Mathematics - 2016-09-27 (page 1 of 4)

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12:11 AM

Hi @TedShifrin

@TedShifrin I don't know if it's homotopy equivalent to $S^4 \times S^5$ or not.

hi @Danu

@MikeM: I don't either, but the topology of the orthogonal groups is well-known.

Long exact homotopy sequence, Bott Periodicity ... put one of your grad students on it.

@TedShifrin I don't think that will work. I suspect it has the same homotopy groups.

ah

I'll ponder later.

Wall has a classification of 3-connected 9-manifolds but I don't really want to do the calculation.

All of their invariants agree except possibly some invariant he calls $\omega \in \Bbb Z/2$. I couldn't figure out how it was defined.

12:25 AM

Anything I can think of differential geometrically won't help with homotopy equivalence questions. Anyhow ...

Sounds like fun... :P

I'm off to drink and eat and get depressed ... See you all later.

@TedShifrin Do all but the last of those!

Sorry, Danu.

@TedShifrin I'm working on union stuff. Seems more productive.

12:34 AM

@TedShifrin Have you changed colors since I last saw you?

Could have sworn you were green.

It's a bug

@Danu A Ted-wide bug? Seems rather specific.

Mike seems the same.

Several others are also different

Huh. Weird.

That from updating the avatar code or something?

I don't know.

12:41 AM

Fair enough.

@Ted I've got some serious misconception going on that's preventing me from understanding the identification of linear homogeneous polynomials with sections of $\mathcal O(1)$ (the higher order case is analogous so once I get this it should be no problem).

What I don't understand is the following: I've got non-trivial transition functions $\psi_{jk}$, but when I restrict a globally defined hom. poly. on $\Bbb C^{n+1}$ to a ray $\ell$, I get one element of $\ell^*$, irrespective of whether I view $\ell$ as lying in the trivialization on $U_j$ or on $U_k$. How do I reconcile this with $s_j=\…

@Axoren I changed and then downloaded and saved my old avatar.

It seems that $s_j$ and $s_k$ cannot agree exactly if $\psi_{jk}$ is non-trivial---but if they come from a globally defined function they must agree (this is all on $U_j\cap U_k$)

This must be really dumb somehow, but I just don't see it.

Hi @Mike

@MikeMiller You like your light blue? :P

@MikeMiller That's a lot of effort to get back the same randomly generated default avatar.

At this point, it's not even randomly generated or default anymore.

It's like you have a completely new avatar.

It's me.

@BalarkaSen I wrote my answer.

Jorge Fernández Hidalgo

12:54 AM

Is anyone gonna watch the debate?

Busy.

1 hour later…

2:06 AM

@MikeMiller I just woke up. Shall I go ahead and post the question now then?

Sounds like some back-room black market answer dealings going on.

"I've got the answer, you got the question?" "Yeah, it's right here."

If you like.

I'm going to be home in like an hour.

Writing it.

Huh. This come up in one of the suggestions for related questions.

2:28 AM

very cool

I was unable to resolve that case

I commented with how I think one could do it

Posted the question, btw.

read the comment; sounds hard.

I actually watched that presidential debate

That looked like a trainwreck for Trump.

2:45 AM

@Balarka Oh, you want examples in which the total spaces ARE the same.

I wrote a post about obstructions. Lol

Lol

That's why you don't write the answer before there is a question :P

That sounds like a fine answer though.

But it doesn't matter much, @MikeMiller: Write your own Q & A!

n

I did mention in the last paragraph that obstructions are welcome too

There, I edited to make it more clear that I want such answers.

2:51 AM

The bit where Trump ranted about how he has "good temperament" while talking over the host of the debate was great.

I am reading the debate. interesting that when asked "how do you bring the jobs back?" he essentially said "we just bring them back".

3:10 AM

At least I had some insight about holomorphic sections now, at 5 am

I think my problem is mostly solved @Ted. Feel free to ignore my earlier comments/question

Does anyone else get annoyed when writing large amounts of latex for an answer and then Ctrl+Z rather than undoing a minor change ends up reverting to like a minute-old state of what you've written?

@Danu I see you're learning blow-ups.

I often just want to change my mind about moving an equation, not erasing the existence of another one.

Cool stuff.

@BalarkaSen the construction for a point is not very spectacular. Then it gets harder...

3:16 AM

Right, of course.

I think blowing up at a point is a very interesting construction though, not just along a subvariety.

The idea is basically to take a m+1 bundle over P^(n-m-1)

In the way that most obviously generalizes the point case

I fail to see why I should care and what it really 'means'

The two must be essentially identical issues

I can tell you the relevance in algebraic geometry.

Again, very little info from huybrechts

Sure

@Danu So, first understand what exactly blowup does. Take a $\Bbb P^2$, blow it up at a point. This is a submanifold of $\Bbb P^2 \times \Bbb P^1$ consisting of pairs $(p, \ell)$ such that $p \in \ell$. Agreed?

$\ell$ is a line through the point I blew up, I should have mentioned.

I am identifying $\Bbb P^1$ with the space of lines through the point I am blowing up in $\Bbb P^2$.

(Perhaps it'd be more intuitively clear to work with $\Bbb C^2$ instead of $\Bbb P^2$ - the latter is just a compactification of the former anyway)

3:32 AM

Huybrechts gives a quite different description. It's just O(-1) with the projection to the complex space

Mhm

These two are equivalent, which is not quite that hard to see

@Danu In any case, I think you should understand the above description to see what's going on. It's much more explicit.

So what's a line in a projective space

Is there an action by C?

A copy of CP^1, coming from a copy of C^2 when quotienting

But just think about it for C^2 instead of P^2 (replace mentally by C^2 whenever you see the latter in the above comments)

Funny. That's not an obvious way to define a line ^^

A line in C^n is a 1-dimensional subspace. When you compactify to CP^n by adding a hyperplane at infinity, a line in there is the compactification of that 1-dimensional subspace (which is a CP^1 of course)

3:42 AM

I wish I could picture C^2

Think of it as an R^2.

That'd be C

Nah, a C is an R

No man :P

It's a good analogy

3:44 AM

I don't approve of this visualization

It doesn't change much, because you can do all this blowup thing with R instead of C

it's called "real blowup"

No its not, especially not when trying to see how to make projective space out of it

You can't make projective spaces out of R???

Never mind it

While it's silly to think C^2 naively as R^2, it's useful in numerous contexts. I can tell you that it has helped me in many contexts.

Hi @Ted.

3:57 AM

Good grief. Isn't it very early morning, @Danu?

hi, @Balarka

@BalarkaSen Danu, he's right.

That's what we all do.

And we draw a cubic curve in the (real) plane and draw the picture of the (real) torus for the real 2-manifold.

Did you figure out that you need to actually look at the trivialization of the line bundle and then see why the transition functions tell you that homogeneous linear polynomials transform exactly right?

@Axoren: Yes, it appears I've been dyed (maybe died, too). Here I see green for my icon, but everywhere else (and everyone else) orange.

Danu - being an ex-physicist - probably is trying to use one of the extra dimensions from the 11 (or was it 10?) of them to draw C^2.

26

Aha, right

I finally posted my question in the main.

Oh, I'll have to look after all the brilliant people answer it

me too

4:03 AM

Has someone told @Danu that blowing up replaces a point with all the tangent directions at it? More generally, replaces a submanifold by all its normal directions ...

I wanted to say that, but he wouldn't listen because I visualize C^2 as R^2

You might indicate in your question that you are aware of that MO question we were discussing.

He's apparently not listening to me, either. I'm going to go watch more debate aftermath.

My favorite way to rephrase that is to say that blowup of a variety at a point replaces the point with the projectivized tangent space of the point and complex topology neighborhoods of the point with open nbhds of the zero section of the tautological line bundle on the projectivized tangent space.

@TedShifrin Good point.

4:20 AM

Hi.

Back home, I suppose?

4:53 AM

yeah

@MikeMiller Thanks for the answer; I'll read it sometime today

Hmm, so presumably your examples, @MikeM, can be chosen to be rank 2 complex v.b. on $\Bbb P^1$?

5:08 AM

@TedShifrin He gives examples of non-homeomorphic vector bundles with same Euler number of the 0 section though, not examples of the question in the title.

(I think)

Wait, we're wanting nonisomorphic bundles with homeomorphic total spaces, I thought.

Yeah, he's answering my side question about obstructions to that happening.

Oh crap.

I'll have to read it carefully tomorrow when I'm awake.

@TedShifrin I see you replied to my diagonal question.

Um, I gave you a diagonal question :)

I asked you to give me the Poincaré dual in general.

5:11 AM

Did you agree with me that diagonal of $T^2 \times T^2$ is not $\alpha\beta + \gamma \delta$?

I had to figure that out for my thesis many years ago.

@AlexW !!!!

Ah. Heh.

Hi @AlexWertheim

!!!!! Good evening, @Ted :)

Hey, @Balarka.

are $\alpha,\beta,\gamma,\delta$ all in $H^1$?

How are you both doing?

5:12 AM

@TedShifrin Yeah, the classes represented by the circles in the 4-torus $T^4 = T^2 \times T^2$.

So-so, @AlexW.

I'm doing decently, @AlexW. I refused to turn on AC, so it's about 88º in my apartment :P

Oh dear @Ted :(

That's much too hot

Oh, @Balarka: I'm on a new antibiotic. She wanted to try anti-inflammatory, but I told her that already hadn't worked.

Why only so-so, @Balarka?

He's sick again. :(

You're missing some terms, @Balarka.

5:14 AM

@TedShifrin Does this new one suit you?

Don't mean to beat a dead horse or assume too much here, but it's good to tend to things other than math now and then so that one can do math. :)

I've only taken one so far, @Balarka. Ask me toward the end of the week?

Regardless, hope you feel better, @Balarka.

@AlexW: I no longer have to do math. I only have to write letters of recommendation for former students.

@TedShifrin So I thought, because $(\alpha\beta + \gamma \delta)^2 = 2$, whereas self-intersection no of the diagonal in $T^2 \times T^2$ is, well, 0.

5:14 AM

How're you doing, @AlexW?

Remember Künneth in general, @Balarka.

I am good, @Ted. I'm listening to Rafal Blechacz and preparing to teach tomorrow, nothing too stressful.

What're you teaching this term?

You still following your algebra/no. theory train?

@AlexWertheim I agree. But it doesn't help when I am sick, have math to do, and pressed with schoolwork. I desperately need the upcoming holidays.

@Balarka: Truly, your health is your first priority, school is your second, and math is your fifth.

fifth? lol

5:16 AM

You just enjoy math too much for your own good.

Nah, it's part of my life now

Or maybe all of it; I don't know.

So you're no longer saying you won't study math when you go to university? :D

@Balarka: fair enough, don't mean to undersell the pressures of life. But I second the advice of Ted - health is #1. That's all :)

@Ted: 2nd semester calculus and abstract algebra.

Yeah, I didn't interpret it as such, @AlexW. I probably am overlooking my health. Need to fix it.

@AlexW: grading for algebra or actually doing some sort of recitation?

5:18 AM

Definitely still on the algebra train, though more on the algebraic geometry side of things than the number theory side of things.

cool beans

@Ted: actually doing recitation, thankfully. Just grading wouldn't be any fun

@TedShifrin I don't know any good actual examples yet.

I've put some effort into constructing some, homotopy theoretic and otherwise, to no avail.

Oh, I thought you had $S^3$-bundle examples on $S^2$. I misread.

@TedShifrin Well, I'll do math nonetheless. It's arguable whether I will (1) pass high school (2) qualify into any uni (3) die in a tragic car accident tomorrow

5:19 AM

Awfully macabre @Balarka, let's hope not.

@Balarka: You will damn well pass high school. You will have letters from various university math professors (you can have one from me if you need one), so I highly doubt you will have trouble qualifying for a uni.

The drivers in India are worse than in LA, @AlexW.

Thanks. I was really joking. But (3) has a higher probability, because of what you said.

Yikes. If that's the case, cross the street carefully... or avoid it entirely. I fear for my life when I run

I don't go out much, @AlexWertheim.

So I'd probably be fine.

@Ted Yeah, it was the opposite. Vector bundles that are not proper homotopy equivalent that you can't distinguish with (compactly supported) cohomology rings.

5:22 AM

OH :(

@AlexWertheim You should stop running on the 405, then.

@Mike: what's life without a little adventure?

Maybe AlexW will run down here to visit.

Also, you joke, but there have been times when I've been trying to get to Santa Monica, and I could run there faster than I could drive there.

Oh, after my visit a few months ago I totally believe that

9 hours in the car to drive about 100 miles in LA

5:24 AM

That would be ambitious, @Ted. When's your next visit around LA?

Well, I'm driving up to the Bay Area and back again, but there's no time to spare to stop in LA ... But I may be driving up to visit a new friend in LA otherwise, so not sure.

@AlexWertheim Assuming you didn't pass out in this heat.

I almost did and that was just from walking back and forth from Wilshire.

@Mike: today was disgusting. Yes.

I walked a couple of miles in the middle of the day (doctor's appointment). I probably could have done a few more.

Today was broiling, @Ted. Hope you drank a lot of water.

5:28 AM

Maybe I should walk a bit today.

Yeah, it was about 100º here.

If you're not too weak from being sick and it won't make you worse, yeah, @Balarka.

@TedShifrin Oh, I think there are more terms of the form (blah)*(blah) involved in the diagonal of T^2 x T^2, blahs being generators of H^1.

Nope

That'd make sense.

Hmm?

Where else does $H^2(X\times Y)$ come from?

5:37 AM

H^2(X) and H^2(Y) certainly.

Right. You need terms.

But classes there can be written as blah*blah, not? I mean, it's an exterior algebra, the cohomology ring.

Oh, because $T^2$ is special. OK.

Right.

I think it's easier to do the general case, actually, or the $\Bbb CP^n$ case, and not get confused by coincidences.

5:39 AM

Evening

heya @Semiclassic ... Did you make it home safely?

Good suggestion. I'll try to do that.

Still going

Aha ... I hope it's all safe and uneventful, @Semiclassic.

Which is about what we expected

Yeah, has been so far

75 miles left, so a bit more than an hour left

5:42 AM

oh wow, that's great.

Wish a safe journey for the next couple minutes.

Main problem right now is that I'm in the backseat and there's enough stuff packed in here that I've got airline levels of legroom

Economy-class

Ugh

I don't like long roadtrips by cars (or anything to do with cars in general) because I have motion sickness.

Not fun, @Semiclassic. Hang in there.

Damn, Balarka. You're just a basket of woes altogether.

Will do

5:47 AM

yeah...

OK ... I'm outta here ...

Bye

6:10 AM

Hi @Anubhav.

Hii @BalarkaSen

From where you are reading this $4$-torus thing?

It sounds interesting

I was just thinking about a problem and it came up. Not reading from anywhere in particular.

Did you see my recent question in MSE?

Perhaps you'd like it.

No...

wait

You mean they are not isomorphic as a bundle map, but they are homeomorphic?

user116211

Is it this @balarka:

user116211

3

Q: Nonisomorphic vector bundles with diffeomorphic total spaces

What's an example of two non-isomorphic vector bundles $E,F$ over the same base such that the total spaces $Total(E), Total(F)$ are homeomorphic? Assume that rank of these bundles is the same as dimension of the base manifold. Notice that the Euler class of the bundle is an obstruction from thi...

algebraic-topology differential-topology vector-bundles

user116211

6:21 AM

?

Yes.

@Anubhav Right.

Sound interesting, I never think,...so let me think

BTW recently I came up with a qes, which I am not able to prove yet...i.e construct a manifold (compact) whose boundary is odd dimensional real projective space

RP^3 bounds a 4-manifold, I believe. In fact every closed oriented 3-manifold bounds a 4-manifold IIRC

Don't know off the top of my head how to construct one. That I have to think.

I've seen this here in some comment

OK, I do know

Take the unit tangent bundle of S^2

Fill in the circle fibers by disks.

6:24 AM

But, Milnor said in his book that it is a good exercise

Aka the unit disk bundle in TS^2.

That won't help, I guess

I tried in that line

but any way, you can think again...

Not sure what you mean. The unit disk bundle in TS^2 bounds RP^3.

What's the big deal?

for general $n$?

Ah, for every odd n, you need an (n+1)-fold bounding RP^n.

I thought for some n. Got it.

6:26 AM

yess

Do you want an answer or do you want to think about it?

think of course, but any slight hint would be helpful @MikeMiller

Understanding the construction for RP^3 might be more helpful than you supposed.

I'll think, now gotta go...I'll msg you if I think something new...thnaks

Interesting question. I'll think about it too I guess.

6:30 AM

Finding bundles with diffeomorphic total spaces is also interesting. I still don't know.

You can't do it for eg $S^2$.

My guess is no sphere will work.

So many questions. Too little time. Soon we'll die.

:D

@MikeMiller How did you come to that conclusion?

I know all the bundles over $S^2$.

Ah. Fair enough.

6:32 AM

I think it's fair to assume you want the base to be the same for both examples.

Yes, I did mention that in the question.

If @Anubhav doesn't mind, can I suggest an approach (which may or may not work) for the RP^n question?

Suggest as in ask if it's doable of course.

Ask elsewhere. I don't think it fair to spoil someone's fun.

True.

1 hour later…

8:02 AM

@BalarkaSen did you solve?

8:20 AM

@Anubhav Yeah.

Mike's hint is a useful one.

1 hour later…

user116211

9:22 AM

Isn't binary operation always closed?

user116211

I.e., $\forall (x,y)\in S\times S: \circ(x,y)\mapsto z\in S\,?$

user116211

Doesn't that mean $(S, ~\circ)$ is always closed?

one may apply the binary operation to subsets, and the subsets may not be closed wrt it

user116211

Could you elaborate @arctictern?

example. N is a subset of Z. we may subtract two things in N, the result may not be in N, so N is not closed wrt the binary operation of subtraction.

user116211

9:27 AM

@arctictern Yes, that's why the minus operation is not a binary operation for $\mathbb N\,,$ isn't it?

no, it isn't. what's your point?

if something somewhere you're reading is bugging you, tell us what it is. don't hide it.

user116211

@arctictern My point is that binary operation is defined to be closed in the concerned set, here, $S\,.$

user116211

So, $(S,~\circ)$ must be a closed algebraic structure.

this conversation seems pointless to me

user116211

But I'm seeing Bourbaki uses the term magma for closed algebraic structure...

user116211

9:30 AM

Not only Bourbaki, even $\mathsf{Pr}\infty\mathsf{fWiki}$ uses magma.

okay. so hypothetically someone could use the word "binary operation" more generally than is standard. that people can choose to use words differently is not earth-shattering. or maybe they just want to re-emphasize closedness, because (again), one may want to check that subsets are substructures in their own right. also please, in the future, don't hide the context of your questions.

user116211

so, does that mean there exists unclosed algebraic structure?

user116211

@arctictern Well, I was writing.... you know I'm not Barry Allen ;))

user116211

@arctictern okay.

user116211

@arctictern Okay re-emphasize; sure.

9:33 AM

A binary operation is a map $S \times S \to S$. So by definition if $S$ is equipped with a binary operation, it's closed (the definition of closed I know is $a \cdot b$ is in $S$ whenever $a, b \in S$ but that has to hold because the range of the map is $S$).

user116211

@BalarkaSen That is the definition in my mind. They are defined to be closed.

Idk, I wouldn't prevent binary operations to not have a result in $S$

@mercio Then that's not really a binary operation on $S$ though.

It is

I wouldn't call it so, at the very least.

9:35 AM

Binary means it takes two arguments

nothing more

user116211

@mercio Maybe that was in the mind of Bourbaki.

in french we say "loi de composition interne"

and the "interne" is there to mean that the output stays in $S$

I guess. Shrug. That's nonstandard terminology in any case.

Nobody calls eg a metric a binary operation

wikipedia seems to agree with you though

then when i click on the french version it drops the "interne" requirement lol

user116211

@mercio ohh.

9:41 AM

also bourbaki's texts were written very long ago

what words meant then and what those same words mean now, may have changed

user116211

@mercio They have some unusual terms, BTW ;)

I wouldn't recommend reading Bourbaki.

I've never actually read bourbaki

user116211

@mercio You need to read the first to get the laters although they hardly used the contents they developed in the first volume in the later texts.

yes

they wanted to firmly establish mathematics from the ground up

because they were concerned about foundations

user116211

9:53 AM

@BalarkaSen Yeh, I know that they are not for the first learners; but I did go in the Set Theory; and it was not a dull venture, though. In the second volume, in the first page, they define magma.

Okay, I don't understand why the preliminary results (2) & (3) in the following article about proving $\zeta(2)$ is irrational... paramanands.blogspot.in/2013/10/…

why what ?

I don't get how you know you can start off with those improper integrals...

They are preliminary results...but how do you connect them to the zeta-function?

well surely he'll use them later ?

True. But how do you actually figure that out?

It is not obvious.

9:57 AM

Because he thought about it for decades and he tried a lot of things and eventually he realized that that integral was useful for his problem ?

you just have to trust him that it will be useful later

10:18 AM

@deostroll It is not always clear from looking at a mathematical article what the author's thought processes were, so some steps might look foreign before understanding why or how those steps interact with the whole big picture. So you'll have to understand those technical steps and move on. You'll eventually understand, perhaps at the end of the proof, why those were important or maybe even why the author thought of those.

Tobias Kildetoft

And sometimes, you will just have to realize that you will never understand how the author got the idea

Very true.

This is a standard issue in reading mathematics most of us face. Solution is to deal with it. How we write is absolutely not how we think.

(on that note, I think Thurston has a soft-question related to this on mathoverflow. I like it a lot)

Here.

Tobias Kildetoft

@BalarkaSen You mean had (sadly)

;w;

its more of asking myself..."what do I need to research before reading that proof?", but I agree with all your points...

10:25 AM

@Tobias The question is still there though, so I do not think past tense is appropriate (is it?).

Tobias Kildetoft

@BalarkaSen Well, Thurston no longer has the question there, but I see your point.

Grammar is always so confusing on the internet. I meant to subjectify the question, not Thurston in that sentence :)

@TedShifrin 26 is only for the bosonic string---the superstring is consistent in 10 only.

@TedShifrin I mostly just thought: Clearly hom poly's are holomorphic in charts, plus clearly they are sections. Therefore, its must work out in trivializations :P

@TedShifrin I fell asleep (you were right, it was *very early :P)

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