So the only things I could reasonably do next year would be analysis-y as far as I know...

Or maybe in a later quarter something like geometric group theory or I dunno?

For MAA Journals is it better to present trigonometric functions using radians or degrees?

@TedShifrin

@Daminark a lot of things in there are kind of relying on a background in basic Riemannian stuff

@Benjamin Do you know of journals which use degrees as the preferred unit?

I'm not being sarcastic or anything, just genuinely curious.

bc there are loads of generalizations for metric spaces that you need to apply when you do ggt

@Benjamin I vote radians because it will make the math simpler most of the time.

Degrees only make the math simpler when you're taking partitions of a circle, because 360 has a lot of possible factors.

It's divisible by $1, 2, \dots, 12$, at the very least.

Actually, nix $11$. Oops.

and 7

Right. Nix $7$, too.

Hmm... How doable is Riemannian geometry?

Well eh

There's a class on it, might do that as more of a DRP

If I can't take it

@Axoren No, but I don't read MAA journals much.

wait is there a class on ggt @Daminark

No on Riemannian geo

oh yeah, the only one is grad

Well yeah, just that I'm not sure if I'll be able to take it or not

@Benjamin If you don't have a preference, radians are a safe bet. Most people consider them to be a trig standard unit due to their relation to arc-length. The benefits of radians outweigh the benefits of degrees.

There is plenty of ggt you can do without Riemannian geometry @Daminark @Eric

@TedShifrin @Axoren Thanks!

hi paul

oh right

:/

@EricStucky Hello anyways

I'm still getting used to this >.<

:)

@Paul what do you need for that?

There's a good whole semester of GGT, probably more, that you can do mostly from scratch, with just Algebra I stuff.

There are plenty of entry points that do not require much of anything.

That sounds pretty dank

basic group theory, and most of the geometry that is important can be picked up on the way.

A bit of algebraic topology would be useful though

like fundamental group stuff, nothing to fancy

Is it going to be a class or were you thinking a reading course @Daminark ?

Reading course, or DRP

(DRP is directed reading program, like with a grad student)

If I uploaded a derivation for a paper, could you check it?

Actually, I retract that question, I just found an error needing ironing out.

@Paul, ah ok, my first exposure was through Riemannian stuff in bridson and haefligers book

Hey @Akiva!

@AkivaWeinberger on Conway's soldiers, what's the minimum amount of moves required to reach the maximum distance? Is it finite?

@SimplyBeautifulArt The image on the Wikipedia article seems to show the minimum amount of pieces required to achieve distances of 1, 2, 3, and 4 units (5 is impossible).

hm, okay.

Each move deletes one soldier, so there is one fewer move than starting soldiers, I guess.

@Eric Even BH doesn't rely a on riemannian stuff, it certainly comes up, but I think an important point of the book is that they discuss a notion of curvature in a way that only relies on being a somewhat nice metric space (length), and no riemannian metric is need. That is actually quite important because a lot of, if not most, spaces which come up in ggt are not riemannian. Plus they introduce the important stuff when it does come up.

@SimplyBeautifulArt

> Image of en:Conway's Soldiers initial positions to reach rows 1 to 4. The soldiers marked A and B denote alternatives.

So I guess you could either have the A soldiers or the B soldiers, but both together are unnecessary.

mhm

I'm aware they aren't, but intuition from the Riemannian stuff motivates a lot of the early results.

@AkivaWeinberger Apparently you can reach the 5th row with an infinite amount of soldiers?

Infinitely many moves, yeah

hehe, interesting

The set of instances during which a move is made is not well-ordered, but it is topologically discrete (I think)

@AkivaWeinberger You don't happen to know the answer to my question above? chat.stackexchange.com/transcript/message/37075881#37075881

It just happens to be a simple trivia that noone bothers to know

@SimplyBeautifulArt Hm, I have no idea

Aw poo-ey, thanks for taking a look :D

So is a Riemann surface just a Riemannian 2-manifold? I learned what they were from my physics TA's complex thing, but in terms of taking branch cuts and gluing them together

Riemann surfaces are not Riemannian 2-manifolds, no :)

And I feel like it should be known and most certainly used. It is quite simple of a relation between functions.

They are distinct concepts

Also why am I not even sleeping right now

shrugs

@BalarkaSen as my friends say, sleep is for the weak.

As my friends also say, that's bad advice, get some sleep.

I was thinking of going to bed soon... @BalarkaSen

thinking, huh?

Well it is only 8pm.

@Simply i think sleep does good to you but hell do i care

i have to decide if i care

@PaulP Ah. only 6 here, AM

I know, that is why I mentioned it

@BalarkaSen build a machine to decide for you.

I thought it was funny

@BalarkaSen ...How many hours have you been awake

@AkivaWeinberger "Perhaps he just got up early." never mind.

@Paul I just realized your interests are geometric group theory :P

:)

@AkivaWeinberger I haven't slept all night man

I was also considering not sleeping too

geometric group theory is cool stuff. I have no idea how it works tho

I guess I'm into it because I know I like group theory, geometry sounds fun, and I kinda like the whole, smashing things together thing?

Like the fact that a complexity theory result, the non-solvability of the word problem in groups, can be used to prove that 4-manifolds are not classifiable... That just blows my mind, you know?

Not to put you off, but why? The complexity theory result is that some things aren't classifiable.

(Well, not classifiable in terms of the fundamental group.)

Well it's more the fact that specifically 4-manifold fundamental groups get you all the groups

It's not super-deep to show every finitely presented group is a $\pi_1$ of some 4-manifold tbh

It doesn't get you all the groups, only finitely presented ones

Oh, that I didn't realize

I heard that it was all groups which is why I was just like, whoa something is very screwy here

Finitely presented is still a pretty huge class anyway, and non-solvability of the word problem is true there

so you're just fine

Ah

i wonder what the fundamental group of non-metrizable 4-manifolds can be

can you get the free group on arbitrary letters for surfaces?

I am pretty sure you can just do that for even nonHausdorff 1-manifolds.

You can up to countable, for non metrizable though I don't know, probably

Generalize the line with two origins to line with continuum many origins

Yeah but I don't want non-Hausdorff.

Wouldn't a "fat" wedge of circles work

Is that a manifold?

you can't embed a wedge of circles in R^4 though, how do you fatten it?

Locally it is R^2

I thought he just wanted a surface

I want uncountably many generators

well, the countable wedge of circles doesn't embed in R^2 even

Oh I see the confusion

it's not 1st countable

I never said it did

I don't see how you fatten it then

I meant to ask about the infinite case. I know you can realize every countable group as the fundamental group of a 4-manifold; I'm interested in what you can do if you drop second countability / metrizability.

So I'm playing around with the surface case

by doing a wedge of open annuli,

It won't be a manifold at the wedge point.

I think what does work is to take the Prufer surface's universal cover and delete an uncountable discrete set. On the other hand since manifolds are first-countable you can't get bigger than $|\Bbb R|$-sized fundamental group.

Can you do your argument for just Long line x R?

I think so

Intervals are not long enough to go through the long line

yeah, in my uncountable discrete set you can walk from any one point to another

(the prufer manifold is path connected)

ah right meh

Maybe there are problems, at the "wedge" point still, but I wasn't thinking of just wedging, I mean something like the hawain earring, except with annuli, and with the weak topology, so each annulus has a common square they all share

maybe it doesn't work though...

I think I would be worried about doing that uncountably many times and getting a manifold

to get it to work you might need some sequence of larger (in actually cardinality) simply connected surfaces for these annuli to connect to

Ah yeah then 100% agreed

I think we have the same idea in different phrasing

If you attach these annuli to other simply connected surfaces I am still worried about first countability...

Ignore that. of course you'd break 2nd countability there (that's the point). I'm not really awake

Ah I was not aware of the prufer manifold.

Makes pun about the proofer manifold

No idea how to go about constructing these arbitrary cardinality manifolds though

I am pretty bad at non-second countable things

You can't do anything bigger than $|\Bbb R|$

It's just that Prufer has a $|\Bbb R|$-discrete set

oh, then you can't do arbirary free group, there are not even enough maps S^1 \to M

Yeah but the point is you can (?) do a free group on |R| letters

yup

Oh I missed that, Mike already said that

what's not obvious to me is how many relators you can get

(I find it plausible, in addition, that there are groups with $|\Bbb R|$ generators and more than $|\Bbb R|$ relators)

can you still triangulate such manifolds?

@MikeMiller That depends on what you mean by $|\Bbb R|$ generators.

If you mean there is a surjection $F(\Bbb R) \to G$.

yes

oh

stupid question

Then you are talking about a subset of $F(\Bbb R)$

which has cardinality $\Bbb R$

(at the maximum)

we can stop now

in such a way that you have a discrete point in each triangle. then after removing it should deformation retract to its one-skeleton.

The ballsy questions always surprise me for some reason math.stackexchange.com/questions/2259798/…

I guess they shouldn't at this point

sighs

Just less subtle than most.

They're not even trying to be subtle at this point

True, Demonark comes into chat and says "Do my homework." :D

?

Well, most of us resist.

There are truthfully a handful of people who post all their differential geometry homework, and most of it gets done for them (not by me, although I used to give some hints).

@TedShifrin but, math.stackexchange.com/questions/2259798/…

"gets done" meaning, they were walked through it or just outright solutions?

How can you not be convinced?

But my rude answer, @GPhys.

Demonark: Other people give outright solutions. My style is to hint and then help in a conversation. But with some of them, I've quit (especially because I've told some of them to look at my notes, where particular things they were asking are done completely and with explanations).

That person literally just joined to post his homework here. Sigh.

@Ted But he doesn't know and has to submit today!

Awww ...

I wonder what time zone they're in. Maybe they have ~40 minutes to submit :(

These comments keep getting better, but I should do my own work

I have laundry to do, and also only some hours to submit....

I am so amused rn

Please Ted it's request

(That acronym is so annoying. It looks like an 'm'.)

Maybe DogAteMy will do the poor person's homework for him.

I would but it isn't 0 almost everywhere so sorry :/

@Akiva Lol I didn't think of that, but yeah

But Ted, it's a request. And he said "please".

I'm an asshole. (Just ask all my former students.)

At least you're not a bum.

Go do your own hw you bum hahahaha

Actually, I am a bum ... a retired bum.

I just seen that and laughed so hard

Faraad will take pity on the poor dear and help him. :P

Ted is roasting people, I should not be enjoying this as much as I am.

BTW, puzzling, interesting question for those who think about multivariable carefully.

Hahaha, its closed. I wouldn't touch that question with a 10ft pole after those comments. I don't want enemies @TedShifrin

Hush, @TimThe.

You could try a 10ft Serb, Faraad. :P

silenced

(That remark was worthy of Demonark or DogAteMy.)

That is one scary Serb.

oh geez hahaha, you are on a rampage today

I invite Faraad, Demonark, @Eric, DogAteMy to ponder that question I just linked.

@TedShifrin I vaguely recall a question like this on an analysis final when I was an undergraduate

except, it was only giving some particular conditions and asking to prove or disprove the conditions were sufficient

(I disproved, with a counterexample)

I would assume it was something more like proving $C^1$ implies differentiable. Oh, differentiable doesn't imply $C^1$ might be what you gave.

I don't know what he means by regularity, but certainly if they are continuous, then $f$ is differentiable since you can define the linearization.

He wants necessary as well as sufficient, @Faraad.

And one needn't have continuous partials — indeed, the partials may not even be defined anywhere else.

I suspect, as I said, that there is no answer.

huh I've never thought about this

this is an interesting question

It'd be real funny if there were a measurability-type answer

I was pondering that briefly, Demonark, but I invite you to investigate.

@TedShifrin: yeah, that was a sufficient condition I gave. I'll keep thinking.

As I gave an exercise in my book, knowing that all directional derivatives vanish, for example, won't come close to telling you differentiable.

@TedShifrin certainly not I

LOL, @MikeM, I didn't page you on it.

So I saw (:

Ah, Dieudonné gives an exercise stating a necessary and sufficient condition, but not in terms of regularity of partials. But I'll go ahead and add that to his question.

@MikeMiller: view it as Ted not trying to call upon stronger forces :)

Certainly not. I'm just having fun.

I know, I'm just kidding

Is it true that any function satisfying the intermediate value property is measurable?

Even in dimension $1$ trying to figure out how to precisely characterize derivatives of everywhere differentiable functions is hard

I think I remember seeing that in 1 dimension, derivatives satisfy the intermediate value property. So maybe that might be something

idk @Daminark but derivatives of everywhere differentiable functions are certainly measurable and have the intermediate value property

they're pointwise limits of continuous functions.

and so they live in the first Baire class or whatever

Yeah

Does that hold in just one or any dimension?

I posted Dieudonné's answer.

Yes, Demonark, that's called Darboux's Theorem.

Right, @Eric.

So being Baire-1 is equivalent to the image of half-lines being $F_{\sigma}$

What do you mean by linear maps satisfying the intermediate value property, Demonark?

I meant more, the function mapping $x$ to $df_x$

But I don't know what the question means.

I mean Darboux's theorem does answer my question, just that intervals are mapped to intervals

But in higher dimensions, you mused ... ?

Oh, I mean, I guess boxes? Or connected sets in general

Boxes is probably wrong...

IVP has a definite "betweenness" in both domain and range. I have no idea what you can do to generalize.

@Ted the condition in that exercise is actually far simpler than I expected that sort of thing to be

It's sort of interesting, @Eric. I suppose we should work the exercise. :P

Speaking of brazen questions, Physics.SE seems to get a lot of a somewhat different character that Math.SE gets more rarely physics.stackexchange.com/questions/329928/…

It's like the "proofs" of trisection of angles with compass and straightedge that we used to get sent to the math department several times a year, @GPhys.

I will think about it once I finish grading. Do you think it's worth it to pick up the volumes of Diedonne's treatise? @Ted

I bought them in France (in French) years ago. They are great resources on everything, with excellent exercises. (There's high-powered analysis, differential geometry, Lie groups ... etc.)

And I did not give those away when I emptied my office out.

@TedShifrin At least it's easy to sort away the emails titled e.g. "EINSTEIN WAS WRONG"

oh that's exciting do you know if there's any good english translation?

@TedShifrin Remember I was talking about a "topology on the topology" a while back?

if not I might pick them up in french, I need to practice anyway

Oh, it's been translated, for sure, Eric.

I think I have a slightly more developed idea

Yeah, DogAteMy.

Instead of the topology (the set of open sets), think of the set of closed sets 'cause it's cleaner.

A few days ago, I noticed that the set of (possibly degenerate) triangles in $\Bbb R^2$ is closed under decreasing intersection, as well as closure-of-increasing-union.

(Triangles include their interior here)

You're allowing infinite triangles?

Yeah

@TedShifrin An online friend of mine told me their public university had a policy of responding to all the mail their department got, and they worked it as a graduate student instead of TAing classes for a while. Apparently it involved responding to quite a bit of bunk.

So let's define a closed family of closed sets in $X$ to be one that's closed like that.

I have not yet checked in this satisfies the axioms for a topology.

Occasionally, faculty responded to the ones we got, @GPhys. But usually we didn't bother.

I do remember responding to an elementary school child who was curious about divisibility tests.

Ah, that's nice :)

DogAteMy: But you're only giving conditions on nested sets?

(I also told that child, as I always told my algebra students when I taught abstract algebra, that I didn't understand the proofs of the divisibility tests, or how to find divisibility tests for 13, 17, etc., until second year of college. One of my classmates in algebra at the time lived down the hall and was a geology/math double major. I went into his room all excited about my "discovery," and he calmly said, "That's good, Ted. I figured that out in 5th grade.".) GRR. :D

I mean, call $\scr F$ closed if, for any sequence $F_n$ of decreasing elements of $\scr F$, we have $\bigcap F_n\in\scr F$, and

if for any sequence $F_n$ of increasing elements of $\scr F$, we have $\overline{\bigcup F_n}\in\scr F$.

Lol I may try to acquire Dieudonne then :P

I always mix up scr and cal.

Test: $\cal F$

Scr is comically fancy

I love scr

OK, you should say decreasing sequence rather than sequence of decreasing elements, etc., DogAteMy. But I'll ponder it later.

OK, sure. I just mean $F_n\subseteq F_{n+1}$ or $\supseteq$.

I know. But you're modifying the sequence :)

Just like we say increasing/decreasing sequence of real numbers ...

True.

Right, OK, that makes sense.

OK, see you all later. Some of you have work to do (or sleep to do). :P

See you around @Ted!

night

Also I think this is an accurate description of what the guy in the other "Please it's request" question is likely going to be feeling right about now: youtube.com/watch?v=3KquFZYi6L0

OK maybe that's not fair but still kek

@Daminark have you talked about the Henstock-Kurzweil integral at all in your analysis class?

Yeah

Marianna brought it up briefly, mentioned that apparently some people want to just do that and drop Riemann/Lebesgue

@Daminark We already knows how he feels because he commented it in emoji

so apparently up to a.e. equality every Henstock-Kurzweil integrable function is the pointwise derivative of a differentiable function in $1$-D

Oh dear. It appears the question was deleted.

Lol I saw it JUST before it was deleted

And @Eric that's pretty dank actually

Well at least the new remade question is narrower in scope math.stackexchange.com/questions/2259846/…

mathoverflow.net/questions/66462/… here's an mo thread with some stuff on it @Daminark

And someone actually gave a hint...

Thanks @Eric!

RIP the hint

I didn't even know this was a badge math.stackexchange.com/help/badges/38/peer-pressure

el oh el

one worn as a badge of dishonor

Do we have a "dishonor on your cow" badge?

So apparently "topological data analysis" is a thing that exists

Yes!

TDA is my favorite thing that I don't understand :D

That sounds like a glorious thing to do in case academia says "Lel no job for you!"

(Or if grad school beats them to the punch)

I am not sure anyone outside of academia believes in TDA. But on the other hand, I am tryharding on the academic front right now and am not as aware of the alternatives as I used to be...

That's a fair point. Though it sounds like maybe the type of thing that AI would be into for sure

Which would be nice because it's still thinking about math

@Daminark At a wild guess: What if the state space for something was like four-dimensional, and the set of states where a certain interesting phenomenon happens was roughly in the shape of an embedding of RP2?

Or: What if, in the state space, there are two interesting data sets, and they're like two linked spheres in 4-space or something

Or two linked whatevers in whatever-space

Losing the teaching part would be sad so I think I'd kinda rather look into academia but eh.

since state spaces can have lots of dimensions (if it's the set of all images, for example)

And yeah @Akiva I think that's the sort of thing this goes into

Knot theory to the rescue!

throws string at people

Please do knot do that

and then, the sentence shortens into the void

Being brief is the soul of being witty or something

I derive quite a lot of amusement from searching meme words in the chat transcript

Such as?

Mostly generic words that are misspelled, or stuff like "kek" and "lel"

Leffing et le-ud

IPA /leud/

question mark appears above head

It's what "lel" should stand for

I don't know how to spell what I'm thinking of with the "leud" thing so if you don't know IPA I can't help you

iseewhatyoumean.jpg

Meeh I always come in here when there is nobody.

Who is zee?

i dunno

but they made a pretty bad first impression :/

I can see that.

Someone was here earlier when we were making memes about topos theory

Ted left bloodstains on the carpet ;)

Well, not memes so much as just talking about it, but point is, Zee was like "Yeah it's nonsense", when he/she didn't actually study it in depth

Bigly

Why the bleep would anybody say that topology is nonsense?

That in itself is nonsense.

Turned into a long-ish argument since Zee thinks that if a subject "doesn't contribute to society", it's not worth studying, etc

Sounds like a doggone alt-righter to me.

Topos theory is different from topology. To my knowledge a topos is a category which is sort of like a category of sheaves but not really?

(At this point I'm just term dropping but yeah)

I wish colorful swear words were not frowned upon in this room sometimes.

@Akiva I see.

Well, even so, category theory is pretty bitchin'.

@ALannister Fucking yellow

kek

@Akiva no worse than that.

$\color{Red}{Cunt}$

HAHAHA

I think that's pretty colorful

yes, Sandor Clegane's favorite insult.

(It says "aunt." The A is just cursive and not closed properly.)

I did actually see a card once that said "To a special aunt" written in a very unfortunate font.

I think we're both thinking of the same one

But yeah, it's definitely one of the subjects which is insanely abstract, we were mostly kinda getting on his case (I'm assuming he because that's what has been said) for talking about things as nonsense without actually knowing what they are

Also, "FLICK" is unfortunate in some fonts as well

Like when republicans talk about global warming?

Anyway, I must confess, I did come in here with ulterior motives.

"It is the mark of an educated mind to be able to entertain a thought without accepting it."

(Which is what I tried to do with Zee's stuff while I could)

Anyway, in most civilized (civilised?) countries, his name is Zed.

And Zed rhymes with Ted.

And Ted make Zed dead.