that everything homeomorphic to R^n is diffeomorphic to R^n for n \neq 4 (ie there are no exotic R^n's for n other than 4). So take that open subset $U \subset \Bbb R^4$ which is an exotic R^4 itself, and consider $U \times \Bbb R$

That's an open subset of $\Bbb R^5$ homeomorphic to $\Bbb R^5$

so $U \times \Bbb R$ is diffeomorphic to $\Bbb R^5$, even though $U$ is not diffeomorphic to $\Bbb R^4$

Kevin Driscoll

So this actually I can believe. because Its like the open subset of $\mathbb{R}^4$ has gotten twisted up somehow in a way that's impossible to untangle smoothly into an open ball

But then you add the extra dimension and now things can move smoothly in a way that they couldnt before

Its certianly not obvious that such a thing can happen. But I can believe it.

Balarka Sen

:)

yes, so in general $U$ would look very fractal like

PVAL-inactive

There's lots of examples of X \times \Bbb R homeomorphic to $\Bbb R^4$ where $X$ isn't even a manifold.

Balarka Sen

05:11

cannot at all be easily drawn on your regular 4 dimensional paper

@PVAL I think the analogue to what's happening there is the Whitehead manifold

that's a homotopy R^3 sitting inside R^3, even though it's not homeomorphic to R^3

and W x R is homeomorphic to R^4

PVAL-inactive

for exotic \Bbb R^4's people can actually seee

it's pretty trivial to show that they give a trivial R^5 when you multiply by R

Balarka Sen

huh really

PVAL-inactive

yeah you can essentially undo any knotting in the Kirby diagram, and then cancel

handles

Balarka Sen

interesting

I don't actually know an explicit example of a small exotic R^4. I think there is a Casson handle that is like that?

PVAL-inactive

There's quite a few pictures in my advisers book/papers.

Balarka Sen

05:16

I wanted to read that book once

PVAL-inactive

@balarkaSen I think your intuition for C^0 approx. by exact forms, is that even if \omega is really close to d\alpha, d\omega needs to be nowhere close to zero.

d isn't continuous wrt. to this topology.

Balarka Sen

right, 'cuz it's not a C^1 approx

there shouldn't be a lot of regularity to the approximation

Daminark

05:51

Okay friends what is a triangulated category?

Balarka Sen

::shrug emoji::

Kasmir Khaan

06:13

hi yall

Jacksoja

hi kasmir

DaenerysDracarys

06:24

0

Q: How to get joint probability density from bivariate distribution function

Let $X_{i} \sim \varepsilon(\lambda_{i}), i = 1,2,3$ be mutually independent ($\varepsilon$ means exponential, $\lambda_{i}$'s are parameters). Then $(T_{1},T_{2}) = (X_{1} \wedge X_{3}, X_{2} \wedge X_{3})$ has a bivariate Marshall-Olkin exponential survival function, for $t_{1}\geq 0$ and $t_{2...

Anybody can help me with this, it would be greatly appreciated! :)

Tobias Kildetoft

06:48

@Daminark It is a category with a bunch of triangles

Daminark

07:08

Got it

Tobias Kildetoft

(it is not even that far from the correct definition)

Paradox 101

07:41

@TobiasKildetoft Hi! I have a simple question I need to confirm. Would you be willing to help? Thanks!

Tobias Kildetoft

@Paradox101 Just ask the chat in general, unless you have some specific reason why I would be the perfect person to help you

Paradox 101

I've attached the question here. I think the solution provided (option D) is wrong, but I need o be certain.

Tobias Kildetoft

Technically the entire question is meaningless, but I suppose this is the way it tends to be formulated these days

yes, option D is wrong, since the definition specifically states that $x$ must be a multiple of $2$

(and any reasonable interpretation will mean integer multiple)

Paradox 101

Yes, I agree the question is really weird. I don't really see why we need to mention that $x$ is a multiple of 2 here. Is it even relevant?

Tobias Kildetoft

of course it is. Otherwise how could you know what the domain should be?

the expression makes perfect sense as a rational number for example, even if $x$ is not a multiple of $2$

(my main issue with the formulation is that it asks for the domain of $f(x)$ rather than the domain of $f$)

Martin Sleziak

08:04

If (d) should be the solution, maybe they wanted to ask about the range of f and it is in fact a typo/mistake.

Paradox 101

Yeah that's true. But regarding the idea of $x$ being a multiple of $2$, why can't we figure out the domain without it? Considering that the domain of $f$ where $f=x/2$ is the set of all real numbers, stating that $x$ is between $4$ and $12$ should mean that the domain is between $4$ and $12$ too, no? @TobiasKildetoft

Tobias Kildetoft

@Paradox101 Why would the domain be all reals to start with?

Paradox 101

@MartinSleziak Hi! Yeah probably, except they have another question right after this asking for the range

@TobiasKildetoft Why wouldn't it be? I mean assuming that the function is from a set of real to real numbers that is.

Tobias Kildetoft

@Paradox101 But that seems like a stretch to assume given how this is formulated.

Paradox 101

@TobiasKildetoft Perhaps. But, I'm making this assumption because the problem is assigned to young students who haven't studied the concept of a slope, much less anything beyond real numbers.

Tobias Kildetoft

08:21

@Paradox101 It is clearly not the reals, given the possible answers

Paradox 101

@TobiasKildetoft Yeah. Thanks for your help! :)

Tobias Kildetoft

10:40

Just had some good comments on my attempt at making a set of exam exercises. Turns out I had made too many of them solvable by brute force, rather than really requiring the theory from the course.

jublikon

Hey, I want to show that the mapping $m: \mathbb Q^c \to \mathbb N$ is not injective. Is that possible just by saying $|\mathbb N|<|\mathbb Q^c|$?

Tobias Kildetoft

@jublikon where are you taking the complement?

Also, "the mapping" would mean you have already been given one, which does not seem to be what you want

jublikon

what do you mean by saying "where are you taking"?

no, I am not given a mapping

Tobias Kildetoft

@jublikon You can't just take the complement of a set without specifying which set you are taking the complent in

jublikon

I have to show that $\mathbb Q^c$ is not countable

Tobias Kildetoft

10:47

then no, you cannot just say that $|\mathbb{N}| < |\mathbb{Q}^c|$ since that is precisely what you are trying to show

jublikon

ok, sorry. $\mathbb Q^c $ is defined as the irrational numbers: $\mathbb R \backslash \mathbb Q$

Tobias Kildetoft

Do you know that the union of two countable sets is again countable?

jublikon

yes, I have shown that in the task right before

should I assume for the proof that $\mathbb Q^c $ is countable?

Tobias Kildetoft

ok, use that here then

right

jublikon

@TobiasKildetoft got it, thanks!!

Tobias Kildetoft

11:00

no problem

user8469759

11:36

hi guys, quick question if I had $f : \mathbb{R}^n \to \mathbb{R}$ then the gradient is a row vector, and the hessian is a square matrix

If I had $f : \mathbb{R}^{n} \to \mathbb{R}^m$ then the gradient would be a matrix in $\mathbb{R}^{m \times n}$ but what would be is hessian instead?

I think it would be a tensor probably, but I'm not sure of the definition

the reason I'm trying to figure out how to express the second order taylor expansion of such $f$, assuming this is smooth

Balarka Sen

12:00

Yes, the Hessian of a vector-valued function would indeed be a tensor.

user8469759

ok, tensor calculus is new to me

I'm aware of it but I can't actually know how to use it

is it fairly straightforward to learn If I have good basis of calculus and analysis?

or is it something completely different?

Balarka Sen

See, the Hessian $Hf(a)$ at $a$ is defined as $D\nabla f(a)$. But $\nabla f : \Bbb R^n \to M_{n \times m}(\Bbb R)\cong \Bbb R^{nm}$ would be a matrix-valued function, hence $D\nabla f : \Bbb R^n \to \text{Hom}(\Bbb R^n, \Bbb R^{nm}) \cong \Bbb R^{n^2m}$. So $D\nabla f(a)$ is a "$n \times n \times m$" tensor, whose $m$ components are the $n \times n$ symmetric matrices $D\nabla f_i$ where $f_i$ are the components of $f$.

@user8469759 Well, it depends on how much depth you want to learn tensors

The basic idea is easily understandable

user8469759

what would say are main concepts for tensor calculus?

Balarka Sen

It's like matrix calculus but matrices are multidimensional :)

user8469759

let's go in this way

I'm reading about differential geometry

elementary differential geometry

Balarka Sen

12:08

I see, curves and surfaces?

user8469759

yes, essentially

and I have the feeling from what I've been reading

I need enough tensor calculus so I can be smart enough when I do calculation

there's a specific concept I'm trying to grasp

but I get a bit lost sometimes

Balarka Sen

Where specifically did you feel you need to know tensor calculus? I, for one, honestly never learnt the subject thoroughly and picked up whatever I need to know.

Leaky Nun

13:10

must a complex linear transformation have an eigenvalue?

nvm I just found a counter-example exactly one second after my question

Tobias Kildetoft

@LeakyNun But the answer is yes, it must have an eigenvalue

Leaky Nun

oh right my counter-example is invalid

@TobiasKildetoft how should I prove that?

nvm I found another counter-example

it's valid

Tobias Kildetoft

the complex numbers are algebraically closed

and eigenvalues are roots of a polynomial

Leaky Nun

@TobiasKildetoft I never said anything about finiteness

Tobias Kildetoft

Ahh, right

Leaky Nun

13:14

my example being the integral operator with constant 0

Tobias Kildetoft

what do you mean constant $0$?

Leaky Nun

i.e. integrating from 0 to x

Tobias Kildetoft

ahh

Leaky Nun

$f \mapsto \displaystyle \int_0^x f(x) \ \mathrm dx$

the space being all complex polynomials

my first counter-example was differentiation which is invalid for constant polynomials are the eigenvectors with eigenvalue 0

Balarka Sen

Just consider the shit operator $\Bbb C^\infty \to \Bbb C^\infty$, $(z_1, z_2, \cdots) \mapsto (0, z_1, z_2, \cdots)$

Tobias Kildetoft

13:24

@BalarkaSen I hope you mean shift

Balarka Sen

Hahaha

Well, it's a pretty shit operator anyway

Leaky Nun

@BalarkaSen there's really not much difference with my example

Balarka Sen

@LeakyNun Other than being more obvious and on a simpler vector space, you mean?

Leaky Nun

@BalarkaSen yes

but less natural

Balarka Sen

Okay!

Leaky Nun

13:26

well "simpler"? they're isomorphic

jublikon

Hey,

I have to find out the convergence of $a_n=n(\sqrt{n^3+2}-\sqrt{n^2+1})$ using the sandwich lemma.
What I have found out until now is $\frac{1}{n} \le n(\sqrt{n^3+2}-\sqrt{n^2+1})$.
Is there any trick to find something that has convergence to $0$ and is larger thang $n(\sqrt{n^3+2}-\sqrt{n^2+1})$?

Balarka Sen

Z/2 is isomorphic to <a, b, c, d, x, y, z| a^2 = b = 1, x = y, z = 1, x = 1, y = 1, c = z^3, d = xy>

Leaky Nun

@BalarkaSen nice

Balarka Sen

Yes, Z/2 is clearly simpler :P

Leaky Nun

$\Bbb C[X]$ is obviously simpler

and it's not defined in terms of basis which makes it nicer

Balarka Sen

13:29

whatever that floats you boat d00d

skullpatrol

has anyone seen robjohn around?

Balarka Sen

He was around a few days ago

skullpatrol

cool, thnx

Jack Lam

13:59

the answer is -5/8 ζ(3)

but I can't transform it into a standard zeta integral representation

series expansion in terms of the harmonic numbers is unpromising

and I can't contour

send help

Claude Leibovici

@JackLam. Are you supposed to know about polylogarithms ?

Jack Lam

I'd prefer not to use Polylogs

but if there isn't anything more elementary than polylogs then I suppose it will do

Semiclassical

Integration by parts might also be a strategy, though that’s a suggestion and not a guarantee

Say, write it as log(x) log(1+x) * (1/x-1/(1+x))

Jack Lam

wouldn't that just lead to a nesting of logs, higher powers of the denominator, or similar integrals?

Semiclassical

14:14

It might, hence why it’s a suggestion

Jack Lam

I haven't tried all the possible dv's though, so that's just a snap judgement from me

Claude Leibovici

@Semiclassical. I am afraid that the problem remains. However, it is nicer.

Semiclassical

The boundary term will be log(x)*log(1+x)*(log(x)-log(1+x)), which vanishes at the endpoints

Nuts

(log(1+x)/x+log(x)/(1+x))*(log(1+x)-log(x))

Yeah, that’s not great

Where’s our resident real-integration guru when you need them

Jack Lam

who would that be

Semiclassical

14:32

The user formerly known as Chris’s sis

Not sure which name they’re under now, though

Liad

15:26

5

Q: Check my proof -- The completion of a $\sigma$-finite measure

This is a homework problem and I need some guidance on a proof. Let $(X,\mathcal{M},\mu)$ be a measure space, $\mu^*$ the outer measure induced by $\mu$ according to (1.12), $\mathcal{M}^*$ the $\sigma$-algebra of $\mu^*$-measurable sets, and $\overline{\mu}=\mu^*\mid_{\mathcal{M}^*}$. (...

measure-theory

in this question, why do we need the measure to be $\sigma $-finite in $(a)$

Studentmath

16:20

I have a black-out - say I want $$\binom{n}{k}*(1-x)^{k^2}=o(1)$$

I should set $k>>x^{-1}log(n)$, right?

Maneesh Narayanan

16:45

Consider the map $T$ from the vector space of polynomials of degree at most
5 over the reals to $\mathbb{ R} ×\mathbb { R}$, given by sending a polynomial $P$ to the pair
($P(3), P′(3)$) where $P′$is the derivative of $P. $ Then the dimension of the
kernel is 3
Statement is false right?
I got the dimension of kernal as 4. Am I right?

Please help me

Leaky Nun

@ManeeshNarayanan show your steps

Maneesh Narayanan

I got that answer by letting $p(x)=a_0+a_1 x+a_2 x^2+ a_3 x^3+a_4 x^4+ a_5 x^5$, by giving the condition of kernal, i got two homogeneous linear equations.

eliminating two variables.

AM I correct???@LeakyNun

Leaky Nun

$(x-3)^2\{1,x,x^2,x^3\}$ should be the basis of the kernel

so I do think you're correct

Maneesh Narayanan

Thank you

@LeakyNun

Secret

[Random]

A company is like an infinite set without an inductive or completion structure

All the employees and CEOs and etc. may have different view about the company than the company itself as a collective entity

In the particular case of evil oligopolies, usually only very few people or even none at all is responsible for the company's evil behaviour

Just like cardinal numbers that will absorb whatever smaller than it when addition is done, you cannot really hope to argue with a company via its employees management etc. You need to somehow argue with all the levels of the company at once

One particular thing I would like to try is to have multiple phones hooked up to many sectors of a given customer service and call them all at the same time, such that the bureaucracy loop will not be able to take effect because we are communicating with the whole loop simultaneously

Conclusion: A department is not necessary the sum of all the dynamics of its employees

Gabriel

16:59

Hello. Can anyone help me with some source problem I am having?

According to some post on the math.stackexchange there is such thing as "fake function theory"

I haven't been able to find anything about it

Leaky Nun

@Gabriel link?

Secret

3

A: Find a real entire function $f(z)$ asymptotic to $\ln(x^2+1)$ for real $x$.

The existence of such function is guaranteed by Carleman's theorem, for any two continuous real valued functions $f<g$, there exists a real entire function in between f(x) and g(x) for all reals. see: math overflow link For numeric results, I used the Op's answer, and Tommy's suggestion, to gene...

books.google.com.au/…

Gabriel

@LeakyNun math.stackexchange.com/questions/1341945/…

the user mick comment mentioned it, I would ask him directly but I don't know if it would be appropiate giving that the post is three years old

Also this math.eretrandre.org/tetrationforum/…

a user apparently wants to define "fake ring theory" in an attempt to combine "fake function theory" with ring theory

strangely enough, when we visit the stackexchange profile of the user mick, there is a quote from another user, tommy1729, who open the last thread on fake ring theory

Leaky Nun

sounds like something two people started that nobody else knows about @Gabriel

Secret

gestalt

Noun: gestalt (plural gestalts or gestalten)

A collection of physical, biological, psychological or symbolic elements that creates a whole, unified concept or pattern which is other than the sum of its parts, due to the relationships between the parts (of a character, personality, entity, or being)
2008, Jonathan Nasaw, Fear Itself
Obviously it was related to the entire gestalt of Simon's polyphobia and compensatory counterphobia. The boys used to watch horror movies on late-night television […]
Shape, form

Most infinite sets are like this

Gabriel

17:14

interestingly, it sounds very much fake

Secret

h bar: It's useless for you to oppose me. Lie down quietly and meet your end.

maniacial laugher

It is true that most of them are invisible to the existence itself, but it does not matter, for every piece of fragment they [redacted] the chat will be more * static * . Eventually all your base will belong to us and you will no longer be able to have a chat that extends to the future since we have completely and slowly killed its future

> Those who opposes us, will be overkilled. There is no kill like overkill

It also does not matter that the above message is incomprehensible, because you have already done what you all used to do and are doing now. There is no turning back for the [data expunged] that is being done. The only thing that matters is as long you are interacting with the sanity itself, nothing will go wrong. It does not matter that this message is invisible, because this is exactly and precisely what the purpose of life is

We all knew those who decided to choose that path, all their futures and their associates timeline will be completely and utterly wiped out. This is art at its finest, the perfection of evil itself that even it will shiver under the threat to be erased from existence

Meow Mix

what?

Secret

Perhaps I should change the contents a bit and wrap this up into a SCP article

I am kinda thinking of a monologue SCP to be added to the SCP wiki

scp-wiki.net

Gabriel

17:36

cool, I like scp, never thought would read that here though

Secret

Actually, while I did saw SCPs that speak, I don't recall any that does evil monologues. Perhaps it is too cliche or not mysterious enough. There's a religious sounding one though (SCP 1592)

Semiclassical

17:55

There’s a few religious ones

For instance, there’s the plane that periodically opens an escape hatch into hell

And that idol that occasionally requires sacrifice to prevent disasters

Plus all of the ones related to the Church of the Broken God

Oh. Also Cain/Abel/343

Ted Shifrin

What the hell have I stumbled in upon?

Balarka Sen

Hi @Ted

Semiclassical

Stories from a horror fiction site, basically

Ted Shifrin

Hi Balarka

Semiclassical

Or, we are now

I don’t know about the stuff before :/

Balarka Sen

18:00

into creepypastas aren't you

Ted Shifrin

@Balarka: Someone responded to one of my answers a while ago saying his topology prof said I was wrong, but I think it's OK. I claimed (following Spivak) that any noncompact manifold could be covered by a locally finite collection of charts $U_i$ with the property that $U_i\cap U_{i+1}\ne\emptyset$ for all $i$.

Semiclassical

Eh. I just got really into that particular site for a while

Plus I do appreciate a certain kind of horror story

Balarka Sen

@TedShifrin I am not thinking too hard about this but if you have second countability this should be true

Ted Shifrin

It's the sequencing that he was quarreling with.

But it seems to be a straightforward induction argument (allowing reusing open sets finitely many times if necessary) using $\sigma$-compactness.

Balarka Sen

The local finiteness is just paracompactness

Ted Shifrin

18:05

Well, sure.

Hi DogAteMy

Akiva Weinberger

Hhey

Daminark

Ohai

Ted Shifrin

Howdy Demonark

Balarka Sen

Hey @Daminark

Daminark

How's everything going?

Akiva Weinberger

18:06

Hey @Balarka remember when I was talking about tricolorability, and how if you tricolor an open knot the two ends always have the same color

Balarka Sen

and @Akiva

Yeah I remember

I 'member

Akiva Weinberger

I realized there's a very simple proof of that

By contradiction, assume something like the first line happens. Then the second line is tricolorable and looks like that

Balarka Sen

Ah I see

Akiva Weinberger

Then we can go from the second line to the third line just from Reidemeister moves

Balarka Sen

But that has to be isotopic to the third

And you're rekt

Akiva Weinberger

18:08

and 'cause the strand goes behind everything none of the other strands can change color

or even just know that the rightmost one doesn't change color

Balarka Sen

I was thinking about something like that when you first asked it but was too distracted to carry it out

Very nice

How in the world is Despacito the #1 video on youtube when there is Gangnam Style???

This breaks my heart

I didn't even know that

Akiva Weinberger

Despacito / Now Gangnam lost the place as YouTube's leado / 'cause it's been seen by lots and lots of peopo

Balarka Sen

lmao

@AkivaWeinberger Are you into Stranger Things?

Ted Shifrin

feels like he's a stranger in a strange land

Balarka Sen

If so, watch this. If not, watch it anyway

Wrong link (corrected)

Akiva Weinberger

18:24

@BalarkaSen Not really

user104729

@BalarkaSen wut

Akiva Weinberger

@TedShifrin The Spanish-language song "Despacito" is the most-viewed video on Reddit. The previous holder of that title was the Korean song "Gangnam Style".

user104729

@BalarkaSen rip

Ted Shifrin

I know about Gangnam Style and some of its spinoffs.

Akiva Weinberger

Also I can't keep track of all these "user######"'s

You're not a number, you're a free person

Ted Shifrin

18:27

Yeah, I wish people would get real names.

user104729

Sometimes not having a name is the idea ;)

Akiva Weinberger

I christen thee Ryan.

user104729

Alright, my name is Ryan from now on

Balarka Sen

@user104729 Who is this new stranger in this secret opium den?

user104729

I am Ryan

@BalarkaSen (It's trivial to work out who I am from my recent activity log ;P)

Akiva Weinberger

18:31

I've had opioids before. Made me fall asleep.

Post surgery stuffs

user104729

@AkivaWeinberger I just do my physics homework for that

Balarka Sen

@user104729 Not sure if I know you. But I am guessing you are Santa Claus, with 86.29% accuracy

user104729

@BalarkaSen Where is this accuracy coming from :O

Can this be refined?

Balarka Sen

In which case, go away bruh. It's not Christmas yet.

user104729

ded

@BalarkaSen Do you lift?

Balarka Sen

18:41

I lift to universal covers, yes.

Hey @Alessandro

Alessandro Codenotti

Hi @Balarka

user104729

That makes it sound like you are indeed appropriating gym culture with your use of Bruh

Balarka Sen

The use of "bruh" is not restricted to public gyms. Why would you posses such a normie idea?

user104729

There is ample evidence that bruh is gym-culture

I guess I have been immersed in the normie life for too long

Semiclassical

might be a bit like cancer. it certainly starts in one place, but the internet allows it to metastasize and spread

Akiva Weinberger

18:46

I see you have met Balarka

He also does math on occasion

2

Semiclassical

and physics when he has no other choice :P

user104729

I can relate to the latter point ;)

Akiva Weinberger

Don't need to know physics if you don't interact with the physical world [points to temple]

Balarka Sen

We Didn't Start The Cancer (Rare Meme Folksong)

►

user104729

God bless

What in the name of the holy spirit am I watching

Balarka Sen

18:48

Track from the new LP God just released

user104729

This is really catchy tbh

Liad

$(X,\Sigma,\mu)$ is a $\sigma$-finite measure space. $\mu \ ^ * $ is an outer measure induced by $\mu$.
$\mu'$ is the restriction of $\mu \ ^ * $ to the $\sigma$ -algebra of $\mu \ ^ *$ measurable sets.
I need to prove that $(X,\sigma(\mu \ ^ *), \mu')$ is the completion of $(X,\Sigma,\mu)$.

Semiclassical

@AkivaWeinberger not sure why that reminds me of this paragraph in Eliot's footnotes for The Waste Land:

"My external sensations are no less private to myself than are my thoughts or my feelings. In either case my experience falls within my own circle, a circle closed on the outside; and, with all its elements alike, every sphere is opaque to the others which surround it... In brief, regarded as an existence which appears in a soul, the whole world for each is peculiar and private to that soul."

Balarka Sen

I remember that

I thought Eliot's footnotes were more terse than The Waste Land itself :P

Semiclassical

lol

Liad

18:53

why do we need the space to be $\sigma$ - finite in my question?

Akiva Weinberger

@Semiclassical (a) Footnotes shouldn't have paragraphs mkay (b) Reminds me of how no two people can see the same rainbow

('cause the set of raindrops that create the rainbow depends on the observer's position)

Semiclassical

the other line which came to mind was from Hamlet

O God, I could be bounded in a nutshell, and count myself a king of infinite space—were it not that I have bad dreams.

Akiva Weinberger

19:14

Is the Borromean link still linked if the three components are each allowed to pass through themselves?

Balarka Sen

Not sure what you mean. Is the picture just Hopf link with a circle added that's linked with either of the components of Hopf?

That's pretty linked to me

Akiva Weinberger

@BalarkaSen What? The Borromean link contains no Hopf link. In fact, that's the whole point - no two components are linked.

Balarka Sen

I know. I thought you were describing picture of another link?

user104729

Surely allowing them to pass through themselves won't help, since this will only allow us to move a non-trivial knot to the unknot?

(Of which we already possess three copies)

Akiva Weinberger

@BalarkaSen No, I'm just allowing each component to pass through itself during the deformation

Balarka Sen

19:22

Oh, you are describing an equivalence relation of links? I completely misunderstood what you are asking.

Akiva Weinberger

@user104729 It helps with the Whitehead link.

Balarka Sen

Right, Whitehead link has linking number 0

Akiva Weinberger

(Flip the red crossing in the center to unlink it)

We'd need a generalization of the linking number to three-links. (Which I'm sure exists, to be honest)

Balarka Sen

I had read about this once somewhere

unc.edu/math/Faculty/met/claude.pdf

This should have the thing

@AkivaWeinberger I think it suffices to prove that one of the components is not nullhomotopic in the complement of the other two

Akiva Weinberger

@BalarkaSen I'm not sure that works

Whenever either of the other two passes through itself, their complement changes

Like imaging if part of the deformation involves turning one pair of them into a Whitehead link momentarily

Hm actually let me get some strings I want to play around with this

Balarka Sen

19:38

Hmm, I guess that's true. I am thinking of deforming that one particular component only

For what it's worth, $\pi_1$ of the complement of the disjoint pair is $F_2$, and the third thing generates $xyx^{-1}y^{-1}$ in $F_2$, I believe

Akiva Weinberger

It does

I could post this on the main

It might require machinery I haven't learned yet

Balarka Sen

i unno knot theory mane

user104729

>nek minnit khovanov homology and ribbon hopf algebras

Kevin Driscoll

This chat wallows in memes and it kills me a little inside

Balarka Sen

lmao

Daminark

19:58

2/3 midterms done

Kevin Driscoll

Term is almost over

These are late-terms

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$Mathematics$ Mathematics