Mathematics - 2016-05-17 (page 2 of 3)

Balarka Sen

14:01

I can't read at the speed of light.

Juan Fran

haha. Well GAGA basically says the following

if you got a variety over C

then you obtain a complex analytic space

(in a functorial way)

and also a correspondence between sheaves

Balarka Sen

what's a complex analytic space?

Juan Fran

essentially a generalization of the notion of complex manifold

sort of "singular" complex manifold

in any case if your variety is nonsingular, its actually a complex manifold

and if it is projective, that complex manifold is compact

and conversly

Balarka Sen

Ah, ok. So this correspondence for smooth manifolds is bijective right? If you have smooth projective complex varieties.

No, that's wrong. Sorry.

Juan Fran

nope

actually the problem

whether any manifold Comes from a variety

is quite deep

it has been solved in a few cases, for instance Riemann surfaces

Balarka Sen

14:04

I was thinking of Chow's theorem. Complex submanifolds of CP^n are varieties.

Juan Fran

i think it is false in general

yeah by chow

if you want an example: let A be the C-Algebra C[X_1,...X_n]/(f_1,..f_m)

then the associated variety is Spec A

Balarka Sen

Well, yeah, I think there are a lot of obstructions. A smooth complex variety can never have free $\pi_1$. In particular $\Bbb Z$ can never appear as fundamental group. Also there's something with the betti numbers I don't remember.

Juan Fran

and the complex analytic space is the set of points p in C^^n with f_i(p)=0 for all i

well surely Z can appear as a fundamental group

I mean the circle

Balarka Sen

Yeah, sure. It's just the zero locus in C^n.

@JuanFran circle is not a complex algebraic variety.

Juan Fran

it is

well wait

Balarka Sen

14:08

Circle is a 1-dimensional manifold. A smooth complex algebraic variety is an even dimensional manifold.

Juan Fran

thats true

Balarka Sen

So obviously a circle can never be a smooth complex variety.

Juan Fran

i was thinking over the reals

Balarka Sen

Yeah, over the reals it's a different story.

Juan Fran

which is from where i got most intuition

Balarka Sen

14:09

Every smooth manifold appears as a real algebraic variety, I think.

This is Nash's theorem.

Juan Fran

a smooth variety can have free pi 1 though

i mean it can be 0

for a K3 surface for instance

but yeah

Balarka Sen

I am not considering 0 as free: there are lots of simply connected complex varieties, e.g., CP^1. It cannot be Z, Z*Z, ...

Juan Fran

it was implicit then

Balarka Sen

Yep.

Juan Fran

so why do you get that the fundamental group cannot be free?

Balarka Sen

14:12

I dunno the proof - something I would like to know at some point. I think you need Morse theory or something.

Note that it's false for singular varieties.

Juan Fran

4

Q: The fundamental group of a complex, quasi-affine variety

Can the fundamental group of a quasi-affine variety over $\mathbb{C}$ be a torsion group?

Balarka Sen

Oh, btw, when I say algebraic variety I mean projective varieties. Not quasiprojectives.

Otherwise C - 0 is a trivial example.

I think it's a big open problem what the fundamental group of quasiprojective varieties are, or so I heard.

So, about GAGA, you said there is a functorial correspondence between the smooth variety and it's underlying manifold. But I suppose there is more to that? Because that's obviously true.

Obliv

what's an easy way to solve for the order/multiplicative inverse of an element in modulo n?

i know to see if it isn't of infinite order or has a multiplicative inverse we can check if it is relatively prime to $n$

Juan Fran

yeah it also gives an equivalence of cateogries for the category of sheaves

isomorphisms in cohomology etc

Balarka Sen

Ah.

Juan Fran

14:17

but it is not even obvious

Balarka Sen

That is interesting.

Juan Fran

that the associated complex space being compact

implies the variety being priojective

etc

it is quite deep actually

Balarka Sen

Oh, that's a good question, whether having a compact underlying manifold implies the variety is projective. Is this true?

Juan Fran

yeah its true

but its nontrivial

Balarka Sen

That's pretty interesting.

Juan Fran

14:20

also the analytic space being smooth (i.e. being a smoooth manifold)

implies nonsingularity of the variety

Balarka Sen

Hmm, I am surprised that is nontrivial.

Juan Fran

i guess this is because you don't have studied schemes

and how nonsingularity is defined in Terms of regular local rings

it is intuitive though

Balarka Sen

I know how nonsingularity is defined in terms of local rings :)

Juan Fran

do you know Serres Theorem on regular local rings?

Balarka Sen

Nope, admittedly not.

I am curious though. Would you like to tell me or shall I google it?

Juan Fran

14:27

well it is a cohomological characterization of regular local rings

in any case the notion of smoothness in algebraic geometry

is much more complicated than one migh think at the beginning

Balarka Sen

I see.

Juan Fran

dependence on the ground field, etc

formal smoothness, geometric regularity etc

in the end this is all just commutative algebra

like most of AG

Balarka Sen

Right, I am not very comfortable with the commutative algebra side of story though.

Mostly because I usually have no intuition for commutative algebra, and most of my intuition comes from the geometry.

Juan Fran

well schemes actually give you intution for commutative algebra

even thought this might be unexpected

for instance modules over a ring

(not so unintuive notion though)

Balarka Sen

are like vector bundles over it's spec, yes

Juan Fran

14:31

should be thought of as quasi coherent sheaves over Spec

etc

well not really vector bundles

but almost :P

vectorbundles are projective modules

Balarka Sen

right, not yet familiar with quasicoherent sheaves.

@JuanFran yep

user 1618033

@Semiclassical are you done with the exams?

Semiclassical

yep.

Balarka Sen

I heard of this a few months back; was quite excited about this analogy.

Semiclassical

finished them yesterday at about noon

Juan Fran

14:32

you probably know the serre swan Theorem right?'

Balarka Sen

yeah

Juan Fran

expected

Semiclassical

and now my advisor has me running calculations as fast as i can :p

Juan Fran

it is quite similar to that

user 1618033

@Semiclassical And how were the tests? They did a good job?

:-)

Balarka Sen

14:33

It was something like vector bundles over a top. space X are in correspondence with projective C(X) modules, right?

C(X) being the ring of continuous functions on X.

Semiclassical

pretty good. they had a harder time with parts of the first problem, but since the parts of that were all largely independent (i.e. not knowing part (a) didn't relate to how you did on part (b)) the distribution was pretty well bell-shaped

Eric Stucky

@Obliv: I don't know of any easy method. But there is a neat trick for almost all elements if $n$ is prime

Juan Fran

yeah

do you know why projective modules are interesting

Balarka Sen

I suppose the way I think of modules over rings as vector bundles over spec of that ring with "varying fibers"

Semiclassical

by contrast, a lot of people aced the second problem

Juan Fran

14:35

form a purely linear Algebra point of view?

user 1618033

@Semiclassical I see.

Balarka Sen

well, they are a generalization of free modules.

Semiclassical

the main thing i had to grade down for there was the way they set up the equations: there's a few ways to set things up which give the same ultimate answer but for the wrong reasons

Balarka Sen

I can use them to build projective resolutions. I don't know of a more fundamental reason for caring about them.

Semiclassical

e.g. you get the same final answer if you had things come in from the LHS instead of the RHS, but the work won't be quite correct

Juan Fran

14:37

well it is perhaps a Little silly

Balarka Sen

I like silly. Please do tell.

user 1618033

@Semiclassical Oh, that is horrible when it happens while calculating integrals and series. It rarely happens but it happens. You do things wrongly, but get the right answer.

Juan Fran

but you most likely know the canonical isomorphisms Hom(V,W)=V^* Tensor W etc for vector spaces over a field

Semiclassical

yeah. here it's due to the way the physics works out

Balarka Sen

Mhm.

Juan Fran

14:38

These are "nontrivial" isomorphisms in the sense that they do not holds for all modules over any rings

Balarka Sen

I agree.

Juan Fran

actually they are true for finite type projective modules

and many more "Nontrivial" isomorphisms

are also true for these Kinds of modules

Obliv

@EricStucky I actually remembered that you can use the euclidean algorithm and re-arrange the terms to produce a linear combination and that gives u the multiplicative inverse

thanks tho

Semiclassical

how i ended up organizing things was to see if they got the right answer, and then check if it was for the right reasons.

Juan Fran

this is why, among other Things, they are useful

user 1618033

14:39

@Semiclassical btw, what is the most difficult concept you met so far in physics, very hard to comprehend?

Balarka Sen

That makes sense. That especially reminds me of the splitting lemma holding when the end term is a projective module.

Juan Fran

perhaps this doesnt sound useful

but i needed it in differential geometry at some point

Balarka Sen

sounds interesting

user 1618033

@Semiclassical I refer to what is taught in the uni classes.

Semiclassical

as a whole, probably quantum field theory

but QFT is more of a grad level course

Juan Fran

14:41

i mean you know that the sections of vector bundles are C^infty(base manifold)-modules

Semiclassical

partly that's hard for me because i don't do a lot with it, though

Balarka Sen

right

Semiclassical

with regards to the math i've seen while doing research, integrable systems encapsulates it fairly well

looooots of crazy stuff there

Obliv

wait nevermind that's to get the multiplicative inverse. doesn't get the order of the element in a multiplicative group modulo n @EricStucky

Juan Fran

something basic is then that S(E Tensor F)=S(E) Tensor S(F)=Hom(S(E),S(F))

user 1618033

14:42

@Semiclassical It sounds attractive then, what is very crazy is also very nice. :-)

Semiclassical

haha

Juan Fran

where S(E) is the sections of the vector bundle E

Obliv

i just have to check the powers of the elements and hope eventually it gets $\bar{1}$

Semiclassical

it gets hard fast

Juan Fran

this follows from what i said

it is useful for interpreting Tensor fields

Semiclassical

14:43

there's also rather esoteric pieces of math which I know of more than understand

Juan Fran

do you know why S(E) is a projective C infintiy module?

Semiclassical

say, K-theory in the context of condensed matter theory.

Juan Fran

balarka?

Semiclassical

it's hard to pick a 'most confusing piece of math' out of all that stuff

Balarka Sen

No, I am afraid not.

user 1618033

14:44

@Semiclassical It's said that math is more about accepting than about understanding. :-)

Semiclassical

because there's a lot of things i've found that i want to know more about, but which are above my head

Juan Fran

well

Semiclassical

sometimes. problem is, using a piece of math without understanding it is always a bit dangerous

Juan Fran

you probably know that if E is a vector bundle

user 1618033

Did John von Neumann say that? I think so (approximately).

Juan Fran

14:44

then there is a vector bundle F

such that E direct sum F is a trivial vecror bundle

now take sections of this direct sum

this shows projectivity

user 1618033

@Semiclassical I know, but this wasn't precisely the point. Sometimes it happens you learn a new concept that requires more time to maturely understand it. So, initially you accept it, and a deep understanding of it comes with time.

Ivan

Hello

I got a question

Juan Fran

in any case i gotta go now

cya

Balarka Sen

Bubye. Nice talking to you.

Hope to see you around more.

Ivan

I have to proof that if a,b,c are sides of triangle, this: https://www.symbolab.com/solver/step-by-step/b%5E%7B2%7Dx%5E%7B2%7D%2B%5Cleft(b%5E%7B2%7D%2Bc%5E%7B2%7D-a%5E%7B2%7D%5Cright)x%2Bc%5E%7B2%7D%3D0/?origin=button
has no real roots? Please help me:(

Semiclassical

14:48

yeah, that's definitely the case

Balarka Sen

I should start working bit by bit now that my non-math works are getting done.

user 1618033

@Semiclassical It's also nice to see that during the time if you return to the same math problem you see it with other eyes, more profoundly by the means of experience.

You simply want to finish it in a more clever way.

Semiclassical

hah, yeah. sometimes your eyes just don't see it initially

Ivan

help?

Eric Stucky

Ivan: So you need to show that $a^4+b^4+c^4-2a^2b^2-2b^2c^2-2a^2c^2<0$.

Semiclassical

14:50

or you don't appreciate what a particular paper is telling you e.g. think it's more obscure than it actually is

Ivan

@EricStucky is that all?

Eric Stucky

You can see in the link you gave me

user 1618033

@Semiclassical Precisely!

Eric Stucky

That if this is true, the solutions are not real.

Semiclassical

i've had a bit of that lately, albeit in a topic to which i'm not likely to contribute anything deep

but it is satisfying, and opens up new conversations

user 1618033

14:52

Indeed. Happily you have access by uni to many papers I suppose ...

Ivan

@EricStucky do I need to do anything after that you showed me?

Semiclassical

it helps, yeah, though a lot of it is from arxiv preprints which doesn't require uni access

Eric Stucky

Ivan, if you do not see why you don't need to do anything else, then you need to prove that what I said is sufficient.

user 1618033

@Semiclassical Yeah, there are many nice papers on arxiv, and good the access is free (so far).

Semiclassical

yeah. i would hope it stays that way

user 1618033

14:55

@Semiclassical btw, I saw some nice integrals on MO

Evinda

Hello @robjohn
If $u \in W^{3,p}(\mathbb{R}^{+})$ I want to construct the reflective extension $Eu$ of $u$ in $\mathbb{R}$ such that $Eu \in W^{3,p}(\mathbb{R})$.

What $Eu$ could we pick?

user 1618033

@Semiclassical mathoverflow.net/questions/200880/…

It's cool.

Obliv

can a multiplicative group $(\mathbb{Z}/12\mathbb{Z})^{\times}$ have negative elements?

Semiclassical

hmm. do the two answers agree? the first one seems to say that the integral goes to zero

Eric Stucky

Obliv, what do you mean by that?

Semiclassical

14:57

whereas the second purports to verify the identity

Obliv

an exercises tells me to find the orders of the following elements of the multiplicative group $(\mathbb{Z}/12\mathbb{Z})^{\times}:\bar{1},\bar{-1},\bar{5},\bar{7},\bar{-7}, \bar{13}$

Eric Stucky

No, the second one also gets zero.

Obliv

oh I misread

Eric Stucky

It says $I=J-\pi^3/6$.

Semiclassical

bah, you're right @EricStucky

i was careless and missed the switch from I to J

user 1618033

14:59

@Semiclassical That paper is about disproving a conjecture by Z. Silagadze saying the answer is about $-\pi^3/12$.

Obliv

it's the multiplicative inverses not the elements in integer modulo 12. the inverses can be negative or larger than 12

Eric Stucky

Yeah I was confused by this the last time she linked it.

Yep, obliv :)

Semiclassical

right

what i wonder is why ZS thought it was that in the first place

the reference in his other question to Landau-Zener sounds neat :)

user 1618033

@Semiclassical Not knowing Landau–Zener formula.

Landau–Zener formula

The Landau–Zener formula is an analytic solution to the equations of motion governing the transition dynamics of a 2-level quantum mechanical system, with a time-dependent Hamiltonian varying such that the energy separation of the two states is a linear function of time. The formula, giving the probability of a diabatic (not adiabatic) transition between the two energy states, was published separately by Lev Landau, Clarence Zener, Ernst Stueckelberg, and Ettore Majorana, in 1932. If the system starts, in the infinite past, in the lower energy eigenstate, we wish to calculate the probability of...

Semiclassical

nor do I, off the top of my head, but i know i've run into it before

user 1618033

15:06

@Semiclassical I said it before, some of integrals coming from physics are damn hard, and they become immediately famous in papers.

Semiclassical

yep

user 1618033

@Semiclassical Like Ahmed integral.

arxiv.org/pdf/1411.5169v2.pdf

Nuclear Physics Division, Bhabha Atomic Research Centre, Mumbai 4
00 085, India

(somewhere I also found details about the way the integral appeared in some stuff related to physics)

Ivan

How to find for what values of a -> x^2-ax+5 = 0 and x^2+5x-a=0 have same roots?

Semiclassical

If they have the same roots, then they both share the same factorization $(x-x_1)(x-x_2)$.

(technically there's the possibility of an overall constant $C$ in front, but $C=1$ since both have leading term $x^2$)

user 1618033

$\frac{A_1}{A_2}=\frac{B_1}{B_2}=\frac{C_1}{C_2}$

This is the answer ^^^ (those are the coefficients)

$a=-5$

Obliv

15:25

anyone know a good way to review sections in a textbook? Is it a good idea to go through all the proofs/exercises again or just the concepts? Also how often should one review material (every 1,2,.. chapters, etc)

user 1618033

By the answer I meant the condition. Well, I suppose everybody knows that.

Anyway.

@Obliv I review almost every day some stuff in my book and it seems I always find something to change, modify.

(all in the idea of improvement)

Some would probably commit suicide if asked to review my book. :-)

Obliv

lol

user 1618033

Have to finish an article. Back a bit later.

Obliv

wish exercises would just incorporate previous material more often in some books so I don't have to review. then again it is hard to think of ways to incorporate previous material all the time

I don't get this proof crazyproject.wordpress.com/2010/01/04/… can someone explain what is meant by this?

wouldn't it make sense to just say there must exist inverses for $xy$ in order for $xy \in G$

user929304

15:46

Hello everyone

Obliv

hi

user929304

tiny question: if I want to ask a question about network flows (in the context of graphs), would math.SE be a good home for that?

Sebgr

15:59

Hi!

If I have a covering map, how is the cardinality of a fiber globally assured?

I mean, Hatcher says it is a local property, which I agree

And the space should be connected for it to be "global"

Balarka Sen

If the base space is connected, cardinality of fiber is indeed constant, globally.

Sebgr

Is this so? I still don't have much experience with examples of coverings. Can I say, without any extra assumptions, that I have an "n-sheeted covering map"?

Balarka Sen

Yes.

You should prove it.

Sebgr

Ok, I see. actually, I'm trying to prove that any two-sheeted covering map has a non- trivial automorphism

So, I wanted to make sure I understood what does it imply to be "two-sheeted"

Balarka Sen

Ok.

Sebgr

16:23

@BalarkaSen For example, the usual covering of the circle has countably infinite "sheets", yet the circle is connected.

@BalarkaSen Is "constant" taken in the sense that it never becomes finite?

Balarka Sen

We're not talking about cardinality in the set theory sense. Infinite things become messy. I mean if cardinality of some point is finite, base is connected, then cardinality of every point have the same cardinality.

Also, yes, it is true that cardinality can never be finite if some preimage is infinite under a covering map. This just follows from the claim above.

Sebgr

Thanks @BalarkaSen

Balarka Sen

I think it's worth knowing a proof of this statement though.

Sebgr

Yes, I'm thinking about a proof right now

Sebgr

16:48

@BalarkaSen What about using the fact that any continuous map $\Bbb{Z}\rightarrow B$ must be constant? and then applying that to the cardinality for each point?

Balarka Sen

1) you've got the arrows wrong, you mean $B \to \Bbb Z$. 2) where did $\Bbb Z$ come from? the cardinality is a finite set

Mike Miller

@BalarkaSen So I don't know how to do the claim I said about deleting a measure zero set and getting a chart in full generality. I can do it for compact manifolds, or at least manifolds that aren't badly noncompact. But eh, that's good enough for me.

Sebgr

@BalarkaSen You're right, I wanted to say that the map that takes the cardinality of the fiber at $b\in B$ to $b$ must be a constant map, because $B$ is connected.

Balarka Sen

Yeah, I figured out later that it was believable. I mean all the examples I can think of has this property.

How did you prove this for compact manifolds?

Mike Miller

Every compact manifold outside dimension 4 has a CW decomposition with one n-cell. Delete the (n-1)-skeleton.

Sebgr

16:53

@BalarkaSen I'll give it more thought.

Mike Miller

I guess I can't literally prove that it's measure zero since I don't know the Lipschitz structure but c'mon.

Balarka Sen

Oh, I did not know this fact about CW decomposition. Then I could have done it too :(

I was just going to say if you have a CW decomposition then it should hold.

Mike Miller

Mm, actually my "but cmon" is a bit unfair. There are Jordan curves with positive measure. What if our homeomorphism to a CW complex does the same for our (n-1)-skeleton?

Balarka Sen

OK, good point.

Mike Miller

Well, I believe it's true. I think I wrote down a proof once after being skeptical of something in a class. But I've forgotten the details.

Balarka Sen

16:55

Those horrible Osgood curve or Oswald curves or whatever those are

@MikeMiller Alright, I am not pushing you to give a proof if you're not interested in it that much :) I am not either. It seems rather doable.

In the sense that it's not very surprising.

Mike Miller

though the proof I gave was definitely in the context of Riemannian manifolds, since I have never ever thought much about Lipschitz or qco manifolds

Balarka Sen

What's a qco manifold, if I may ask?

Mike Miller

quasiconformal

this is weaker than Lipschitz, I think my officemate told me. Donaldson and Sullivan proved that Donaldson's instanton theory extends all the way to quasiconformal manifolds, hence proving 1) not every topological 4-manifold admits a qco structure 2) homeomorphic qco 4-manifolds needn't be qco-homeomorphic

Balarka Sen

sounds cool, but probably out of my reach right now

thanks.

Mike Miller

it's out of my reach, and my officemate's

he apparently tried to read the paper and got frustrated with it and gave up

he's a qco person

maybe we'll try to fight through it together at some point, since i'm familiar with the instanton theory

Balarka Sen

17:09

Hmm, oh wait, so 1) tells me in particular that not every topological 4-fold admits a Lipschitz structure?

Mike Miller

yes

Balarka Sen

cool. is there an explicit example?

Mike Miller

there's no such thing as "explicit example" of non-smoothable topological 4-manifold. you invoke Freedman's theory, which says that there are exactly one or two topological 4-manifolds with given intersection form; one if the form is odd (that is, $q(x) \neq 0$ for some $x$, working here over $\Bbb Z/2$), two if the form is even

Donaldson's original theorem was that smooth 4-manifolds with definite intersection form have diagonal intersection form. this says that qco 4-manifolds that have definite intersection form have diagonal intersection form

Sebgr

@BalarkaSen It's getting kind of messy, am I at least going in the right direction by considering a continuous function $B\rightarrow \{ 0,1 \}$?

Mike Miller

so you use Freedman's machine to say that there is a topological 4-manifold with intersection form $E_8$

Balarka Sen

17:12

You're thinking too hard, @Sebgr.

@MikeMiller ah, ok.

Mike Miller

similarly Freedman's machine implies that smooth 4-manifolds with the same intersection form are homeomorphic. there's no way of doing this "by hand", it's always through Freedman's highly inexplicit process

Balarka Sen

I see. So since it's hopeless to "see" a nonsmoothable 4-manifold, you're saying in particular it should be hopeless to "see" a nonLipschitzeable (ugh) 4-manifold?

Mike Miller

it's not that it should be hopeless... the class of manifolds we can show are non-smoothable is the same class of manifolds we can show to be non-qcoable

a (known) example of one is an example of the other

of course, it could be true that qco =/= smooth

Balarka Sen

ah.

interesting stuff.

Mike Miller

I think there's some people who believe that they shouldn't be the same. But nobody's actually made serious progress on it. I don't know if there's anybody who really still thinks much about it.

So what you're calling "high school" is two years? And to get into a good one of those is tough?

Sebgr

17:43

It's amazing how @BalarkaSen is so young yet he knows already so much math!

Sebgr

17:55

@BalarkaSen So, this is what i've got: If for a $b \in B$, a trivializing set $U \subset B$ has a fiber of cardinality $n$, then any other set with a different cardinality must not intersect $U$. If we try to cover $B$ with trivializing sets with fibers of each cardinality we would eventually cover $B$, but since it is connected, tht's impossible. So, there's only one trivializing open set: $B$ itself.

Semiclassical

i like this problem: math.stackexchange.com/q/1789282/137524

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