I think I actually got a 30min chat suspension a year or two ago for calling someone a troll

ah, the translation actually has a meaning which is close enough.

never heard of it used in this context

My intuition for your integral curve question is opposed to yours, I'd say (guess) it is maximal.

Not trolling, so please give examples that prove I am idiot.

Hmm, I guess you are right.

It still can be made minimal if you throw in a negative sign.

:)

Yes, but now what worries me is whether either is true.

I think not

Definitely maximal

I think it's kinda like saying that, if you keep going downhill, you'll get to the lowest spot on Earth, when in reality, you'll just end up in a nearby valley. I don't think I know enough to make the analogy more accurate

Along the lines of $\mathbf v\cdot\mathbf v\ge \mathbf v\cdot\mathbf w$

@AkivaWeinberger I don't know how that's an analogue.

The length of the curve is fixed though @Akiva

I suppose the physics formulation would go like this: Suppose I have a force field such that a field line connects two points. Is the work required to move along this line the minimum possible?

*maximum, based on @Karl's observation.

have to be a bit careful whether one is talking about the work done on the particle

@Semiclassical I am pretty sure you need to minimize among the paths of fixed lengths.

minimize/maximize.

i doubt that. if i can write the force as the gradient of a scalar function---i.e. if there's a suitable potential energy---all that matters are the beginning and end points

not every field is conservative

Oh, Akiva's observation is good.

It translates into a counterexample with some effort methinks.

Yay!

@balarka sure. i just mean that 'length' won't necessarily be the relevant functional.

Talking incredibly cryptically works!

perhaps. I haven't thought about it nearly enough seriously to give any opinion right now.

Stone-Cezh compactification

I was thinking about ways to make length irrelevant, seems @Semiclassical has the same sort of problem in mind

@Mambo What about it

If my vector field is conservative there is nothing to prove at the end.

@AkivaWeinberger Do you have any idea what is it ?

Also, I think I have a counterexample.

@Mambo Vaguely? (No.)

what counterexample you want?

my intuition is that, if i have a very light particle, then it'll move along the force field line by itself. (if it's not very light, its inertia would come into play)

What do you know about it?

what do you want to know?

@Semiclassical yes, that's what the definition of integral curve says.

and so if i want to force it along another path, i'll have to apply an external constraint. and while the particle might do work on me at times, my feeling is that the net work i do in keeping it on that trajectory is positive.

@Mambo

For instance, what is Stone-Cech compactificaton of $Bbb{Z}$?

How is it related to $\Bbb{Z}$?

the caveat here, though, is that what i have in mind for 'nonconservative force' in that case is by including some kind of dissipation i.e. friction

its a crazy space

well, it contains $\mathbb Z$ as a dense open subspace

How will you construct it?

rather than, say, a force field defined on $\mathbb{R}^2\setminus \{0\}$ which lacks a potential

I take back what I said on having a counterexample.

by ultrafilters

Oh, that's the space of ultrafilters, right?

yes

I don't actually know what the Stone-Cech compactification of anything else is.

its very big

and points are not accessible in $\omega$-many steps

What is an ultrafilter?

even there, though, a potential energy should exist locally. so it'd need to be some global property of the force field.

Remind me what accessible means?

@Mambo Awesomeness

But, um, also complicatedness?

I mean it is not first countable at points in the remainder

I forgot what first countable means

so not metrizable

I'm a terrible topologist

countable basis at a point

Ahh.

@Semiclassical @Karl @AkivaWeinberger Can we have this discussion later? I am sorry to jump out of my own question but I have work to finish up before thinking about this. :(

@Mambo you first learn about filters

i have to head out soon anyways, so sure

@ForeverMozart Not really

and then use Zorn's Lemma to extend a filter to a maximal filter

speaking of: "compute faster, mathematica, i want to know this before i have to leave"

i'll ping you if i get the time to think it through, @SemiC.

sure.

I am very afraid of Zorn's lemma

$\Bbb Z$ is a dense subset, despite the fact that every point in $\Bbb Z$ is isolated. (Kinda like how, in $\{0\}\cup\{1/n:n\in\Bbb N\}$, the nonzero points are a dense, isolated subset)

It's more afraid of you than you are of it @Mambo

Disclaimer: It is very late here so I am spouting nonsense

this would be a nice ted question :/

lol

it originated from one of ted's exercises ;)

@AkivaWeinberger yes it is first countable at points in $\mathbb Z$, but not at the added points

I think I can sort of explain ultrafilters, but it's late so that makes it harder

So, um, suppose we play Guess My Number

lol

You think of a number in $\Bbb N$, I try to guess it

not great odds for that game

Can you please tell me an element apart from points of $\Bbb{Z}$?

and I can ask yes-or-no questions, infinitely many times

@Mambo They're not guaranteed to exist without the axiom of choice!

those will be the free ultrafilters @Mambo

(We're assuming choice)

but, um, let me finish, because this is relevant

So, you think of a number

5

And I can always win at this game, right?

no, because i thought of sqrt 2

I just ask, "Is it 0?" "Is it 1?" "Is it 2?" until I hit it

lool

attempts to smack Balarka

ok, continue I am enjoying this story

but, the thing is, you can cheat at this game!

ok

like i did

Don't actually think of a number! Keep on saying "no" to each of those questions!

@BalarkaSen NO

no not like you

damn it i have to get work done.

Do work!

you are messing up the rules

@BalarkaSen

lol

Apply forces to masses!

well, it has the same effect (always saying "no" to cheat and always saying "no" because you didn't listen to the rules)

Fd or something

see ya'll. i'll continue trolling @Akiva after i get back

flagged for trolling

$W=\int_C \mathbf{F}\cdot d\mathbf{r}$

So, um

You can't actually always say "no"

'Cause, like, I can ask, "Is it even?" and later I can ask, "Is it odd?"

and you can't say "no" to both of these

Then

So, an ultrafilter is kinda like a game of this, where you may or may not be cheating

and the principal ultrafilters are where you're actually thinking of a number

I'll explain.

ahahahahah mathematica proved me right

Please go ahead

So, an ultrafilter is a family of sets of numbers.

Put $A\subseteq\Bbb Z$ in the ultrafilter if you answer "yes" to the question "Is your number in $A$"?

and don't if the answer is "no"

What is $A$? any subset

If $A$ is not in the ultrafilter, then $\Bbb Z-A$ is, because if you answer "no" to "Is your number in $A$" then you'd have to answer "yes" to the question "Is your number in $\Bbb Z-A$"

@Mambo Yeah

So $\Bbb{Z}$ is always there

Yeah.

And the empty set is always not there.

@Akiva Did you come back yet?

Also:

@MikeMiller ?

Oh, no, still in Israel, gonna leave in a few hours

Union is also there

If $A$ and $B$ are in the ultrafilter, then so is $A\cup B$ and $A\cap B$

@Huy is my favorite

In fact, superset is always there. If $A$ is in it and $B\supseteq A$, then $B$ is in it

@ForeverMozart: why

cause you say cool things

^_^

yeah, true

see I can't do that

So, example of an ultrafilter:

Set of everything containing $6$

So, set of all supersets of $\{6\}$ (including $\{6\}$)

containing as in "has it as a digit"?

oh

This is a set of sets

This corresponds to you not cheating, and legit thinking of the number six

Yes

So, um, formal definition time, I think

An ultrafilter $\scr U$ on a set $X$ (we've been using $\Bbb Z$ but let's generalize) is a set of subsets of $X$ satisfying four things:

1) $\varnothing\notin\scr U$

2) If $A\in\scr U$ and $A\subseteq B$ then $B\in\scr U$

It is almost like a associating a measure to $\Bbb Z$ fixing a point $x \in Z$ and giving weight if the set contains $x$ ; otherwise $0$.

3) If $A,B\in\scr U$ then $A\cap B\in\scr U$

4) If $A$ is any subset of $X$, either $A\in\scr U$ or $X\setminus A\in\scr U$

And you are considering set of all measurable set of non-zero measure

Yeah, pretty much

Another example of an ultrafilter:

On $\Bbb R$, the set of neighborhoods of a point

Wait, no

That's just a filter

Sorry

free ultrafilters are difficult to construct

you can do it inductively

A filter is something that doesn't satisfy #4. #4 is what makes it "ultra."

wonder why 'ultra.'

I mean, doesn't need to.

superfilter? megahyperfilter?

Funny. I'm in NY. Will probably gone by the time you're back.

Aw :(

Not that we'd be able to meet anyway, to be honest

Also: I can't take you for drinks.

I probably won't touch alcohol even for years after I turn 21

Doesn't seem appealing to me

It's tasty.

So, um, a filter is like a game of Guess My Number where we didn't finish

@MikeMiller I'll still with my soda, thanks

For the set of neighborhoods of a point, why would any superset be contained?

So, like, we might have neither $A$ nor $X\setminus A$ in our filter

Also, there are plenty of coffee bars.

(We can't have both, because then their intersection would also be in it, but that's the empty set, and we can't have that in a filter)

Don't like coffee either. Probably should, though, seems useful

I didn't ask about that

Oh, and, uh, filters have to be nonempty. Forgot to mention. A filter is a nonempty set such that yadda yadda yadda.

@Mambo Right, yeah, sorry

Uh, isn't a neighborhood of a point defined to be any set containing an open set containing the point?

Then okay

So any superset of that still contains that same open set containing the point, and thus is a neighborhood

So, the ultrafilter lemma

says that every filter on $X$ is the subset of some ultrafilter on it

This cannot be proven in ZF.

It, however, is weaker than choice.

So, it's kinda somewhere between.

If you're familiar with IST, than IST without choice is essentially equivalent to ZF+ultrafilter lemma

Don't like coffee? Weird kid

(IST is kinda like ZFC but with infinitesimals and infinitely large stuff)

(It's fun)

Does it stand that $(p_1\cdot p_2)\circ f=(p_1\circ f)\cdot (p_2\circ f)$ where $p_1,p_2,f$ are polynomials?

So, um, did my sleep-deprived, about-to-be-on-a-plane self explain adequately well? @Mambo

@MaryStar Yeah, pretty sure

But why? Hpw could we show that? @AkivaWeinberger

(Not sure why being about to be on a plane would make my brain function worse. Blame the sleep-deprived part of me for writing that)

Of course, you did a great job there

@MaryStar Plugging in an arbitrary value $x$ would be a nice place to start

Yay!

I would be happy to see you speak more about it @AkivaWeinberger

Actually I am studying a very famous theorem of Gelfand and Naikmark

So, um, ZF doesn't prove that nonprincipal ("cheating") ultrafilters exist, but ZFC does. Zorn's lemma was mentioned earlier —

I think you use it (equivalent to choice) to show that they exist in ZFC

I see

IST's proof of it is really awesome, though. But you have to know about IST (and trust that everything provable in IST is provable in ZFC) to understand it

(I don't know how to prove that IST and ZFC are equivalent like that.)

Well, no problem. Take your time. I would read this slowly.

I can't explain IST :P

Not from the beginning

lol okay Do you know any good reference?

Side note: IST is one way of doing something called "nonstandard analysis", which has the nice acronym "NSA"

Not really. I have bad references

I know someone who's writing a paper on it, though, aiming to explain it

@GPhys

So, when that's done, it would be a good reference

Isn't he in MSE?

I think so

I couldn't find his profile

To do the IST proof without explaining any of the words: You take a nonstandard element $x$ of $X$, and take the set of all subsets of $X$ containing $x$ (the principal ultrafilter generated by $x$); call it $\scr U$. Then take the unique standard set $^{\Large\sf s}\!\scr U$ containing all standard elements of $\scr U$. That's the nonprincipal ultrafilter.

Difficult

The nonstandard elements of $\Bbb Z$ are the infinitely large ones, which I said existed in IST

IST kind of breaks induction

It's weird

'Cause you can only prove induction for properties that can be described in the language of ZFC, but "not infinitely large" is not describable in the language of ZFC

I just solved 2 more problems in a row which I thought were hard initially. Funny how that works.

Time to send Ted an e-mail.

Good job

Hi @BalarkaSen. Do I still need to look at something?

Hi doc

Nope, @MikeMiller, my argument was fine. Thanks.

who's doc

Could you check out this bump for me?

So, now I know that pulling back forms and feeding them vectors is the same as feeding them the pushforward of those vectors. Cute.

Thank you very much @AkivaWeinberger.

that's the definition of a pullback form ;)

yay you're welcome

It wasn't quite how Ted defined it. It was an exercises that they were equivalent.

Not a very hard one, but an exercise nonetheless.

Sure, I agree. I just meant that it's the correct definition in broader generality than you need right now.

Oh, that makes sense. Given a smooth map $f: M \to N$, and a diff form $\omega$ on $N$, I guess I can define it's pullback on $M$ coordinate-freely by defining it to eat a vector field and spit out what $\omega$ spits out after eating the pushforward of $X$ by $f$.

yeah, since a differential form is just something that eats vectors.

right, argreed.

well, no, i meant $\omega$ is a 1-form up there. for k-forms i have to eat k vector fields multilinearly.

Sure.

@Mambo Oh, and, uh, presumably you can put a topology on the set of all ultrafilters on $\Bbb Z$. Don't remember what it is, though.

Lots of thunderbolts outside. I hope it's going to rain.

weird kid

i take that as a compliment. in my not so defense it's 43 degrees.

converts to $^\circ\!$F

Wow

Argh. Just there I was typing out the joke if it is C or F.

It's Kelvin.

So neither.

Good luck with rain and 43 Kelvin

liquid nitrogen?

and all the super-fluidity.

@KarlKronenfeld to cold even for that AFAIK

damn

Maybe hydrogen.

No, still too hot for that.

Let's ask on chem.se which substance is liquid at 43 K.

This discussion is quickly becoming too technical for me.

converts from degrees to radians

("let us discuss topological quantum field theories instead")

Start by naming one. :)

Admittedly I don't know anything other than the definition, but that was meant as a joke ;)

Eventually I will learn something about this type of mathematics. But maybe not today.

I guess I better start doing what I actually need to do till next week.

See you!

Byes.

@quid Different people have different tastes. I know some people who study TQFTs in general but would have trouble naming them. I study one particular thing which is not quite a TQFT.

I think I and they would have a hard time talking.

Do they still at least give relevant information about particular TQFTs?

hm

@MikeMiller yes, I somehow knew that it is quite diverse. And in part very abstract.

@KarlKronenfeld I think the trouble is that there are not many particular TQFTs. I think the general theory is quite interesting but not always entirely helpful, since you need to calculate something eventually.

(I also think the fact that my thing fails some of the TQFT axioms is part of why it's useful, maybe? Less restricted.)

But I am slightly ignorant of all this. :)

I am a bit curious about what is the thing you study which is not completely a TQFT. But I shouldn't be.

I've said the name before, I'm sure.

hi guys I have a doubt in combinations.

Admittedly I don't remember if you said it to me when I was reading Lurie. But I have calculus to do!

Oh, but I have already done the calculus I had to do today.

say I have to choose 2 teams of 5 from 10 ppl. How many ways can I do it?

So I guess it's time to think.

the answer is (10 C 5)/2

I don't understand the division by 2.

I probably didn't say it then. But I've definitely said the name of the thing I do plenty.

@AbhishekBhatia you do not distinguish between selecting team A and then B and selecting team B and then A.

Hmm, I'll have to decide what I want to think about first though. There's a bunch of things, but not sure which one is the most juicy thing to ponder on.

10C5 is the number of ways to choose a set of 5 ppl, but you want to choose pairs of sets (the set and its complement). Notice that 10C5 will count for a set and its complement separately

depending on your search terms, either 9 times or ~30 times, i guess

@KarlKronenfeld how will it count the complement?

say I want to choose a team 6 and 4. The answer is (10 C 6) or (10 C 4) in that case.

I'm guessing Seiberg Witten theory.

it matches with the statistics.

@quid okay, but how does that result in a division by 2.

@BalarkaSen Close, but not quite.

@AbhishekBhatia Because the complement of a set with 6 elements has 4 elements, so the complement will not be counted in 10C6.

@AbhishekBhatia Let us take a simpler example. You have 4 players w,x,y,z and want two teams. You have 4 C 2 ways to chose one team of 2 players.

but why would it counted in 10 c 5?