Mathematics - 2015-06-24 (page 2 of 5)

Balarka Sen

07:02

@Paul Thinking about anything interesting?

Paul Plummer

@AlexClark It sort of seems like you are just taking a course for taking a courses sake. If you want to study beforehand, because you want to, that is fine, but I would rather spend my on something that I really want to learn/understand and save courses for when I actually take courses.

@BalarkaSen Mildly, reading 1.1 of hatcher and doing some problems. Looking at showing that a path in a locally star shaped space is homotopy equivalent to a PL curve

Which reminds me @MikeMiller I feel like I remember you saying you didn't care for Morse Theory by Milnor, are there any other sources you recommend?

Balarka Sen

ok, cool. ps : there are some nice exercises in 0 you shouldn't skip. the join of spheres is a pretty interesting exercise.

the contractibility of $S^\infty$ is another one.

as a motivation, the first one gives you a way to see hopf fibrations :)

Paul Plummer

:D I don't plan on skipping them I am sort of doing a back and forth thing, especially when a concept from ch0 comes up, or I feel like I missed something.

Balarka Sen

I don't recommend skipping, but most of ch0 (exception : cell complexes, and what's before them) you can come back to when you need to learn them and skip for the time being.

Mike Miller

@PaulPlummer: Morse theory is a fantastic book. It's his + Stasheff's chadacteristic classes I dislike.

Balarka Sen

07:09

just pointing out which exercises you shouldn't skip :)

Paul Plummer

Fallibility of memory... I came across some applications of PL Morse theory in some geometric group theory, and it looked quite cool, I am not 100% sure if one really needs "classic" Morse theory, but I figured it wouldn't hurt to have a nice source around for whenever I get to that point @MikeMiller

Hello. You left in a hurry! @AlexClark

user147690

@PaulPlummer Oh no, don't get me wrong, I really enjoyed my previous semester courses, and I want to get a feel for what I like about next semesters courses.

user147690

Yes you were stressing me out

Paul Plummer

Sorry @AlexClark

user147690

I spent all day trying to organise how I am going to study, I don't know what's wrong with me haha

Paul Plummer

07:17

Well you spend all day organizing how you are going to study :P

Tobias Kildetoft

@AlexClark Still need references for certain topics?

user147690

@TobiasKildetoft Well I did some research and apparently there is no such textbook that covers the range of topics

Tobias Kildetoft

@AlexClark That would be my guess also, as it seemed like you were looking for something rather broad

user147690

Yep the course hasn't changed much, and I found some information from 2010 that said you would need to have access to:

1. J.E. Humphreys, Introduction to Lie algebras and representation theory,
QA3 .G7NO.9 1980
2. B. G. Wybourne, Classical groups for physicists, QC174.17.G7 W9 1974
3. J.F. Cornwell, Group theory in physics, QC20.7.G76 C67 1984-
4. V. Chari and A. Pressley, A guide to quantum groups, QC20.7.G76 C48 1994
5. G. Lusztig, Introduction to quantum groups, QC20.7.G76 L88 1993

user147690

If you wanted to find textbooks to cover the course haha

user147690

07:19

@PaulPlummer I was diagnosed with anxiety last year but I haven't been able to force myself into a doctor to get re-diagnosed and get a script or something haha

Tobias Kildetoft

@AlexClark I like Humphreys for Lie algebras. I never read Lusztig's book on quantum groups (I think I learned about those from some notes of Andersen)

Balarka Sen

@AlexClark Just pick up a book you'd like to and start studying.

user147690

@BalarkaSen Okay haha, I'll do some tropical geometry first

Tobias Kildetoft

But I don't really know anything about applications in physics nor to knot theory (apart from some somewhat technical things)

Balarka Sen

@AlexClark Sure. Anything you'd like to. That's the best solution to anxiety I've seen.

user147690

07:21

@TobiasKildetoft That's alright, I found a course notes from 2012 that seem to cover most of what I need

Paul Plummer

@AlexClark I believe it!

Tobias Kildetoft

though I actually think that some of the things I have been working on recently have applications to knot theory (at least we have generalized something that was used to give an alternative formulation of Khovanov homology)

Paper actually appeared on the arXiv today

Mike Miller

@PaulPlummer I think PL Morse theory is probably quite different than the standard theory.

Tobias Kildetoft

@AlexClark BTW, I have all of those books if you need some of them

Paul Plummer

I am sure it is different, but at least the bit I read it seemed like a decent amount motivation and ideas (and probably intuition) carry over, although I guess I wouldn't know @MikeMiller

But I also had a feeling it was not too necessary, to have a deep knowledge of the standard theory

maybe even no meaningful knowledge would even be necessary

user147690

07:25

@TobiasKildetoft Thanks very much, but I don't know how many will be relevant to me since the list was from 2010, and I think I'll get a proper course notes to cover everything at the start of the sem

Tobias Kildetoft

@AlexClark Ok

I would recommend Humphreys anyway, since I find that quite nice

user147690

I'll check them all out for sure though, no doubt

Balarka Sen

I've never plucked up the courage to go through the computation of the kernel of the van Kampen map :(

Tobias Kildetoft

(the follow-up to it is a lot harder to read though)

but the follow-up is probably not that relevant for someone doing mathematical physics anyway

user147690

Well it's the only math phys course that I expect to take

Tobias Kildetoft

07:28

@AlexClark What sort of direction are you heading for then?

Balarka Sen

@MikeMiller Can you give me a walkthough/direct me to a proof that every group is fundamental group of a 4-manifold?

user147690

Honestly, no idea, I really enjoy mathematics in general, and I found functional analysis I much more fun in the end than I expected, although I found complex analysis rather boring

Tobias Kildetoft

@AlexClark That might point towards operator algebra then

Balarka Sen

I've heard about it and take that fact as granted, but I don't really know the proof.

user147690

I also did group-theory, ring-theory, and really enjoyed it(except at the end when they pushed me a little too fast)

user147690

07:34

@TobiasKildetoft The person you published with has a heap of papers

Tobias Kildetoft

@AlexClark Yeah, he is absurdly active

user147690

I just saw the reference list on your paper with him

Tobias Kildetoft

And it is no surprise to me after having worked with him for a while. He has a way of thinking about things that just produces results really efficiently

He has almost 150 publications on MathSciNet (and he is not even that old. He had his first publication in '93)

user147690

Wow, do you know how exactly he attacks the problems? Or is he just talking to you and things just hit him

Tobias Kildetoft

@AlexClark He has a very visual way to think about modules, which works very nicely for figuring out what sort of things one needs to do to them

He is also quite good at explaining his thought process, so I am learning a ton by working with him

Balarka Sen

07:40

I'm curious : how does he think about modules visually?

Tobias Kildetoft

@BalarkaSen It is a bit tricky to explain without drawing stuff. But basically, he draws the module as a big oval, then the socle is a smaller oval at the bottom, and stuff just above that is things in the socle of the quotient by the socle. Similarly for the head

Paul Plummer

I think Shelah is catching up to Erdos in terms of number of papers... sort of (I think it is like 500 more or so)

Tobias Kildetoft

@PaulPlummer But 500 more is still a 50% increase to the current number

@BalarkaSen I don't think I am explaining the ideas so well here. They also involve "distance" between submodules when there is a grading

Balarka Sen

I am not going to pretend I understand that analogy well. But interesting nonetheless!

Paul Plummer

True, hence the "sort of" @TobiasKildetoft

Tobias Kildetoft

07:46

@BalarkaSen He also has a nice way to visualize the cells in finitary 2-categories (and Coxeter groups) which includes left-, right- and twosided cells at the same time

Balarka Sen

what're cells in 2-categories?

Tobias Kildetoft

@BalarkaSen a generalization of the ones in Coxeter groups introduced by Kazhdan and Lusztig

(though actually I am not quite sure if one can get the ones for arbitrary Coxeter groups, i.e. ones that are not Weyl groups, in terms of 2-categories in a nice way)

Ohh, scratch that last. One just considers the 2-category of Soergel bimodules

(I am just more familiar with the construction for Weyl groups as that is basically the topic of that paper from today)

Balarka Sen

all over my head, but very fun.

Tobias Kildetoft

@BalarkaSen But it has applications in knot theory

Balarka Sen

oh?

I'm trying to understand some of these stuff from here, which seems reader-friendly.

Tobias Kildetoft

07:56

@BalarkaSen Yeah, I think I read that some time ago, it is quite nice

Balarka Sen

cool.

Paul Plummer

See you guys later, going to go to bed

Sorry for stressing you out @AlexClark

Mike Miller

@Balarka: Take a 4-ball (0-handle). Add 1-handles for generators, 2-handles for relators. Pass to the double.

Balarka Sen

By adding 1-handles, you mean attaching cylinders/connected summing with S^1 \times S^1?

or do you mean summing with S^4 \times S^1?

Mike Miller

We're in 4-manifolds, I hope I'm not summing with a 2- or 5-manifold. The above is in the language of handlebodies; look it up. It's just gluing on disks in a certain way.

It's the CW complex idea but built for manifolds.

Balarka Sen

08:07

yikes. that definition might take a good many hours for me to digest.

Mike Miller

Alternatively, avoiding handlebodies (poor move; handlebodies are good), take $\#_k S^1 \times S^3$, where you've got $k$ relators. Pick loops representing the relators. Do surgery on these loops (delete a tubular neighborhood $S^1 \times D^3$, reglue a $D^2 \times S^2$ back in).

Balarka Sen

ah, that looks less murky. (I can barely visualize 3-manifolds, so seeing 4-manifolds are about hopeless for me)

Mike Miller

Handlebody diagrams for 4-manifolds are essentially inherently 3-dimensional. This is why they're a powerful visualization tool.

(This is in the same sense that a Heegaard diagram is a 2-dimensional representation of a 3-fold.)

Balarka Sen

oh.

$\pi_1(\# S^1 \times S^3)$ is clearly the free group on $k$ generators. Now picking loops according to the relators and doing surgery along these make these path-homotopic to the basepoint, I presume.

Mike Miller

Don't presume, prove.

Balarka Sen

08:19

Yeah, I guess it does : contract bits of the loop through the disk $D^2$ the copy of $S^1$ bounds and contract bits of the loop through the sphere $S^2$ (which should be no problem, as the sphere is simply connected)

Right, I am trying to visualize the situation before trying to prove it.

Mike Miller

OK. The point is more that the thing that used to be an $S^1$ factor - that is, the loop representing your knot - now bounds a $D^2$ factors.

Balarka Sen

right

that's an interesting technique.

but, as usual, I am seeing surgeries as a systematic way to build spaces with given homotopy/homology groups. I guess there's more to it?

Mike Miller

Every 3-manifold is obtained by surgery on a link in $S^3$. It is a bit reductionist to say that the point is to build spaces with given invariants.

(Good luck controlling the higher homotopy groups, in any case.)

Not to mention very related to cobordisms of 3-manifolds.

Balarka Sen

@MikeMiller Um, probably I am being silly, but what does knowing that every 3-manifold is obtained from a surgery on a link tell us?

@MikeMiller ah, true, I don't expect those beasts to be easily controllable.

Mike Miller

It tells you that there's a lot more going on than just changing some homotopy groups? You can completely specify the topology of a 3-manifold by a link.

Also, we're talking about two different things here. Dehn surgery is the thing with neighborhoods of tori. "Surgery" is this construction I'm doing with $D^3 \times S^1$ and $S^2 \times D^2$.

(And more general versions thereof.) It corresponds to a Dehn surgery with surgery coefficient 0.

Tobias Kildetoft

08:33

@BalarkaSen Well, it tells us that to get 3-manifold invariants, we can instead consider link invariants

Balarka Sen

yes, I get it. but what in particular about the topology do we get to know from that?

ahh, @Tobias.

Mike Miller

@TobiasKildetoft: How? we have to carry the surgery coefficients with us, which links normally don't, and there's no reason your link invariant will give the same invariant for different links that surger to the same 3-manifold

@BalarkaSen: The point was that your statement that "Surgery is a systematic way to do something with chosen homotopy/homology groups" is not really accurate. That's all.

Tobias Kildetoft

@MikeMiller Right, we may lose faithfulness of the invariant (and of course we need to specify it as integral surgery to make it work)

Balarka Sen

no, I didn't expect it to be accurate (thus the question "I guess there's more to it?"). I am just trying to coax more facts out of you :P

interesting, all in all.

Mike Miller

OK, then you're just taking invariants of framed links, which is reasonable data to carry around. But I still don't understand that we may lose faithfulness. If invariant means "thing associated to a 3-manifold", it's not a well-defined invariant.

Balarka Sen

08:37

thanks for the construction of the 4-manifold with a given fundamental group.

Mike Miller

I guess you could say it's "set of things associated to a 3-manifold."

Where the set is just the set of invariants I get from the set of framed links that give me the 3-fold.

Tobias Kildetoft

@MikeMiller By invariant I mean a map from 3-manifolds to some other set which sends homeomorphic (or diffeomorphic or whatever) objects to the same thing

Mike Miller

Fine, I don't understand what your other set is.

Sorry, I think that sounded hostile. I didn't mean that.

Tobias Kildetoft

@MikeMiller It could be whatever. The idea is that an invariant of links (meaning a map respecting whatever is the obvious thing for links, I forgot) gives an invariant of 3-manifolds, by realizing the 3-manifold as surgery and applying the link invariant to the corresponding link

(I am not actually familiar with how the details go that allows this to be done completely)

There is some result about the relation between links that produce the same 3-manifold on integral surgery

Mike Miller

I see, it's that last bit I was hunting for. I think you're referring to Kirby's theorem.

Surgery on two different framed links gives homeomorphic 3-manifolds iff the links are equivalent under isotopy and a couple extra moves called handleslides.

Tobias Kildetoft

08:43

Possibly. As I said, I am not really familiar with the details, I have just seen some talks where they discuss various link invariants

Mike Miller

So you need your link invariants to be stable under taking these moves. OK, thanks :)

Tobias Kildetoft

usually ones related to attaching quantum representations to the components

Mike Miller

I was just worried about having two links with different invariants that both surger to the same 3-fold.

How do you promote the Jones polynomial to a framed link invariant? Do you just forget the framing?

Balarka Sen

Milnor's theorem is nice.

unexpected, actually.

Mike Miller

which one? lol

Balarka Sen

08:48

haha

I agree he's got a lot of theorems

Mike Miller

no but that wasn't a joke

Balarka Sen

@MikeMiller "Two closed, oriented manifolds are cobordant iff one can be obtained from the other by a sequence of surgeries"

Tobias Kildetoft

@MikeMiller Not actually sure how to do that

Mike Miller

Is that really due to Milnor?

Balarka Sen

this guy says so

Mike Miller

08:50

I thought it was Cerf. Thanks.

Huy

@BalarkaSen: $\mathbb{C}\mathbb{P}^1 \cong S^2$, even diffeomorphic, right?

Balarka Sen

yes

Huy

@BalarkaSen: If I have a 3-sphere through the origin and I look at its intersections with the coordinate planes, e.g. $x_3 = x_4 = 0$, what object would that be? An ordinary circle?

Looks like it.

And if I only choose $x_4 = 0$ I'd get a 2-sphere.

Balarka Sen

3-sphere sitting where?

$\Bbb R^4$?

Huy

Yeah.

Balarka Sen

09:03

It's a 2-sphere

Huy

?

Balarka Sen

The intersection with the coordinate plane is a 2-sphere

Huy

How so?

Coordinate plane being $x_3 = x_4 = 0$ for example.

Then we still have $x_1^2 + x_2^2 = 1$, so a circle, no?

Balarka Sen

oh, yikes, I thought you were intersecting with $x_2 = x _3 = x_4 =0$

Then it's a circle, sure.

Huy

Oh, it seems that's basically a cotangent.

Didn't know that. :P

Balarka Sen

09:06

@Huy Think of a 3-sphere as $S^2 \times [0, 1]$ with top $S^2 \times \{0\}$ and bottom $S^2 \times \{1\}$ pinched to a point (i.e., where you have meridians as spheres instead of circles). That being said, intersection of the plane with this thing intersects the greatest sphere at the equator -- which is a circle. Of course, you can just plug in $x_3 = x_4 = 0$ and prove it algebraically.

@Huy I am not willing to do all the problems for you :P

Huy

@BalarkaSen: I'm just thinking about the Hopf fibration $S^3 \to CP^1$.

Balarka Sen

oh. don't think about that algebraically.

I mean, you can write down the fibration, but it wouldn't help you to "see" it

Huy

How do you think about it then?

Balarka Sen

The fibration is just $(z_0, z_1) \mapsto z_0/z_1$, where $(z_0, z_1) \in S^3 \subset \Bbb C^2$, iirc.

@Huy The way I imagine it is a bit too topological. D'you know about join of topological spaces?

Huy

Unfortunately, no.

Alyosha

09:12

Sanity check: is the following true?

Any ordinal $\alpha$ is expressible uniquely as $\alpha= \beta^ {\alpha_1} \gamma_1 + \cdots + \beta^ {\alpha_k} \gamma_k$ where $k$ is a natural number, $\beta < \gamma_i$, and $ \alpha_1 > \cdots > \alpha_k$.

I assume the proof is identical to the usual Cantor Normal Form theorem, but I ask just in case there are any complications.

Balarka Sen

Anyway, the gist is that $S^3$ can be decomposed into two solid torii $\cong S^1 \times D^2$ pasted at the boundary via identifying longitudal circle of one to the meridianal circle of the other. Now there's a natural $S^1$-action on the boundary torus which, after quotienting by it, gives you back $S^2$.

@Huy I could have explained it to you better if I wasn't in a hurry right now. Sorry for that.

Huy

No worries.

Alyosha

Oh, Balarka! Do you have any idea what Lang, in Algebra 3rd ed. p.215, wants the answer to be in part b of the Davenport exercise?

Huy

I think we're doing something similar here, @BalarkaSen. We're partitioning $S^3$ into sets $T_r := \{(z,w): |z| = \cos r, |w| = \sin r\}$ for $0 \leq r \leq \pi/2$. Then $T_0$ and $T_{\pi/2}$ are great circles of $S^3$ and Hopf fibers. For all $0 < r < 1$ we get equidistant tori and $T_{\pi/4}$ is the Clifford torus.

Balarka Sen

oh, ps, you have to vary those boundary torii and look at S^1-action over the class off them. i forgot to mention that

@Huy ah, that's precisely it. try to see the picture instead of caring about the algebraic formulas, though.

Huy

09:22

@BalarkaSen: I try. :)

Balarka Sen

also, interesting exercise to ponder on : Hopf fibration is closely linked (no pun intended) to the Hopf link. how?

@Alyosha I don't have Lang with me

OK, now I absolutely need to leave. Good problem you're thinking about, @Huy.

But make sure you get to the bottom of it. Hopf fibrations are elusive.

Huy

I will try to make the best of it.

nullgeppetto

09:41

Hi everyone. I need your help please! I want to prove that the Riemannian manifold $\Bbb{S}_{++}^n$, i.e., the cone of symmetric positive definite $n\times n$ real matrices is isometric to some Euclidean space $\Bbb{R}^n$. Could you help? Thank you very much!

nullgeppetto

11:00

If you want to help, you may want to answer to this question. Thank you very much for your help!

Jasper Loy

11:46

Hello @Chris'ssistheartist. The doctor gave me 5 tablets a night for the next 8 weeks, lol.

Chris's sis the artist

@JasperLoy Hi. Maybe it's the proper dose for you.

Hippalectryon

@JasperLoy ಠ_ಠ wot

What are the tablets for ?

Jasper Loy

@Hippalectryon For my mental illness, lol.

Hippalectryon

Does that even exist ? (medicine for mental illness)

Jasper Loy

Depends on the definition of mental illness.

Chris's sis the artist

11:49

I wanna address myself a bit to @TedShifrin and the rest of those that starred another sick message. You should learn to calculate some elementary integrals, series and limits before talking anything about me.

And many middle school problems after all.

Jasper Loy

@Chris'ssistheartist He might come in and write a novel. =)

user147690

@SohamChowdhury How are you man

Jasper Loy

@AlexClark Hi, bro.

user147690

Hey @JasperLoy how are you?

Jasper Loy

@AlexClark I just went to see the doctor, lol. Taking 5 tablets each day.

user147690

11:53

Oh wow, all at the same time?

Chris's sis the artist

@TedShifrin An elementary solution to it? Because in my class you wouldn't pass without that. $$\sum _{n=1}^{\infty } (24 \psi ^{(-3)}(n)-24 \psi ^{(-3)}(n+1)+24 \psi ^{(-2)}(n+1)-12 \text{log$\Gamma $}(n+1)+4 \psi ^{(0)}(n+1)-\psi ^{(1)}(n+1))$$ $$=6 \log (A)+3 \gamma+\zeta (2) -6 \log (2 \pi )-\frac{3 \zeta (3)}{2 \zeta (2)}$$

Jasper Loy

Yes. 4 tablets of type A and 1 of type B.

user147690

lmao @Chris'ssistheartist

Jasper Loy

@Chris'ssistheartist Don't provoke him, lol.

Chris's sis the artist

@TedShifrin In fact, I changed my mind, if you only tell me a single clever step in such a solution, I consider you solve the whole problem.

Anyway.

I have to finish some solutions to my book.

BBL

user147690

11:56

Cya later @Chris'ssistheartist have a good night

Jasper Loy

@AlexClark See you in your dreams.

Chris's sis the artist

@AlexClark You too.

user147690

@JasperLoy Oh I'm staying haha

Jasper Loy

@AlexClark Maybe your exgf will start missing you soon...

user147690

Shouldn't talk about such things attached to my real name

Jasper Loy

11:58

Your real name is CTAC, LOL.

user147690

:P

Chris's sis the artist

@JasperLoy Well, he's probably a great professor but this doesn't mean I'm going to allow him to talk like that to me. It's available for all.

BBL

Jasper Loy

@Chris'ssistheartist I think you should ping Thomas instead. I think he is more mean to you.

Chris's sis the artist

@JasperLoy I only wanna do mathematics, no need for fights.

Jasper Loy

@Chris'ssistheartist I only wanna get well and marry Laura Ramsey.

2

Chris's sis the artist

12:03

@JasperLoy Great! :-)

Jasper Loy

@Chris'ssistheartist You can publish your book and marry Monica Bellucci, lol.

Chris's sis the artist

@JasperLoy Monica Bellucci ;)

@JasperLoy she's a great woman! :-)

Jasper Loy

@Chris'ssistheartist Just like you!

Chris's sis the artist

2

:-)

Perfect!

Jasper Loy

@Chris'ssistheartist Just like me, lol.

Chris's sis the artist

12:06

@JasperLoy :D

@JasperLoy I set that picture as my desktop background. It inspires me a lot in my work.

BBL

evinda

Hi @Huy

Jasper Loy

The most mature community on SE is TeX. There are almost no downvotes there.

Hippalectryon

Doesn't mean it's more mature

evinda

Hi @Hippalectryon

Hippalectryon

@evinda o/

evinda

12:18

@Hippalectryon What does this mean?

Hippalectryon

Oh, it's someone waving to say hello

evinda

@Hippalectryon A ok...

@Hippalectryon How are you?

Hippalectryon

Fine, and you ?

evinda

Me too @Hippalectryon

Jasper Loy

@evinda Why no hi to me?

evinda

12:21

Hi @JasperLoy :)

@JasperLoy What's up?

Jasper Loy

Nothing is up. Everything is down.

Hippalectryon

JasperLoy.reverse()

evinda

@JasperLoy The sun is up. Or isn't it sunny today?

@Hippalectryon Java?

Jasper Loy

@evinda It is. I walked 1 hour to the hospital and 1 hour back home.

Hippalectryon

Python

evinda

12:25

@Hippalectryon A ok

@JasperLoy How often do you go there?

Soham Chowdhury

I was actually thinking that Laura Ramsey was somehow related to Frank Plumpton Ramsey. :P

Jasper Loy

@evinda Once a month or two.

evinda

@JasperLoy Does it help?

@SohamChowdhury Hi

Soham Chowdhury

codepen.io/anon/pen/xGPgZo

Jasper Loy

@evinda Maybe...

evinda

12:27

@JasperLoy What do you do there?

Jasper Loy

@evinda See the doctor and take more meds.

evinda

@JasperLoy Aha...

Jasper Loy

I dislike people commenting on my answer when they should be posting another answer. It is as if they are teaching me and not the asker.

evinda

@JasperLoy So you have started answering again?

Jasper Loy

@evinda Yes, like 1+1=2.

Soham Chowdhury

12:32

@evinda Hello, what are you up to?

evinda

@SohamChowdhury I have finished with my exams.. What's with you?

Soham Chowdhury

Oh, we had a debate in school today and I was forced to take part in it. It went semi-well, so I can do a little math now.

Du (?) bist Deutscher, ja?

evinda

@SohamChowdhury Halb deutsch

Soham Chowdhury

Haha, das ist genug.

evinda

@SohamChowdhury :D Lernst du im Moment Deutsch?

Soham Chowdhury

12:35

(My German is weird :P)

evinda

@SohamChowdhury Ok

Soham Chowdhury

@evinda Nicht im Moment. Ich besuchte einen Deutschkurs in meiner Schule fuer 2 Jahre.

(My cases are all over the place)

evinda

@SohamChowdhury Siehst du? Geht doch!!!

Soham Chowdhury

Studierst du in Deutschland?

evinda

@SohamChowdhury Nein, ich studiere in Griechenland.

Soham Chowdhury

12:39

Ah. Wo wohnt deine Familie? In Griechenland auch?

Jasper Loy

Eine kleine Nachtmusik

My favourite German movie is Sommersturm

evinda

Ja, wir wohnen auf Kreta. @SohamChowdhury

Soham Chowdhury

@evinda Cool. Es gibt im Moment viele Wirtschafts-Probleme, oder?

evinda

@SohamChowdhury Genau...

Soham Chowdhury

Aber der Finanz-Minister hat ein tolle Bike. Es gibt viele Fotos in letzten Woche in die Zeitungen. :P

Ich weiss fast nichts ueber wie Griechlenland im Moment ist. :(

evinda

12:47

@SohamChowdhury Du meinst wie es hier aussieht?

Huy

Moin @evinda.

evinda

Wie geht es dir? @Huy

Huy

Ganz ok, bisschen DiffGeo lernen. :)

Jasper Loy

There is French and German in this room, time to bring in some Russian.

evinda

@JasperLoy как дела?

@Huy Gut :)

Jasper Loy

12:52

@evinda You know Russian? I only know English. =(

Hippalectryon

nin hao :3

evinda

@JasperLoy No, just the elementary.

Jasper Loy

@Hippalectryon My Chinese is worse than your Arabic, lol.

Hippalectryon

:c

evinda

@Hippalectryon You speak german too, right?

Hippalectryon

12:54

@evinda Nope

evinda

But your brother can a little. Or do I remember wrong? @Hippalectryon

Hippalectryon

Yeah, he can. I can't.

evinda

@Hippalectryon Which languages do you speak?

Hippalectryon

French (native) , English and bit (little bit) of Chinese

Jasper Loy

@Hippalectryon Why did you learn Chinese? Waste of time, lol.

evinda

12:57

@Hippalectryon Tell us something in Chinese!

Hippalectryon

@JasperLoy Not at all, it's very interesting as a language. very different from French and English

Jasper Loy

@Hippalectryon Is it because you have a Chinese girl?

Hippalectryon

@JasperLoy I don't have any kind of gf

Jasper Loy

@Hippalectryon I see. Maybe bf? LOL.

Hippalectryon

@evinda Urm... 你好吗 ? :P

@JasperLoy ಠ_ಠ plz no

evinda

12:59

@Hippalectryon How is this pronounced and what does it mean?

Jasper Loy

@Hippalectryon What do you use to type Chinese?

Hippalectryon

@JasperLoy chine-nouvelle.com/outils/editeur.html

Jasper Loy

@evinda Ni hao ma = How are you

Hippalectryon

^

evinda

很好，谢谢。

Hippalectryon

13:02

:-)

user147690

@SohamChowdhury

evinda

@Hippalectryon 你会在夏天去度假？

Hippalectryon

I most likely will stay in France. I don't travel a lot @evinda

evinda

@Hippalectryon 行 :D

@TobiasKildetoft Are you from Germany?

Jasper Loy

I wish to be born in Germany my next life.

evinda

13:08

@Hippalectryon How old are you?

Hippalectryon

@evinda 17

@TedShifrin o/

Ted Shifrin

Salut @Hippa, guten Tag, @evinda, hi, @AlexC and @Jasper

evinda

@JasperLoy What's with you?

@TedShifrin Guten Tag!!! Wie geht es Ihnen?

user147690

Hey @TedShifrin

Ted Shifrin

Ganz gut, danke, und Ihnen?

evinda

13:09

Ganz ok... @TedShifrin
Was gibt es so neues?

Ted Shifrin

Ich mach' mich bereit zu beziehen ...

Jasper Loy

@Hippalectryon I will most likely stay in my room.

Ted Shifrin

@Jasper: I understand your point, but I occasionally do try to comment on answers if I think something in an answer is unclear or misleading. Usually, the original answerer modifies his answer and improves it.

Jasper Loy

@TedShifrin OK. Anyway, don't fight with @Chris'ssistheartist, lol.

evinda

@TedShifrin Wann ist es soweit? :D

Ted Shifrin

13:12

Interestingly, I downvoted someone over a week ago who had answered the wrong question, and in a somewhat glib style. I pointed this out to him and, when he didn't change it, I downvoted. (I had already answered the question before he did any of this.) He still hasn't done anything about it.

Jasper, I'm not fighting. She keeps going on and on and on, and I'm fed up.

es gibt einen Monat, @evinda.

evinda

@TedShifrin Freuen Sie sich schon?

Ted Shifrin

Ein Bißchen :P

Soham Chowdhury

13:44

^ You didn't attend the spelling reform meeting? :P

@evinda ah, danke.

Ted Shifrin

I'm an old dog, @Soham; I don't care what Huy says :P

Soham Chowdhury

@TedShifrin Hahaha

What timezone are you in?

Ted Shifrin

eastern time zone of the US, for another month

Soham Chowdhury

so it's 9:50?

Ted Shifrin

correct ...

Alyosha

13:53

What can I do to intuit fractional ideals?

Is there a good set of exercises somewhere?

Ted Shifrin

@Alyosha, you might want to ask @anon that question. I have no idea.

Alyosha

Okay, thanks.

@anon, do you have any good idea?

(See a couple of lines above).

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$Mathematics$ Mathematics