Mathematics - 2015-07-26 (page 2 of 3)

9:23 AM

@BalarkaSen why is the space obtained from $S^1$ by attaching two 2-cells by degree 2 and degree 3 maps respectively homotopy equivalent to $S^2$?

@iwriteonbananas I asked this to Mike a long while ago. he doesn't know how to do with without some machinary.

hmm ok

what machinery?

whitehead theorem

i see

yeah, i think this was it

let $X$ be that cell complex

$X \to S^2 \vee S^2$ be the map obtained from pinching the equator. compose with $S^2 \vee S^2 \to S^2$ obtained from wedging the identity map on both.

this map $X \to S^2$ clearly induces identity on $\pi_1$. it also induces isomorphism on $\pi_2 = H_2$.

9:35 AM

oh

and i guess higher homotopy groups are 0. how do you show it incudes an iso on $H_2$?

hmm. higher homotopy groups are not 0. that's a huge problem, lol.

there ought to be a solution that doesn't use the whitehead theorem

recall that \pi_3(S^2) = Z

i never knew that fact until now :P

really?

9:38 AM

whitehead's theorem seems pretty impractical then

but what do i know... :)

computing higher homotopy groups of spheres is the biggest open problem right now in algebraic topology

interesting

sounds like such a simple proposition

I read your message back up. Quite interesting.

oh, there is a homological version of whitehead's theorem

corollary 4.33 in Hatcher

weird, didn't know it existed.

anyway here is an easier proof.

Isn't it simply because any 2-vector in $\Bbb F_p$ can be taken to another by multiplying it by a $2\times 2$ matrix with coeffs in that field?

(not a proof, I know)

the other direction would be harder, I guess.

9:42 AM

ah

well, an aut of (Z/pZ)^2 is determined by a 2 \times 2 matrix with entries in F_p.

so there you go

isn't that what I said?

leastways, that's what I meant.

@BalarkaSen neat solution

my nose and lungs are acting up again :'(

brb

9:45 AM

@SohamChowdhury yikes

I just sneezed 17 times

not fun

;-;

i understand, i went through a month of sinuisits

still not completely cured

@Soham I made a typo, btw. Gal(Q^alg/Q) acts on p-torsion points of E(Q^alg)

not E(Q), of course.

ok

where did you learn all this?

ANT course?

lectures of a very good number theorist i met.

Do more sports, then you'll get sick less often, @BalarkaSen @SohamChowdhury ;)

9:51 AM

asthma, @Huy.

@SohamChowdhury: I have asthma too.

mine is worse, I assure you. but you're right about the sports.

@Huy I guess I understand now that people say that for a reason.

Sports still helps me, even though I have asthma. I just have to not go over the top.

@BalarkaSen: I'm pretty sure at some point in your life you'll wish you would have done more sports in your youth. I do it already, even though I did sports about 4 times a week.

yeah, I played tennis for two years. used to feel somewhat better as long as I didn't overdo it.

9:53 AM

@SohamChowdhury: Did you play it because you wanted to or because your parents did? :P

I walk a lot. But I guess that's not useful enough.

@BalarkaSen: I'd say it's better than to not move at all all day.

I didn't want to grow up having played nothing, and I like tennis, so . . .

:D

When we move to our new apartment in two months, I'll be able to cycle again. I love cycling, but haven't done much of it in years because the area I live in now doesn't have nice places for it.

9:55 AM

Are you moving to a more rural place then?

heh, I don't even know how to cycle. don't think I'll ever need to.

Semi-suburban, you might say. But definitely in the middle of the city.

@SohamChowdhury: If you come here, we can cycle around the lake. I live next to it. :)

@Huy My parents have urged me to get tutors or join a coaching centre ever since I was in 6th grade. I always refused -- today I'm the only one in my entire class who doesn't do any of that. ;)

@Huy ah, that'd be cool. If I only could. :)

@SohamChowdhury Cool, I don't have any tutors either.

takes up good math time

9:58 AM

Yeah, all those "tricks" for "guaranteed success" are painful.

Any time is good math time.

Jul 15 at 13:48, by Balarka Sen

rock climbing is useless. everything excluding math is useless.

that was a joke, btw.

I wasn't sure.

:P

(that was one, too)

you know everything in the book, more or less?

roughly, i think so.

10:01 AM

cool

Chris's sis the artist

@robjohn I found a new $\zeta(2)$ integral representation.

by "knowing", i mean in the sense that the book says etale cohomology has a natural action of Gal(\bar Q/Q), and i "know" that. but i'd rather prefer to understand that :P

which would reduce to understanding what etale cohomology is

10:15 AM

meh @Soham, they don't tell you about how the galois representations arise from modforms either.

that's another thing i wanted to know :(

@SohamChowdhury: Coaching centre? What does that include? Is that so common in your country?

Very.

Why?

There are a few (apparently) "elite" institutions called the Indian Institutes of Technology.

snorts @ elite

10:19 AM

Most Indian parents want their kids to get into one of those (or into a certain medical school called AIIMS).

@BalarkaSen the air quotes were there for a reason

i know

The idea is that you get a B.Tech, pass out and get into a firm where you earn boatloads of money, @Huy.

Aha.

It's similar in France, except people are mad about studying pure math there in ENS and Mines and whatnot.

I wish . . .

What do you wish?

10:21 AM

And so kids go to prep schools from fifth grade onward here, where they learn "tricks" to solve the problems on the entrance exams to get into those institutes. It's disgustingly like a game.

I wish people were as mad about math here.

@SohamChowdhury to be in France? you don't wanna be a french algebraist, trust me

My parents are like, "why do you want to study useless abstract things?"

@BalarkaSen lol, why? ultra-abstract stuff like Arnold once said?

you don't know the "what's a variety" joke?

integral scheme of finite type over a field?

:P

@SohamChowdhury: Yeah, I remember that some Indian told me in India it's all about engineering/cs and people think maths is useless.

10:23 AM

exactly right

haha. dunno why I remember.

I can already appreciate one thing about schemes, though, with my limited knowledge: they help differentiate the variety corresponding to $x = 0$ from $x^2 = 0$, or something similar, right?

and then there's the grothendieck riemann roch theorem joke

what?

@SohamChowdhury I have already told you what an affine scheme is

@Huy remember that profile I linked a few days back?

@BalarkaSen what's the joke?

10:25 AM

@SohamChowdhury: Yeah, pretty confirmative.

I should go back and check your explanation. I knew even less then.

It's amazing: your knowledge increases (say) exponentially, but the amount of things you know you don't know increases superexponentially. :P

@SohamChowdhury so true

but what's the joke?

I need a hit of Seroflo powder, brb

I'm searching for it.

ah, I feel much better.

10:33 AM

oh well, can't find it. it went something like this :

?

a phd students enters the classroom with his professor, preparing to talk about his recent researches. all the while the student talks about what his research about functional analysis, the professor only half-listens and dozes off during long proofs. the students fears that his work might not be interesting enough. however, just when he wrote down is main theorem, the professor jumps upright and says "that's exactly like the Grothendieck-Riemann-Roch theorem!!!".

the student is confused, because he doesn't understand what's the connection. however, believing that the professor is seeing something he isn't, he asks "how so?". the professor says "you have a left hand side, and you have a right hand side, and you have an equality between them! that's exactly like the Grothendieck-Riemann-Roch theorem!!!"

oh, this one.

I saw it on MO.

the professor was a french algebraist, apparently.

but of course

Hippa and Gato and Agawa will have our heads off (?)

10:37 AM

lol

what are you doing now? (don't be stupid)

also, the functional analysis was a nice touch :P

@BalarkaSen: mathoverflow.net/a/53228

nothing, i have to go to do schoolwork in about a second or so.

I do, too.

bad schoolwork. always getting in the way of math.

I learned a bunch of things today. :)

> "At last, we come to the heart of this book, or at least the pericardium."

:P

where are you in right now?

10:40 AM

@BalarkaSen: math.stackexchange.com/a/1160938/3787 Do you know this one?

@Huy that was it, thanks.

just hit representations.

good

will see that later, @Huy. I have to run fast now.

bye, everyone

bye

I really liked that one, but only one other person voted it up. :(

10:51 AM

wooooooow

representations are cool!

@Huy, if I have a representation $A_4\to \rm GL(3,\Bbb R)$, what does the $\det$ of the matrix represent?

@SohamChowdhury: I don't remember what a representation is. :(

oh, okay.

I never did more than just basic algebra up to some Galois theory.

aren't you interested in algebraic geometry at all?

Not too much, and I hardly have spare time anyways. I'd very much like to revise my complex analysis and algebra at some point though, because I don't remember as much of it as I'd want to.

10:58 AM

oh.

what are you interested in learning?

Mostly mathematical physics, and subjects needed to do that. For example I did functional analysis to study quantum mechanics, and right now I'm still studying differential geometry to understand general relativity better.

I'll probably do PDEs next semester because I have no knowledge about them whatsoever.

So I guess I'm more on the analytic side than on the algebraic.

cool.

But I really liked algebra when I had to learn it for exams, in fact it was my best exam at uni so far.

(gradewise)

I used to be mad about mathematical physics once. I still like it. Someday, when I know enough diff geo and everything, I'll learn some QFT.

Yeah, I want to take QFT next semester even though my QM knowledge is only basic.

I want to see what it's all about. :D

11:01 AM

everything is a perturbation of a field.

what a powerful idea.

There was also a course about string theory last year at my uni, but unfortunately I didn't have time to take it. :(

Did you do mechanics already, like Lagrangian and Hamiltonian?

I know a tiny little bit about Lagrangians. Don't have enough time, unfortunately.

:(

I find the ideas tremendously elegant.

:(

I really liked those. Yeah, it makes all those messy problems really easy to solve.

11:04 AM

I have a splitting headache.

(not a Galois theory joke)

Then you should take a break from your PC.

Indeed. Ciao for now.

Laters.

@MikeMiller: I found two versions of covering maps, one says it's continuous and surjective, the other says it's open and surjective. Are they equivalent for covering maps? I mean they are only locally homeomorphisms so I would have suspected it not to be equivalent, but maybe I'm misunderstanding.

Mats Granvik

12:20 PM

Functional equations look like fluffy nonsense, but I am starting to believe that they contain some solid substance.

Chris's sis the artist

I rarely found such impressive results as today!!! What an incredible day!

I'm completely overwhelmed. Am I dreaming???

A M A Z I N G

@Huy covering maps aren't just local homeomorphisms.

the official definition of covering maps is : $E \stackrel{p}{\to} X$ is called a covering map if for pt $x \in X$, there is an nbhd $U$ around $x$ such that $p^{-1}(U) = \coprod V_i$ where $V_i$ are open sets in $X$ such that $p|_{V_i}$ are homeomorphisms of $V_i$ with $U$.

@BalarkaSen: In the definition on mathworld it is worded as if continuous surjection that is a local homeomorphism implies the other properties: mathworld.wolfram.com/CoveringMap.html

yes, i think that is also a valid defn

but you need the surjection part.

But in this definition, do we need open or continuous or does it not matter?

12:35 PM

a local homeomorphism is continuous by defn.

a local homeomorphism is also open.

does that resolve your issue?

Yea.

ok, cool.

Ben Dover

1:29 PM

@BalarkaSen the official definition should be: fibre bundle with discrete fibre

1:39 PM

haha

Hippalectryon

@Chris'ssistheartist :D Finally I have the internet here

Chris's sis the artist

@Hippalectryon Welcome back! :-)

Hippalectryon

For some reason, Chrome still displays the chat with an ugly monospaced font though :(

Btw does anyone here have an unlimited mathematica online (cloud) account ? I'm computing quite a lot of data and with my bad computer it takes about 20 minutes or so, if someone could run it for me on the cloud it would be way faster

2:05 PM

No, there are definitely continuous, open (local homeomorphisms are open!) surjective maps that aren't covering maps. You want to throw in the word 'proper' to conclude that they're all covering maps.

ah, ok.

Pick your favorite surjective local homeomorphism $\mathbb R^2 \to S^2$ for a counterexample.

I'm back, thankfully. I have not had such a horrible headache in my life.

@MikeMiller this might or might not (this one with greater probability) interest you : if you look at the galois theory-covering spaces analogy closely, finite separable extensions of a field $k$ are treated as finite covers of $X$. even the grothendieck fundamental group of complex algebraic varities is inverse limit of fundamental group of finite covers. so why this finiteness? one answer is that infinite covers might not correspond to separable extensions.

that said, here's a probably promising idea : $k^{alg}$ is direct limit of separable extensions of $k$. so dually, one should think about inverse limit of finite covers of the base space $X$. so "infinite covers" should be some version of "solenoids".

@Balarka, what is the significance of the $\det$ of a representation $A_4\to {\rm {GL}}(3,\Bbb R)$?

2:18 PM

what is a det of a representation?

every element gets mapped to a matrix, no?

yes

what does the det of that matrix say about the element of $A_4$?

oh, i see.

(replace with trace if possible)

2:19 PM

doesn't seem as if something interesting can be said without knowing what the representation is. no idea.

makes sense

can you check the pdf? page 173?

essentially, first project the tetrahedron onto a unit sphere.

embed an xyz coordinate system into the sphere. move the sphere around so that the vertices of the tetrahedron are mapped to themselves. let the new positions of x, y, z be x',y',z' (row vectors).

now, the representation of this rotation is $(x', y',z')^T$.

> "That is enough examples. This is already looking too much like a textbook on Galois theory."

hahahaha

@SohamChowdhury how is that a 3 x 3 matrix?

three 3-vectors make a 3x3 matrix.

oh, i see, so you're putting the position of the vertices on each coloumn of the matrix

yes.

2:33 PM

not sure if det of that says something meaningful

probably it doesn't

ah, ok

I'll ask Tobias to be sure.

it has nothing to do with representation theory

ask Ted

well, if you say so.

@BalarkaSen: Sorry, this just doesn't excite me.

hmm. i think the det of that matrix represents det of the linear transformation that switches the vertices of your tetrahedra. i'm not entirely sure.

sure, @MikeMiller. if you don't want to think about it, don't bother.

@Soham representation theory is subtler than just fiddling with homomorphisms into matrix groups (of course, I don't know any more about it than you). a representation G --> GL_n(V) is just an everyday homomorphisms. how can it tell you something interesting about G? representation theory deals with that.

2:39 PM

you've done Artin toh.

he has character tables and stuff.

not all of artin.

?

I didn't do representation theory, probably won't for the time being.

ah.

user2004685

4:16 PM

Hi

I was wondering how Golden Proportion is related to prime numbers.

Any theories which relates Golden Proportion to the Prime Numbers or Prime Numbers Distribution?

4:30 PM

Hi, on p. 391 here (math.ucdavis.edu/~hunter/book/ch13.pdf), how does the author get the inequality 13.17 from the preceding two?

user2004685

It's simple

Multiply both sides of second inequality by -1

That will change the sign

Now add that to the first inequality

Since it's mod every term after addition will become positive

Sorry, but that's not correct. The t in the second inequality is between -1 and 0

user2004685

In first inequality, a<=b

In second one, c<=a

If I multiply the second inequality by -1 then

-a<=-c

Now, there are two terms on RHS on 1st inequality

a<=b - x

Also, -a<=-(c - y)

You'll see the second term of RHS of both these inequalities becomes same

i.e. -|u(x)|^p

The domain of t in the two inequalities is not the same

hi @samuel

4:45 PM

they're being pretty terse about that second inequality, which doesn't help the reader

Samuel Yusim

hey @mike

what's up

@PhilipHoskins: i mean, that first one is just a rearrangement of the convexity bound given just above. but the second one looks to involve a slightly different convexity argument

yes

using the range $-1<t<0$ in order to get $|u(x)-h(x)|^p$ instead

not much

have a bit of a headache but when it calms down it's reading time

4:49 PM

@mike I searched for the 3-manifolds appearing as branched covers over S^3 branched over a link fact. couldn't find it. probably will try again tonight.

Hallo.

Hello

hi @Pedro

@Balarka: projecteuclid.org/download/pdf_1/euclid.bams/1183536038

If you can figure out what the branched covers in question are you can reconstruct the proof.

What's new?

4:51 PM

@Semiclassical Do you see an argument to get the second inequality for t in (0,1)?

the guy refers to [5], which is apparently not written in english

no, i really don't

It's by Alexander. Of course it is.

oh, [1].

Oh, he's referencing the wrong paper. In any case, the paper I just linked sketches a proof. Some details are left to you.

4:53 PM

oh, i see. ok, let me read it.

oh, though

i don't think you need to. since the signs of the $t$'s in the two inequalities are different, the claims made are slightly different as well

@Pedro how's it going?

I'm hungry.

I also need to see that something is a quasi-isomorphism.

I thought I had it pinned down yesterday but apparently I didn't.

i don't even know what a quasi-isomorphism is.

Here's a surgery proof, apparently. I haven't read it. projecteuclid.org/download/pdf_1/euclid.bams/1183535815

4:54 PM

@Semiclassical You would need it for user2004685's argument to work

i mean, if the numerator of that different quotient is positive, then in the first inequality the LHS is positive. but then the RHS of the second inequality is negative

Something that is seemingly an isomorphism.

yes, you would. hence why i'm not going by that argument :)

@Semiclassical I know how the argument goes for the second inequality with -1<t<0, but I don't know how to get the final inequality by combining the two

i'm thinking more in terms of proof by cases

4:56 PM

Hmm, maybe it is that simple

i think it may be simpler than it looks: pick one set of signs for the numerator and denominator, and confirm that the inequality holds

there may be a better way, mind, but that brute force casework should give it

Then there would be four cases to check

hence why it's brute-force :/

oh, ok. google tells me it's a chain map which induces an isomorphism of homology groups.

Samuel Yusim

someone hype me up for cell/cw complexes. I started reading about them a couple days ago and I haven't picked my book up since then. I figure I'm letting myself down here but I need some extra motivation

4:58 PM

one might be able to get away with just doing $t\gtrless 0$, though

@BalarkaSen It's a morphism of complexes that induces an isomorphism in homology.

isn't morphism of chain complexes the same as chain map?

@SamuelYusim: Name your top 10 favorite spaces.

One can define what a weak equivalence in a model category. Apparently, the category of chain complexes admits a model structure, and these are the suitable weak equivalences.

right

4:59 PM

Something similar happens in the category of topological spaces.

doesn't interest me, as maps between spaces which induces isom on homology is a htpy equiv

:)