Mathematics - 2015-06-03 (page 2 of 4)

Soham Chowdhury

07:00

nope, I was going to ask you a question when anon when offline, and then you did too

@iwriteonbananas evil grin

what are you up to?

iwriteonbananas

@SohamChowdhury trying to understand why this commutes: i.imgur.com/Qp3ypvc.png

and proving exactness

Balarka Sen

what commutes?

Soham Chowdhury

as in, faar above my pay grade

and the man is here.

iwriteonbananas

everything in that diagram apparently

Balarka Sen

oh, the diagram took ages to load

iwriteonbananas

07:03

i mean, the bottom two rows are just LES of the pair (X,A) and (P,P) resp.

P is the one point space btw

Balarka Sen

chuck out everything above the two rows. the resulting diagram commutes by naturality

so all you have to do is to verify this for the hat-like diagram above

iwriteonbananas

yeah, if we ignore the top two terms, everything commutes by naturality. so far so good

what exactly are the maps $H_1(X,A) \to \tilde{H_0}(A)$ and $\tilde{H_0}(A)\to H_0(A)$?

Balarka Sen

the little square inside the hat containing $H_0(A), H_0(X), \tilde{H_0}(A), \tilde{H_0}(X)$ also commutes.

iwriteonbananas

why does that square commute?

yeah, i guess so

Balarka Sen

recall what the maps are

you can verify this by hand

@iwriteonbananas the first map is the snake map in the reduced long exact sequence, apparently

iwriteonbananas

07:07

well, i think that's sort of my problem. im not entirely sure what the maps are

i dont know the snake lemma

and i dont know the reduced long exact sequence

Balarka Sen

but you do know the reduced long exact sequence, right?

iwriteonbananas

no, im trying to prove its existence

that's why im looking at this diagram

im trying to show that the reduced LES is exact

Balarka Sen

oh. but anyway, $H_1(X, A) \to \tilde{H_0}(A)$ is the map obtained from taking a class $[z]$ and mapping this to $[\partial z]$, the homology class in $H_0(A)$ of it's boundary

iwriteonbananas

ok, yeah. that's what i figured

Balarka Sen

@iwriteonbananas ? why d'you need a buckload of commutative diagrams for that?

iwriteonbananas

07:09

how do you suggest i prove it?

i mean, i guess we dont really need the bottom row of that diagram to prove exactness

Balarka Sen

write down the short exact sequence of triples, apply $H_\bullet$ functors and extend to long exact sequence by snake lemma, and then set $(X, A, B) = (X, A, pt)$

you can check long exactness by hand

iwriteonbananas

short exact sequence of triples?

Balarka Sen

ack, you're not working with singular homology, are you?

oh noes.

iwriteonbananas

:( nope

i know LES for triples

wait, cant we just use that

with $pt \subset A \subset X$

Balarka Sen

LES for triples holds for arbitrary homology?

iwriteonbananas

07:14

yes

Balarka Sen

@iwriteonbananas that's what I told above

@iwriteonbananas ok, then, you're lucky this time

iwriteonbananas

hahah

$H_i(X, pt) \cong \tilde{H_i}(X)$, right?

Balarka Sen

mhm

iwriteonbananas

perfect

Exam in 5 days

Can someone give me some simple topological objects(for product spaces) like $S^1\times I$ is a cylninder and $S^1 \times S^1$ is a torus

iwriteonbananas

07:17

what do you mean

Balarka Sen

lots of product spaces out there

Exam in 5 days

I just wanted some more to think about

$I\times I$ is a cube I suppose

Balarka Sen

it's a square

Exam in 5 days

Oh indeed

$I^2 \times I$

What is $S^1\times S^1\times S^1$?

Balarka Sen

it's the 3-torus

Exam in 5 days

07:19

As in genus 3?

Balarka Sen

hard to visualize, so not good for thinking-purpose

@Examin5days no, it can't be embedded in $\Bbb R^3$.

it's something in $\Bbb R^4$

iwriteonbananas

(orientable) genus 3 surface refers to the connected sum of 3 tori

Exam in 5 days

Well can I have some simple examples now that you probably know what I mean

Are there other important letters to know for one thing

I know intervals $I$ and circles $S^1$ spheres $S^2$ and so on

iwriteonbananas

you should know what $D^n$ is, too

Balarka Sen

in Homotopy Theory, 8 hours ago, by Tyler Lawson

sometimes it's just a bunch of long exact sequences that get together and have a party

hahaha

Exam in 5 days

07:22

@iwriteonbananas What is $D^n$?

iwriteonbananas

@BalarkaSen loool

@Examin5days it's the disk bounded by the (n-1) sphere

Exam in 5 days

Hmmm

So $D^n$ is the disk bounded by $S^n$?

iwriteonbananas

e.g. $D^2$ are all the points of $\mathbb{R}^2$ that have norm $\leq 1$

@Examin5days no $S^{n-1}$

Exam in 5 days

Oh sorry I thought the n-1 sphere is $S^n$ since a sphere is $s^2$

@iwriteonbananas Standard euclidean norm I imagine?

iwriteonbananas

sure

Balarka Sen

07:28

it seems that spectral sequences on galois cohomology are serious tools in class field theory.

ok, i got learn this stuff thoroughly.

iwriteonbananas

i dont know a lot of those words. where are you learning this from?

Balarka Sen

what, spectral sequences?

iwriteonbananas

yeah

Balarka Sen

picked some up from chow's article, but now learning from hatcher's "spectral sequences in algebraic topology"

iwriteonbananas

cool

Exam in 5 days

07:31

I have heard Hatcher is a bad text, and to use an alternative text, one sec, I will try to remember what the alternative text was called

Balarka Sen

what you heard is crap

and anyway, I am not talking about Hatcher's alg top book

Exam in 5 days

It was William Fulton's text

Balarka Sen

Fulton had an algebraic topology text?

Exam in 5 days

A first course in Algebraic Topology

Apparently it is a much better text

Balarka Sen

weird. but anyway, hatcher is a good book.

iwriteonbananas

07:32

i doubt it, hatcher is amazing

Exam in 5 days

(based on anecdotal evidence of a very intelligent friend)

Balarka Sen

don't judge a book by it's cover

ok, i gotta go

Exam in 5 days

I will borrow out both of them and compare them directly when I have time.

iwriteonbananas

study point set topology before you study algebraic topology

Exam in 5 days

Point set? This refers to metric spaces? I have studied them decently

Or this refers to general topology, like $T^n$(or $T_n$ if you like) spaces?

iwriteonbananas

07:37

General topology

In mathematics, general topology is the branch of topology that deals with the basic set-theoretic definitions and constructions used in topology. It is the foundation of most other branches of topology, including differential topology, geometric topology, and algebraic topology. Another name for general topology is point-set topology. The fundamental concepts in point-set topology are continuity, compactness, and connectedness: Continuous functions, intuitively, take nearby points to nearby points. Compact sets are those that can be covered by finitely many sets of arbitrarily small size. Connected...

i mean all of those things

Exam in 5 days

Yes all of them are fine except perhaps the countability axioms. I am restudying product spaces and the Kolmogorov classification now actually.

Soham Chowdhury

@iwriteonbananas I've jumped in at the deep end. :)

Bredon, specifically. Slowly working through the intro point-set chapter.

iwriteonbananas

@SohamChowdhury so, what are you studying?

cool, bredon is decent

Soham Chowdhury

Doing all the exercises

iwriteonbananas

pats on shoulder

Soham Chowdhury

07:45

Actually, the point-set chapter is short, which was the draw. :P

iwriteonbananas

'atta boy

Soham Chowdhury

Haha.

iwriteonbananas

yeah, point set topology can be hairy

Soham Chowdhury

Hairy?

iwriteonbananas

yes, interpret that as you may

Soham Chowdhury

07:46

uhmm.

Exam in 5 days

He is referring to the hairy ball theorem.

iwriteonbananas

hahah

Soham Chowdhury

this way?

damn.

Exam in 5 days

To be fair, you linked it :P.

iwriteonbananas

hairy ball theorem is neat

Soham Chowdhury

07:48

hm. have you done much from Bredon?

Exam in 5 days

Why choose Bredon? Have you done comparisons with other similar textbooks?

iwriteonbananas

not at all, only parts of the axiomatic homology section

i only use it because my professor uses parts of it

Soham Chowdhury

well, I didn't like Munkres very much. and it seems like the point-set intro is quite short and covers all the alg-top prerequisites.

(aaaand Balarka is here)

will you be online now, @Balarka?

Balarka Sen

spectral sequences are so awesome that i can't study them calmly.

Exam in 5 days

@SohamChowdhury You didn't like Munkres? Nor do I - but Armstrong also seems very verbose.

Balarka Sen

07:55

@iwriteonbananas have you studied cellular homology yet?

iwriteonbananas

@BalarkaSen no, not really

Balarka Sen

you know what they are?

iwriteonbananas

i will once i have time

nah

but im looking at that huge diagram in hatcher now

Balarka Sen

well, recall that we discussed previously how $H_n(X^{(n)}, X^{(n-1)})$ is the free abelian group generated by the $n$-cells in $X$.

iwriteonbananas

yes

Balarka Sen

08:01

cellular homology tries to arrange these in a chain complex and computes the homology of that.

iwriteonbananas

compute the homology of what exactly?

Balarka Sen

for example, $C_n(X)$ are free abelian groups generated by the simplices in a simplicial complex $X$, and $H_\bullet(\mathcal{C})$ is a good topological invariant for the simplicial complex. just like that, $H_n(X^n, X^{n-1})$ are free abgrps generated by cells in a cell complex $X$, and $H_\bullet$ of the chain complex of $H_p(X^p, X^{p-1})$ also gives us a topological invariant for our space

the interesting bit is that we don't need a simplicial structure anymore : just knowing a good cell structure (which is generally by far easier) does the trick

iwriteonbananas

ok, i see

Balarka Sen

@iwriteonbananas of the chain complex $\cdots \to H_p(X^p, X^{p-1}) \to H_{p-1}(X^{p-1}, X^{p-2}) \to \cdots$

iwriteonbananas

yeah, i get it now

Balarka Sen

08:06

of course, you have to prove that such a thing exists.

but all of that is in hatcher

iwriteonbananas

yeah

Balarka Sen

so cellular homology is like "homology of homologies"

spectral sequences goes one step higher and does "homology of homology of homologies" and so on and so forth

iwriteonbananas

yeah, that's what confused me a second ago

what the hell lol

Balarka Sen

it's super-cool stuff

spectral sequences are like books, where the first page is about homology, second page is about homology of homologies, third is about homology of homology of homolgies and so forth.

and that's what makes them good tools : you have pages of informations about something instead of just one big nasty long exact sequence

but of course, this is what i understand, and i haven't even read about serre spectral sequences.

the theory so far is intellectually satisfying

Soham Chowdhury

sigh

my second time around, i understand free groups okay

Exam in 5 days

08:13

@SohamChowdhury Free groups just refers to strings of letters concatenated and reduced?

Soham Chowdhury

just like number theory is the study of terminating sequences of letters from the alphabet $\{0,\cdots,9\}$

:P

Exam in 5 days

Is there pedagogical value in learning free groups early?

iwriteonbananas

@BalarkaSen that's awesome, i gotta study those some day

Exam in 5 days

@SohamChowdhury :P.

Soham Chowdhury

@Examin5days yeah; later, the author uses them to talk about subgroups

Agawa001

08:56

i had a question yesterday wthout any lead :( , are these expressions equivalent?

$\sum_{i=1} ^\inf {(\frac{1}{2})^i (a+\frac{1}{2}+...+(\frac{1}{2})^i)^2}$ and $\int_a^{a+1} x^2 dx$

oh wait

$\sum_{i=0} ^\inf {(\frac{1}{2})^i (a+1+\frac{1}{2}+...+(\frac{1}{2})^i)^2}$

@Emrakul

Alec Teal

09:26

@BalarkaSen you never answered my questions

user61230

@Agawa I'm not totally sure what you're trying to do. Are you trying to create a Riemann sum expression for the integral of $x^2$?

Andrew Thompson

09:41

What does "adapted charts" mean?

Alec Teal

What context?

Andrew Thompson

Jet bundle

In differential geometry, the jet bundle is a certain construction that makes a new smooth fiber bundle out of a given smooth fiber bundle. It makes it possible to write differential equations on sections of a fiber bundle in an invariant form. Jets may also be seen as the coordinate free versions of Taylor expansions. Historically, jet bundles are attributed to Ehresmann, and were an advance on the method (prolongation) of Élie Cartan, of dealing geometrically with higher derivatives, by imposing differential form conditions on newly introduced formal variables. Jet bundles are sometimes called...

"Adapted coordinate charts"

Masacroso

ey, this seems interesting... I never tried do it @Emrakul... a riemann sum that express a integral need another internal infinite sum, isnt?

Andrew Thompson

In particular, what would the dimension of the jet manifold be?

Masacroso

forget it :p

evinda

10:30

0

Q: The functional is not continuous in respect to the strong norm

Let $V=C^1([a,b])$. If $J$ is a continuous functional for the norm $\|\cdot\|_\infty$ then it is continuous for the norm $\|\cdot\|_1:= ||y||_{\infty}+||y'||_{\infty}, y \in V$. But the converse is false. In my book there is the hint that we can use the functional of arc length, in order to show...

Do you have an idea for the above?

Agawa001

11:04

yes @Emrakul

using geometric series

not arithmetic series

Masacroso

11:36

@Agawa001 you surely want to see that

or check this

Agawa001

@Masacroso this expression is it correct $ \lim_{x\rightarrow 0} \sum_{i=0} ^\inf {(x*2^i) (x+x*2+x*4+....+x*2^i)^2} = \int_0^{inf} x^2$ ?

Masacroso

Idk, sry... I cant help you with this now, my knowledge of calculus is so limited... well, my knowledge of every topic on maths is very limited :)

Agawa001

ur link helps

thx

in fact it helps much

Masacroso

oh, no... well, I just wanted destroy your interest on the topic @Agawa001 (joking, Im pleased it help you)

Agawa001

oh its @robjohn :D he was present when i ws begging fo help

8

A: Exactly expressing integral as a sum

We can break an alternating sum into the difference of the non-alternating sum and twice the sum of the even terms: $$ \begin{align} &\sum_{m=1}^{2^n}(-1)^{m-1}(b-a)2^{-n}f(a+m(b-a)2^{-n})\\ &=\sum_{m=1}^{2^n}(b-a)2^{-n}f(a+m(b-a)2^{-n}) -2\sum_{m=1}^{2^{n-1}}(b-a)2^{-n}f(a+2m(b-a)2^{-n})\\ &=\un...

The Artist

11:54

@Chris'ssistheartist ^_^

Masacroso

@robjohn is the best nick that I know (we dont know if exist a person, an IA or whatever under it xDDD). He is my best and bigger ultrafriend :D (metafriend is legit too)

Agawa001

do u mean theres chatbots arround ?

u can know whether u talk with a bot when u ask it "Would you lie if i ask you whether are you human ?"

just try it @Masacroso xD

Remember me

12:28

Hello

robjohn

@Agawa001 I was? When was that?

Remember me

Why do we call the definition of continuity in topology more natural??

I feel the definition of continuity is more better in analysis

robjohn

@Rememberme It is applicable in settings other than only in analysis.

"more better"?

Remember me

I didnt get you @Robjohn

I mean the definition in analysis is more intuitive

robjohn

@Rememberme perhaps the question is whether intuitive and natural are the same thing.

Remember me

12:43

In analysis it is just like saying that if i take two roads which are supposed to meet at a point and do actually meet then we will call that whole road continuous at that point@robjohn

Agawa001

@robjohn i was jkin , i know u are ready to help anyone who doesnt offer any bounty for assistance :D , maybe my question was poorly formulated

robjohn

@Rememberme actually the topological definition implies the "analysis definition" once you apply the idea of a metric space topology.

@Agawa001 did you pose a question to me?

Agawa001

no

i was nt precise

Remember me

But in topology we think in terms of open balls which are basically collection of points and if the value of the function in that ball is the subset of a ball with the center which is the function value at that point@robjohn

Then it is continuous

robjohn

@Agawa001 Ah... just because I have a chat window open (which puts my avatar on the avatar bar) does not mean that that window is on top (so that I can see all questions). And if I see a question that is not posed directly to me that I can't answer at the time, I usually won't butt in :-)

@Rememberme Yes, and that is just the same thing as in the analysis definition. They are the same once you define the balls using a metric.

Remember me

12:48

If I have to state it then
$$f(B_\epsilon(c)) \subseteq K_\delta(f(c))$$ @robjohn

Agawa001

i said i was jking :/ and everyone here knows how much beneficent are yu :)

Remember me

robjohn's the man!!!

robjohn

@Agawa001 okay... I just don't want someone to think that I was ignoring them just because I didn't respond.

@Rememberme That is pretty much the topological definition (inverse image of an open set is an open set) written in a form that resembles the analysis definition.

Remember me

Okay @robjohn I know the definition of an ball using a metric. So basically
$B_\epsilon(c)=d(c,c_0)\leq \epsilon$

Soham Chowdhury

The topological defn. implies the analysis defn. once you use, say, the Euclidean metric. Am I missing something, @Rem?

Munkres explains this, AFAIK.

Remember me

12:53

@Soham Pls no abbreviations..... they make me feel dyslexic

Soham Chowdhury

As far as I know.

Also, "Pls". :P

Remember me

please :p

I want to know how does it imply

can you explain it @Soham??

Soham Chowdhury

Bredon covered this very well. I'm shocked how comfortable I am with continuity and $\epsilon-\delta$s now.

@robjohn is here, I won't even try.

:)

Remember me

So @robjohn Can you explain me how does it imply?

robjohn

@Rememberme which direction? (they are equivalent in the metric space setting)

Remember me

12:55

Definition of continuity implies definition of real analysis

How^?

Soham Chowdhury

@robjohn topology $\implies$ analysis.

robjohn

@Rememberme so that the topological definition implies the epsilon-delta definition.

Remember me

Yup

Hello@Balarka

robjohn

@Rememberme Suppose that for all $\epsilon\gt0$ there is a $\delta\gt0$ so that if $|x-x_0|\le\delta$ then $|f(x)-f(x_0)|\le\epsilon$. Suppose that we have an open set $U$. Pick a point $x\in f^{-1}(U)$. That is $f(x)\in U$. Since $U$ is open, there is an $\epsilon\gt0$ so that $|y-f(x)|\lt\epsilon$ means that $y\in U$. Choose the $\delta\gt0$ we have assumed for this $\epsilon$, then if $|x-y|\le\delta$, $|f(x)-f(y)|\le\epsilon$. That is, $f(y)\in U$ and therefore, $y\in f^{-1}(U)$. QED

That is, all points within $\delta$ of $x$ are in $f^{-1}(U)$ so $f^{-1}(U)$ is open and $f$ is continuous by the topological definition.

Remember me

13:11

Ohh.. Okay I get this now...

A huge thanks!!@robjohn

Soham Chowdhury

@Balarka have you ever seen Hatcher's little intro to Farey sequences and everything?

Agawa001

interesting book @SohamChowdhury

Remember me

Hatchers looks like a great book.... But
Never judge a book by its cover..... :)

Agawa001

lol i think such cover isnt sthing attractive :D

$user image$

it looks medieval

Alec Teal

13:30

NO

NO MORE HATCHER BOOKS

3

evinda

Hi @DanielFischer
Could you explain me why the existence interval of $ \tan x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$ and not the domain of the function that is $\mathbb{R}$ ?

Daniel Fischer

@evinda $\cos$ has zeros at $\bigl(k+\frac{1}{2}\bigr)\pi$ for $k\in \mathbb{Z}$, so $\tan$ has poles there. So if you want an interval in $\mathbb{R}$ on which $\tan$ is finite, it must be contained in $\Bigl(\bigl(k-\frac{1}{2}\bigr)\pi, \bigl(k+\frac{1}{2}\bigr)\pi\Bigr)$ for some $k\in \mathbb{Z}$.

$\tan$ is only defined on all of $\mathbb{R}$ if you let it take values in the real projective line $\mathbb{R}\cup \{\infty\}$ [note that $\infty$ is glued to both ends of $\mathbb{R}$, unlike $+\infty$ and $-\infty$].

evinda

@DanielFischer With : $\tan$ has poles there , do you mean that there are values of x for which $\tan x$ is not defined?
Also, is $\tan$ finite in intervals, at which it has no poles? Or have I understood it wrong?

Remember me

7

Q: Are there infinitely many N^3 (especially for prime N) that cannot be expressed as a sum of three positive cubes?

The sequence A023042 on the OEIS website shows that a large percentage of $N^3$ are a sum of three positive cubes. OEIS lists only N<1770, but we can extend that:$$\begin{array}{|c|c|} N&\text{%}\\ 2000&85.8\text{%}\\ 4000&89.8\text{%}\\ 6000&92.1\text{%}\\ 8000&93.3\text{%}\\ 10000&94.2\text{%}\...

nt.number-theory

Now atleast someone answer this question

Daniel Fischer

13:46

@evinda Recall $\tan x = \frac{\sin x}{\cos x}$. When $\cos x = 0$, you have a problem.

evinda

@DanielFischer I see... So are we looking for an interval at which it doesn't hold that $\cos x = 0$ for any $x$?

@DanielFischer If so, then the interval of existence is the smallest interval of the form $\Bigl(\bigl(k-\frac{1}{2}\bigr)\pi, \bigl(k+\frac{1}{2}\bigr)\pi\Bigr)$ , $k\in \mathbb{Z}$ so that $\cos x$ never gets the value $0$. Or am I wrong?

Daniel Fischer

@evinda These intervals are all equally long. It is convenient to consider the one containing $0$, however, so unless specified otherwise, that one is assumed.

Masacroso

a quick question, I need confirm (or refutation if I did something wrong) that this diophantine equation have no solutions: $$a^2+b^2=c^2+d^2\ ,\ a\ne b\ne c \ne d$$

evinda

@DanielFischer So at any of these intervals, $cos x$ does never get the value 0, right?

Semiclassical

@Masacroso must all four variables be nonzero? if not, then i see an immediate refutation

Masacroso

13:56

just natural numbers @Semiclassical

Semiclassical

alright, just making sure

Masacroso

no zero allowed :)

Daniel Fischer

@Masacroso $8^2 + 1^2 = 7^2 + 4^2$

Masacroso

I get a "solution" that says is impossible, cause is a equality between a odd number and a even one... but Im not sure if I did correctly

Daniel Fischer

@evinda Right.

Masacroso

13:58

so I failed, as I expected after all xD

evinda

@DanielFischer Great... Thanks a lot :)

Semiclassical

what that makes me curious about (along the lines of how one finds pythagorean triples) is how to construct such solutions

Masacroso

this means that you cant caracterize different vectors only with euclidian norm... at least vectors of 2 dimensions (vectors formed only with natural numbers)

I was wanting a way to characterize isomorphic vectors that are composed of entire quantities on its coordinates

but, maybe, if I use a norm of dimension greater than 2 it may work (or not :p)

anon

14:26

I have no idea what you just said @Masacroso

robjohn

@Masacroso note that $a^2+b^2=c^2+d^2\iff(a-c)(a+c)=(d-b)(d+b)$

Masacroso

I dont have idea either @anon, LOL

Semiclassical

try saying it in plainer language, then :P

Masacroso

yes, I knew that @robjohn but

forget it @Semiclassical... i dont remember the context, just have the diophantine equation

user253008

hi

Mary Star

14:39

Hello!! Could someone of you take a look at my question:

0

Q: Streamlines - Pathlines

Construct and draw the streamlines of the velocity field $u=az-bt, v=\frac{b}{4}z-cy, w=2(a-1)$. Calculate $c$ (as a function of the constants $a$, $b$) such that the flow field $\overrightarrow{u}=(u, v, w)$ represents the flow of an incompressible fluid. Find also the pathlines of the element o...

physics classical-mechanics fluid-dynamics

?

user253008

I have a question about Clique Percolation Method ?

0

Q: Clique Percolation Method calculation

According to Clique Percolation Method as shown in Wikipedia I could not understand why the result of Graph1 is Graph2 if we assign k to be=2, According to the definition of CPM, then the cluster resulting from cliques must be 4 cliques not only 2, because also :{2,4,12} and {8,9,13} is also max...

discrete-mathematics graph-theory computer-science

؟

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$Mathematics$ Mathematics