Mathematics - 2019-05-21 (page 1 of 2)

Ted Shifrin

00:12

@BalarkaSen LOL ... Forsaking topology?

Balarka Sen

Why not both? Hi, @Ted.

GFauxPas

good evening Ted

Ted Shifrin

You said it was unhealthy, Balarka, not me :P

hi @GFauxPas

GFauxPas

can you help me with ODE's or is that not your department of expertise?

Ted Shifrin

certainly not my department of expertise

Balarka Sen

00:15

I am not being able to comprehend how this is garnering so much upvotes. There are obvious counterexamples $S^1 \to \Bbb R^2 \setminus \{0\}$.

GFauxPas

I'm doing practice problems for my exam

alright then, hopefully Semi will have time to help me tonight

thanks anyway

Ted Shifrin

@Balarka: Is your point that it may not longer be a weak homotopy equivalence?

Balarka Sen

To the image, no.

Ted Shifrin

I mean, it could be a homeomorphism to the image ...

Balarka Sen

Well, that's precisely the source of counterexamples. Take an immersion of S^1 in R^2\0 as a figure eight with one lobe wrapping around 0

This is a fine (strong!) homotopy equivalence, but the map to image is a map S^1 -> S^1 v S^1

Ted Shifrin

00:20

Oh, I see your point. I was missing it.

I don't think it's a horrible question for a newbie to ask (I have to say that cuz I missed your point entirely).

Ryan Unger

That's a cool counterexample. The point is that the image has more holes than the codomain.

In hindsight obvious...

Balarka Sen

Pondering on whether to post as an answer to gain cheap upvotes, or whether to comment to ruin other's chance of answering and gaining upvotes

Ted Shifrin

I think you should just post the answer, a @Balarka.

Ryan Unger

theres an answer

Ted Shifrin

or post a hint as the answer.

Balarka Sen

00:22

Great, I have to do neither!

Ted Shifrin

Your answer is way more interesting than his ...

Contractible things can wrap and make circles. Duh.

Balarka Sen

Ok fine I'll post a comment

@Ted By the way, every simplicial complex is indeed a "simplicial quotient" of a simplex, in the sense that there is a quotient map $f : \Delta^n \to K$ from a triangulated $\Delta^n$ which takes $i$-simplices to $i$-simplices (this is stronger than saying $f$ is a simplicial map).

Ted Shifrin

Crazy.

I guess we never were taught this because ... who really cares?

Balarka Sen

Here's the idea: take the dual graph of $K$, joining midpoints of every face, and take a maximal tree $T$ of it. Mark the codimension 1 faces of $K$ which doesn't intersect the edges of $T$, and cut $K$ along those faces. You get a new simplicial complex $K'$ which deformation retracts to a tree, so is topologically a ball.

So $K$ is a quotient of $K'$ obtained by gluing some faces, and $K'$ is just a triangulation of $\Delta^n$ as promised

Ted Shifrin

What a strange picture.

Balarka Sen

00:31

This is basically "cutting a cube into a cross of squares", formalized :D

Ted Shifrin

Not that I understood that.

Balarka Sen

There are those pictures of polyhedra unwrapped, so you can glue the edges back to get the polyhedron again, right? Just making that idea precise

Ultradark

Can it be shown that integration is "harder" than differentiation on average

Balarka Sen

Anyway just a random question with a random answer on my behalf

Ted Shifrin

a @Balarka: I agree it is interesting, but I don't see why we should care, since the whole premise of homology is that we do chain complexes.

Balarka Sen

00:34

I agree there's no major point to it. The only thing you could say is perhaps that nice spaces we see in topology are all ball quotients but that's easier: every connected finite dimensional finite CW complex is quotient of a $D^n$

Ted Shifrin

I don't know that that is a helpful way to visualize the CW complex, though.

In fact, I know it isn't :)

Anyhow, the right way to think about the fundamental class, obviously, is the way you explained ... not in terms of a singular singular simplex.

Balarka Sen

Haha I agree

Ted Shifrin

But thanks for teaching me something :)

Balarka Sen

Nah just wanted to share a proof of something which is useless but partially creative maybe

Ted Shifrin

I think it's an interesting fact ... which I would have been (and was) loath to believe.

But it's useless nonetheless, unless we find some cool application.

Balarka Sen

00:38

Yup

Ted Shifrin

I think it's a great exercise.

Balarka Sen

00:50

this is a very interesting theorem

Ted Shifrin

Well, knot so interesting to me.

Balarka Sen

cheap joke Ted

Ted Shifrin

I was about to leave to go cook dinner.

Balarka Sen

Hahah

Ted Shifrin

So it was my parting gesture.

GFauxPas

01:00

Ted's favorite area of math is knot theory, and neither is it mine

Ryan Unger

@BalarkaSen oh that reminds me, I wanted to say that Freedman had a proposal to characterize unknots by computing a certain nonlocal energy over an isotopy class

Balarka Sen

o shit tell me more

Ryan Unger

jstor.org/stable/2946626?seq=1#metadata_info_tab_contents

the first couple pages are readable

the problem is that the functional is really badly behaved

there's been some recent progress

as far as I know that's the only real geometric problem people have tried to tackle with nonlocal techniques

Balarka Sen

I see pages of integrals my dude

Ryan Unger

well yeah it's an energy

were you expecting topology?

Balarka Sen

01:04

sigh i guess

I've had good experiences with Freedman in the past. Lots of pictures in his Bing topology notes, and some proceedings on low dimensional topology I pilfered from the library, ...

Ryan Unger

now one could look at the gradient flow of this thing

but that's getting into the realm of things that are too complicated

Balarka Sen

geometric analysis is too hard

Ryan Unger

it's definitely hitting a steep difficulty barrier

Someone told me he wouldnt take any more PhD students because Ricci flow is too hard for grad students these days

too much work for very small results

Balarka Sen

we should just all do algebra

motivic stacks is where the results are at

Ryan Unger

that's what I've been saying

but then you look at what these ppl are doing and its like jesus

Balarka Sen

01:08

i have no clue how people do algebraic geometry

i cant even read chapter 2 of hartshorne

how do they just do it

Ryan Unger

there's still open problems about harmonic functions in R^3

why do we need sliced stack infinity categories

when we can just study simple differential equations

it was good enough for Lagrange

Subhasis Biswas

You guys didn't sleep?

Ryan Unger

it's 9PM...

Balarka Sen

it's 6:38AM!

Ryan Unger

well I have no evidence that you're a real person

Subhasis Biswas

01:10

Oof

Ryan Unger

you could just be a google ai

would make a lot of sense actually

Balarka Sen

alexa jones meme here

Subhasis Biswas

Did he fail the captcha test?

Ryan Unger

have you read the kirby calculus book

it's on my list of books I want to read but never will

Balarka Sen

Gompf-Stipsicz?

Ryan Unger

01:12

sounds right

Gompf is retiring! I was sad to find that out when I visited UT Austin

Balarka Sen

I want to. I remember reading a little once; I learnt the proof of representability of homology classes of $H_2(M)$ by embedded surfaces for a $4$-manifold $M$ from there

Travis R

Sorry to interrupt any legit conversations going on. I'm not a mathematician and it's possible that my question is boringly simple to most participants on this exchange, but I don't have any reputation to put up for bounty and the meta exchange suggests using the chat as another way to get attention for an unanswered question, so that's what I'm doing. Thanks in advance to anybody who can help: math.stackexchange.com/q/2896723/500494

Ryan Unger

oh that's in there?

hmmmmm

Balarka Sen

They do some strange proof, using duality to get a map $M \to K(\Bbb Z, 2) = \Bbb{CP}^\infty$, homotope that down to $M \to \Bbb{CP}^2$ and then make it transverse to $\Bbb{CP}^1$, take it's preimage. That dude will represent the class

Ryan Unger

what the hell lol

Princeton doesn't have an algebraic topology class so I'll never actually learn topology

It doesn't even have a number theory class but somehow there's lots of number theorists

Well there's that one I showed you

Maybe that's a good intro :P

Balarka Sen

01:17

I wish you luck with anabelian p-adic Hodge theory, which is of course grad number theory 101

Ryan Unger

introduction to perfectoid spaces

Érico Melo Silva

@RyanUnger they in fact do offer one

Ryan Unger

@ÉricoMeloSilva this fall?

Subhasis Biswas

Okay okay hol' up.

Érico Melo Silva

they offer it in springs

Subhasis Biswas

01:18

Noob here

Ryan Unger

Oh cool

Érico Melo Silva

yeah at least this spring they had one

Subhasis Biswas

How one goes about learning abstract algebra. Step by step?

Ryan Unger

Oh I didn't rember that

oh yeah there is it, whoops

"Topics in Algebraic Geometry:
Introduction to \ell-adic etale
cohomology"

@BalarkaSen explain

Balarka Sen

vOv

u learn Weil conjectures

thats obviously the right place to start doing alg geo

because historically it was :D

Ryan Unger

01:20

the right place is probably mirror symmetry

it's at least somewhat analytic

Balarka Sen

I bought a book with my contingency grant this semester

"Enumerative Geometry and String Theory"

by Katz

Subhasis Biswas

@RyanUnger @BalarkaSen

Balarka Sen

it has mirror symmetry in it

Ryan Unger

@SubhasisBiswas probably take a class unless you're very disciplined

Joe Shmo

hi chat

Balarka Sen

01:22

@RyanUnger There are direct proofs if you have a smooth 4-manifold; any cohomology class can be represented by a map from a simplicial complex which is a manifold away from codimension 2 (this is not obvious but not hard), so in this case you get a map $K \to M$ from a simplicial $2$-complex which is a manifold away from a bunch of points. "Blow it up" at those points to get a map $S \to M$ from a surface. Homotope that to a immersion with clean double points

Now you have an immersed representative

Ryan Unger

@BalarkaSen I've seen maybe half a dozen mirror symmetry talks, only two defined what it is (and it was the same guy twice)

Balarka Sen

then at the double points just "desingularize it"

Subhasis Biswas

Okay. I have to school myself. I missed two semesters of work.

Ryan Unger

but I don't think he defined the Fukaya category

Balarka Sen

embedded representative viola

Ryan Unger

01:23

@BalarkaSen does the blow up make $S$ noncompact

Balarka Sen

Nah

you just remove a cone nbhd of the singularity and cap off balls on two sides

Ryan Unger

then what do you mean by blow up

two sides? so $S$ is disconnected?

Balarka Sen

think of transitioning from a cone to a hyperboloid of one sheet

$S$ may possibly be disconnected

Ryan Unger

@BalarkaSen ok sure

do you do the Whitney trick for the double pts

Balarka Sen

Maybe. I am not very familiar with the Whitney trick

The local model at the double points is $x = y = 0$ and $z = w = 0$ in $\Bbb C^2$

Two transverse 2-planes in R^4 intersecting at a point

Ryan Unger

01:27

well when you say desingularize a double point for something half-dimensional ,that's literally the whitney trick

Balarka Sen

The link of that point is a Hopf link in S^3, so just cut out the interior and glue a Siefert surface of the Hopf link back

this makes it an embedding at the cost of increasing genus

@RyanUnger Hm ok

I thought it was changing an immersed half-dimensional disk to be embedded

Ryan Unger

yes

is that not what you're doing here

maybe I'm misunderstanding what kind of singularity you've got here

Balarka Sen

Not really I suppose. I am replacing two transverse 2-disks by an embedded surface, leaving the boundary intact

Anyway all of this requires a smooth structure which misses the point lmao

Ryan Unger

manifolds are smooth m8

Balarka Sen

that's why Gompf-Stipsicz do it in that weird way

Oh but they are also making things transverse I suppose

Ryan Unger

01:31

well just general position right

you can triangulate the surface

Balarka Sen

$M^4$ might not be triangulable/have a smooth structure

So I don't know how to play general position games there

Ryan Unger

but you want the 2-dimensional guy to intersect itself transversally, no?

it admits a triangulation

Balarka Sen

That's true, but so what? The triangulation will be messed up in the rangle unless the map is simplicial

but to get a simplicial map you need a triangulation on the range as well innit

Ryan Unger

can you triangulate a neighborhood

Balarka Sen

that's local flatness

IDK homie too hard im leaving

Ryan Unger

01:34

ok good morning

Balarka Sen

I am a $C^\infty$ man

Ryan Unger

3 mins ago, by Ryan Unger

manifolds are smooth m8

hypersurfaces, on the other hand...

hah, Freedman's name is misspelled in vol 2 of Yau's collected works

Ted Shifrin

@BalarkaSen Not if you do Whitney stratifications :P

loch

@BalarkaSen exciting

Ted Shifrin

Sad that Gompf is retiring. He's way younger than I am :P

Ryan Unger

01:40

you retired too!

Ted Shifrin

Yes, but I'm more than 4 years older :P

Balarka Sen

@TedShifrin I do Thom-Mather $C^\infty$-stratified spaces

Ted Shifrin

Actually, it seems it's exactly 4 years. But he got his Ph.D. 5 years after me (he started at Berkeley just as I left).

But the spaces aren't smooth, @Balarka :P

Ryan Unger

@BalarkaSen what the hell is that

Ted Shifrin

LOL

Balarka Sen

01:43

@loch I was reading Schubert calculus from that, although that's prolly not your flavor of enumerative geometry

There's no schemes tho lol

@TedShifrin But it's the next best thing!

Stratumwise smooth structure, and glues smoothly along the frontiers

Ted Shifrin

Schubert calculus is great.

Well, @Balarka, algebraic/analytic varieties are even nicer :)

Balarka Sen

True, all of those were motivating examples for Whitney/Thom/Mather/MacPherson/Goresky et al to study stratified spaces of course

I should know the algebro geometric side of the story one day

loch

@BalarkaSen oh Schubert calculus is great!

Ryan Unger

where can I learn about branched conformal immersions and what that has to do with Riemann Roch

Ted Shifrin

What are your domain and range, @Ryan?

Ryan Unger

01:50

$\Sigma^2_g\to S^2$ or $\Bbb RP^2$

Balarka Sen

That's a branched covering my dude

Riemann-Hurwitz formula :3

Ted Shifrin

Mapping to $\Bbb RP^2$ it won't have anything to do with Riemann Roch.

If you're studying holomorphic maps, then Riemann Roch will be relevant.

Ryan Unger

@BalarkaSen yeah when I google what I wrote, I just get back the paper

what you said makes more sense

Ted Shifrin

Branched immersion suggests you're mapping into a higher-dimensional $\Bbb CP^n$.

Ryan Unger

Well upon closer inspection Yau doesn't seem to say RR has to do with the RP^2 piece

He just wants there to be a conformal map to S^2 with degree <= g+1

in the orientable case

Ted Shifrin

01:54

Yeah, so if you put the corresponding complex structure on the torus, you're doing holomorphic maps and so RR applies.

Balarka Sen

You can construct one explicitly I believe. There's a degree g-1 unbranched covering map Sigma_g -> Sigma_2 and then there's a degree 2 branched map Sigma_2 -> S^2 which is the "pillowcase map"

Reflect Sigma_2 about an axis cutting through the hole; that's a Z/2-action (this can be done holomorphically as well)

Total degree g-1+2 = g+1

Ted Shifrin

reflection doesn't sound holomorphic.

Balarka Sen

It won't be that map, I think, but it can be done. You do reflection along an axis cutting through the hole in T^2 in R^3 to get a branched map T^2 -> S^2, the holomorphic realization is complicated: it's the Weierstrass $\wp$-function

Ted Shifrin

You need degree $\ge g+1$ to get a nontrivial holomorphic map.

Balarka Sen

I think you'll need a modular form here

Ted Shifrin

01:59

Yeah, but $\wp$ doesn't generalize to $g>1$.

Ah.

I don't know that stuff.

Ryan Unger

So...it does follow from RR if you know it but there might be an explicit construction too?

Daminark

Hello

Balarka Sen

Hi nerd

Ted Shifrin

Linear systems are pretty explicit, but you won't have a topological picture.

Ryan Unger

hi

Ted Shifrin

02:00

heya Demonark

Bye for now

Daminark

Did I drive you away that quickly?

Ted Shifrin

I would have left sooner if I'd anticipated your arrival !

I have to go cook dinner, in truth.

Balarka Sen

Cya, Ted

Enjoy cooking

Daminark

Haha, bon appetit!

Balarka Sen

I have to sleep

will I?

Ryan Unger

02:03

no, start reading the 4D poincare paper

Balarka Sen

oh yeah what happened to that paper

Ryan Unger

Yau talks about it in his new autobio

@BalarkaSen when trying to applying RH here, how do you compute the ramification index

Balarka Sen

I don't know the context of your calculations but give a holomorphic branched covering of Riemann surfaces $f : M \to N$, around a ramified point $x \in N$ the local picture is $f^{-1}(x) = \{x_1, \cdots, x_m\}$ where $x$ has an nbhd $U$ in $N$ and $x_i$ has an nbhd $V_i$ in $M$ such that $f|V_i : V_i \to U$ is $f(z) = z^{n_i}$ in local holomorphic coordinates

$n_i$ is the ramification index of $x_i$

Ramified point is what happens when the preimages upstairs get clubbed togather, but they might club togather in groups; ramification index measures the size of those groups

(Indeed, $n_1 + \cdots + n_m = \deg(f)$)

Ryan Unger

So for now I'm happy to put this in a black box for when I take a Riemann surfaces course

But the point is that RR and RH give you such a branched covering?

You get a meromorphic function with g+1 simple poles

Balarka Sen

I guess so, yeah

I can try to come up with a modular form that gives a degree 2 covering $\Sigma_2 \to \Bbb{CP}^1$ if you want

That's enough

Ryan Unger

02:18

I don't know what modular forms are

Balarka Sen

Meromorphic functions on the upper half plane with special symmetry relations bro

"Special symmetry relations" will help it descend down to $\Sigma_2 = \Bbb H^2/\pi_1(\Sigma_2)$. You only care about "meromorphic functions"

Ryan Unger

ugh i thought you werent an algebraic geometer

Balarka Sen

02:33

@RyanUnger I have example.

Ryan Unger

ok

Konformist Liberal

Hey, a small question, and a targeted reference would be enough - What's infinite dimensional complex projective space and orientation have to do with each other?

Balarka Sen

$\text{SL}_2(\Bbb Z)$ acts holomorphically on $\Bbb H^2$ by Mobius transformations; let $\Gamma(n)$ denote the kernel of $\text{SL}_2(\Bbb Z) \to \text{SL}_2(\Bbb Z_n)$ also known as the congruence subgroup of level $n$. Consider $X = \Bbb H^2/\Gamma(n)$. This is a Riemann surface.

Ryan Unger

This is algebraic geometry.

Explicit examples are even worse than Ch 2 of hartshorne!

Which surface is it?

Balarka Sen

Now $X$ might be noncompact, it'll have certain "cusps" which are the ends of $X$ which is what happens when the fundamental domain associated with the action of $\Gamma(n)$ has some corner going off towards the ideal boundary in $\Bbb H^2$

But the since the ends are cusps, the conformal metric can be compactified, i.e., there is a compact Riemann surface $\overline{X}$ inside of which $X$ can be biholomorphically, densely embedded, and $\overline{X} \setminus X$ is a finite number of points in bijection with the ends

Basically the conformal metric on $X$ comes from the hyperbolic metric on $\Bbb H^2$ which decays nicely towards $\partial \Bbb H^2$, so near the cusp the surface becomes thin

That's why the metric extends

Now look at a modular form on $\Bbb H^2$ which is $\text{SL}_2(\Bbb Z)$-invariant, for example something like the j-invariant. This preserves the $\text{SL}_2(\Bbb Z)$-action, hence in particular the $\Gamma(n)$-action, hence descends down to $X = \Bbb H^2/\Gamma(n)$

Extend that over to $\overline{X}$ holomorphically by extending over isolated singularities, so this defines a holomorphic function $\overline{X} \to \Bbb{CP}^1$

According to this list, if $n = 23$ then $X$ has genus $2$. I am sure that this is the desired branched covering.

Ryan Unger

03:08

@BalarkaSen lol

Balarka Sen

I couldn't compute the genus so I looked it up. OEIS fam

Maybe $X \to \Bbb H^2/\text{SL}_2(\Bbb Z)$ is a branched covering and you use R-H on that to compute genus

Oh who am I kidding that's enough, $\Bbb H^2/\text{SL}_2(\Bbb Z)$ compactifies to $\Bbb{CP}^1$. So $\overline{X} \to \Bbb{CP}^1$, a branched covering.

Ryan Unger

I'm lost

what is X now

Balarka Sen

$X = \Bbb H^2/\Gamma(n)$

Ryan Unger

with n=23 still?

Balarka Sen

Yeah, that seems to be the scenario when this is genus 2

Konformist Liberal

03:48

(Again - ) hey, a small question, and a targeted reference would be enough - What's infinite dimensional complex projective space and orientation have to do with each other?

Balarka Sen

04:02

@KonformistLiberal A choice of orientation on a $2$-dimensional real vector bundle $E$ over a manifold $M$ gives a reduction of the structure group from $\text{GL}_2(\Bbb R)$ to $\text{GL}_2^+(\Bbb R)$. So oriented $2$-dimensional real vector bundles are in 1-1 correspondence with homotopy classes of maps $M \to B\text{GL}_2^+(\Bbb R)$ but $\text{GL}_2^+$ deformation retracts to $SO(2) = U(1)$, so this is in bijection with homotopy classes of maps $M \to BU(1) \cong \Bbb{CP}^\infty$.

More precisely there is a bijective correspondence between oriented rank 2 bundles over $M$ and complex line bundles on $M$: if $\omega$ is a fiberwise $2$-form on $E$ establishing an orientation, then define $J : E \to E$ by $\omega(v, Jv) = 1$; note that $\omega(J^2 v, Jv) = -1$ which implies $J^2 v = -v$ i.e., $J^2 = -I$ i.e., $J$ is a complex structure on $E$

Conversely, every complex structure $J$ on $E$ gives a fiberwise orientation by declaring $\{v, Jv\}$ to be a positively oriented basis

But now there is a bijective correspondence between complex line bundles on $M$ and homotopy classes of maps $M \to \Bbb{CP}^\infty$, namely, the one obtained from pulling back the universal tautological bundle $O(-1)$ on $\Bbb{CP}^\infty$

In general a rank $2$ real bundle has classifying map $M \to \text{Grass}(2, \infty)$, but there is an inclusion $\Bbb{CP}^\infty \hookrightarrow \text{Grass}(2, \infty)$ given by the complex 1-dimensional subspaces of $\Bbb R^{2\infty} = \Bbb C^\infty$, and we're demanding that the classifying map factors through that

In summary, oriented rank 2 real bundles must have it's classifying map taking values in $\Bbb{CP}^\infty$. That's the relation

Konformist Liberal

@BalarkaSen Thanks, I was hoping/looking for a lower level explanation-run through

But thanks, will try to understand

Adam

04:39

So how much credibility does Princeton have in the mathematics department? This was strange to see youtube.com/watch?v=5tu32CCA_Ig

Balarka Sen

05:00

@Adam It's like one of the top 5 mathematics departments in the world

Interesting video and seems totally believable to me

Ryan Unger

I doubt Princeton math had anything to do with such a study...

Balarka Sen

yo @Ryan, Gabai is the dep head right?

that's a topologist there

Ryan Unger

there's multiple topologists

the people who invented Heegard-Floer homology are there

and John Pardon

who's probably one of the few people naturally smarter than you

or maybe you're working on some huge conjecture

Balarka Sen

Oh yeah Pardon

Secret

07:49

440

A: The staircase paradox, or why $\pi\ne4$

This question is usually posed as the length of the diagonal of a unit square. You start going from one corner to the opposite one following the perimeter and observe the length is $2$, then take shorter and shorter stair-steps and the length is $2$ but your path approaches the diagonal. So $\sq...

In other words:

Mar 19 at 14:40, by anakhro

I would highly recommend just being more formal with how you define your faux-finity, and then move on from there.

Even the intuition that I can just use polygons to approximate a circle and hence somehow define an infinite object in a finitistic fashion is flawed

back to the drawing board

Adam

@BalarkaSen yeah I was being a dic I was working of their post grad textbook in analytic number theory but it got torn in half by whoever stole my wallet n a bunch of um.. meds n stuff

i know they are very well respected it's just that what the you tube link's premise is isn't a very popular thing i would say

Secret

now that makes me wonder, if a diagonal cannot be coverged with a series of horizontal and vertical zigzags, what does a diagonal actually consists of. Is it just an uncountable set with no horizontal nor vertical intervals?

in particular, a diagonal line interval is lebesgue measurable, meaning it has to contain some kind of intervals somehow

Alessandro Codenotti

The Cantor set is Lebesgue measurable, but it'll be hard to find an interval in it

Secret

ah right...

For something that generalise the notion of "volume", I think the least intuitive thing about measures is its dependence on the ordering of the points in the set, at least for the uncountable case

Since finite sets (and even countable sets) cannot do that, it is hard to find an intuitive example of that

user131753

08:09

chat.stackexchange.com/rooms/pingable/36

user131753

Wow!

user131753

Very interesting.

Akiva Weinberger

@user170039 What is this

user131753

@AkivaWeinberger Did you click on it?

Akiva Weinberger

@Secret Reminds me of this thing I wrote once

https://math.stackexchange.com/a/1390907/166353

on surface area

@user170039 Yeah

user131753

08:16

@AkivaWeinberger This basically gives the number of pingable users in a room. I thought that it is interesting and so shared it.

user131753

@BalarkaSen: I recently found out this. Is there any pdf of the talk available?

Akiva Weinberger

@user170039 What do the numbers mean

user131753

@AkivaWeinberger The room number (for this room it is "36"). Change it and see the fun.

Akiva Weinberger

Not that number

Secret

> There is a way to save this. If 𝐴A and 𝐵 B are both convex, then we do have that 𝐴 A has a larger surface area than 𝐵
B, and our approximations do work. And we can use limits to get the volume exactly. However, I'm pretty sure your union of cylinders isn't convex.

Akiva Weinberger

08:19

> [124967,"Akiva Weinberger",1558426680,1558426557]

Those numbers

Secret

Hmm, somehow concavity makes things divergent

Subhasis Biswas

intersections are guaranteed to be convex

but union?

Secret

[134719,"Secret",1558426699,1558425710]

From what I can understood, the 1st is your user id in number form

user131753

@AkivaWeinberger "124967" is your user id I guess. About others I don't know.

Secret

(one can see that when trying to use the chat search bar and see how it turns into a number)

Akiva Weinberger

08:22

@SubhasisBiswas Easy to find counterexamples

Secret

[134719,"Secret",1558426912,1558426848]

the numbers changed

[134719,"Secret",1558426983,1558426983]

[134719,"Secret",1558427009,1558427009]

It seems to change for each new posts by that user made

Akiva Weinberger

Strange

Secret

but I cannot pinpoint it to anything, they are not message ids as far I am aware

but they should be related in some way, since they are always identitical, similar to how message ids in the permalink always have the form <id>#<id>

Adam

08:48

I've found that unless you are in some kind of warzone, pathological searching for number sequence - meaning connections can land you in a padded cell pretty quick

nothing frightens people more than numbers and an unsolved situation or number of. I think that was what they were trying to warn people with the movie the beautiful mind, utilize the advantages of the computer age and don't get too artsy or leer at a flock of pigeons for too long, even if they are disgusting little sacks of every which virus that's airborne

PM 2Ring

09:19

@Secret But you can use polygons to approximate a circle. The core problem with the staircase construction, as TCL points out, is that while the points on the staircase converge to the desired function, the slopes of the staircase definitely do not converge to the derivatives of that function.

Secret

true

In fact, I think the limit of the staircase function is differentiable nowhere

@Adam Deal with the pigeons with pigeon hoes, it always works, even if there are inaccessibly many of them

(disclaimer: do not include sex workers, else you will be banned for hate speech)

Adam

10:05

Well absolutely I mean hookers have taken childhood trauma and chosen to lower the sexual assault rate with the total opposite of gratitude from society, so yeah I do not want to go down the road of being the type of idiot that looks down on those people.

i mean other people's childhood trauma manifests itself as they themselves becoming sexual predators, so in that light it should be easy enough for you to see the virtue in the sex workers character

but no i thought this would get a ban tbh lol youtube.com/watch?v=5tu32CCA_Ig&t=4s

Secret

10:27

Video -> USA, Brexit and Australia if you are not conservative

Adam

12:45

oh dear really, sounds like a very broad rhetoric of someone without any rational retort or counter argument to make against that implied the content of said video but OK

Secret

13:20

@ Semiclassical: I seriously think that Adam is trolling most of the time, but his troll pattern is something I never seen before, as it has a strange sawtooth wave profile

Semiclassical

shrug

Secret

I have a strange political position in the political spectrum such that the left think I am too right the right think I am too left the centrists think I am too revolutionary the revolutionary think I am too reactionary

GFauxPas

Hey @Semiclassical

Wanna help me with ozez

Odes?

PM 2Ring

13:49

Shall I compare thee to a Summer Glau?
Thou art more lovely and more temperate:
Worse art thou at ballet, and thine brow
Has never broken any reaver's pate
Thou art less morbid, yet more creepifying
No horde of nerds profess their love for thee
Thou art not a term'nator undying
Thou dost not travel 'pon Serenity
Thou dost not consort with Angels fair
And never hast thou shared in Echo's strife
And yet, thy blow doth cause me to declare
"I've never been so turned on in my life"
So long as men can breathe or eyes can see

That's a sonnet, but I guess it's also an ode. :)

GFauxPas

Boo

Thorgott

14:46

Hey, does anyone know a simple example of a differentiable function $f\colon U\rightarrow\mathbb{R}^m$, where $U\subseteq\mathbb{R}^n$ is open and path-connected, s.t. $f$ is not Lipschitz, but $\mathrm{d}f$ is bounded?

Secret

[Random]

$\frac{\pi}{7-\pi}=\frac{p}{q}$

$7p-\pi p = q\pi$

$7p = (q+p)\pi$

contradiction

therefore, $\frac{\pi}{7-\pi}$ is irrational

$\frac{\pi}{P(\pi)} = \frac{p}{q}$

$q\pi = p P(\pi)$

$Q(\pi) = 0$

Contradiction

$\frac{\pi}{P(\pi)}$ is transcendental for all polynomials $P$

Secret

15:27

$P([e]_n)= Q( \pi)$

Contradiction

$Q(\pi)$ is transcendental

Ryan Unger

@Thorgott Is the issue there that $U$ is not convex?

Secret

$P(e)=Q(\pi)$

Thorgott

Yeah, that's necessary. Such an example can't exist for convex or even quasi-convex $U$.

Adam

15:50

@Semiclassical everyone seems to introduce you unnecessarily into stuff, kinda like u are the same guy that is tagging you but he wants to make it a numbers game as is standard for his brand of sad. anyway thanks for this meme it really made my day

2

Transcript for

$Mathematics$ Mathematics