Mathematics - 2017-09-28 (page 2 of 2)

Semiclassical

18:06

Ehhhh

Infinite dimensions is worse :)

Leaky Nun

hi everyone from London

Abcd

@LeakyNun Are you in London right now?

Leaky Nun

@Abcd yes

Ted Shifrin

Leaky, your accent has changed.

Abcd

@LeakyNun And you are generally talking from Hong Kong if I am not wrong?

Tobias Kildetoft

18:15

@LeakyNun Have you started the final exercise yet?

Leaky Nun

@Abcd I'm staying in London

@TobiasKildetoft not really

@TedShifrin heh?

Abcd

@LeakyNun ohkay.

Ted Shifrin

Well, now you type with a British accent.

Leaky Nun

@TedShifrin lol

@Secret hmm, radius must be integer (and can be infinity)

Semiclassical

18:39

@TedShifrin I actually have a bit of that myself, and I know why: the online text RPG I was into for years uses the British spellings

So artefact instead of artifact, honour instead of honor

Perturbative

Hey everyone!

Semiclassical

I mostly don't do that now, but it does show up every so often

PVAL-inactive

howdy yall

my avatar is different again.

Daminark

Hey everyone!

TastyRomeo

ohi

Daminark

18:43

How's it going?

TastyRomeo

Academic year has started, so yeah, preparing first classes now and hopefully fit in some research

Daminark

Nice. Same here, we're just about done with first week

@PVAL so is it gonna cycle between the three now?

Perturbative

Yo! @Daminark

Daminark

Oy @Perturbative!

Perturbative

@Dami I was Googling some stuff on Cobordism Theory and stumbled my way into here : math.uchicago.edu/~may/REU2016 :P

Daminark

18:48

Lol, found my paper? :P

Balarka Sen

cobordism theory is le shit

Krijn

Heyoo

Perturbative

@Dami, Just found it :D

Balarka Sen

oy

Krijn

@Balarka, explain the genus of a curve to me

Assume I know nothing, though

That's not too bold of an assumption I hope

Balarka Sen

18:57

I dunno, there are a lot of definitions. How about half the dimension of the 1st cohomology group as a free $\Bbb Z$-module?

That sounds like a definition you'd love and adore you algebraist

Krijn

I need to explain my thesis in 45 minutes, which I can' t, so I need to explain what the genus of a curve is in like, 5 minutes.

No, I literally mean, assume I know nothing

Balarka Sen

Oh

Krijn

And that presentation is aimed at people who barely know what a curve is

Balarka Sen

Oh man

Hm

So the first step is to convince them a complex algebraic curve is actually a surface, not a curve :P

Perturbative

@Daminark What TeX templates did y'all use?

Krijn

19:03

Well, yeah, over $\Bbb C$, but all my images are over $\Bbb R$ of course

Soooo

Balarka Sen

From which point the job is easy because "by classification of surfaces", every surface is like a torus with multiple holes.

And the count of those holes is what we call the genus

Which I imagine can be explained just by drawing a picture

TastyRomeo

@Perturbative Looks like amsart ?

Krijn

@BalarkaSen Counting the holes is my approach at the moment

But then, explaining why $\mathcal{E}: y^2 = x^3 + x + 1$ is a torus...?

Perturbative

@TastyRomeo Ahhh

Balarka Sen

@Krijn Ah, ok, I can do that.

So the first step is

PVAL-inactive

19:05

I mean it's a torus with a hole in it/

Balarka Sen

Draw them the graph of a generic elliptic curve in the Cartesian plane

PVAL-inactive

the projectivization is a torus.

Balarka Sen

That's, like, the usual picture of two disconnected components, one circle-ish and the other unbounded

Daminark

Peter May mentioned the font to use somewhere on the website

Balarka Sen

Then explain them that if you "add a point at infinity" to the unbounded component the true plot actually becomes like two disjoint circles

And then tell them that if you actually consider the "plot of $\mathcal{E}$ over the complex numbers", then the resulting surface you'd obtain would have that as a section

Which the torus does!

A section of the torus is two disjoint circles.

Krijn

19:08

Ah, I see

Never thought of it in that way

PVAL-inactive

I think if I wasn

t allowed to use RRT/(degree, genus)

I would compute the singularities of f(x,y)=|x|^2 +|y|^2

and work from there.

I guess that's like f(x)=|x^2|+|x^3+x+1|

Balarka Sen

@Krijn $y^2 = x^3 + x + 1$ is not a great example to demonstrate what I said because it has connected graph

It's one of the "bad sections" of the torus

$y^2 = x^3 - x$ does the trick

Krijn

Yeah, but I already made most of the presentation with that curve :(

I might switch up though

I'll see

Balarka Sen

kk

When is the presentation?

Krijn

Dunno actually

I'm waiting for feedback on some chapter of my thesis

So doing this in the meantime

Balarka Sen

19:14

Aha

Well you have a lot of time to decide what to do then

Krijn

I do

Reading a lot of drama too

Balarka Sen

actual drama this time?

Krijn

Have you solved the Riemann Hypothesis already?

Yeah, proper drama

PVAL-inactive

@Balarka Wanna read this and explain it to me arxiv.org/pdf/math/0506523.pdf ?

Mike Miller

I read that at some point but don't remember it. It was a good paper.

PVAL-inactive

19:18

It kind of gets a little carried away with the graphs...

Mike Miller

But a corollary of the first sentence is that I do not want to explain it to you

Balarka Sen

Oh man I'd love to

But I wonder if I will understand shit

PVAL-inactive

It should be all basic 3-manifold topology +thurston black boxes.

Mike Miller

Ya

Balarka Sen

Ok, I'd be down for it.

PVAL-inactive

19:19

I'm supposed to explain theorem 4.18

so read that

Mike Miller

My job this qr's seminar is to explain the relationship between contact and symplectic topology I think

Balarka Sen

I should learn those things at some point

Mike Miller

should you?

PVAL-inactive

The answer should be that level sets of the natural real-valued functions on symplectic manifolds are contact manifolds.

Job done.

Mike Miller

Oh, typo. -topology +homology.

Balarka Sen

19:27

@MikeM Well I thought that was our vague goal of the foliations readings

Oh

Mike Miller

which i think is harder.

PVAL-inactive

Not sure how well defined those things are.

contact homology should correspond to psh curves in the symplectization which limit to Reeb orbits.

I don't know a definition of "contact homology" which doesn't directly involve psh curves in the symplectization.

Mike Miller

i think that's still what they're doing but there's some Gysin sequence for symplectic homology of something w contact boundary

it's the second to last talk of the quarter

-Gysin

I confuse myself too easily. Ignore me.

PVAL-inactive

I'm talking the day before the day after tomorrow.

Semiclassical

19:51

@PVAL-inactive I see what you did there.

Balarka Sen

confusion intensifies

@PVAL I am making a wild attempt at reading section 2, let's see if I can understand what's happening

(I actually don't know what the JSJ decomposition is)

Oh lord, this is technical, but I probably see the picture.

Semiclassical

20:07

I'm amused at how the Wiki page for the JSJ decomposition includes two "warning" paragraphs

Balarka Sen

Prop 2.2. is interesting

Semiclassical

@BalarkaSen hmm, this question seems up your alley: math.stackexchange.com/q/2449201/137524

Balarka Sen

The little Picard theorem is about entire functions so it's not clear to me how the OP wants to mash that up with the Casorati-Weierstrass to conclude the big Picard, both of which are statements about holomorphic functions with one essential singularity.

Doing any business around a neighborhood of the essential singularity would render me paralyzed to apply the little Picard at all

The question mostly sounds like sketchy shit to me tbh

Semiclassical

The lack of the word 'entire' from their statement of the little picard theorem does make me a bit dubious

PVAL-inactive

probably as fruitful as trying to prove Fermat's last theorem from Fermat's little theorem.

Balarka Sen

20:18

lol

Semiclassical

well, at least in this case there's a definite parallel: both theorems involve the ranges of functions in the complex plane

PVAL-inactive

both of fermats theorems involve the powers of positive integers.

Semiclassical

so I think comparing it to Fermat is rather a stretch. (doesn't mean what they do can be made to work, mind)

Balarka Sen

I like the analogy

Semiclassical

and both are questions in elementary number theory

PVAL-inactive

20:22

Pretty certain any non-polynomial entire function has a singularity at infinity.

Semiclassical

anyways, the disanalogy between little and great Picard is that one's about functions which are well-behaved everywhere in the complex plane, whereas the other is about functions which behave really really badly at one point

PVAL-inactive

so big picard implies little picard. (has an essential singularity)

Balarka Sen

yeah

PVAL-inactive

though this certainly doesn't follow from Cas-Weir as that's really strictly a weaker statement than big picard.

Semiclassical

I'm not sure how one would 'disprove' the question, though.

PVAL-inactive

20:26

I mean I'm pretty certain little picard is some clever appeal to Liouville. I don't know the std proof of big picard off the top of my head, but people who did analysis seemed to think it was difficult (I think they skipped it. For reference they did not skip uniformization.)

Balarka Sen

@PVAL I think little Picard is proved by constructing a holomorphic cover $\Bbb D \to \Bbb C - \{0, 1\}$

If $f : \Bbb C \to \Bbb C$ misses two points, then the image goes inside $\Bbb C - \{p, q\}$

So you can lift this map

Ok, yeah, Liouville :P

ALannister

20:53

0

Q: Determine the polar cone of the convex cone

After showing that the set $K_{1} \in \mathbb{R}^{n}$ defined as $$ K_{1} = \{ x = (x_{1}, \cdots , x_{n})^{T} \in \mathbb{R}^{n} : x_{1} \leq x_{2} \leq x_{3} \leq \cdots \leq x_{n} \}$$ is a convex cone, which I've done, I need to determine its polar cone. In other words, for $x \in K_{1}$, I ...

Raptor

21:06

Could someone please hint how I may show $f,\hat f\in L^1(\mathbb R)$ implies $f\in L^p(\mathbb R)$ for all $p\in[1,\infty]$?

mixedmath

21:21

@Danu I understand congratulations are in order!

Congratulations!

Raptor

I think I got it using the Fourier inversion formula.

Balarka Sen

Hi @Ted

Ted Shifrin

Hi @Balarka

You're messing up your sleep cycle yet again?

Balarka Sen

So it seems :(

ALannister

Hi @TedShifrin

Ted Shifrin

21:34

One of your remarkable talents, Balarka.

Hi @ALannister.

ALannister

0

Q: Set of all probability distributions supported on $\Omega$ is a convex set.

Let $\Omega$ be a set of $n$ elements. I need to do the following two things: Describe the set $\mathcal{P}$ of all probability distributions supported on $\Omega$. Show that $\mathcal{P}$ is convex. Now, the second part is probably very easy once I've figured out the first part. For the fi...

probability probability-theory vectors convex-analysis convex-optimization

Ted Shifrin

I would assume probability density functions (at least in the continuous case); for the discrete case, it's just the probability as a function on the discrete measure space, I guess.

Convex linear combinations of positive things that add up to 1 are again positive and add up to 1.

ALannister

Well, I just emailed my prof and this is what she said: "It means that \Omega as a sample space contains finitely many simple events to which we assign probabilities. Thus, any distribution is completely described by stating what are the probabilities of those simple events. "

So then, how do I express them as vectors?

Ted Shifrin

For the points $\omega_1,\dots,\omega_n$, take $(p(\omega_1),\dots,p(\omega_n))\in\Bbb R^n$?

ALannister

For one thing, they're not even all going to be of the same length

Ted Shifrin

21:38

Length?

ALannister

will they?

Ted Shifrin

You have a finite sample space, with $n$ points.

ALannister

I mean, there won't all be $n$ elements in each

Ted Shifrin

Sure there will.

ALannister

Oh, and we're looking at the probability of each sample point

Okay.

Ted Shifrin

21:39

You're fixing the $\Omega$ to start with.

But this even makes sense in the continuous case, in the space of functions. But never mind that ... :)

ALannister

Then, it's just a question of taking two vectors, say $p$ and $q$ and then showing that the convex combination is also a distribution?

Ted Shifrin

Right.

ALannister

Why the hell did she have to make the wording so damn complicated?

Communication is hard :(

Ted Shifrin

I just told the father of a teenager that it's language issues that make primary/secondary math hard for a lot of kids. (Assuming they can do basic arithmetic.) Mathematics is language.

user685252

but language is not math

:)

Ted Shifrin

21:45

the structure of language is very much mathematical, according to Chomsky

user685252

opinion

Ted Shifrin

well, I'll take one of most famous linguists to live over you

user685252

:(

Ted Shifrin

And, I had to do some (computer) math modeling work for a linguist colleague of mine at the university in his language analysis of Croatian, as well.

user685252

learning is intimately related to language

Balarka Sen

21:47

Language is a virus from outer space. - William S. Burroughs

ALannister

I like Chomsky but I find the sound of his voice irritating.

Ted Shifrin

He's old, ALannister. Give him a break.

user685252

math is learning

ALannister

I think it's misonophobia

Ted Shifrin

That's a new word for me.

user685252

21:49

me too

ALannister

It's like when you have a really strong negative reaction to the sound of someone chewing

user685252

Ewww

ALannister

Actually I spelled it wrong

Ted Shifrin

misophonia?

ALannister

Yes

Ted Shifrin

21:50

hatred of sounds?

user685252

tsk tsk :)

ALannister

Misophonia

Misophonia, literally "hatred of sound", was proposed in 2000 as a condition in which negative emotions, thoughts, and physical reactions are triggered by specific sounds. It is also called "select sound sensitivity syndrome" and "sound-rage". Misophonia has no classification as an auditory, neurological, or psychiatric condition, there are no standard diagnostic criteria, it is not recognized in the DSM-IV or the ICD-10, and there is little research on how common it is or the treatment. Proponents suggest misophonia can adversely affect ability to achieve life goals and to enjoy social situations...

Ted Shifrin

You'd really hate me, then, ALannister. I'm not a silent teacher.

ALannister

I so have this.

There used to be another TA I shared an office with who constantly slurped.

Ted Shifrin

We all do things that get on other people's nerves. It's part of being human.

ALannister

21:51

Other sounds that bother me: Hungarian accents.

I found that one documentary about Erdos particularly hard to watch without closed-captioning and the sound off.

user685252

me too

ALannister

@user685252 woooow! You're the first person I've ever met who felt the same way about it! Maybe I'm not completely screwed up in the head after all.

Ted Shifrin

twiddles thumbs

user685252

:-D

ALannister

@TedShifrin LOL

user685252

21:55

Accents can be impenetrable.

As shown by Peter Sellers in the pink panther :)

"Do you have a room?"

Ted Shifrin

a rhume

user685252

exactly

Ted Shifrin

well, English is a totally f***ed up language, anyhow. Back to my G.B. Shaw example: GHOTI. Pronounce that.

user685252

I'll pass, thanks.

@TedShifrin was he one of your students?

Ted Shifrin

No, no, I've never met the kid.

But I've had 40+ years of math teaching experience ...

Balarka Sen

22:04

Gee each owe tea eye smells fish

Ted Shifrin

Perfectly "logical" English @Balarka

spells, you meant

Balarka Sen

Nah Joyce wrote smells

Ted Shifrin

oh

Balarka Sen

It's a multilevel pun

Ted Shifrin

We've had this argument before :P

Balarka Sen

22:05

lol yep

Ted Shifrin

smells fishy :P

Go to sleep, Balarka.

ALannister

FISH!

Or, if you're a hippie: PHISH!

Ted Shifrin

maybe user68525 is a hippy. I dunno.

user685252

nah

ALannister

So, Let $X$ be probability distribution supported on $\Omega$. Then, for $\{\omega_{1}, \omega_{2}, \cdots , \omega_{n}\} \in \Omega$, let $x:= (p(\omega_{1}),p(\omega_{2}) , \cdots , p(\omega_{n})) \in \mathbb{R}^{n}$. Then, since $x$ is supported on $\Omega$, $p(\omega_{i})>0$ $\forall i$, and since $\sum_{i=1}^{n} p(\omega_{i})=1$, $p(\omega_{i})<1$ $\forall i$. Is that enough of an analytic description?

Ted Shifrin

22:08

I'm annoyed by how many questions I've answered on main only to get no response whatsoever.

Balarka Sen

Ptashion, where P stands for Ptashion

Ted Shifrin

No, @ALannister. $p(\omega_i)\ge 0$.

ALannister

I didn't know you still did that @TedShifrin

But it's supported on $\Omega$, doesn't that mean that it can't be zero there?

Ted Shifrin

No, it means it's 0 outside $\Omega$.

ALannister

But it still could be 0 inside $\Omega$?

Ted Shifrin

22:09

Of course.

ALannister

Oh.

Then, what's the analytic description, then? just that $0<p(\omega_{i}) < 1$ and

$\sum p(\omega_{i}) = 1$?

Ted Shifrin

still wrong inequalities

$p(\omega_i)\ge 0$ and $\sum p(\omega_i)=1$.

ALannister

Ooooh.

But that's what is meant by an "analytic description"?

Ted Shifrin

Seems fine to me.

ALannister

It sounds so official and complicated...

Ted Shifrin

22:12

Don't ask me. Ask your prof.

ALannister

Just did. We'll see what she says.

ALannister

22:58

Okay, she answered me back "no"

Never mind that was for something else chuckles

Daminark

23:17

Hello!

ALannister

I am going to fail this class

ALannister

23:33

I do not know how to find polar cones

And when questions get posted on MSE about them, nobody answers them!

I'm quitting. I'm not going in tomorrow. I can't do this. I don't know why I thought I could.

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$Mathematics$ Mathematics