Mathematics

@AndrewThompson If you have a compact Hausdorff topological space $X$, and it's ring of real-valued cont. functions $C(X)$, then you can recover $X$ as maximal-spectrum of $C(X)$.

Mike Miller

@robjohn: woo, made it.

Andrew Thompson

@BalarkaSen Really?!

Balarka Sen

That is, $X$ is homeomorphic to $\text{mSpec} \, C(X)$, with the later getting subspace topology from Spec C(X).

@Andrew Indeed.

Andrew Thompson

7:20 PM

That's supercool.

Balarka Sen

(what this says is that you can recover the topological space from the sheaf of functions on it, in a fancier language)

Mike Miller

Really? Inside the prime apectrum? Strange.

Balarka Sen

@Andrew A-M leads you through a rigorous proof.

Andrew Thompson

Is that the exercise?

Balarka Sen

Anyway, this thing actually generalizes to varities.

yeah, @MikeMiller.

Mike Miller

7:20 PM

"Generalizes" is the wrong word...

different direction

Balarka Sen

ok, analogizes, then.

@AndrewThompson that's the statement of the problem.

So the thing with varities is actually equivalent to Nullstellensatz.

Andrew Thompson

I'll have a look at some point, thanks!

Where are you in your studies?

Balarka Sen

No problem, I thought you'd like to know about it.

@AndrewThompson Not very far from you. I am doing singular cohomology and commutative algebra too :P

Andrew Thompson

That's almost demotivating if the difference in our knowledge is as it seems. Undegraduate or graduate?

Balarka Sen

None.

Nah, you'll catch up with me soon. I am generally very slow.

Andrew Thompson

7:24 PM

Oh, just hobby?

Balarka Sen

You can say that.

Mike Miller

A different direction: $C(X)$ is a normed algebra; indeed if you consider complex functions a $C^*$-algebra. then the statement is "The category of unital commutative $C^*$-algebras is opposite to the category of compact Hausdorff spaces." The equivalence is by mapping to the maximal spectrum as a subset of the dual space.

And hence, noncommutative topology is the study of unital $C^*$-algebras.

Balarka Sen

interesting, but what is so noncommutative here?

(I'm not very familiar with C^*-algebras)

Mike Miller

I dropped the commutativity hypothesis. :P

If a compact space is a commutative unital algebra, then clearly a unital algebra is just a noncommutative compact space.

Balarka Sen

oh, I see.

I didn't see that.

@MikeMiller How do you define a noncommutative space?

@AndrewThompson Have stumbled upon anything interesting while studying cohomology?

Andrew Thompson

7:35 PM

Not started yet. It starts in the middle of the semester, half the credits of a normal course.

Mike Miller

@Balarka: It's a unital $C^*$-algebra!

Andrew Thompson

Its essentially chapter 3 of Hatcher.

Mike Miller

Oh, I guess you want to drop compactness? You're in trouble. You're always locally compact; non-unital means locally compact but not compact.

Balarka Sen

Oh, I just googled. A noncommutative space doesn't exist, but you think of noncommutative C* algebras as algebra of functions over the space.

Indeed, that sounds exactly like that thing Alain Connes would do.

:P

@MikeMiller hmm, ok.

Mike Miller

It's a nontrivially interesting idea. Very deep concepts from topology can be given interpretations in the more general setting of $C^*$-algebras.

Balarka Sen

7:39 PM

I was not trying to imply it was trivial, just that Alain Connes is reputed to work on things which exist only hypothetically, e.g., motives, $\Bbb F_1$. Of course, I don't think any of them are trivial - I wish I knew more mathematics so that I could understand what the $\Bbb F_1$ business is about.

@MikeMiller It certainly sounds fun.

Mike Miller

That's bullshit. Makes it sound like he doesn't do math.

(I know you weren't implying this.)

Balarka Sen

Grothendieck hypothesized about motives... and surely he did math!

Mike Miller

I just don't like the sentence "is reputed to work on things that only exist hypothetically".

Balarka Sen

@MikeMiller oh, ok. I guess it does sound silly.

Mike Miller

They exist; whether or not they're the right notion of motive or $\Bbb F_1$ is up for debate.

Balarka Sen

7:44 PM

@MikeMiller Fair enough. My english isn't that good.

@MikeMiller Last time I heard, there was no known definition for $\Bbb F_1$.

Did they finally find something interesting?

Mike Miller

Plenty of people have proposed definitions. You can't really do serious math that gets published without definitions.

The question is whether this definition has the desired properties.

columbus8myhw

Hi

Balarka Sen

ok, I didn't know that.

columbus8myhw

You're talking about that $\Bbb F_1$ thing?

I know nothing about that. From what I've heard, there are somewhat sensible ways to define it, but none sound like the "right" definition of it.

Balarka Sen

I thought people only used to study $\Bbb F_1$ as taking it as a blackbox, and see what kind of mathematical structure lives over $\Bbb F_1$.

columbus8myhw

7:49 PM

That makes sense.

But I feel kind of uneasy about it.

Has this happened before?

With some object other than $\Bbb F_1$?

Balarka Sen

I don't know nearly enough to feel uneasy about it.

columbus8myhw

@BalarkaSen That's probably what I should be feeling, too

You know how, when people first invented calculus,

they defined the derivative of $x^2$ to be the value of $\dfrac{(x+h)^2-x^2}h$ when $h=0$?

Which makes some kind of sense, since expanding gives you $2x+h$.

Mike Miller

Sure, maybe. I don't really see why that's much different than knowing the object itself.

It's varieties over $\Bbb F_1$ we care about anyway.

columbus8myhw

Hm. I should probably back off now, since this is clearly headed into territory I am unfamiliar with.

Balarka Sen

Yeah, varieties over $\Bbb F_1$ are number fields, that's why :D

columbus8myhw

7:53 PM

$^{^\text{what's a variety}}$

morphic

first lecture in topology we covered something called the universal properties and it was so confusing

columbus8myhw

Hello, @morphic!

morphic

hi @columbus8myhw

Mike Miller

The problem with $\Bbb F_1$ isn't that people don't do rigorous mathematics with it, it's that chatrooms that can't absorb the rigorous mathemarics are obsessed with it.

3

(Of course, this includes me.)

Balarka Sen

@columbus8myhw pick a bunch of polynomials in $k[x_1, \cdots, x_n]$. Look at their common zero locus in $k^n$. This is called a (affine) variety over $k$.

@MikeMiller +1

columbus8myhw

7:57 PM

You know what there's no rigorous definition for?

Mathematics.

:P

@morphic What type of topology? I just looked up "universal properties" and it seems to be related to category theory.

(A subject I want to learn but there are other things I want to get to first)

Balarka Sen

Some of Ted's exercises are pretty challenging.

columbus8myhw

Ted who? Shifrin?

morphic

@columbus8myhw Point-set

Mike Miller

"Mathematics is the smallest subject satisfying the following:
• Mathematics includes the natural numbers and plane and solid geometry.
• Mathematics is that which mathematicians study.
• Mathematicians are those humans who advance human understanding of mathematics."

columbus8myhw

I learned point-set topology and never ran into universal properties.

(And by "learned point-set topology" I mean "read through a tiny introduction")

(So maybe you shouldn't listen to me)

useanalias

8:01 PM

Hello all! Could someone take a look here (math.stackexchange.com/questions/1414254/…) and quickly check my logic? I just want to make sure I'm not posting a bad answer to my own question on SE - thanks!

columbus8myhw

That book was this thing, by the way

I'm not sure I should have called it an introduction; it was pretty thorough.

Balarka Sen

@MikeMiller Sorry, but that's not a rigorous definition. To say "smallest", you have to define what the partial order on the set of all subjects is.

morphic

One of the supplementary texts is Awodey's Category Theory

Mike Miller

It's not my definition, and also the correct partial order is obvious.

columbus8myhw

Hm. So you

're probably learning them both, or something.

@MikeMiller It seems to be a recursive definition, then.

Mike Miller

8:04 PM

There are only a couple universal properties worth caring about. That of the product and disjoint union are by far the most important. In some cases there is a universal property for the mapping space $X^Y$ that's useful.

That's all I can think of.

morphic

we did the mapping one

Mike Miller

It's not really recursive. That's one way to come up with the smallest subset, but this definition just says that math is the smallest one.

columbus8myhw

You start with natural numbers and plane/solid geometry. At every step, you give it to a few mathematicians to see what they do to it. This gives you a slightly larger subject. What you can get in a finite number of steps is called "math."

morphic

oh wait no, we did the disjoint union universal property

we have to draw some commutative diagrams for homework

columbus8myhw

aaaaah

Balarka Sen

8:10 PM

@MikeMiller What about the universal property of quotient maps?

Mike Miller

Ok, that too.

columbus8myhw

Is this at all related to the universal quantifier $\forall$?

Semiclassical

@Chris'ssistheartist my tastes being what they are, what i find interesting isn't so much the question "is the stated bound valid" but rather "by what margin is that bound satisfied"

Chris's sis the artist

@Semiclassical interesting. It's an IMO question, and for what they asked one doesn't need very complex stuff.

Semiclassical

and rates thereof, etc.

sure.

Chris's sis the artist

8:17 PM

@Semiclassical Better say, it's a IMO shortlisted question, but still it was proposed for IMO.

Semiclassical

i do like the question, to be sure. but i like it on two levels; on one hand, the question about whether a bound exists is a sensibly challenging one for an IMO problem

on the other, one can turn it into more of a research question by asking questions about how different the two sides have to be

useanalias

Off topic, but would someone mind looking at the link I posted? I'm sorry to double-post like this, but I'm confident someone will probably know how to solve it and it'd be a huge help.

Semiclassical

there are presumably proofs which don't touch the latter, and that's fine for a contest context. but it's the latter which i find more interesting.

Chris's sis the artist

@Semiclassical after a while the problem reduces to a telescoping series.

Semiclassical

I'll play around with it some more and see if I find an explanation I like.

Mike Miller

8:23 PM

@useanalias: I'll comment when I get to my office.

useanalias

Awesome - thank you @MikeMiller!

morphic

I hate chat on mobile because I have to manually refresh the page in order to get chat updates

columbus8myhw

@morphic My phone doesn't seem to have that problem

(It's an iPhone)

morphic

Maybe I should switch to an iPhone

Mike Miller

@useanalias: You are correct that in the case of Euclidean spaces $df$ and $\grad f$ are essentially the same thing, and that $M \hat i + N \hat j = \grad f$ is essentially the same equation as $Mdx + Ndy = df$, so that the first equation is true for some $f$ iff $Mdx + Ndy$ is exact.

Two notes. 1) The thing that one uses simple connectedness for is that, in a simply connected domain, $Mdx + Ndy$ being closed (integrals are path-independent) implies it's exact. Integrals are path-independent for any exact form; but the converse is not necessarily true in a non-simply-connected domain.

This might be what you were thinking of.

2) When one day you move on to manifolds, $\grad f$ is not something that will make sense without a lot more structure (something called a "Riemannian metric"; basically something that lets you pretend you're working in Euclidean space). The object that's always defined is $df$. One is just lucky in the Euclidean case that $\grad f$ is defined.

In general, a Riemannian metric gives you a way to identify vector fields and differentials, and $\grad f$ will end up being defined as "the thing that's identified with $df$ under this identification". It's sort of cheating, but it ends up being the correct thing to do. Anyway, this is all probably a bit too much for right now

Semiclassical

8:46 PM

@Chris'ssistheartist as for that integral you posted earlier---the one Jack tackled with symmetric homogeneous polynomials---the tool which i really want to be useful in that context is Feynman parameterization

as a way of making things even more symmetric, albeit at the cost of an increase in dimensionality

Chris's sis the artist

@Semiclassical Ah, I see. Still the answer seems far away, it's an evil integral. Well, if you find some way to tackle it then post an answer on main.

Semiclassical

certainly

Chris's sis the artist

OK

Semiclassical

i say that as more of "i want that approach to work"

rather than a certainty than it does in fact

Chris's sis the artist

@Semiclassical Yeah, I know.

Anthony

8:51 PM

@MikeMiller!!!

Mike Miller

Morning.

Anthony

Morning.

Chris's sis the artist

@Semiclassical I don't know if you got my point above, some messages above, I wanted to show you the closed form of that cubic integral I was talking about yesterday.

Semiclassical

don't think i did

i probably missed it amongst the chatter

Chris's sis the artist

@Semiclassical it's not simplified yet.

Semiclassical

8:54 PM

sweet merciful crap that's a lot of logs and polylogs

2

Chris's sis the artist

@Semiclassical Yeap, it's a very long answer. :-)

Semiclassical

though the factors of $\pi$ seem a tad superfluous. just push it into the LHS and say you're looking for the average value :P

Chris's sis the artist

:D

Semiclassical

i say that with at least some seriousness, since that interpretation lends itself well to asking what the rest of the Fourier coefficients would be

Chris's sis the artist

The amazing thing is this: I know how to develop a way of calculating that integral for any $n\ge1$.

:D

@Semiclassical Yeap.

Semiclassical

8:59 PM

hmm!

Chris's sis the artist

The amazing thing at the tool I developed is that it allows me to calculate that integral for any $n$. Although for the tedious integrals involved, but that can be calculated by hand (if needed), I prefer to use Mathematica.

Semiclassical

more generally, one could ask things like: Given a 'nice' function $f$, what can be said about the Fourier coefficients of $f(\sin\theta)$? That's something I find rather intriguing

Chris's sis the artist

Hard to answer the question in this form.

Transcript for

$Mathematics$ Mathematics