00:00 - 16:0016:00 - 00:00

12:12 AM
@BalarkaSen LOL ... Forsaking topology?

Why not both? Hi, @Ted.

good evening Ted

You said it was unhealthy, Balarka, not me :P
hi @GFauxPas

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

certainly not my department of expertise

12:15 AM
I am not being able to comprehend how this is garnering so much upvotes. There are obvious counterexamples $S^1 \to \Bbb R^2 \setminus \{0\}$.

I'm doing practice problems for my exam
alright then, hopefully Semi will have time to help me tonight
thanks anyway

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

To the image, no.

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

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

12:20 AM
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).

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

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

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

or post a hint as the answer.

12:22 AM
Great, I have to do neither!

Contractible things can wrap and make circles. Duh.

@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).

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

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

What a strange picture.

12:31 AM
This is basically "cutting a cube into a cross of squares", formalized :D

Not that I understood that.

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

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

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

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.

12:34 AM
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 $D^n$

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.

Haha I agree

But thanks for teaching me something :)

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

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.

12:38 AM
Yup

I think it's a great exercise.

12:50 AM
this is a very interesting theorem

Well, knot so interesting to me.

cheap joke Ted

I was about to leave to go cook dinner.

Hahah

So it was my parting gesture.

1:00 AM
Ted's favorite area of math is knot theory, and neither is it mine

@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

o shit tell me more

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

I see pages of integrals my dude

well yeah it's an energy
were you expecting topology?

1:04 AM
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, ...

now one could look at the gradient flow of this thing
but that's getting into the realm of things that are too complicated

geometric analysis is too hard

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

we should just all do algebra
motivic stacks is where the results are at

that's what I've been saying
but then you look at what these ppl are doing and its like jesus

1:08 AM
i have no clue how people do algebraic geometry
i cant even read chapter 2 of hartshorne
how do they just do it

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

You guys didn't sleep?

it's 9PM...

it's 6:38AM!

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

1:10 AM
Oof

you could just be a google ai
would make a lot of sense actually

alexa jones meme here

Did he fail the captcha test?

have you read the kirby calculus book
it's on my list of books I want to read but never will

Gompf-Stipsicz?

1:12 AM
sounds right
Gompf is retiring! I was sad to find that out when I visited UT Austin

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

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

oh that's in there?
hmmmmm

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

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

1:17 AM
I wish you luck with anabelian p-adic Hodge theory, which is of course grad number theory 101

introduction to perfectoid spaces

@RyanUnger they in fact do offer one

@Ã‰ricoMeloSilva this fall?

Okay okay hol' up.

they offer it in springs

1:18 AM
Noob here

Oh cool

yeah at least this spring they had one

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

Oh I didn't rember that
oh yeah there is it, whoops
"Topics in Algebraic Geometry:
cohomology"
@BalarkaSen explain

vOv
u learn Weil conjectures
thats obviously the right place to start doing alg geo
because historically it was :D

1:20 AM
the right place is probably mirror symmetry
it's at least somewhat analytic

I bought a book with my contingency grant this semester
"Enumerative Geometry and String Theory"
by Katz

@RyanUnger @BalarkaSen

it has mirror symmetry in it

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

hi chat

1:22 AM
@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

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

then at the double points just "desingularize it"

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

but I don't think he defined the Fukaya category

embedded representative viola

1:23 AM
@BalarkaSen does the blow up make $S$ noncompact

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

then what do you mean by blow up
two sides? so $S$ is disconnected?

think of transitioning from a cone to a hyperboloid of one sheet
$S$ may possibly be disconnected

@BalarkaSen ok sure
do you do the Whitney trick for the double pts

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

1:27 AM
well when you say desingularize a double point for something half-dimensional ,that's literally the whitney trick

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

yes
is that not what you're doing here
maybe I'm misunderstanding what kind of singularity you've got here

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

manifolds are smooth m8

that's why Gompf-Stipsicz do it in that weird way
Oh but they are also making things transverse I suppose

1:31 AM
well just general position right
you can triangulate the surface

$M^4$ might not be triangulable/have a smooth structure
So I don't know how to play general position games there

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

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

can you triangulate a neighborhood

that's local flatness
IDK homie too hard im leaving

1:34 AM
ok good morning

I am a $C^\infty$ man

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

@BalarkaSen Not if you do Whitney stratifications :P

@BalarkaSen exciting

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

1:40 AM
you retired too!

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

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

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

@BalarkaSen what the hell is that

LOL

1:43 AM
@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

Schubert calculus is great.
Well, @Balarka, algebraic/analytic varieties are even nicer :)

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

@BalarkaSen oh Schubert calculus is great!

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

What are your domain and range, @Ryan?

1:50 AM
$\Sigma^2_g\to S^2$ or $\Bbb RP^2$

That's a branched covering my dude
Riemann-Hurwitz formula :3

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.

@BalarkaSen yeah when I google what I wrote, I just get back the paper
what you said makes more sense

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

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

1:54 AM
Yeah, so if you put the corresponding complex structure on the torus, you're doing holomorphic maps and so RR applies.

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

reflection doesn't sound holomorphic.

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

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

I think you'll need a modular form here

1:59 AM
Yeah, but $\wp$ doesn't generalize to $g>1$.
Ah.
I don't know that stuff.

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

Hello

Hi nerd

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

2:00 AM
heya Demonark
Bye for now

Did I drive you away that quickly?

I would have left sooner if I'd anticipated your arrival !
I have to go cook dinner, in truth.

Cya, Ted
Enjoy cooking

Haha, bon appetit!

I have to sleep
will I?

2:03 AM
no, start reading the 4D poincare paper

oh yeah what happened to that paper

Yau talks about it in his new autobio
@BalarkaSen when trying to applying RH here, how do you compute the ramification index

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

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

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

2:18 AM
I don't know what modular forms are

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"

ugh i thought you werent an algebraic geometer

2:33 AM
@RyanUnger I have example.

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?

$\text{SL}_2(\Bbb Z)$ 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.

This is algebraic geometry.
Explicit examples are even worse than Ch 2 of hartshorne!
Which surface is it?

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.

3:08 AM
@BalarkaSen lol

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.

I'm lost
what is X now

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

with n=23 still?

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

3:48 AM
(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?

4:02 AM
@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

@BalarkaSen Thanks, I was hoping/looking for a lower level explanation-run through
But thanks, will try to understand

4:39 AM
So how much credibility does Princeton have in the mathematics department? This was strange to see youtube.com/watch?v=5tu32CCA_Ig

5:00 AM
@Adam It's like one of the top 5 mathematics departments in the world
Interesting video and seems totally believable to me

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

yo @Ryan, Gabai is the dep head right?
that's a topologist there

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

Oh yeah Pardon

3 hours later…
7:49 AM
440

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 @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 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 The Cantor set is Lebesgue measurable, but it'll be hard to find an interval in it 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 user131753 Wow! user131753 Very interesting. @user170039 What is this user131753 @AkivaWeinberger Did you click on it? @Secret Reminds me of this thing I wrote once on surface area @user170039 Yeah user131753 8:16 AM @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? @user170039 What do the numbers mean user131753 @AkivaWeinberger The room number (for this room it is "36"). Change it and see the fun. Not that number > 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. 8:19 AM > [124967,"Akiva Weinberger",1558426680,1558426557] Those numbers Hmm, somehow concavity makes things divergent intersections are guaranteed to be convex but union? [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. (one can see that when trying to use the chat search bar and see how it turns into a number) 8:22 AM @SubhasisBiswas Easy to find counterexamples [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 Strange 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> 8:48 AM 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 9:19 AM @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. 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) 10:05 AM 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 10:27 AM Video -> USA, Brexit and Australia if you are not conservative 2 hours later… 12:45 PM 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 1:20 PM @ 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 shrug 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 Hey @Semiclassical Wanna help me with ozez Odes? 1:49 PM 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. :) Boo 2:46 PM 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$is bounded? [Random]$\frac{\pi}{7-\pi}=\frac{p}{q}7p-\pi p = q\pi7p = (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$3:27 PM$P([e]_n)= Q( \pi)$Contradiction$Q(\pi)$is transcendental @Thorgott Is the issue there that$U$is not convex?$P(e)=Q(\pi)$Yeah, that's necessary. Such an example can't exist for convex or even quasi-convex$U\$.

3:50 PM
@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

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