Mathematics - 2018-05-23 (page 3 of 3)

Ted Shifrin

7:00 PM

If it's an $SU(2)$-bundle, @Semiclassic, on a $4$-manifold, there's only one of interest.

Semiclassical

hmm

Ted Shifrin

Yeah, it's the 3-volume book of Greub-Halperin-Vanstone I was thinking of. I recommend looking at it, too.

I haven't looked in years, but I had vague good feelings about it.

Oh, and @Balarka, "welcome back." :)

Balarka Sen

I'm back with a notepad lololol

@TedShifrin Thanks!

Mary Star

I tried it again:

$\frac{2-n^2}{-4n}=\frac{n^2-2}{4n}>\frac{n^2-2}{4n^2}=\frac{1}{4}-\frac{1}{2n^2}$

For $n>N$ we have that $n^2>N^2\Rightarrow 2n^2>2N^2\Rightarrow \frac{1}{2n^2}<\frac{1}{2N^2}\Rightarrow -\frac{1}{2n^2}>-\frac{1}{2N^2}\Rightarrow \frac{1}{4}-\frac{1}{2n^2}>\frac{1}{4}-\frac{1}{2N^2}$

Is this correct?

loch

one thing i like about chern classes is e.g. it gives a nice proof in showing that a smooth cubic is expected to contain 27 lines. (then there's a bit more to show that you do actually get 27 distinct lines) - where the essence is you're computing the top chern class of some bundle over the grassmannian G(2,4) (projective lines in $\mathbb{P}^3$), where a generic section corresponds to some cubic polynomial,

and its vanishing locus on the grassmannian gives you the points (i.e. lines in $\mathbb{P}^3$) s.t. your cubic polynomial vanishes, i.e. the line lies in the cubic.

Ted Shifrin

7:05 PM

Your algebra is totally wrong, @MaryStar. Keep in mind that $n^2-2>n^2$. Now divide by $4n$. (Simplify as much as possible. You keep complicating.)

Eric Silva

@Semiclassical pointwise a.e. you can get close to this idk what their norms will be tho

Ted Shifrin

Chern classes are all about enumerating things, @loch. I've written numerous papers on that. :)

You can see very beautiful and subtle combinations of algebra-geometric and differential-geometric loci in terms of vector bundles and then count with Chern classes.

Mike Miller

@BalarkaSen mathoverflow.net/a/462/40804

Semiclassical

Weirdly, I'm finding that Gaussian convolution is preserving the L1 norm

Eulb

@BalarkaSen How many topologies would it take to trade you for your current favourite atmospheric black metal album?

Semiclassical

7:07 PM

I can't quite believe that's true though

Mary Star

Ahh, so we have: $\frac{2-n^2}{-4n}=\frac{n^2-2}{4n}>\frac{n^2}{4n}=\frac{n}{4}>\frac{N}{4}$
So the sequence is not bounded from above and so it douesn't converge, right? @TedShifrin

Mike Miller

@loch cute

Eulb

@BalarkaSen Also, remember hatcher's quick point set notes you linked me before? Would that be helpful to read along too? Unsure if related to the course I'm taking.

loch

@TedShifrin i might come across your papers at some point then! I am also going into enumerative geometry side of things (assuming all goes well)

Mike Miller

@Eulb yes those are great

Ted Shifrin

7:08 PM

@MaryStar: That will show that the sequence tends to infinity, so doesn't converge. Right.

Eulb

@MikeMiller I'm taking intro topo with munkres though, how much overlap is there?

Mike Miller

complete and total but Hatcher has good exercises

I just think that one covers more topology in a usual course than necessary

Eulb

Okay that's nice.

Ted Shifrin

@loch: Not that I'm trying to coerce you, but I can send you .pdfs of a few if you're interested. Just email me at the address in my profile if you decide you want 'em.

Mike Miller

@loch I wish you luck

It's not an easy path, and not one entirely determined by our force of will

Balarka Sen

7:10 PM

@Eulb Hatcher's notes are great! Also the answer to your first question is about 42

@MikeMiller Thanks let me have a look

Ted Shifrin

I sure hope Balarka is keeping up with all his tasks. :)

BBIAB.

loch

@TedShifrin That's great! I'm quite busy recently with some other stuff but I'll email you nonetheless :)

Eulb

@BalarkaSen Hey that's my shoe size in EU. I guess "Remember to look up at the stars and not down at your feet" doesn't apply if the answer to the universe lies below.

Mary Star

@TedShifrin Ok! Thank you!!

Balarka Sen

LOL

Hm, the external cup product is the map $H^*(X) \times H^*(Y) \to H^*(X \times Y)$ given by pulling back the classes from $X$ and $Y$ to $X \times Y$ by the projections, and taking the cup there?

I'd forgotten that.

Mike Miller

7:20 PM

I hadn't thought of that

Balarka Sen

That description in Eric's answer is sick

I don't completely understand why the classifying map factors through the fibered product though

Oh, it's packaging that the classifying map is homotopic to $f \circ \text{flip}$, and that double of that homotopy is homotopic to the identity homotopy, and subsequent coherences of infinite order

Is it clear why the second sentence is true? (The first is true because both maps classify $a \otimes a$)

Yes, I think. The homotopy $h$ between $f$ and $f \circ \text{flip}$ is a map $X \times X \times S^1 \to K(\Bbb Z_2, 2n)$, so lives in $H^{2n}(X \times X \times S^1, \Bbb Z_2)$. In fact the element that it represents is $a \otimes a \otimes 1$. $2h$ represents the element $a \otimes a \otimes 2 = a \otimes a \otimes 0$.

So $h$ is homotopic to the "identity homotopy", which is just constant on the $S^1$ factor.

I see, so that's how they make this big ass map $f : X \times X \times_{\Bbb Z_2} S^\infty \to K(\Bbb Z_2, 2n)$, by putting all these informations togather cell-wise.

Daminark

7:43 PM

Hello

Balarka Sen

helo

Mike Miller

@BalarkaSen Indeed. That is precisely what this says.

Balarka Sen

Very very cool

Mike Miller

A map X -> Y factors through the homotopy quotient (X x EG)/G iff the G-action is 'coherent trivial'

And a factorization is basically such a coherent trivialization

Balarka Sen

Ahh. So (1) the map is G-equivariant upto homotopy (2) That homotopy is G-equivariant upto homotopy (3) etc etc

Mike Miller

7:48 PM

Yup

Balarka Sen

Very fucking cool

Anderson Felipe Viveiros

8:12 PM

Hello, guys. Let $M$ be a three-dimensional compact manifold (without boundary). A orientable surface $\Sigma$ in $M$ is Heegaard spliting if $M\backslash \Sigma$ has to open components homeomorphic to handlebodies (i.e. a "solid" (3d) ball with "solid" handles).

For any such $M$, is there always a Heegaard spliting? (I guess that the answer is "no").

When does it exist?

Balarka Sen

The answer is yes.

Anderson Felipe Viveiros

Really?

Balarka Sen

Yup. Well, when $M$ is connected and orientable anyway.

Anderson Felipe Viveiros

Any references?

Hahaha, ok $M$ is always connected

*and orientable also, sorry

Balarka Sen

Take a triangulation of $M$. Take an $\epsilon$-neighborhood of the 1-skeleton of the triangulation. This is a handlebody $H_1$. $M \setminus \text{cl}(H_1)$ is an $\epsilon'$-neighborhood of the 1-skeleton of the dual triangulation, so is again a handlebody $H_2$.

Well, their closures are.

So $M$ can be written as union of two handlebodies glued along their boundary surfaces.

(Dual triangulation = 0-simplices on the centroid of each 3-simplex of the original triangulation, edges transverse to each 2-simplex of the original triangulation, 3-simplices centered at each 0-simplex of the original triangulation)

Anderson Felipe Viveiros

8:21 PM

Oh, I thank you @BalarkaSen I think that's right

Balarka Sen

It's one of my favorite proofs of Heegaard decomposition

Of course, you need to know that 3-manifolds can be triangulated... that's not an easy theorem. :P

Moise, I think, proved that?

Anderson Felipe Viveiros

And that when $M$ is orientable compact, this is triangulation is finite, right?

Balarka Sen

Yep.

If it's compact, the triangulation is finite. Orientability is to ensure that the epsilon-neighborhood of the 1-skeleton doesn't become a "nonorientable handlebody"

GFauxPas

9:03 PM

Not all manifolds have a triangulation?

Nope. :)

Surprising

In mathematics, the E8 manifold is the unique compact, simply connected topological 4-manifold with intersection form the E8 lattice. == History == The E8 manifold was discovered by Michael Freedman in 1982. Rokhlin's theorem shows that it has no smooth structure (as does Donaldson's theorem), and in fact, combined with the work of Andrew Casson on the Casson invariant, this shows that the E8 manifold is not even triangulable as a simplicial complex. == Construction == The manifold can be constructed by first plumbing together disc bundles of Euler number 2 over the sphere, according to...

GFauxPas

Compact and s.c. But still no?! D:

That's uncomfortable

Delete it

Mike Miller

Simple connectivity shouldn't really play a role in any case. But try to think how you would construct a triangulation

You triangulate as much of one chart as you can, and you pass to a neighboring chart

Anderson Felipe Viveiros

9:07 PM

@GFauxPas Reeeeally uncomfortable haha

Balarka Sen

@MikeMiller I really like that perspective

Mike Miller

In this neighboring chart you see what the old triangulation became and try to extend jt

ÍgjøgnumMeg

Evening

Mary Star

Does someone of you have an idea about my question: math.stackexchange.com/questions/2793512/… ?

Mike Miller

If it becomes an actual decomposition of R^n into simplices then all is well, but perhaps the simplices got really crinkly

And you don't know how to glue more simplices to the boundary of what you already have

GFauxPas

9:09 PM

Sadface

Mike Miller

The reason smoothness has a place here is that it essentially means "linear subspaces essentially go to linear subspaces". Up to a reparameterization, you can straighten out what you have already built

But for something not locally linear, who is to say?

Balarka Sen

(The conclusion is that smooth manifolds are always triangulable, just to write that down)

@MikeMiller So, if you can say that your manifold admits a locally flat structure (i.e, transition functions are locally flat diffeomorphisms of R^n), then it's triangulable.

By locally flat homeomorphism of R^n I really mean that it's restriction on linear subspaces are locally flat embeddings.

(In dimension 2 Jordan-Schoenflies guarantees that)

Not diffeomorphism, homeomorphism.

Mike Miller

I see what you mean. I am a little skeptical

That sounds more like a foliation I guess

Balarka Sen

Yeah.

I was gonna say

Mike Miller

I guess you're not demanding it preserves slices...

Just a regularity condition on all linear subspaces

Balarka Sen

9:15 PM

Right.

Mike Miller

No this is tautological

The fact that it's a self-homeo of R^n means that the things are locally flat

I think you're more worried about crinkly intersections with the "boundary" of your small open set

Balarka Sen

Oh, duh.

I was thinking that under the transition function one of the faces of the triangulation on one chart becoming like the surface of Alexander's horned sphere and then I'm screwed

I don't understand what else can happen

@MikeMiller Yeah I see. The boundary of the open set is the issue

I don't know how to write down exactly what I want the transition functions of my atlas to be for your procedure to work, and I'm interested in doing that

B. Mehta

Perhaps I'm missing something obvious, but I'm struggling to see why $X_n \to X$ almost surely implies $\mathbb{E}(\|X\| I_{\|X\| \geq M}) \leq \mathbb{E}(\liminf_{n \to \infty} \|X_n\|I_{\|X_n\|>M})$

Mike Miller

@BalarkaSen I have never thought seriously enough about this to write out a proof. It should be something like "linearization at infinity".

Balarka Sen

I want the transition function to be a self-homeomorphism of the open n-ball which extends in the domain to the closed n-ball and it's a locally flat embedding at the boundary, or something.

Mike Miller

9:25 PM

You need to know that every simplex is locally linear, and that you can do this in a compatible way for everything near the boundary of the chart

Think of the 2-chart case eg for the most obvious worry

Balarka Sen

I think we have similar pictures in mind.

I like "linearization at infinity"

Mike Miller

There's not a good enough reward mechanism in place to have anybody write this proof out

You won't get better Geometry scores for it

Balarka Sen

lmao

I was wondering if you can recover the problem of having a simplicial structure as an obstruction class from this interpretation. Like, you're demanding a topological atlas to lift to a specific atlas. Or rather, the tangent microbundle to have structure group lift from TOP(n) to ???

Idk

Mike Miller

I would work through the paper of Galewski-Stern where they construct the obstruction class with you, but I would not do it alone

Balarka Sen

Lmao

I'd love to do that after I learn some classical obstruction theory

Mike Miller

9:30 PM

I think it should be a very clean paper but I have not spent time on it

Ted Shifrin

Wow, those are names of people dear to me, @MikeM, @Balarka (one deceased about 10 years).

Mike Miller

I did not know that Galewski had passed away, nor that he worked at UGA.

Ted Shifrin

Yup to both.

He came to UGA in the 70s.

Balarka Sen

Very cool

Alessandro Codenotti

9:59 PM

Hi @Ted

Ted Shifrin

hi demonic @Alessandro

How goes it?

Alessandro Codenotti

Guess who got his driving licence today? Pay attention when you cross the road ;)

Ted Shifrin

Oh hell.

Balarka Sen

Oh god

Ted Shifrin

Now I can never visit Europe again

Balarka Sen

10:01 PM

Alessandro is officially Arturo Beneditto Giovanni Giuseppe Pietro Archangelo Alfredo Cortafelli da Milano now

Ted Shifrin

The standards on driving licenses in Italy are even slacker than the standards on mathematics diplomas. :(

LOL, you left out a few, Balarka.

Eric Silva

how long does it take to get a license there

Ted Shifrin

depends how many time you run over the examiner and fail.

Mike Miller

@BalarkaSen yeah the Assassin's Creed games have really gone downhill

Ted Shifrin

Oh, Eric. A question for you and @0celo. ...

Balarka Sen

10:05 PM

I have heard of ass creed but never played it

Eric Silva

@MikeMiller they went uphill again!

@TedShifrin i would look if my life werent a living hell of work rn

Mike Miller

I was just making a long Italian name joke

@Ted Interesting, seems plausible

Eric Silva

i get it

Alessandro Codenotti

@TedShifrin mathematics diplomas where?

Ted Shifrin

I dunno, @MikeM. I don't see why the star doesn't mess it up.

Alessandro Codenotti

10:08 PM

@EricSilva a few months if you don't have to the exam three times like me

Ted Shifrin

I said in Italy @Alessandro :P

Alessandro Codenotti

pfff

Eric Silva

i think the total time i spent getting my license was collectively under 2 hours lol

Alessandro Codenotti

we have to take some mandatory driving lessons (6 hours), but I actually took waaaaaaaaaaaaaay more

Mike Miller

@Ted I don't see any way to prove it but his intuition seems plausible to me

Alessandro Codenotti

10:09 PM

Well anyway now that this is done I can think about maths

Ted Shifrin

In my day, we had driver training in high school ... it was probably a dozen hours of experience.

Balarka Sen

i prefer walking

Eric Silva

i waited in a line and parked and was done

but i lived in florida the land where you can be blind and get a license

Ted Shifrin

You didn't bother passing the written test, Eric? That still petrifies me more than any math exam I ever took.

Eric Silva

i took no written test

Ted Shifrin

10:11 PM

Say what?

Every state requires a written test, I thought.

Alessandro Codenotti

I should have gone to Florida to take my test

Mike Miller

Presumably someone should just write down a smooth map from R^n to R^n with some metrics on each and write down the PDE "preserving bi-closed" gives

Ted Shifrin

And those don't transfer. I've had to take those in every state I've lived in (and redo it in CA, obviously).

Eric Silva

oh it was on the comp

Ted Shifrin

Right, now it's on the computer.

@MikeM: Or think about Poincaré duality. Didn't we talk once about John Millson's paper on harmonic P.D. on R.S.?

Eric Silva

10:12 PM

it was reeeeeaaaal easy in FL

Alessandro Codenotti

@TedShifrin There is a probability strictly bigger than 0 that I'll be in the US for a week or so in December

Ted Shifrin

I never remember how far from a hydrant you must park or how far ahead one must signal (not that 90% of the drivers do) ...

Oh yeah, @Alessandro, you coming to visit?

Mike Miller

I don't remember it but my brain is mush

Eric Silva

no clue

Mike Miller

He wants to work locally, mind

Alessandro Codenotti

10:13 PM

A friend from Italy who now lives in the US is getting married

Ted Shifrin

Oh @MikeM ...

Eric Silva

oh it looks like in FL at least you dont have to retake it when you get your license if you just have a learners permit youve gotten recently

Ted Shifrin

I wonder if this is answered on MO.

Where is the wedding, @Alessandro?

Alessandro Codenotti

Salt Lake City

or somewhere in Utah

Ted Shifrin

Oh, Mormon land.

Mike Miller

10:14 PM

@AlessandroCodenotti Ah, so he has two countries to be frustrated with!

Eric Silva

what a weird state

Alessandro Codenotti

@TedShifrin yep

Ted Shifrin

Well, if you wanna come visit CA, I will probably be here.

And, unlike you, I won't threaten to run you over.

Alessandro Codenotti

Wow that's much further from Utah than I thought

Ted Shifrin

Italy is smaller than a number of our states.

Alessandro Codenotti

10:18 PM

I don't really understand how big the USA are

Eric Silva

the us is like 30 times the size of italy

Ted Shifrin

Does that count Alaska and Hawaii?

Balarka Sen

USA is thrice the size of India. And India is very fucking huge

Ted Shifrin

Very.

Alessandro Codenotti

and how many people are there in the US compared to Italy?

Ted Shifrin

10:20 PM

We're over 325 million.

What's Italy?

60 million.

Eric Silva

im p sure it counts alaska @Ted

if u throw in hawaii nothing changes

Ted Shifrin

Wow ... Only 5 times the population. Italy is crowded.

Balarka Sen

Apparently India is 1 billion

Rip

Ted Shifrin

Yeah, and China?

Balarka Sen

1.3 billion. India is 1.2 I think

Eric Silva

10:22 PM

the US has a very low pop density away from the coasts

Ted Shifrin

Right ... we won't discuss education density ... :)

Balarka Sen

Asia (without Russia) is quite overcrowded I think

Alessandro Codenotti

Apparently the surface are aof the US is close to that of the whole of Europe wtf

Ted Shifrin

I might have thought more, actually.

Balarka Sen

2 big

Eric Silva

10:24 PM

The US is massive dude

it would take like months to hitchhike across if you walked a decent chunk of the way

Ted Shifrin

But less than 3 days to drive across if you drive nonstop.

Eric Silva

yeah dont do that it's horrible

Ted Shifrin

I think it's not hard to drive from the southern tip of France to the northern in one day.

Balarka Sen

how is that possible

Ted Shifrin

I didn't quite do that after grad school, Eric. The car died a few times and I stopped in Chicago for a day.

Balarka Sen

10:26 PM

what's the total length you'll be driving

Mike Miller

My uncle used to drive from NY to CA for work

Eric Silva

ive made the drive from florida to chicago nonstop like 7 or 8 times and it makes me wanna die every time

Mike Miller

He liked to pump red bull and get it all done in one push

Ted Shifrin

about 3000 miles, Balarka. Average 60 mph that's only 50 hours. :P

Right now, San Diego to San Francisco non-stop is all I'm willing to do.

But I'm old.

And LA makes it miserable. (Nothing person to Mike.)

Balarka Sen

@TedShifrin Oh ok. The diameter of India is like 2000 miles and I've heard stories of people driving through the national highways for two days to get across.

So that makes sense

Mike Miller

10:28 PM

LA traffic is not good. I think it hubris for any other city to think they won't see this in 10-20 years

The urban sprawl is the limiting model

Ted Shifrin

The Bay Area is almost as bad now, Mike, and Atlanta is horrific.

Balarka Sen

@EricSilva y

Mike Miller

Yup

Balarka Sen

do you have the mOtIoN SiCknEss

Ted Shifrin

San Diego can be bad at certain times, too. I'm lucky I can choose to avoid that for the most part.

Mike Miller

10:29 PM

Symptoms of the market floor's successful war on mankind.

Eric Silva

@BalarkaSen i did a bunch of jobs that involved making the drive when i was broke in hs

Balarka Sen

Ah. Ugh.

Eric Silva

plus i drove when i moved here so that was 1 of them

@MikeMiller snaps fingers in agreement

Balarka Sen

Cars will eat transportation but I can't digest cars because I have motion sickness - some tankie rapper

Daminark

"I can't digest cars"

I would hope not

Ted Shifrin

10:39 PM

Hi Demonark

Daminark

How's everything going?

Ted Shifrin

You missed our comparisons of sizes of countries and populations.

Eric Silva

i would hope you can digest cars

that's a super cool ability

Daminark

Yeah that's true

@TedShifrin anything surprising discovered?

Balarka Sen

i eat lamborghinis with peanut butter as breakfast - lyrics taken from the same rapper's repertoire

Ted Shifrin

10:43 PM

People who aren't in the US don't realize how humongous it is.

Eric Silva

people who are the us dont realize how small other places are either

Ted Shifrin

Well, lots of US people also have very bad attitudes — say, regarding languages people speak. We won't even mention such things.

Eric Silva

oh you dont gotta tell me that

Balarka Sen

American is a language of plebians

Ted Shifrin

I still love telling the story about the Texans who came into a cafe in Paris after I was sitting there sipping my pastis. It was a hot summer day (back in the 80s). They asked for ice tea in their drawl. The waiter didn't understand (or perhaps feigned not understanding). I called him over and asked him (in French) to bring a pot of tea and all the ice in the kitchen. He laughed, but he did close to that. The Texans doffed their hats at me.

Well, Balarka, we can't all speak the Queen's English.

Balarka Sen

10:48 PM

Neither can the British

Eric Silva

i got in trouble in my elementary school for speaking portuguese on the phone once

p bad

Ted Shifrin

Verily.

Leaky Nun

@EricSilva Portuguese is such a beautiful language

Balarka Sen

Only Irish and Scottish people can speak English

Ted Shifrin

Well, now we have a president and representatives who can't even speak American.

Balarka Sen

10:48 PM

LOL

Eric Silva

@LeakyNun i agree

Mancala

Does a complete properly embedded surface mean that the parameterization is its proper map? That is, what inverse image of compact is compact by parameterization?

Ted Shifrin

There needn't be a global parametrization, @Mancala.

But if there is one, yes, it'll be a proper map.

Mancala

Thanks! @TedShifrin

Balarka Sen

11:19 PM

@Eulb 'Sup

Mancala

11:33 PM

Could someone tell me a good introductory book to study hyperbolic space?

GFauxPas