Hi @JasperLoy @Incurrence I'm so sleepy

@JasperLoy Yep

@ᴇʏᴇs Bart.

@ᴇʏᴇs I am too. I have had 6 coffees today though

Wow, my 2 friends appeared at the same time, I am so happy.

I have done 8 hours of study today though, so that's alright

8 hours not enough

Must do more @Incurrence

Well I will be doing another 3, so I'll hit 11 hours :)

I lost a few from deleting them from my study records(for being too inefficient)

Once I get well, I will study 12 hours a day.

12 a day is essentially impossible

Unless you have no social life and you are efficient with daily activities

It is possible if you have no work or school.

My only social life is this chat

Truly Bart?

Yea lol I haven't made any friends at school yet

@ᴇʏᴇs @Incurrence @Chris'ssis are my best friends in this chat. =)

@ᴇʏᴇs Why not?

@JasperLoy No time

yo

Hi @JC574.

@TedShifrin sorry, didn't see your message, hi!

how's it going?

@JasperLoy

@JC574 What is your favourite algebra text?

@JasperLoy ummm

@JasperLoy I have a really good one on representation theory and character somewhere

@JasperLoy I can't remember the name

@JC574 I mean a general one. I am doing a mass survey.

I wish I did less courses so I could focus more on each of them

a general one... hmm

I wish I can reverse time.

I mean I like Dummit + Foote

but I haven't read it in depth

OK.

I find Artin's book nice to read

@JasperLoy amazon.com/Basic-Algebra-II-Edition-Mathematics/dp/048647187X

@iwriteonbananas Yes, just take a non-abelian group and its abelianization

i deleted the question out of embaressment

haha

No need to be shy.

We are all stupid here.

We all ask questions that turn out to be trivial. That is what chat works so well for

yeah no judgement. I spend most of my time asking trivial questions

i used to ask a few trivial questions before

but everything i ask now are trivial questions.

i'm doing exercise 6:

@Bal pls

we attach n-cells, $n\geq 3$, to a path connected space $X$. Then the map induced by the inclusion $X \hookrightarrow Y$ is an isomorphism

ah

it makes sense intuitively

can we prove it easily with homology?

yep : try to formalize your intuition

@iwriteonbananas yes, you can use cellular stuff.

This sounds like biology now

@iwriteonbananas map induced on what? Fundamental groups?

yeah, @Tobias

@BalarkaSen not sure how...basically $S^2$ is simply connected

But what if the original space is a point and you attach a solid torus?

a solid torus is not a cell

well, attach cells forming a solid torus. Or I suppose these will not attach to the same point.

cells are by definition homeomorphic to D^n

a solid torus is not

@TobiasKildetoft ah, but then you're not attaching a 3-cell.

right, I need to also attach some smaller cells

right

been too long since I did cell complexes

should i follow the hint and proceed like in proof of prop 1.26 or is there a cooler way?

i cant think of a better way atm

the map induces a surjective map on the $\pi_1$ level. you just have to show that kernel is trivial.

not much machinery is needed for this.

recall what an attatching map does, and proceed with that.

hmmm

ooh i think i did this the other day

when you do this and consider the case n= 2 it allows you to build complex for a specific finite presentation of a group i think

it does't hold for n = 2.

i know

but you get a specific change right

or something

i guess you're referring to Cayley/presentation complex?

it quotients by the image of the attaching map

yeah think so

@Chris'ssis?

@Chris'ssis ?

@robjohn ?

damn it i cant think straight

when n=2

and a loop is trivial in Y, it needn't be trivial X...for example if the loop encloses a hole in X and we attach the boundary of a 2-cell to cover the hole

right. the place you attatch your 2-cell onto might have noncontractible loops

but we cant cover any holes by attaching a sphere

@iwriteonbananas that's not the correct logic.

your intuition is wrong

ok

well, what i wrote above for n=2 no longer works if the boundary we're attaching is $S^2$

right : why doesn't it?

$A\cap B$ is simply connected if $n\geq3$

wheree A, B are as in the proof of prop 1.26

@BalarkaSen what is the answer you have in mind?

@iwriteonbananas right, there you go.

and now you can van kampen it

@BalarkaSen I cannot tolerate the MSE corrupted atmosphere anymore!

@BalarkaSen I asked a question, it got 5 downvotes in less than 2 minutes!

@BalarkaSen yeah the rest is an the proof

hey guys!

I have a question regarding game theory...

if you had to choose between winning 100 points with probability 1/100 or 1000 points with probability 1/1000 which should you choose?

which one*

@BalarkaSen how would you prove it? can you elaborate on your intuition?

are they equally good because they have the same expected value?

@iwriteonbananas prove what?

@BalarkaSen the claim from the exercise

well, i'd have done it by proving that no noncontractible loop of $X$ on the place your cell gets attatched to is contractible on $D^3 \sqcup_\varphi X$.

it's essentially equivalent to van-kampen-ing.

@BalarkaSen how would you proceed in this? (if u feel like it)

@FreeMind did you see my reply?

i gotta go

@FreeMind if you have any questions, see the Alternating Series Test.

@FreeMind However, I am sure you know that the Harmonic Series diverges.

@FreeMind When a series converges, but does not do so absolutely, it is called conditionally convergent. The Riemann Series Theorem says that the order of the terms is important for a conditionally convergent series.

Hey @ted

@robjohn math.stackexchange.com/questions/1237353/…

@FreeMind Aren't you well aware of the 'show no work, get a downvote' policy on MSE?

@Incurrence Do I have to write a wrong attempt for just beating around the bush?

@FreeMind Actually - yes

@Incurrence Why don't they change the stupid policy of pretending?

Because trying to answer the question should be your job

If you can't answer it, surely there is something you failed on. Put that there, and they know what to help you with

@Incurrence Well, it is but I don't have anything special to show as an attempt when I have not gotten any kind of clue!

@FreeMind Then atleast put down where you got it from, and why you wan't to solve it

@Incurrence Honestly, it seems like a political atmosphere dominating here instead of the freedom of asking questions which leads to absolutely a better community, at least that is what I think. I am not sick to ask a question that I have not made any kind of attempt on, I may be sloppy and forgetful but I am absolutely not that kind of guy.

@FreeMind Unfortunately noone really knows that, and the main answerers deal with so many people it would be unlikely they remember specifically you until after a long time. I find it unlikely that you would truly be stuck without being able to try anything - I can normally try more than 5 things for any given problem.

Greetings

Greetings

How is it going?

I delete my account right now, shut the door of coming to this site. Thanks all for helping me.

@FreeMind I didn't mean to offend you

@FreeMind Life is often tough. Learn to deal with it.

@FreeMind I was just trying to give incite into the politics of having your question well-received

@FreeMind Try my suggestion

hmmm, I need to buy some food for my pets. BBL (in 30-60 min)

@robjohn How do I show that the p-norm has $d(f,g)=0 \iff f=g$?

On the space of continuous functions $[a,b]$

I take $d(f,g) = ||f-g||_p =\left(\int_a^b |f(x)-g(x)|^p \,\mathrm{dx}\right)^{\frac1p}$

$\left(\int_a^b |f(x)-g(x)|^p \,\mathrm{dx}\right)^{\frac1p}- \left(\int_a^b |g(x)-f(x)|^p \,\mathrm{dx}\right)^{\frac1p}= 0$

Should I start with $||f-g||_p^p$?

yes

I think so

If $f \neq g$ then there exists subinterval where $|f-g|^p \ge \mu $ for some $\mu> 0 $

you can use that to show the integral is nonzero

Is that $\mu$ a measure?

no sorry

it's just a positive constant

Oh that's fine haha. I don't know pretty much any of measure theory, but usually $\mu$ is for measures(and is on the wiki page for this)

like, take $x_0$ where $f(x_0) \neq g(x_0)$

@JC574 Thanks, I'll give your suggestions a go

Just gotta go eat something haha

then on a small interval about $x_0$ we know $|f(x)-g(x)| \ge \frac{1}{2} |f(x_0)-g(x_0)| $ for example

the integral over the whole of $[a,b]$ is going to be at least the integral over this small interval, since $|f-g|^p$ is non-negative

Could anyone pls have a look at math.stackexchange.com/questions/1233811/… ?

I offered a bounty but still drew no attention...

Offer more bounty :p j/k

@ᴇʏᴇs Bart.

@Incurrence this is wrong

@Incurrence Can you show that if $f$ is continuous and $\int_a^b|f(x)|\,\mathrm{d}x=0$, then $f(x)=0$ on $[a,b]$?

@Incurrence but we do have: $\int |f| d\mu = 0$ if and only if $f=0$ $\mu$-almost everywhere

@JC574 Ah, I see you are showing that :-)

@iwriteonbananas He is assuming that $f$ is continuous

f measureable of course

if $f$ is continuous

oh ok

almost everywhere is everywhere

well not that

but you know what i mean

sorry i didnt see that :P

How can you intuitively tell if a function is probably measurable

By using a ruler.

if X and Y are independent and identical this implies E(X^2Y^2)=E(X^2)E(Y^2)? right?

@user60887 Using covariance, yes

i dunno, @Mike, the proof looks okay to me, and it's not really that long either [most of the proof goes into proving basic stuff, like contractibility of $S^\infty$]. Do you have a different proof in mind that doesn't use homology/homotopy?

i've never seen a proof of non contractibility of $S^2$ using elementary stuff.

@BalarkaSen I need to bark to you about algebra

Hi @Chris'ssis!

Cliffford ones to be exact

i dunno any clifford algebra

Consider every 2x2 matrix of the form (a, b, kb, a)

under matrix mult

okay

This is the same as a+bi where i^2=k

So k=-1 gives complex, k=1 gives split complex, k=0 gives dual etc.

If a, b are from some module R

right, okay

then k=1 gives the direct sum of algebras of R + R

k=-1 gives the complexification of R

you're losing me, alizter

Where don't you understand

tell me if i need to define stuff for you

what's a complexification?

complexification is the tensor product with V x C, where C is complex numbers, V is a vector space over some ring R and tensor product is with respect to R

in other words

weird stuff.

basically v in V

all v_1 + i v_2

where i^2 = -1

i get it.

Choosing k = 0 gives dual numbers

which are wierd

so what's the point of all this?

I am going to eat.

@Alizter right, extension by idempotents.

I think as far as I understand clifford algebra construction covers these

but why should i care about them?

Hi @Ali

@BalarkaSen Good question, don't know, i find them interesting

that's not interesting enough for me :P

I am trying to see how different combinations of these constructions on top of eachother can form corresponding groups of their units (basises)

and perhaps algebraic structures arising from treating theese constructions as operators and algebras as elements

i am really thinking about topology these days, so that may not be a quite so interesting to me

Well Lie theory joins us up

This stuff is related to Lie groups as well

ah?

Thats how I got into it

Constructions that product lie groups

or have subliegroups maybe interesting

I might be overmy head right now

you know a bit about algebraic topology, @Alizter?

@BalarkaSen depends prob not much

i have been trying to compute fundamental group of some toy lie groups for some time.

which ones?

I havn't studied fundemental groups in detail but I can look at the lie groups

$SL_2(\Bbb R)$, say.

hi @Alizter @Balarka

hello

@TedShifrin :)

The topology in question is easier if you know some linear algebra :P

it can be done by just using linear algebra?

and there i was cooking up a suitable fibration

@TedShifrin consider (a, b, kb, a) under matrix mult. This is the same as a+b i with i^2=k what do you call these things? For example k=-1 is complex nums, k=1 is split comp and k=0 is dual.

@BalarkaSen I am having a doubt understanding generators and relations. ......I know what both of them are but I can't understand how they are used to define a group?

Hi @TedShifrin professor

sorry, @Sayan. told ya that i won't help you understanding groups in depth. you just don't have the mathematical maturity yet.

try doing some linear algebra

Hi @Sayan. Damn, @Balarka: You're being more obnoxious than I was to you :D

I don't know, @Alizter.

@TedShifrin :P

Can't both be done simultaneously

don't think so. but i won't stop you if you want.

just not gonna help you.

Ok.....

ok, i have to run

Hi @TedShifrin

How can one gain enough mathematical maturity to win @Bal's favor

@ᴇʏᴇs No need to win his favour.

@JasperLoy But what if I want his help in algebra :(

@ᴇʏᴇs @Incurrence @Chris'ssis are my 3 best friends in this room. =)

Me @JasperLoy

@SayanChattopadhyay Hello.

hi mr eyeglasses

I am not your friend?????? :p

@SayanChattopadhyay We already told you not to study groups yet, you just don't listen.

What's the matter in groups....

If you don't listen to us, we will not advise you anymore.

Why isn't @SayanChattopadhyay allowed to study groups?

You must understand the context @ᴇʏᴇs.

@JasperLoy Hi. Pretty busy at the moment. I'll be more free later on. ;)

Before he studies X, he will ask about Y. And then before he studies Y, he will ask about Z, and it never ends. And in the end he still has not studied X.

So before scolding me, please know the context @ᴇʏᴇs.

I think if you know sets then you can easily understand groups.....and also I am doing both simultaneously no prob in that right @JasperLoy? :p

@SayanChattopadhyay We already gave you a long list of books to study which you have not completed.

@JasperLoy How should I scold you

@ᴇʏᴇs ~!@#$%^&*

@ᴇʏᴇs Yes, you ~!@#$%^&*

Which books @JasperLoy I finished apostol just a few days ago so what I am aiming is algebra both linear and abstract

@SayanChattopadhyay Are you sure you finished Hammock and Apostol? That's 3 very long books.

@SayanChattopadhyay Wow you did all the problems in Apostol

Why they are not hard except last parts of integration @ᴇʏᴇs

@SayanChattopadhyay If you finish 3 books in such a short time, you are Superman. I wonder how you finish them.

@SayanChattopadhyay Do you know every definition, every theorem and every proof by heart?

@JasperLoy I did hammock when I first saw it because hammock is a lot similar to my 11 grade so I was able to do it very quickly and the last 5 months I have just been doing apostol and aylast I have been able to finish it without a question left

@SayanChattopadhyay Are you sure you understand all the material in all the books?

Yes I do....... You or anyone can ask me questions if they want proof @JasperLoy

@SayanChattopadhyay OK, now finish Petersen first before we talk about groups, that is all.

But why aren't both of them algebra....?

Are you going to ask about algebraic geometry next?

@Chris'ssis I am going to sleep now, see you in my dreams, pray that I succeed next month, good night.

Never I have no idea what it is @JasperLoy

@BalarkaSen: The proof is not elementary, and uses tools that Lipschitz uses in his research (symmetric products). It's an amusing way of proving that $S^2$ is non-contractible, but it's far from elementary. It is secretly - but not out loud - doing homotopy theory. The point of my comment on the question is that the problem is not at all elementary.

If you want an "elementary" way, prove that a smooth manifold is contractible iff it's smoothly contractible; define the degree of a map $S^n \to S^n$ by picking a regular value and counting its (oriented) preimages (if you don't want to deal with orientation, count points in the preimage mod 2); show that it's a smooth homotopy invariant.

@TedShifrin I just realised

It is $R[X]/(X^2-k)$

So $R[X]/(X^2-k)\cong \left\{\begin{bmatrix} a & b \\ kb & a\end{bmatrix} :\ a,b\in R\right\}$

@MikeMiller Hello

off topic : What is roll back ?

Hi people can you clear me one thing: Is Z[i] a principal ideal domain (PID)?

@BalarkaSen

@HKLee when an older edit is put in place

@zed111 What is your definition for PID?

@Alizter Every ideal is principal ideal <a> = ra: r \in R

Z[i] is a eucledean domain

and every euclidean domain is a priniciple ideal domain

so show that Z[i] is a eucledean domain

Yes, @Alizter, I believe you're right. I have an exercise like that (with rational coefficients) in my algebra book. :)

Does @MikeMiller still exist?

yes, @zed111. You can give an argument exactly like for $\Bbb Z$.

@Alizter For ED, which function \phi gives that property?

zz*

Wait a second

@TedShifrin Cayley-Dickson Construction $R_n\to R_{n+1}$ is just $R_n[X]/(X^2+1)\cong R_{n+1}$

starting from $R_1=\Bbb R$

I don't know anything about that, @Alizter ...

or Complexification as its called

Getting from $\Bbb R \to \Bbb C \to \Bbb H \to \Bbb O \to \Bbb S \to \cdots$

@TedShifrin What is your favourite lie group?

Hello :)

@evinda guten Abend :)

@Alessandro Wie geht es dir?

@ Alizter Thank you for your reply. But I cannot understand. Older edit is the edit which I did someone's question or answer ?

@evinda gut, und dir?

Ganz ok.. @Alessandro Was hast du heute so gemacht?

@evinda nichts besonderes, ich bin in die Schule gegangen :( Ich werde nächste Woche die TELC B1 Prüfung schreiben

@Alessandro Und bist du schon fit für die Prüfung?

@Ted Hey Ted. Do you want to take a look at today's Ecole Polytechnique math exam ?

@evinda Ja, ich habe keine Angst davor! Ich habe viele Probeprüfungen gemacht und sie mit guten Ergebnissen bestanden

Salut, @leDodo :)

@TedShifrin marocprepa.com/temp/2015_conc/xens/2015_xens_b.pdf (and keep in mind I only got 4 hours and proofs have to be flawless)

@Alizter: Don't have a favorite, but have used $SO(n)$ and $U(n)$ a lot in my math life.

@Alessandro Gut!!! Und musst du danach noch mehr Prüfungen in Deutsch schreiben? Oder ist es die letzte?

I have office hours now, so students are here. I'll check back later.

@HKLee Somebody rolled the edit back to an older revision. You can see all the revisions and authors of the revisions of a post

okay serious problem. rotational rigid motion makes total sense with calculus and algebra. this is something i don't get though. we have a trapezoid with coordinates (-1,1),(-1, 4),(3,4), and (3,-3). the rotational isometry at R(1,0) is (0,-1). The rotational isometry at (0,1) is (1,0). Why?

@Ted alrighty, please ping me your feeling about it :)

i'm so confused

@evinda Ich muss die DHS Prüfung schreiben, an die Uni zu anmelden

Aha! Also geht der Unterricht dann weiter? @Alessandro

@evinda ja, mindestens bis Juli

oh wow, i'm dumb

i got it

@Alessandro Achso

@Alitzer : Thank you. I get an cleanup

@MikeMiller i am interested : how would you interpret the proof homotopy theoretically?

i find it to be a fun proof.

hi again, Ted.

@LeGrandDODO: On blague? I doubt if I could do half of that in 4 hours. How much did you do?

hi again, @Balarka

guten Abend, @Alessandro

we were talking about this proof, @Ted

LOL ... Obviously the most direct proof possible, @Balarka. However, the symmetric product is an important idea for all sorts of higher math.

but i like this Sym functor. wonder what else one can do with it.

Ah, very important, for example, in studying the Abel-Jacobi map in algebraic geometry.

Unordered $k$-tuples of points in a space is often an important thing to understand.

yikes flees in fear

just be careful of fleas ...

that's not a good joke, @Ted.

I specialize in not-good jokes.

Ah, the exercise that $\text{Sym}^n(\Bbb CP^1) = \Bbb CP^n$ is super important. Sadly, for manifolds of dim > 1, the symmetric product has singularities.

really? should i try that exercise?

Yes, that's very cool. :)

So symmetric products of curves (Riemann surfaces) are complex manifolds. Not true for higher dimensions ...

ah. sounds very fun.

@BalarkaSen: He's proving that $\Bbb{CP}^infty$ is not contractible, and doing so by calculating that $\pi_2(\Bbb{CP}^\infty) = \pi_1(\Omega \Bbb{CP}^\infty)$; and by demonstrating a nontrivial covering space of the latter space he shows that this is nontrivial.

My phone vibrates whenever you ping me. I'm trying to work right now. Please ping me less frequently.

okay : sorry for disturbing you. i'll try to figure what's happening by myself. thanks!

@TedShifrin Ah, I got it.

Any thoughts?

Let's pick a point $z \in \text{Sym}^n(\Bbb C - \{0\})$ : this corresponds to $n$-tuples in $\Bbb C - \{0\}$ which are permutations of each other (let there be $k$ of them). Consider the tuple $(a_0, a_1, ..., a_k, 0, ..., 0) \in \Bbb C^n$ which are coefficients of the polynomial the roots of which are the $k$ of the tuples that correspond to $z$. This coefficient tuple is never zero, as we are considering $\text{Sym}^n(\Bbb C - \{0\})$, so this is in $\Bbb C^n - \{0\}$.

How do I make my phone vibrate when someone pings me

Obviously $(\lambda a_0, \lambda a_1, ..., \lambda a_k, 0, ..., 0)$ is the same thing as $(a_0, a_1, ..., a_k, 0, 0, ..., 0)$ (i.e., same in the preimage of the map Sym^n(S^2) --> C^n - {0} we constructed) as the polynomial obtained from this coefficient tuple has the same roots, the ones corresponding to $z$. Thus, $\text{Sym}^n(\Bbb C - \{0\}) \cong \Bbb CP^n$.

@Bib I missed that, sorry

@Danu Would it work to translate the bra-ket notation to typical math notation and post on MSE?

This wasn't hard.