@BalarkaSen depends on what language

Interesting, @Danu.

I meant English of course

Oh that sucks @Ted :( Selfish neighbours are the worst

I think the hardest letter to kill is the "v"

@TedShifrin Have you brought the issue up to your neighbor?

@Danu So you're trying to calculate the cohomology of $\widetilde{\text{Gr}_k}$?

"v" like ...?

vonderland imo

Hi @Ted and every one I haven't seen yet

@MikeMiller No

Of $G_2/SO(4)$

I thought you just told Ted it was an oriented Grassmannian :)

Do you know what the cohomology of $G_2$ is?

Nope, that's a space that fibers over it

Nope, I don't know nothing

@s.harp I googled. Chivvy = chivy

I've complained before, Balarka, but ultimately it's my own fault. I just blame them for making my life difficult.

@Mike Hiya

Hi

hi @TedShifrin

hi PVAL

hi @MikeMiller @BalarkaSen

h

@TedShifrin show them who's the boss, no time to say sorry

I know nothing about cohomology of $G_2$.

@TedShifrin can I talk to you on facebook for a sec?

Not now.

okay

apparently the integral cohomology of $G_2$ is $\Bbb Z[x_3, x_{11}]/(x_3^4, x_{11}^2, x_3^2x_{11}, 2x_3^2)$.

@TedShifrin Are now people wasting your time with a question Robert Bryant can answer?

@BalarkaSen i have never heard of either of those words xD

@MikeMiller Great ^^

and googling is cheating

LOL, PVAL, Robert can answer most everything.

@PVAL-inactive You know Bryant?! I think he knows the answer to all the problems I'm facing :P

@Danu Run the Serre spectral sequence (sorry!) on the fibration $SO(4) \to G_2 \to G_2/SO(4)$. With mod p cohomology, p not 2, the cohomology of the fiber is trivial (except in degree 0). So you get that the mod p cohomology of $G_2$ agrees with that of $G_2/SO(4)$.

Same thing works with R.

Cry every time

that's a difficult order @Danu

You're the one that wants to calculate cohomology, I'm just using a calculational tool :D

Deep sigh

laughing is healthy

Learn it from Hatcher's notes man

Yup, Mike wins. I even needed spectral sequences in my thesis for similar reasons.

My supervisor did tell me that "it is known" that there is no odd order torsion in the cohomology of $G_2/SO(4)$.

That follows from what I said (sorry!)

Yeah, I'm sure it does

@SteamyRoot Sorry, I'm really slow with Algebra. So you said I can write $a=a'\gcd(a,n)$ with $\gcd(a',n)=1$. I'm guessing we can't have $\gcd(a',n)>1$. For the sake of proving this, say $k=gcd(a',n)>1$. Somehow, I should be able to derive from this that $a=a'\gcd(a,n)$ is a contradiction then. All I can write down is $k\mid a'$ and $k\mid n$. So I guess we could write $a=rk\gcd(a,n)$, for some $k\in\mathbb Z$. But I'm kind of stuck...

So now other than understanding the difference between Z/2 and Z/4 you have the cohomology ring of G_2/SO(4)

Cohomology rings of (oriented) Grassmannians are well known.

but black-boxing that for now, can I hope to get all the cohomology from just the $\Bbb Z_2$ and $\Bbb R$ coefficients?

@TedShifrin Right; but this space is not the oriented Grassmannian.

@ShaVuklia what are you doing ?

That's $G_2/U(2)$

Almost. You won't be able to tell the difference between Z/4 and Z/2 summands.

@MikeMiller What about other powers of 2?

@Astyx I've done what you said with the prime decomposition, but now I'm also trying to do SteamyRoot's approach

(why can I tell those apart?)

Hey @TedShifrin , do you know something about curvature in normed vector spaces that are not euclidean? ie how people talk about such a thing concretely?

@ShaVuklia Suppose that $\gcd(a',n) = k > 1$. Then $k$ divides $a'$; hence it divides $a$, and it divides $n$

I think everything I know about the cohomology of Grassmannians has to do with the difference between Z/4 and Z/2.

@Danu You can't. I was just writing down the first two.

@Sha, I haven't been following, but your question is one I always assigned in algebra and lcm needn't appear ...

@MikeMiller D'oh

@Astyx glad you came here :)

@PVAL-inactive Is that supposed to tell me I will know nothing? :-)

Right, I'll let you to it then :p

I sincerely doubt there is torsion higher than order 4 in the cohomology of G_2.

Thanks @Alucard, how are you by the way ?

I don't mean the summands.

I think I'm gonna go ahead and assume for now there is n't even any order 4 and see what I come up with

That's probably false

@Astyx i am fine, i hope so are You!

@ShaVuklia What you're looking for might be this: $\gcd(a / \gcd(a,n) , n) = 1$

I was trying to make a joke about the Bockstein homorphism being the "difference between Z/4 and Z/2"

@Steamy right, I'm going to write it out!

@MikeMiller Let's hope I run into a contradiction then :D

"If you divide $a$ by all factors it has in common with $n$, it doesn't have any common factors with $n$ left"

I'm also fine for my part, thanks

@PVAL-inactive I don't get it :P

Is there anything to be learned from that joke? I'd love to hear it

No

LOL ... What's a Bockstein among friends?

Hi, DogAteMy.

That was an information-packed burst of math chat.

Thanks all

In fact let's just run the spectral sequence. The bottom row is the cohomology of G_2 (with Z coefficients). The second row is the cohomology of G_2 with Z/2 coefficients. You can find the relevant calculation in 2.1.4. here. So the only possibly nontrivial differential is on the $E_3$ page. This is a multiplicative spectral sequence, so it suffices to determine what the differential does on $E^{0,2}$.

But since we know the calculation of the cohomology of $G_2$, and there's no 3-torsion in degree 3, this must vanish identically

I don't agree with my previous claim that it suffices to determine the differential on that one spot. But whatever.

Hm @SteamyRoot Say we work with $\gcd(\frac{a}{\gcd(a,n)},n)=k>1$. Then there exists $m_1,m_2\in\mathbb Z$ such that $a=\gcd(a,n)km_1$ and $n=km_2$. But we can also write $n=m_3\gcd(a,n)$ for some $m_3\in\mathbb Z$. Ehm... maybe another hint?:(

Oooh, that paper has significant info in it, @MikeM.

I suspect the differential says $x_3$ to $x_3^2$ and $x_5$ to 0.

@ShaV: If the gcd of $a$ and $n$ is $k$, what is the order of $\bar k$ in $\Bbb Z_n$?

You need to resolve extension problems at the final page but that's as easy as spectral sequences get.

I see no reason to be able to get anywhere close to a calculation without using spectral sequences.

@TedShifrin That the craziness of Chicago's curriculum generating stuff like that.

@ShaVuklia Then $k \gcd(a,n)$ divides both $a$ and $n$

Is that person an undergraduate, @PVAL?

Yes.

But it's an REU.

Pretty sure its an undergraduate working under May

those guys at the military went totally nuts, they recruit absolutely everyone, no matter what

But $gcd(a,n)$ is the largest integer dividing both $a$ and $n$ by definition

Ah. Makes sense.

or someone working under someone working under May

@PVAL-inactive the latter

In the dictionary order topology on the unit square I^2, the set {<0,0>} is closed, since I^2 is Hausdorff (all finite sets are closed), and its complement is (<0,0>, <1,1>] is open, implying that {<0,0>} is clopen. Hence, I^2 is disconnected. But evidently I^2 IS connected. Where did I amiss in my reasoning?

Might not even be a UC undergrad, if it's an REU.

@Ted I'll have to think about that! @SteamyRoot: Yea but, from what I've written, it's not apparent that $k\gcd(a,n)$ divides n.

Where did you get that the set is open, @user193319?

@PVAL-inactive lol

How are you doing awesome user @ShaVuklia?

I just remember that paper having something to do with May and an undergrad.

@Givemeabreak Hi! Doing great, and you?

@Balarka: There are a lot of people buried under May.

@ShaV: I always recommend that people try concrete examples. Try actual numbers. See if you can get an idea of a proof from concrete examples.

@PVAL-inactive It was really directed by one of May's grad students, who has either graduated or will soon, Zhuli Xu.

@MikeMiller Thanks for all the information. I'll try to get back to it... when the time comes.

He's working on exotic spheres stuff. No exotic structures in dimension 61.

@TedShifrin Are you referring to the set (<0,0>, <1,1>]? I am pretty certain Munkres defines this as basis element.

No, @user193319, I meant where did you get the idea that $\{0\times 0\}$ is an open set?

@SteamyRoot I only have that $k\mid n$ and $\gcd(a,n)\mid n$. I don't have yet that $k\gcd(a,n)\mid n$, as far as I see

@MikeMiller I think that means Adams SS stuff.

Ya it does.

He's really good at Adams and ANSS.

Yeah, wait, you're right.

@ShaVuklia Glad for you. Not that bad. Working hard on something.

@TedShifrin I never claimed it was. I first began with that it is closed, and then deduced that it was open.

(you should listen to Ted's suggestion, though. Try it with some actual numbers and see if you get an idea of the proof from there)

How did you deduce it was open?

Ah. In order to show that {(0,0)} is open, I need to show its complement is CLOSED, not open.

Which it isn't :)

I watched some talk by Hopkins online a few years back (resolution of the Kervaire invariant 1 problem I think)

@TedShifrin Yup! Thanks!

Anyway, back to work.

Sure thing :)

and he made the Adams SS seem like this sort of incomprehensible behemoth of mathematics.

@Givemeabreak ah okay! @SteamyRoot that's not going to work here, because as soon as you know the $\gcd$ has value $x$, you automatically know what the other factor has to be, and there is no point in checking that this factor has $\gcd =1$

Oh, I saw that talk

The one where he says the pictures look like porn to him

LOL, what?

Or rather, refrains from saying it because there's a camera :-)

Yeah he didn't say it

@TedShifrin The spectral sequence things---no idea what they mean

I think YouTube likes pushing that video onto people watching math ^^

well, I suppose lots of arrows can look porn-like.

I don't think I watched it on youtube.

Oh well

an arrow-fetish? that's new

He says a lot of funny stuff there.

Pretty nice talk

Only if i understood it

@PVAL-inactive It's not incomprehensible but it is a behemoth.

I think I listened to it several times; never understood anything but his voice is pretty soothing.

@MikeMiller I don't mean the work is incomprenhensible

I think the construction of the spectral sequence is pretty easy but I don't have the yoga like those people do, even a little bit.

I mean like understanding the thing is really hard.

Oh sorry I agree.

Like the object itself can't be comprehended

@Danu I only ever understood the beginning

I like the anecdotes ^^

after the weird spectral sequence pictures it flies out of my head

Right, he talks about this nice geometric stuff

Pontryagin's stuff. Cool!

Yep

@Ted Huh. I tried out an example ($n=12$, $a=8$), and now I have that $\gcd(a/\gcd(a,n),n)=\gcd(2,12)=2$

Wasn't the point to prove that $\gcd(a/\gcd(a,n),n)=1$ ?

Wait, @ShaV. The gcd of $8$ and $12$ is $4$. I'm asking you what the order of $\bar 4\in\Bbb Z_{12}$ is.

I haven't been following all the contortions that you and Steamy went through.

I didn't know we were working in $\mathbb Z_n$?

the original question took $\overline a\in\mathbb Z_n$ such that $1\leq a\leq n$

Your original question was about the order of $\bar a$, right?

but all the arithmetic had to do with $a\in\mathbb N$

oh, right...

in my syllabus they wrote $a$

That's a mistake.

that got me confused

It's very important to keep track of where things live :P

Yeah, they should be smacked for that :P

calls for DogAteMy or Demonark or Astyx to administer a smack

aha.....

So let's follow through with this example.

lends a smack without knowing to whom or why

(the writers of my Algebra syllabus @Astyx)

bing

wait wait! @Ted

I think my syllabus is right on the notation:

I know, it is Dutch

but as you see, they work with both $\overline a$ and $a$

No, but that should say orde($\bar a$).

so they make a distinction, and question was dealing with the part where they use $a$

it says so later on!

They left out the bar in two places that I see.

Does "door" mean "mod" ?

No

Note that they started with orde($\bar 1$).

"door" = "by"

yes

I'm officially bad at Dutch then :)

confuzled

wait

Anyhow, let's get back to the order of $\bar 8\in\Bbb Z_{12}$. First, what is the order of $\bar 4$? Why? And then why is the order of $\bar 8$ the same?

I just realised I'm stuck on a problem that I already know one solution for (the prime decomposition)

the order of $\overline 4$ is 3

RIGHT, sorry.

hahahaha :P

Can we prove that?

Well, yes

you can show it's not 1 and not 2 :P

And can we prove that in general if $k|n$, then the order of $\bar k$ is $n/k$?

so 3 is the smallest one for which we get $\overline 0$

OK, and now in general, do you have an argument for what I just said?

@PVAL-inactive I like that symplectomorphism question you essetnially answered.

@Ted I can see that it's indeed a factor that gives $\overline 0$

but I would have to prove that it's also the smallest positive number

Correct.

An inequality does that in a second.

If $0<\ell<k/n$, what can you tell me about $\ell k$?

@MikeMiller Me too. There's a few more hard interesting questions on the I. Smith exam.

ah!!!

that's smaller than $n$

:)

okay, so that's settled

Cool. Now you just have to explain to me why the order of any nontrivial multiple of $\bar k$ is the same as the order of $\bar k$.

well that's kind of easy

wait

let me write that out neatly

it has to do with the fact that $\overline 0$ is multiplied

Why no smaller, though?

let me think for 1 sec

:(

@PVAL-inactive I saw this question you asked about diffeomorphism invariants of Stiefel-Whitney numbers... Could information on the invariance of Chern numbers be helpful for that at all?

uhm

So we have $\operatorname{order}(\overline k)=r$, and $\operatorname{order}(\overline{mk})=l$, where $l<r$. Also, $m\neq 1,0$.

@Danu sure but realize that question is asking something weird

its asking about invariants of the diffeomorphism of the total space

So we have $l\overline {mk}=\overline 0$ and $r\overline k=\overline 0$ @Ted

oh, right

OK, so you're observing that $\bar a$ is a multiple of $\bar k$, @ShaV.

Sorry, where does $\overline a$ come from?

Is it also true that $\bar k$ is a multiple of $\bar a$? :) $\bar a$ is where we started!

What has $a$ to do with $k$? All we know is that $1\leq a\leq n$ and $k\mid n$.

@MikeMiller Which reminds me, I was saying to arctic a few days ago that the Milnor's fibration S^3 --> S^7 --> S^4 and Hopf fibration are non-equivalent. That gives two nonisomorphic fiber bundles with homeomorphic total spaces, right?

$k=\gcd(a,n)$.

So you said that $\bar a$ is a multiple of $\bar k$.

oh right, sorry

I'm asking if we can be sneaky and deduce that $\bar k$ is likewise a multiple of $\bar a$. It works for your example. Try a few more examples.

Hey @Ted

yes, sorry, I need to reread our conversation partly

Hi @Ali.

@ShaV: We're almost at the end, believe it or not. There's just one ingredient missing.

Does anybody know of any good literature on concordance groups?

concordance groups of what

of S^3 mostly

I meant, of knots/links or ... ?

@Ted oh think I see it now

Oh, @Balarka, we're back to that question from ages ago? Cool.

$a$ is a multiple of $k$

so we have that order $a$ = order $k$

(I have to think about the bars)

@BalarkaSen sorry, let me fix my internal definition quickly

Well, order makes no sense without the bars, @ShaV.

@TedShifrin Yeah, only that was for vector bundles. Mike gave examples for that in the question I asked

That's true