Mathematics - 2016-03-13 (page 1 of 2)

TheGreatDuck

01:06

are there any moderators or advanced users here. i have a question about the site.

MickLH

@TheGreatDuck Did you check the FAQ? Did you check Meta? Did you post on Meta?

TheGreatDuck

no because it's not worthy of such a thing

i was just going to ask why im all of sudden question banned even though my last question was a week ago

Mike Miller

Seems deserving of meta.

TheGreatDuck

ok

Mike Miller

Probably you got another downvote on a question, or one of yojr old questions was closed.

TheGreatDuck

01:13

nope

the only thing i did do was remove some old garbage

as it was months old with no answers

(and frankly i already had the information so it was useless to keep as it was a very specific question about a specific formula)

possible i got a downvote

it doesnt even make sense

the only ones with downvotes is one post with -1

and another with -8

out of 14

and the -1 one is listed as someone's favorite...

Mike Miller

04:56

Hi @PVAL.

PVAL

Hi

Mike Miller

How's it going

PVAL

Alright

I'm on spring break which is nice

But I have to probably work everyday to prepare my candidacy talk

Mike Miller

Yea I get that

We're on break too which means a week to get work done.

Aditya Dev

06:47

I was going through this question: math.stackexchange.com/questions/255/…

Isnt the series $\sum\frac{1}{r}$ when n tends to infinity the same as -ln(1-x) when x is 1?

DeNiSkA

? it's unclear

Aditya Dev

oops. sorry. nvm

Martin Sleziak

09:07

@TheGreatDuck I might be wrong, but my guess is that you have better chance that moderators notice your message if you post in the office. (Or maybe both here and in the office, to have improve your odds.)

user322355

12:02

well hello everyone!

I'm new here, made an account today and just managed to get a reputation of >20 points

Andrew Thompson

12:32

@TobiasKildetoft Regarding the Loday-thing: it was wrong in the text. Just got an expression that works.

Simply Me

12:45

Does anyone have any ideea how to prove $$\lVert A\rVert_\infty =\sup_{v\in C^m \setminus\{0\} }{\lVert A_v\rVert_\infty \over \lVert v\rVert_\infty} = \max_{i} \sum_j |a_{ij}|$$? It's regarding the 2-norm, proving it is equal to $sqrt(\rho(AA*)).

De Moivre

What is an example of a non-abelian group where some elements are their own inverses?

hhh

Is every complex number algebraic over reals?

SoumyoB

@DeMoivre an example could be the set of all $n \times n$ matrices with multiplication of matrices as the group operation? And the elements that are their own inverses would be all symmetric orthogonal matrices?

hhh

13:02

Yes true by 46, maths.manchester.ac.uk/~mprest/commalg6.pdf, understood :)

"Every vector space is isomorphic to the vector space corresponding to some field extension." -- can you think about any counterexample to show that this is false?

De Moivre

13:28

@SoumyoB Cheers. Do symmetric orthogonal matrices form a subgroup?

De Moivre

13:41

They do, in fact.

I'm looking for an example of a non-abelian group $G$ such that the subset $H = \left\{g^{-1} \in G: g = g^{-1} \right\}$ is not a subgroup.

SoumyoB

@DeMoivre I don't think symmetric orthogonal matrices form a subgroup. I'm looking for counter examples but I don't see how $A$ and $B$ being two symmetric orthogonal matrices implies that $AB$ would be symmetric, though of course it is orthogonal.

Mike Miller

14:06

@SoumyoB: Try to prove it (what does it mean in terms of the transpose?) and you'll see how to get a counterezample.

Tobias Kildetoft

@MikeMiller Depends on what you need to know about root systems.

De Moivre

I think I found my example in the Dihedral group.

Evinda

Hey @robjohn !!!
Let $u$ and $v$ be two distributions on $\mathbb{R}^n$, at least one of which has compact support. I have to show that $supp(u \ast v)=\supp u + supp v$.

But does the equality hold? Or does it only hold that $supp(u \ast v) \subset \supp u + supp v$ ?

Mike Miller

14:43

@Tobias I want to be able to parse this paper; the representation thwory starts in 2.1 (page 10) and continues for quite some ways

Mary Star

14:57

Hello @TobiasKildetoft !! I am looking again at the exericse:
Let the finite group $G$ act transitively on the set $\Omega$. Then the action of $G$ and on $\Omega\times\Omega$ is defined as follows $(a,b)\cdot x=(a\cdot x, b\cdot x)$.
Let $a\in \Omega$.
Show that the number of orbits of $G$ on $\Omega\times\Omega$ is equal to the number of orbits of $G_a$ on $\Omega$.

You said the following:

2 days ago, by Tobias Kildetoft

@MaryStar Given such two, they have the form $(x.g,y.g)$ and $(x.h,y.h)$ where $x.g = x.h = a$. This should tell you how to find an element in $G_a$ that sends $y.g$ to $y.h$

I haven't really understood why $x.g = x.h = a$. Could you explain it to me?

JC574

15:11

hey anyone here?

Mike Miller

sure

JC574

I'm looking at the projective variety/ planar algebraic curve $xy = 0$

trying to do some cohomology stuff

quite inexperienced at actually working things out

Mike Miller

I am not your man

JC574

dang

just singular cohomology?

SoumyoB

I wish I could help too but I'm also not very familiar with cohomology advanced topology

Mike Miller

15:15

Sure, but that would just be mayer-vietoris

JC574

yeah that's all i'm after

some

confirmation etc

so the intersection is just a point

and has cohomology $\mathbb{Z}$ in degree $0$ and $0$ elsewhere

?

Mike Miller

This is over $\Bbb C$?

JC574

just a sec :)

Mike Miller

Projective varieties are unlikely to have 0 cohomology...

JC574

yes over complex numbers

umm

ugh god I suck at this

OK

$xy = 0$ is the union of $x=0$ and $y=0$

oh derp

intersection is not a point is it

oh it is

$[0,0,1]$

Mike Miller

15:19

The intersection is a point, but you've misidentified the things you're intersecting.

JC574

I'm confused here

you mean it's not those $x= 0$ and $y = 0$

Mike Miller

It is. And what are those, topologically?

JC574

ohh

so those are circles topologically

right

no um

daaamn i'm rusty

aah

no in the real case they are circles

here they are spheres (?)

Mike Miller

yes... when you projectivize a line you get $\Bbb{CP}^1 = S^2$

JC574

yep sorry

it's just the wedge of two spheres?

Mike Miller

15:29

yes

JC574

i am being so stupid today

thanks

Gridley Quayle

Could someone please explain this answer to me? I'm not too sure what it means to put the 'basis vectors together' and generally don't quite get how it works:math.stackexchange.com/questions/546728/…

GGG

15:50

How do you prove that $\left({\mathbb{R}^{+}, \times}\right)$ is not cyclic?

Mike Miller

Tell me anything about cyclic groups.

GGG

Busted!

I know that if $H$ is cyclic, then there's an element $x$ in $H$ such that $H = \left\{x^{\ell}: \ell \in \mathbb{Z} \right\}$

Mike Miller

So, is there such an element in your group?

GGG

Nope.

I can't come up with such $x$.

s.harp

you could look at what kind of cardinality a cyclic group can maximally have

GGG

16:11

Is it not possible to do by contradiction? Assume that there's such element $x$, then derive a contradiction...

Mike Miller

Yes, it is. Have you tried?

Evinda

16:39

Hey @DanielFischer !!!
I have a set, let C, over $\mathbb{F}_2$ and I want to show that $C_0$ is a linear subspace of $C$.

I took $x,y \in C_0$ and $\lambda, \mu \in \mathbb{F}_2$ and I have shown that $\lambda x+ \mu y \in C_0$.

So does it now remain to show that the set is non-empty? Or do I have to show also something else?

Daniel Fischer

@Evinda Yes, if you have shown that $C_0$ is closed under linear combinations, it only remains to verify that it is nonempty.

Typically, the easiest way for that is to show $0 \in C_0$.

Evinda

Nice, thank you!!! @DanielFischer

Ted Shifrin

Guten Tag, @DanielF, @evinda.

Evinda

Guten Tag @TedShifrin

Daniel Fischer

Good day @Ted.

The Kind Cat

16:58

@DanielFischer Hello, I was trying to use something from an answer of yours but I don't know javascript

7

A: How to calculate coefficients of polynomial using Lagrange interpolation

Well, you can do it the naive way. Represent a polynomial by the array of its coefficients, the array [a_0,a_1,...,a_n] corresponding to a_0 + a_1*X + ... + a_n*X^n. I'm no good with JavaScript, so pseudocode will have to do: interpolation_polynomial(i,points) coefficients = [1/denominato...

Evinda

@DanielFischer We have that $||x|| \equiv 0 \mod 4$ and $y \not\equiv 0\mod 4$. How could we show that $||x+y|| \not\equiv 0 \mod 4$ ?

Or isn't it like that?

The Kind Cat

I need to calculate the LI interpolation polynomial for cos(x) with one thousand points

do you think that JS code is the way to go? @DanielFischer

Evinda

What does it have to hold so that $||x+y||=||x||+||y||$ ? @DanielFischer

Tobias Kildetoft

@MikeMiller At a quick glance they just recall the basics, so probably any source would do. Humphreys has a fine chapter on root systems for example. Though a source that also deals with Lie groups might be better as it seems some normalizations are done differently (they have an equality that in my world would be an integer on one side and not on the other).

L33ter

hi @TedShifrin

Ted Shifrin

17:06

hi Karim

L33ter

@TedShifrin I am learning to ask questions to both prof and his assisant now

a lot of stuff makes sense after doing that

he has like 3 graduate students so I should use that or else I am stupid.

Ted Shifrin

That's what professors are for, you know, as long as you're working hard.

L33ter

yeah

The Kind Cat

Yeah, some professors have the ability to make everything clear with a single drawing or phrase.

L33ter

yeah

Ted Shifrin

17:09

And some of us are horrid.

L33ter

my prof is very good actually

like my topology professor

I found actually like I am able to totally depend on myself with algebra, but topology I need professor's help.

to understand some stuff.

especially stuff that has geometrical intuition in it.

I haven't really taken geometry courses even in high school, so maybe that is why.

Ted Shifrin

I would like to believe that my students got more value from me than they would just from reading a book.

Shame on you for avoiding geometry.

L33ter

yeah

I like it actually much more than algebra, but this summer I am dedicating all my time to it and topology.

I will be working with my topology professor, so hopefully during that summer I can gain all the geometry machinary in my head.

I found also Massey book is quite good !

I am gonna use with hatcher's book.

The Kind Cat

I'm working with a proffesor which used to be a teacher for a class I took. His class wasn't bad but I could have learned most of it from a book. But now I'm working with him he has been awesome, he is really good at conveying his insight and years of experience, talking a couple of minutes gives me more clarity than a couple of hours staring into a blackboard.

Daniel Fischer

@TheKindCat With one thousand points, you should probably spend some time on thinking about minimising calculation errors, and optimisation. Can't you use a library that implements Lagrange interpolation? Using code "seriously tuned by someone else" is usually the best way.

@Evinda What is $\lVert x\rVert$?

L33ter

17:16

yeah I agree

The Kind Cat

All I know about programming is solving contest problems in c++

ok, I'll look for a library, thanks

Ted Shifrin

@DanielF: Someone actually posted an undergrad diff geo problem a few days ago that took me over an hour to solve. I was ashamed of myself. :) Easy, of course, once I realized how to approach it. Usually when problems are assigned, students know what they've just been taught :D

Sort of cute, though. If we have a convex plane curve, it boils down to the fact that any breadth (distance between parallel tangent lines) must be at least $2/M$, where $M$ is the maximum curvature.

Daniel Fischer

@TedShifrin Yup, that's often important knowledge.

Dinnertime, back later.

Evinda

@DanielFischer It's the support of x...

Ted Shifrin

Bis später. Guten Appetit.

DogAteMy !

How's that history paper going?

Akiva Weinberger

17:19

Good, mostly

I also was scheduled to tutor someone today but they never showed up

user19405892

hi

Ted Shifrin

That'll happen to you plenty of times in your life, DogAteMy.

hi @user19405892

user19405892

the rearrangement inequality for sequences ≤ is the same for < right?

Ted Shifrin

not sure what you're referring to

user19405892

on wikipedia here they seem to make a distinction

Rearrangement inequality

In mathematics, the rearrangement inequality states that for every choice of real numbers and every permutation of x1, . . ., xn. If the numbers are different, meaning that then the lower bound is attained only for the permutation which reverses the order, i.e. σ(i) = n − i + 1 for all i = 1, ..., n, and the upper bound is attained only for the identity, i.e. σ(i) = i for all i = 1, ..., n. Note that the rearrangement inequality makes no assumptions on the signs of the real numbers. == Applications == Many famous inequalities can be proved by the rearrangement inequality, such as the arithmetic...

Ted Shifrin

17:22

That even starts with a sentence that is not a sentence. Pretty pathetic.

Akiva Weinberger

?

Ted Shifrin

@user19405892: The actual article assumes $\le$ but makes a stronger statement if you know $<$.

user19405892

they say if the numbers are different but i don't see the reason to make a distinction since it works the same for ≤

Ted Shifrin

No, they assume $\le$. They make the stronger statement that if $<$ holds, then only one permutation gives you the lower bound.

Interesting: I've gotten through my whole mathematical life and never seen this before.

Gridley Quayle

It comes up in contest mathematics quite frequently

Ted Shifrin

17:27

I believe that. Interestingly, I cowrote dozens of (hard) math tournament exams and I don't recall our ever using it there, however.

Quite neat.

Gridley Quayle

A simple example question is proving : a^2+ b^2+c^2 ≥ ab + bc + ca.

Huy

@TedShifrin: can you help me out with this here?

Ted Shifrin

@Gridley: Yeah, that's basically geometric/arithmetic or Cauchy-Schwarz, sure.

GGG

@MikeMiller Assume there's such $x$ such that $(\mathbb{R}^{+}, \times) = \left\{x^{\ell}: \ell \in \mathbb{Z}\right\}.$ Then in particular we would have say $x^{n} = 2$ suggesting that $x = 2^{\frac{1}{n}}$. Trying this on $3$ we have $3 = 2^{\frac{m}{n}} \implies 3^n = 2^m$, which is impossible, right?

Ted Shifrin

@Huy: All that sort of stuff is far more easily done in the context of differential forms. I wrote a handout exercise for my physics students in my multivariable math class. Shall I forward it to you?

That looks good @GGG.

Huy

17:30

@TedShifrin: sure, but I'd still like to have a local expression for the divergence operator

Ted Shifrin

Sure. Just saying that's far easier to do in the context of forms. You just need $d$ and the star operator.

Huy

I know that one too

GGG

Cheers @TedShifrin

Ted Shifrin

So just do it, @Huy :P

Huy

well it leads to the same local expression, doesn't it?

and then I end up with the same question

Ted Shifrin

17:31

It should. So what's the precise question.

Huy

@TedShifrin: I plug in the metric of the upper half-plane model for $\mathbb{H}$ and don't really get what I expect

Gridley Quayle

Any tips on solving recurrences like T(n)=6T(n/2) + BigTheta [n^2 * log(n)] or T(n) = T[n/log_2 (n)] + BigTheta[n]? Master's theorem doesn't work and when I try and solve recursively I get a a divergent sum for the first one at least.

Ted Shifrin

@Huy: You're not being very helpful here. What do you want from me? Do you want me to calculate it myself?

Huy

@TedShifrin: no, I thougt maybe you know what the result should be and then I could use it to find out where my mistake is

Gridley Quayle

So for the first I get $$T(n)=6^k \cdot T(\frac{n}{2^k})+ \sum_{i=1}^{k-1}6^{i-1}\cdot(\frac{n}{2^i})^2log(\frac{n}{2^i})$$

Ted Shifrin

17:36

So the best thing to do is work with an orthonormal basis or cobasis, @Huy. If you start with $f e_1 + g e_2$, where $e_i$ are orthonormal, then I get that the divergence is $f_1+g_2-g$, where $df = f_1\omega_1+f_2\omega_2$. Here $\omega_1=dx/y$ and $\omega_2 = dy/y$ are the orthonormal coframe. @Huy

user67081

Hi all, I really need a push in the right direction in showing this trivial bound on the L2 error of a K-sparse approximation of a vector, math.stackexchange.com/questions/1695406/…

Gridley Quayle

where $$6^{i-1}\cdot(\frac{n}{2^i})^2=4n^2\cdot \left (\frac{3}{2}^i \right )$$

Ted Shifrin

@Huy: So $e_1=y\,\partial/\partial x$ and $e_2 = y\,\partial/\partial y$.

Gridley Quayle

I'm only trying to solve these asymptotically (big theta)

L33ter

hey @TedShifrin do you have like access to midterms and questions solved in algebraic topology ?

Ted Shifrin

17:41

No, Karim, not really.

I've never taught the course. I've just written several topology qualifying exams.

Huy

@TedShifrin: thanks, that's helpful

L33ter

could I have them ?

Ted Shifrin

You can download those and lots more at the UGA math department website. Go to graduate students and you'll find tons of old quals, Karim.

L33ter

oh ok

Ted Shifrin

When you want differential geometry or multivariable analysis type stuff, then I have things.

L33ter

17:44

that will be next things I study after I finish with algebraic topology this year

I want to do topology and differential geometry

The Kind Cat

@DanielFischer I found one in java! commons.apache.org/proper/commons-math/apidocs/org/apache/…

but I'm not sure how to use it

user67081

anyone have an idea on my question?

PVAL

17:59

@TedShifrin math.stackexchange.com/questions/1695556/… This user posted a nearly identical question, which I gave a big hint on and then they deleted it and then copy-pasted my hint as their own "thoughts".

L33ter

haha

Gridley Quayle

Any help with my problem? chat.stackexchange.com/transcript/message/28219168#28219168

user19405892

18:22

@TedShifrin So the lower and upper bounds for $<$ and $\leq$ are different?

Vrouvrou

I have a question: to prove that a sequence has no adherent value is it sufficient to prove that $(f_n)$ is not a Cauchy sequence

?

Daniel Fischer

18:33

@Evinda Err, weren't we talking of vector spaces over $\mathbb{F}_2$? What is $x$, what's the definition of the support of $x$ here?

Vrouvrou

@DanielFischer have you an idea about my question please

Evinda

The number of non-zero components of x. @DanielFischer

Daniel Fischer

@Vrouvrou What's the definition of an adherent value of a sequence? An accumulation point?

Vrouvrou

@DanielFischer yes

Daniel Fischer

@Evinda Aha. Then we have $\lVert x+y\rVert = \lVert x\rVert + \lVert y\rVert$ if and only if there is no component where both of $x$ and $y$ are nonzero.

Evinda

18:42

@DanielFischer This doesn't hold in my case... I will rethink about it and tell you...

Daniel Fischer

@Vrouvrou Then it's neither sufficient nor in general necessary that $(f_n)$ is not a Cauchy sequence. $f_n = (-1)^n + 2^{-n}$ has two accumulation points, but is not a Cauchy sequence. And a non-convergent Cauchy sequence (which exists if the space is not complete) has no accumulation point.

The Kind Cat

@DanielFischer I installed this

commons.apache.org/proper/commons-math/apidocs/org/apache/…

but I don't know how to ask it to compute the polynomial

here is my code

Vrouvrou

@DanielFischer let $E=\mathcal{C}([0,\pi],\mathbb{R})$ and $$d(f,g)=\sqrt{\int_0^{\pi} (f(x)-g(x))^2 dx}, ~\forall f,g\in E$$ how to prove that $(f_n)$ has no adherent values please where $f_n(x)=\sin(nx), n\in \mathbb{N}$

Daniel Fischer

@TheKindCat From the docs, I'd say PolynomialFunctionLagrangeForm pol = new PolynomialFunctionLagrangeForm(xValues, yValues); and then you'd query pol.value(z) to get the value at some point.

The Kind Cat

oh, thank you

Vrouvrou

18:45

@DanielFischer can you help me please

The Kind Cat

@DanielFischer that worked !!

Daniel Fischer

@TheKindCat But check that the implementation isn't too inaccurate or too slow before committing to use it.

@Vrouvrou To prove that sequence has no accumulation points, show that it doesn't have any Cauchy subsequence. Compute $d(f_n,f_m)$.

Vrouvrou

Why @DanielFischer a Cauchy subsequence please ?

Daniel Fischer

@Vrouvrou Because if a sequence in a metric space has an accumulation point, then it has a subsequence converging to that point. And that subsequence is, since convergent, a Cauchy sequence.

Michael Albanese

Hi.

Mike Miller

19:01

Hi @MichaelA. I'm trying to get my office mate to use SE.

Michael Albanese

Hi Mike.

I've had that discussion.

Tobias Kildetoft

@MikeMiller BTW, I was somewhat surprised that you needed a reference for root systems. I would have thought it was impossible to do anything with Lie groups without at some point learning about those

Mike Miller

@Tobias: I know very little about Lie groups. I basically only work with U(2), SU(2), and SO(3).

And I guess subgroups of those.

Tobias Kildetoft

@MikeMiller Ahh, I suppose restricting that far down might make it easier to not do the general theory

Mike Miller

Thanks for the reference. I should probably just read Humphreys. (I'm coping with that paper by restricting to one feop.)

group*

Tobias Kildetoft

19:04

@MikeMiller But note that there are probably some normalizations that are different in Humphreys

Mike Miller

@MichaelA: Yes, this inspired by the fact that you got your officemate to.

Tobias Kildetoft

at least related to the various inner products and the Coxeter numbers

Mike Miller

@Tobias: Like the $2\pi$s everywhere?

Tobias Kildetoft

(which partly comes from how one identifies the coroots with elements of the ambient vector space, which is not really necessary and not always desirable in other contexts)

@MikeMiller I didn't think that much about those $\pi$'s

Mike Miller

Oh so those weren't the problem

Anyway I don't even know what a root is so tjay seems like a good place to start

GGG

19:07

"An integer $k$ in $\mathbb{Z}/n\mathbb{Z}$ is a generator of $\mathbb{Z}/n\mathbb{Z}$ iff $\gcd(n, k) = 1$." Is this just for the multiplicative group, or for the additive group as well?

Tobias Kildetoft

Root systems are just collections of vectors with certain nice integrality properties with respect to some inner product

plus some restrictions to make sure you don't have too many vectors

@GGG Only for the additive

@GGG It clearly fails for the multiplicative, as otherwise $1$ would be a generator

Mike Miller

(I have the recollection that in that paper when they use SU(N) as their Lie group they just end up recovering sl(N) Khovanov-Rozansky for N>2, so SU(2) is somehow still the most exciting one

GGG

Cheers. @TobiasKildetoft

Tobias Kildetoft

@MikeMiller Root systems just somehow encode all the information one needs for a semisimple Lie algebra (or Lie group/algebraic group or quantum group or whatever)

Mike Miller

Sure, just somehow data involving the root system shows up in their dimension formulae

Tobias Kildetoft

19:12

@MikeMiller Anyway, I don't really know anything about the rest, though I have done some work on stuff that generalizes stuff that was previously used to do stuff related to Khovanov homology (or at least some stuff like it)

yep, lotta stuff

Michael Albanese

@MikeMiller: I have a question for you.

Frank Science

Hello, all.

Mike Miller

Yeah, this is some gauge theory related to Khovanov stuff. A related instanton homology was used to prove Khovanov detects the unknot, by proving it detects the unknot, and that there's a spectral sequence from Khovanov to it, or something like that.

@MichaelA: Oh dear.

Michael Albanese

...?

Frank Science

Suppose projective plane curve $C$ has only a nondegenerate double point.

Mike Miller

19:15

It's probably too hard.

Michael Albanese

I don't think so.

Mike Miller

Ok.

Frank Science

Is it true that after a small perturbation, $C'$ also has only a nondegenerate double point?

By perturbation, I mean, a perturbation of the defining equation of $C$.

Vrouvrou

@DanielFischer i calculate $d(f_p,f_q)=\sqrt{\pi}$

but this means that $(f_n)$ is not Cauchy

Michael Albanese

Suppose $X$ is a closed topological $n$-manifold, and $E \to X$ is a rank $k$ vector bundle with $k > n$. Is there an easy proof that $E$ admits a nowhere-zero section?

If $E$ and $X$ are smooth, there's an argument using transversality.

Tobias Kildetoft

19:19

@FrankScience what do you mean by "only"

Daniel Fischer

@Vrouvrou It means more than that. Much more. It means that $(f_n)$ has no Cauchy subsequence.

Frank Science

@MichaelAlbanese $X$ is a CW complex, and apply the canonical induction on the dimension of $X$.

Vrouvrou

@DanielFischer sorry i don't understand

how ?

Mike Miller

@FrankScience: I believe this should follow from compactness/projectivity and the fact that detwrminant is algebraic, and detwrminant of Jacobian or something like this tells you if you're smooth.

Michael Albanese

@FrankScience: What is "the canonical induction"?

Mike Miller

19:20

@Frank: If you can prove that topological 4-manifolds are all CW complexes you will probably get a nice job somewhere.

Frank Science

@MichaelAlbanese Induction on skeleton.

Daniel Fischer

@Vrouvrou Because if it had a Cauchy subsequence, for every $\varepsilon > 0$ there'd be $m \neq n$ with $d(f_n,f_m) < \varepsilon$.

Frank Science

@TobiasKildetoft After perturbation, it could be smooth, but if not, then at most a singular point, which is nondegenerate. Sorry, I was inaccurate.

Vrouvrou

but this is the definition of not Cauchy sequence @DanielFischer

Daniel Fischer

@Vrouvrou No. You're taking the wrong order of quantifiers.

Frank Science

19:24

@MikeMiller Okay, but I guess the classification of vector bundles only depends on the homotopy type of the base space?

Mike Miller

@MichaelA: Maybe pick a cover such that the elements of the cover and all their intersections are contractible, and then extend one by one over the whole thing.

Michael Albanese

That sounds plausible.

Vrouvrou

@DanielFischer we say that $(f_n)$ is Cauchy iff $\forall \varepsilon>0,\exists n_0\in \mathbb{N}, \forall p,q\in \mathbb{N}, p>q\geq n_o\Rightarrow d(f_p,f_q)<\varepsilon$

that's the deffinition i have

Mike Miller

@Frank: But when I change the homotopy type, mightn't I also change whether or not there's a nonvanishing section? I guess not, but since j mihjt be adding higher-dimensional cells, I'm worried.

Frank Science

@MikeMiller It's equivalent to the existence of a section of the sphere bundle?

Daniel Fischer

19:27

@Vrouvrou Right. And now negate that to get the definition of "not a Cauchy sequence".

Michael Albanese

I have never seen a proof of the fact I asked about, but it should be not too bad. I want to avoid obstruction theory, etc.

Mike Miller

@Frank: You're right, so it's well-defined under pullback. But let's take the following situation. M is a topological manifold and the only CW complez equicalent to it has higher dimensional cells.

Now why does it still have a section?

Vrouvrou

$\exists \varepsilon>0,\forall n_0\in \mathbb{N}, \existsl p,q\in \mathbb{N}, p>q\geq n_o~\text{and}~ d(f_p,f_q)\geq\varepsilon$ @DanielFischer

Daniel Fischer

@Vrouvrou Okay. And now, can you write down what "doesn't have a Cauchy subsequence" would be?

Vrouvrou

@DanielFischer i don't know

Daniel Fischer

19:33

@Vrouvrou Yes, that's a bit complicated. Perhaps we better let that be for the moment. But, looking at e.g. the sequence $(-1)^n$, can you see that "is not a Cauchy sequence" is much weaker than "doesn't have a Cauchy subsequence"?

Danu

Hi guys

Does anyone have some trivial examples of sheaves, to build intuition?

Tobias Kildetoft

@Danu What examples do you already know?

Danu

So far, the only ones I found reasonably intuitive were sheaves of continuous/holomorphic/meromorphic functions.

Tobias Kildetoft

Well, that is the general intuitive idea they are supposed to be a generalization of

Danu

But for instance the one of locally constant functions is less nice

Mike Miller

19:38

Why?

Danu

Less intuitive, for me.

So the way I think of a sheaf so far is like this: A collection of objects that are determined by the objects they contain (so what they restrict to)

Mike Miller

Why?

Tobias Kildetoft

@Danu Note that one half of the sheaf requirement (above being a presheaf I mean) will be automatic as long as you are looking at functions with some specified property

Daniel Fischer

@Danu When you say sheaf, do you think $p \colon S \to B$, or $F(U)$?

Danu

@TobiasKildetoft You mean that they're always determined by their restrictions?

Tobias Kildetoft

19:40

@Danu Right, so the only thing that could fail is the existence part

Danu

@DanielFischer I'm thinking about what the book I'm reading (Forster - Lectures on Riemann Surfaces) calls $\mathscr F(U)$

@TobiasKildetoft Is it fair to summarize the existence part as "even-more-local information always builds up to something larger"?

Tobias Kildetoft

@Danu Another important examples comes from affine schemes, but those might be a bit too technical for this purpose

@Danu Hmm, I would not have any idea what you meant by that

Mike Miller

@Danu: I still can't say anything unless you say why it's unintuitive. Is it because you think it should be just the "constant sheaf"?

Danu

@MikeMiller I can't really explain why it's less intuitive (yet) :\

@TobiasKildetoft That if you have "compatible" information from smaller subsets, you get something on the bigger subset (that restricts to the compatible chunks of info)

Tobias Kildetoft

@Danu Right, that is what the existence means

aras

19:43

Hi, I have a question. Is it ok English to say 'Let $n$ be an integer of the form $2m$', or would I have to say 'Let $n$ be an integer with the form $2m$' or something else? Thank you

Tobias Kildetoft

note that this is the only thing preventing you from just picking something trivial for the global sections without doing so for the rest.

Danu

So, for instance, why is $\mathscr F(\varnothing)$ not empty?

Tobias Kildetoft

@Danu What are these sheaves over?

@aras The latter. The former would be considered incorrect

Danu

@TobiasKildetoft Topological spaces

Daniel Fischer

@Danu Because it contains the empty function.

Tobias Kildetoft

19:44

@Danu I highly doubt it

aras

OK, so 'Let $n$ be an integer with the form $2m$'. Thank you!

Tobias Kildetoft

@Danu By "over" I mean what sort of sections

Danu

@TobiasKildetoft Eh?

Frank Science

sheaves of abelian groups

sheaves of sets

Danu

Right, abelian groups

Sorry

Tobias Kildetoft

19:45

@aras Woops, I missed which part you were referring to.

Danu

I didn't know that "over" was the terminology for that.

Tobias Kildetoft

@aras It should be "of", not with (I saw it as having two $m$'s in the first version)

Danu

But the author emphasizes several times that it doesn't really matter what objects one has

aras

@TobiasKildetoft Oh ok then. Thank you

Tobias Kildetoft

@Danu Yeah, that was not the right word (they are sheaves of X on Y)

@Danu The only reason it does not matter is that one could do it for any category. It really does matter for certain things (for example once they are abelian groups, it should be clear why it cannot be empty)

Danu

19:48

@TobiasKildetoft That last remark makes no sense to me :P

Tobias Kildetoft

But anyway, even for sheaves of sets, you should show that if the global sections are not the empty set, then no other section can be the empty set.

@Danu I meant why the sections of the empty set is not themselves the empty set

Danu

Can the presheave still be empty?

Tobias Kildetoft

You should also show that assigning the empty set to every open set does define a sheaf of sets on any topological space.

Frank Science

The original statement I wanted to check is whether, the set of projective plane curves with only one nondegenerate double point is open in the space of singular projective plane curves.

Tobias Kildetoft

@Danu No, it is not possible to have an empty section even for a presheaf, unless all sections are empty

Danu

19:50

What do you mean by an empty section?

Vrouvrou

@DanielFischer yes i see

Danu

You mean $\mathscr F(U)=\varnothing$?

Tobias Kildetoft

@Danu Yes, I refer to the value at an open set as the sections of that open set

@FrankScience So essentially that the double point cannot turn into another type of singularity without passing through a non-singular curve?

Frank Science

@TobiasKildetoft I don't think there's anything to do with passing through a nonsingular curve?

Daniel Fischer

@Vrouvrou Good. And if there is a $c > 0$ such that $d(f_n, f_m) \geqslant c$ for all $n \neq m$, then the sequence doesn't have any Cauchy subsequence. Hence it has no accumulation points.

Tobias Kildetoft

19:54

@FrankScience I mean when perturbing the curve

@FrankScience It was not meant as any particularly precise statement

Frank Science

@TobiasKildetoft Sorry, I ignored the condition that the degree of the curve is fixed, say, $d$.

So the space of curves of deg $d$ forms a projective space, and endowed with classical topology (I guess there is nothing different if we consider the Zariski topology)

Axoren

What's a good general topology book? Is there an industry standard text?

Frank Science

For general topology, you can try Dixmier's book. It's very thin.

Danu

@Axoren There are many books

I have a small list:

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