Mathematics - 2016-03-20 (page 2 of 3)

Mike Miller

16:00

quote from MSE: "the proof is very, very long. Continues over two pages."

Semiclassical

haha

Mike Miller

debating whether or not to bring my laptop on this trip. it's a bit bulkt

Semiclassical

@anush anyways, i hope that wasn't too much information overload :)

where are you heading? @MikeMiller

Balarka Sen

@MikeMiller maybe sell your current laptop and buy a lighter one.

Mike Miller

conference in the bay area

my current laptop is old enough that it's got essentially no value. I'm going to buy a chromebook (again) sometime

time cube appears to be permanently offline. rip

luckily everything on the Internet is archived

Semiclassical

16:13

in honor of that kind of crazy, a conspiracy theory: Wikipedia says the domain expired in August 2015, in the midst of the enduring crazy train that is Donald Trump

Ted Shifrin

@MikeM: Don't you know a one-line proof of the Riemann Mapping Theorem? Geez.

oh, Balarka is still alive ...

Albas

Hello@TedShifrin

Balarka Sen

@TedShifrin Hello.

Ted Shifrin

hi, @Semiclassic, DogAteMy, @Albas.

Balarka Sen

Yes, but only barely.

Mike Miller

16:15

@Ted Yes, of course, assuming uniformization.

Ted Shifrin

Talk about putting the carriage ahead of the horse, @MikeM.

Semiclassical

obviously, then, Time Cube was a secret NWO plot to subconciously prepare us for the coming of the Trump, and had thus served its purpose.

Logan Toll

Hey everyone!

Semiclassical

#TrumpCube

Mike Miller

@Ted: I try.

This reminds me of my favorite construction of the Lebesge measure.

Ted Shifrin

16:16

hi @Logan

Semiclassical

(this is one of those instances where i am far more amused by my joke than anyone else.)

Ted Shifrin

I find the whole thing way too depressing, @Semiclassic.

Semiclassical

fair enough. farce comes after tragedy, after all, and i doubt we've reached rock bottom yet :/

Ted Shifrin

No, we surely have not.

Logan Toll

What exactly is time cube? I heard about it a while back but was too lazy to spend any time looking at it.

Ted Shifrin

16:18

@Balarka: Have you been sequestered in an asylum?

Mike Miller

Riemann integration defines a positive linear functional $C_c(\Bbb R^n) \to \Bbb R$. So by the Riesz representation theorem, this is given by integration against a regular Borel measure. :)

@Logan: I linked to the webpage.

Ted Shifrin

Yes, that would be a good carriage, as well, @MikeM.

Semiclassical

i do like the phrase #TrumpCube, though

Logan Toll

@MikeMiller Oh, thanks!

Ted Shifrin

Reminds me of Nissan Cube.

Balarka Sen

16:20

@TedShifrin Much worse. I have been semi-forced to study something I don't think I have the mathematical maturity for. Namely, Shafarevich part I.

Mike Miller

@Ted: It's probably not circular. :)

Ted Shifrin

Forced? @Balarka

Obviously, there are people who think it's more important to learn second-year graduate courses before learning the important undergraduate material. I'm not one of those.

Balarka Sen

Well, told to, at least. But the stuff's hard.

Mike Miller

Is the implied claim that Shaferevich is second-year graduate material?

Logan Toll

@TedShifrin I don't know I find myself seeking the higher level stuff out because I think its more interesting; I'm in high school and it's not that I find the curriculum easy but it's so god damned boring.

Ted Shifrin

16:23

Assuming one takes graduate algebra and commutative algebra first year, yes, @MikeM.

Mike Miller

How does anyone in algebraic geometry get a PhD in five years?

Ted Shifrin

@Logan: I don't think Balarka has been talking to us about high school math in, say, 5 years.

Logan Toll

@TedShifrin Oops, read it out of context!

Pedro Tamaroff

I have a problem for you guys.

Ted Shifrin

I don't see that as impossible, @MikeM. But not everyone can go to top research universities as an undergrad and take two years of graduate courses as an undergrad.

mr @Pedro !

Pedro Tamaroff

16:26

Give $S_n$ uniform distribution. The expected number of $k$-cycles in a permutation is $1/k$ for $k\leqslant n$.

Ted Shifrin

@Balarka: In all seriousness, it's no harder than the recondite topology you've been wrestling with. But I still don't like it.

Balarka Sen

@TedShifrin On a brighter note, I'll probably be learning real analysis (1st Rudin then Stein Shakharchi or something) from an analyst after this is over (I don't think I can cope with Shafarevich anyway, probably would drop out soon).

I don't like it either, to be honest.

Mike Miller

Given any elliptic linear operator $D$, does D-flow ($d/dt-D$) always produce smooth solutions? I guess so, but it's not obvious why.

Ted Shifrin

You still should finish what we were doing, @Balarka, and I don't see any reason you can't do differential topology, finally. There's not that much left in Rudin for you to learn. Skipping multivariable and his crummy chapter on Lebesgue, that leaves the first 8 chapters. You might need to do a month of exercises, but that's it.

Mike Miller

Topology qual here is in a few days.

Ted Shifrin

16:29

Excitement.

I'm confused, @MikeM: On what Banach space are we working?

@Pedro: I find that hard to believe. Are you talking about a minimal representation, for starters? I can put in an awful lot of 2-cycles.

Pedro Tamaroff

I mean, a permutation can be written uniquely as a product of cycles: order the cycles first in size and then putting the largest element of the cycle first, order the cycles.

Mike Miller

I dunno, whatever space of functions you want. If one needs help at the start of bootstrapping up, the minimal $L^2_k(\Bbb R^n \times \Bbb R)$ such that functions are continuous.

Ted Shifrin

@Pedro: Uniquely? Really?

Pedro Tamaroff

@TedShifrin Yes.

Balarka Sen

@TedShifrin I actually peeked at G-P when I got stuck and frustrated with Shafarevich. The first 6 chapters looked nice, I like the material: the exposition runs quite smooth. I can probably do a couple exercises too.

arctic tern

16:32

(disjoint cycles)

Pedro Tamaroff

I mean that this representation is unique, Ted.

Balarka Sen

Speaking of, I looked and Rudin's multivariable exercises are easy.

Ted Shifrin

But why can't I throw in arbitrarily many $(1\,2)(1\, 2)$?

Balarka Sen

Admittedly most of the exercises are in your book.

Albas

Balarka the exercises in the special function chapter are very difficult. I saw a few of them

Ted Shifrin

16:34

His treatment of multivariable sucks, @Balarka. And you're learning a lot of analysis if you do the exercises I've asked you to do.

Pedro Tamaroff

@TedShifrin Disjoint cycles.

Ted Shifrin

Ohhhhhh, disjoint cycles :P

Pedro Tamaroff

Write $\sigma$ in disjoint cycles, and order them by length first and then by the largest element inside the cycle.

Ted Shifrin

Still very strange that the expected number of cycles should be that tiny. So what is the expected cycle-length, then?

arctic tern

calculate the # of permutations with a given k-cycle and count the number of k-cycles, multiply and divide by n!

Balarka Sen

16:35

@TedShifrin I probably did most of the exercises you asked me to do, give or take a few. I hope I did learn analysis.

Anush

@Semiclassical thanks very much!

Ted Shifrin

Well, I haven't heard anything from you about the chapter 7 stuff, @Balarka, and there's still plenty of chapter 8.

Balarka Sen

I didn't get the chance to talk. Admittedly I did not study chapter 8.

Ted Shifrin

Since we spend more time and effort on you than these other people, @Balarka, why are they giving you all the orders?

Balarka Sen

Well, they spent some time teaching me some topology, so I can't say right away that I don't want to study Shafarevich right now whereas they want me to learn it. But I don't think I need to, as it's not going well. I keep getting stuck. Probably they'll agree that it's too early for me and put me on differential topology or something.

Or at least that's what they said when I said it's too hard for me.

arctic tern

16:43

@TedShifrin assuming I've understood correctly, $\frac{1}{n!}\sum_{\sigma}\sum_{\rm cycles}{\rm size(cycle)} = n$ because the inner sums are $n$, and the total # of cycles in the list of all cycle representations of all permutations will be $\sum_k n!/k$, so the expected cycle size of a cycle drawn from the list of all cycle representations of all permutations uniformly at random will be $n/H_n$ where $H_n=1+1/2+\cdots+1/n$.

which is about $\pi(n)$. there are analogies between prime statistics and random permutations, qiaochu has a blog post or two about it, and andrew granville I believe wrote a thing on it

Ted Shifrin

Well, expected cycle length should be $\sum k P(\text{length }k)$? Am I missing something?

arctic tern

what do you mean by P(length k)?

Ted Shifrin

the probability that a given permutation has a length $k$ cycle in it?

I freely admit this is stuff I never think about ... unlike Granville, my former colleague and good friend :)

arctic tern

so would two permutations with different #s of k-cycles >1 contribute the same to kth summand there?

Ted Shifrin

oh, what I wrote isn't right.

Balarka Sen

16:47

@TedShifrin Granville was your colleague? Cool.

arctic tern

the formula will of course depend on how we are drawing our k-cycles. I've pretended we've written down the disjoint cycle representation of all permutations on paper, then circled all the k-cycles present, and drawn from that collection uniformly at random.

Ted Shifrin

Yup, for many years before he left UGA to escape the gun culture of the US. And now look ...

arctic tern

cool

Balarka Sen

When did he leave UGA?

Ted Shifrin

Hmm, over 10 years ago, @Balarka. I don't remember precisely, but one can find it on the web easily enough.

Balarka Sen

16:49

Ah, ok.

Pedro Tamaroff

@TedShifrin Yes. The point is to obtain a combinatorial proof.

Ted Shifrin

Now the whole question is confusing me, so I'm not going to think about it any more ... :)

Mike Miller

@TedShifrin: Here's the point of the question, rephrased. Heat flow is a smoothing operator (that is, solutions to the heat equation for $t \geq 0$ with boundary value at $t=0$ are smooth for $t>0$). Put more elementarily but with less power, solutions to the heat equation on $\Bbb R^n \times \Bbb R$, the second term the time variable, are smooth.

Is this special to the heat equation? Or is it just because it's the flow of the Laplacian, which is elliptic in ever sense?

Ted Shifrin

@MikeM: Surely what works for the Laplacian should work for any elliptic operator. Not that I remember my PDEs from grad school ...

Mike Miller

@TedShifrin: That's how I feel. Unortunately I also can't remember why heat flow is smoothing, other than "I can construct the unique sol'n"

Maybe something something convolution with heat kernel.

Akiva Weinberger

16:52

Hola

Ted Shifrin

Yes, I was about to say convolve with the fundamental solution.

Akiva Weinberger

I got a gig tutoring people. I just taught someone the Pythagorean theorem and how to prove it

Ted Shifrin

Presumably you can use the symbol of the elliptic operator to tweak it.

DogAteMy: How many different proofs? :D

Semiclassical

doesn't heat flow have a conserved quantity?

Akiva Weinberger

Just one, but I mentioned that there are at least four hundred

arctic tern

16:53

@AkivaWeinberger which proof?

Mike Miller

So, what's special to ellipticity here? The existence of a fundamental solution?

Semiclassical

though i may well be confusing that with something else. my formal PDE background isn't great either.

oh, @ted, did you see the horrible PDE i cited earlier? :P

Mike Miller

I guess someday I should just sit my ass down and read Evans. Sigh. Too much to do.

Ted Shifrin

No, @MikeM, you always have a fundamental solution in the appropriate sense, don't you? Ellipticity gives you nice exponential decay or something.

No, @Semiclassic.

Akiva Weinberger

This one: images.pcmac.org/SiSFiles/Schools/FL/GadsdenCounty/…

Mike Miller

16:54

I don't know much about the theory of fundamental solutions. I'l believe you.

Akiva Weinberger

@arctictern (Nice name, by the way)

arctic tern

you familiar with the joke?

Akiva Weinberger

Who's Line reference?

Semiclassical

2 hours ago, by Semiclassical

\begin{align} \partial_t h - \partial_{xx} h &= p-\frac12 (\partial_x h)^2,\\ \partial_t p+\partial_{xx}p &= -\partial_x(h\partial_x p),\\ p(x,1)&=\Lambda \delta(x),\\p(x,0)&=\Lambda \delta(x)-2\partial_{xx}h(x,0)\end{align}

arctic tern

@AkivaWeinberger :D

Ted Shifrin

16:55

DogAteMy: I guess I'm fondest of the similar triangle proof. But I don't know the age/level of your tutee.

Semiclassical

it kind've sucks :/

Balarka Sen

@TedShifrin Garfield's proof is cool.

Semiclassical

i'm not trying to solve it exactly, mind, just numerically in mathematica

Ted Shifrin

@Semiclassic: This looks like the PDE you wrote down a week ago re Mathematica. I have no clue.

Semiclassical

it is, yeah

Akiva Weinberger

16:56

We went over similar triangles just last week… I wouldn't want to use something that she might still have problems with

Mike Miller

@Semiclassical: Oh my, I hadn't realized without TeX on just how nonlinear your equation is.

Semiclassical

to make life easier i'm working with periodic BCs in the $x$ direction but still

Mike Miller

Are yuou sure that's legal in the state of Michigan?

Balarka Sen

I actually put that in my answer sheet on the board exams. It asked to give a proof of Pythagoras' theorem. I put up the similar triangle proof, and for extra credits, Garfield's. :P

Semiclassical

don't know, but I don't live in Michigan

Ted Shifrin

16:57

He's in Minnesota, @MikeM.

Semiclassical

I'm fond of the 'visual proofs', though they're really not

Mike Miller

Knew it started with Mi.

Akiva Weinberger

This one's an interesting one:

Semiclassical

main thing I need to remind myself is how certain things look when I periodize them, e.g. gaussians

Akiva Weinberger

(Though it's substantially less rigorous and intuitive than the others)

Semiclassical

16:59

which amusingly is related to the Riemann Theta function discussion i had earlier, since perioidization is basically just poisson summation

Mike Miller

Whatever floats your boat, buddy.

Semiclassical

well, if i periodize them, that both gets rid of any discontinuities in boundary conditions and gives me some knobs to play with

how many fourier modes i use, what size interval i restrict to, etc.

Balarka Sen

@MikeMiller That has become a signature idiom of yours.

Akiva Weinberger

(I refer to goats, myself)

arctic tern

@Huy sorry for not being around. there is a 2-1 (hence "spin") homomorphism ${\rm SL}_2(\Bbb K)\to{\rm SO}(1,n+1)$ for $\Bbb K=\Bbb R,\Bbb C,\Bbb H$ and $n=\dim\Bbb K=1,2,4$ respectively (plus probably $\Bbb O$ too with definition shopping). ${\rm SL}_2(\Bbb K)$ acts on hermitian matrices $h_2(\Bbb K)$ by similarity transformations $A\cdot M:=AMA^\dagger$, preserving the determinant form of signature $(1,n+1)$: $$\det\begin{pmatrix} t+x & \bar{u} \\ u & t-x \end{pmatrix}=t^2-x^2-|u|^2 $$

Akiva Weinberger

17:05

But whatever floats your goat

Balarka Sen

If that floats your goat, that's fine by me.

Semiclassical

whatever floats your goat on a boat in a moat

Balarka Sen

That's less funny, because it doesn't make too much sense.

Semiclassical

pfff, you people and your "jokes actually have to make sense"

Vrouvrou

hello, who can help me on path connectedness

?

Mike Miller

17:09

@Semiclassical This is far from Balarka's position.

Balarka Sen

I don't think I have made nonsensical jokes, if not good ones.

Semiclassical

mine really wasn't a joke so much as pointless rhyming for my own amusement

Vrouvrou

@TedShifrin pouvez vous m'aider sur la connexité par arcs ? s'il vous plait

Balarka Sen

17:23

@MikeMiller As a side note, I remember I annoyed you by saying I understood why injective proper immersions and embeddings once when I didn't really understand it. I believe I do now, along with the first few chapters on Guillemin-Pollack. (just to make up for that).

I meant first few sections, not chapters, oops.

Mike Miller

I believe borh statements.

Balarka Sen

There's one thing that bugs me though. I remember G-P mentioning there are smooth manifolds in $\Bbb R^n$ which are not globally cut out by smooth real functions $g_1, \cdots, g_k$. What would be an example?

Obviously locally this is true, as inclusion map $M \to \Bbb R^n$ of any manifold is a immersion and thus locally at each point is of the form $(x_1, \cdots, x_k) \to (x_1, \cdots, x_k, 0, \cdots, 0)$ upto setting the correct parametrization by immersion theorem, so given by $x_{k+1} = \cdots = x_n = 0$.

Vrouvrou

someone knows path connected set ?

Balarka Sen

The reason this statement bugs me is that I vaguely remember hearing that smooth manifolds are all real algebraic varieties. I am worried whether I misremembered the theorem.

Adeek

17:47

hi

@BalarkaSen would you like to see my elliptic curve project ?

Balarka Sen

Sure.

Mike Miller

@BalarkaSen: There's two points here. One is subtle, the other is not.

Adeek

Mike Miller

1) Pick a smooth manifold that's not closed as a subset of $\Bbb R^n$. This is the not subtle one.

Adeek

1 sec

I will upload it

speedyshare.com/ZTAZX/final-elliptic-curve-beamer.pdf

speedyshare.com/THtsG/Algebraic-elliptic-curve.pdf

one is beamer file presentation

and the other is the write up

let me know what do you think

Balarka Sen

17:51

@MikeMiller You're right. I was not looking for those.

Mike Miller

2) The subtle part is when something can never be cut out transversely by smooth functions, even if you re-embed. So suppose $M \subset \Bbb R^n$ is cut out transversely by $k$ smooth functions; then its normal bundle is trivial of rank $k$. $TM \oplus NM$ is trivial, so we see that $TM$ is stably trivial. In particular, it can't be orientable; so eg it's not possible for $\Bbb{RP}^2$.

Think too much about this you start thinking about framed bordism. I'm not going to do that here.

So your result, where things are algebraic varieties (IIRC it's actually just that they're components of smooth algebraic varieties? you may not be able to get something by itself, I think), they're being cut out by the equations, but not transversely. $x^2+y^2=0$ type stuff.

Balarka Sen

Being cut out transversely essentially means $Dg_1, \cdots, Dg_k$ are linearly independent, correct? Just to confirm.

Semiclassical

so degenerations of algebraic varieties?

Mike Miller

You have the wrong number there, $n$ instead of $k$. But yes.

Semiclassical

or something more subtle than that

Mike Miller

17:56

It means they span the tangent space of the image, aka that you're a regular value.

Balarka Sen

Thanks.

Right.

Mike Miller

@Semiclassical: I gues so.

Balarka Sen

@MikeMiller Ah, alright. I see now. Good point.

Mike Miller

It's extremely gross.

Semiclassical

though i guess the example you cited doesn't quite fit with the canonical example i have in my head (namely the legendre family $y^2=x(x-1)(x-\lambda)$ as $\lambda\to0,1,$ or $\infty$)

that'd be stuff like $y^2=x^2(x-1)$.

Mike Miller

17:59

@Semiclassical: This is a very, very bad type of degeneration. The dimension of what I cut out is smaller than what I expect it to be.

Semiclassical

gotcha

Mike Miller

Think again $x^2+y^2=\varepsilon$.

Balarka Sen

@MikeMiller I don't have the background to understand the reasoning there (probably I would when I get to ch 2), but that's certainly very interesting. Thanks.

Mike Miller

It's sickening, I'm telling you.

Semiclassical

$y^2=x(x-\lambda)(x+\lambda)$ is probably closer to the mark, but hrm

probably still not enough.

Mike Miller

18:01

@BalarkaSen: Take it as an exercise to prove that if $M$ is a regular value, its tangent bundle is stably trivial, using what's above.

Balarka Sen

Why so?

Mike Miller

Smooth manifolds being cut out non-transversely?

Balarka Sen

I mean, why is it sickening?

Semiclassical

what i find interesting in general is the question of how the periods of that algebraic variety behave. as an example

Mike Miller

Since when have you ever cut out a smooth manifold non-transversely?

Balarka Sen

18:02

Oh. Heh. Ok.

I get you.

@MikeMiller I have to probably get more familiar with the tangent bundle before I can prove that. I take it as a future exercise.

Semiclassical

suppose i take the family of algebraic varieties $y^2+x^2=\epsilon$ for $\epsilon\in\mathbb{C}\setminus\{0\}$ and consider the integral $\oint_C \dfrac{dx}{y}$ over some nice cycle on the surface

there we go

Mike Miller

No! No $\Bbb C$!

Semiclassical

most definitely $\mathbb{C}$

Mike Miller

$x^2+y^2=0$ is still 1-dim there!

Semiclassical

in $\mathbb{C}\setminus\{0\}$?

i'm specifically excluding the point where things become bad.

Balarka Sen

18:06

@Semiclassical Mike means $x^2 + y^2 = 0$ is a 1-dimensional variety over $\Bbb C$.

Indeed, $\Bbb C$ is algebraically closed, so nothing bad can happen. Degree and dimension doesn't behave badly.

Akiva Weinberger

It's a good thing that I'm doing these for free, since there's no way anyone would pay to get tutored by me.

Semiclassical

i'm not following you.

Balarka Sen

$x^2 + y^2 = 0$ has a buckload of solutions on $\Bbb C$.

While the real locus is just a point.

Semiclassical

yes, but i'm specifically not talking about $\epsilon =0$.

Balarka Sen

That's all we are saying.

Akiva Weinberger

18:08

There's really a big difference between knowing something and being able to teach it… It's a lot easier if they come in with a sheet of exercises, because you know what to do, and more-or-less what they have and haven't learned

Semiclassical

alrighty then.

Mike Miller

@Semiclassical: I mean, you're now talking about something unrelated to what I was talking about before. I was talking about how you can get smooth manifolds as real algebraic varieties by cutting them out with fewer polynomials than their codimension. This isn't something that could happen over $\Bbb C$.

Balarka Sen

@Semiclassical I am confused. So you're talking about something unrelated, then?

Akiva Weinberger

The last two people who came in just had textbooks with them

One was learning exponential growth/decay, his brother was learning piecewise functions

Mike Miller

That's all I was saying. Anything discussed now is an unrelated phenom.

Exponential decay is still a mystery to me.

Akiva Weinberger

18:10

(I assume they were brothers)

Semiclassical

where i was going was: suppose i pick some $\epsilon>0$ and compute that integral. i then take $\epsilon\to \epsilon e^{2\pi i}$, smoothly deforming my cycle of integration along the way to avoid any jumps

Akiva Weinberger

$a(1-r)^t$, that sort of thing

Semiclassical

ugh, this connection

Adeek

Hey @BalarkaSen the reason we have $Im j_{*} \subset ker(\partial)$ is because in the definition of the map $\partial$ we have $c \mapsto b$, $b \mapsto \partial(b)$ and $a \mapsto \partial(b)$. In the definition though of the image though we must have that particular b being element of the $ker$, so we have $a$ gets mapped to 0, so the equivalence class of $[a] = 0$

Semiclassical

i'm talking about something related. namely, how does that period behave if we smoothly change $\epsilon$ in such a way that it returns to its initial value but encircles the point $\epsilon=0$

Adeek

18:12

I am talking about the proof of the exact sequence $H_n(A) \rightarrow H_n(B) \rightarrow H_n(C) \rightarrow H_{n - 1}(A) \rightarrow H_{n - 1}(B)$ etc

@BalarkaSen

Semiclassical

@MikeMiller ah. hrm.

Balarka Sen

@Adeek I have no idea what you're talking about, but that's fine.

Semiclassical

to the extent that i have any comfort with algebraic varieties, its in the complex setting not the real one

Mike Miller

Real algebraic varieties are very strange.

Semiclassical

yeah.

Balarka Sen

18:15

@MikeMiller Do you want to elaborate on that?

If not, I understand. Otherwise I am interested.

Semiclassical

well, i take the $x^2+y^2=\epsilon$ example as indicative of that

in the complex setting, $\epsilon\to0$ gives you $x/y=\pm i$ (a pair of Riemann spheres, i should think)

Balarka Sen

$x^2 + y^2 = 0$ phenomenon is nothing special. It happens in many non-alg closed fields, not just $\Bbb R$.

Semiclassical

eh, fair enough.

Balarka Sen

I think Mike had something else in mind when he said they are strange.

Semiclassical

I'd consider that an example of their weirdness relative to complex algebraic varieties, at least.

Mike Miller

18:20

@Balarka: It's the only field you can visualize it happens in. ;)

Balarka Sen

...

@SemiC wins.

Semiclassical

wooooo-wait what did i win at

Balarka Sen

That $x^2 + y^2 = 0$ was what Mike was referring to when he said weird, apparently. I was looking for a better and exciting explanation, but oh well.

Semiclassical

ahh.

Mike Miller

Don't be greedy.

I should have gotten a haircut before traveling. Oh well.

Ephemeral

18:41

I'm probably being short-sighted, but in a metric space, how can I show that a singleton set is closed?

I'm trying to prove that if we have a general metric space $(M,d)$, a subset $F$ of $M$ being finite means that it's closed.

Mike Miller

What does it mean to be closed?

Ephemeral

That the complement of the set is open!

Mike Miller

So what does it mean to be open?

Ephemeral

For every point in the set, to be able to find an open ball within the set that's centred at that point with radius that depends on the point in question.

Mike Miller

Ok, so pick a point in the complement of x, right? And you want to show there's an open ball around that point that doesn't contain x.

Ephemeral

18:47

Oh, so for a point $r \in M\setminus \{x\}$, we can just choose a radius that's less than $d(x,r)$.

So $r$ being arbitrary means that the complement is open and so the singleton is closed?

Mike Miller

Well, we just proved the complement is open using the definition, so yea.

Ephemeral

Awesome!

Thank you, @MikeMiller!

Evinda

Hey @DanielFischer
Which set does $L^1(\mathbb{R}^n \times \mathbb{R}^m)$ represent?

Mike Miller

Thank the definition, @Ephemeral - it did all the work for us.

@SemiC I went back through it a bit and I think the key detail in understanding the time cube is that the four corners of the four-cornered day are the vertical edges of the cube.

That's the essential detail.

Semiclassical

really? i figured the essential detail was that the guy was batshit crazy

Mike Miller

18:59

Wait, you don't believe in the four cornered day?

Semiclassical

nah. five corners, all the way

Mike Miller

A time dodecagedron? Don't kid around.

Balarka Sen

what is the time cube?

Semiclassical

psh, you sound like my parents. no respect for the Natural 20.

Daniel Fischer

@Evinda When no measure is specified, for subsets of an $\mathbb{R}^d$ the Lebesgue measure is implied (except if a different measure is implied from the context). So it's most likely the set of equivalence classes (modulo being equal almost everywhere) of (Lebsegue) integrable (with respect to the Lebesgue measure) functions on $\mathbb{R}^{n+m}$.

Mike Miller

19:06

@Balarka: I have a link somewhere above. Just search time cube, possibly archive if that doesn't work.

Essentially, it (as of the time of its writing) was a new way of understanding the passage of time. It's now long-accepted.

@SemiC You got an audible laugh out of me

Balarka Sen

What the ...

Semiclassical

i'm trying to think of what some good cultish devotionals would be for the time dodecahedron. " 'May the Natural 20 be with you.' 'And also with you.'"

Balarka Sen

This is even worse than the Cicada 3301 webpage.

Semiclassical

i still like the conspiracy theory i came up with earlier

oops

3 hours ago, by Semiclassical

in honor of that kind of crazy, a conspiracy theory: Wikipedia says the domain expired in August 2015, in the midst of the enduring crazy train that is Donald Trump

3 hours ago, by Semiclassical

obviously, then, Time Cube was a secret NWO plot to subconciously prepare us for the coming of the Trump, and had thus served its purpose.

#TrumpCube

oddly enough, trumpcubes are actual things apparently

Evinda

19:51

A ok.. I have to formulate Fubini's theorem for $L^1(\mathbb{R}^n \times \mathbb{R}^m)$ functions and use it to calculate the integral $\int_0^{+\infty} \frac{\sin x}{x} \left( \frac{1-e^{-x}}{x}-e^{-x}\right) dx$.

There is a hint that one should apply Fubini's theorem for the calculation of $\int_{(0, \infty) \times (0,1)} y \sin x e^{-xy} dx dy$.

Why do we have to calculate the last integral?

Also it holds that $\int_{(0,\infty) \times (0,1)} y \sin x e^{-xy} dx dy=\int_0^{\infty} \int_0^1 y \sin x e^{-xy} dx dy=\int_0^1 \int_0^{\infty} y \sin x e^{-xy} dy dx$, right?

Daniel Fischer

@Evinda What happens if you calculate that integral? Use both orders of integration.

Evinda

@DanielFischer I found that $\int_0^{\infty} y e^{-xy} dy=\frac{1}{x^2}$. Am I right?
If so, then the integral gets $\int_0^1 \frac{\sin x}{x^2} dx$.

How can we continue?

wolframalpha.com/input/?i=int+sinx%2Fx%5E2+,+x%3D0,+x%3D1 @DanielFischer

So have I done something wrong? @DanielFischer

Daniel Fischer

20:09

@Evinda The integral with respect to $y$ extends only over the interval $(0,1)$.

Balarka Sen

I wonder if there's a less horrid way to show that Grassmannians are projective algebraic varieties.

Shafarevich does it by realizing $Grass(r, n)$ as a subspace of $\Bbb P(\bigwedge V)$ where $V$ is an $n+1$ dimensional vector space, by sending a $r$ dimensional subspace of $V$ to $e_1 \wedge \cdots \wedge e_r$ where $e_1, \cdots, e_r$ is some basis of that subspace, and noting this is well-defined as any change of basis multiplies that thing with a nonzero constant (namely the det of the basechange matrix).

And then he writes out an equation for the subset using a complicated operation he calls "convolution". Convoluted!

Mike Miller

Realize it as the space of matrices that are rank r orthogonal projections.

Oh, wait, that's the Stiefel manifold. Hmm.

rorty

Let $F^*$ be the multiplicative group of nonzero elements of the field $F$. Consider the subgroup $H=\{x^2:x \in F^*\}$. Why is it that any element $x^2 \in H$ is the square of only $x$ and $-x$? Why can't $x^2=y^2$ for some $y \notin \{x, -x\}$?

Mike Miller

I dunno, figure out of my idea is workable and if so fix it.

Balarka Sen

Thanks, I'll try.

Vrouvrou

20:32

@DanielFischer hello, can you help me on path connectedness ?

please

Balarka Sen

@MikeMiller I haven't thought about it, but I am not sure if it's helpful to realize them as subset of some matrix space. Because then you only know that it's an affine variety: how can you guarantee it's cone over some projective variety? This is all hypothetical blabber though, I haven't done any computation at all.

Evinda

@DanielFischer You mean that x goes from 0 to $+\infty$ and $y$ from 0 to $1$ ?

Daniel Fischer

@Evinda That's what the hint says.

Mike Miller

@Balarka: Oh, projective algebraic variety.

Ask Ted.

Balarka Sen

Hmm, I guess I will. Thanks though.

Evinda

20:50

Oh yes, right...Then $\int_0^1 y e^{-xy} dy=-\frac{e^{-x}}{x}-\frac{e^{-x}}{x^2}+\frac{1}{x^2}$, right?

And then the integral gets $\int_0^{\infty} \sin x \left( -\frac{e^{-x}}{x}-\frac{e^{-x}}{x^2}+\frac{1}{x^2} \right)$.

How do we continue? @DanielFischer

Balarka Sen

There are quasiprojective varieties with non-finitely generated global section of the sheaf of regular functions to $k$! Ugh!.

Mike Miller

yeah

Balarka Sen

Guess that's why Shafarevich calls this ring of regular functions to $k$ as $k[X]$ but doesn't give it a name for non-affine things. Not at all the right coordinate ring.

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