Mathematics - 2016-07-26 (page 4 of 4)

9:00 PM

@Krijn: and even more importantly, the drummer looks happy when he pulls it off, so to sum up, the teacher made the drummer happy.

Krijn

That's how this abuse is supposed to work

That's why it's mental abuse!

Huy

but then I don't see the problem of it. it is only abuse in the very moment and for the non-involved observer. in the end, the abused drummer is happy.

Alessandro

ok, that's true, I see your point @Huy I still think the movie condemns the method even though it shows good results, for those who survive it at least

Krijn

Most people in 1984 are also quite happy, and even Winston loves Big Brother, but the observer clearly sees that we should not go down that road!

Huy

@Krijn: but why is the observer supposed to be right? why does he have to intervene with a different person's life just because he thinks his way of life is so much better, even though the other person is completely happy?

Krijn

9:05 PM

Ah, that's a fine philosophical point, @Huy, and one that we seem to disagree on.

Balarka Sen

heh, i googled: they put the j. jonah jameson guy as the abusive teacher? excellent fit

Huy

@Krijn: I could just never understand that point of view, because if I'm happy with my life, I don't feel like someone else should tell me "hey, you're doing it wrong, my way makes you way happier"

(and then forces me to do it his way)

user 1618033

Back.

Balarka Sen

"being happy with life" might be a controversial concept in itself

user 1618033

@Huy Hope you think the same about different attitudes you see here (or elsewhere) toward mathematics. That's perfectly fine.

Huy

9:08 PM

@user1618033: of course! :)

user 1618033

@Huy Cool! ;)

Krijn

@Huy I get your point, I do!

Huy

I've always been the "whatever makes you happy as long as you don't hurt anyone else" kind of guy

user 1618033

@Huy What if your mathematical performance becomes annoying for a bunch of guys? Maybe they feel hurted by you.

Huy

@user1618033: ignore them and tell them to ignore you too, if they're bothered

user 1618033

9:10 PM

Of course, you struggle so much to be kind, but look, you don't manage to be (because some decided so).

@Huy Right! :-)

Huy

@Krijn: so are drums your only instrument?

Krijn

@Huy No, my main instrument is a vibraphone. Drums are more of a side thingy (which, to be fair, I have neglected for some time).

Huy

@user1618033: just one thing: if you think some people are not mature enough yet to understand, just accept it and think "he/she is just a kid, one day he/she will understand". if you say it out loud, people will likely get offended and more drama will result.

@Krijn: oh, very interesting. for how long have you been playing the vibraphone?

Krijn

@Huy I'd say at least 6 years, but I'm not entirely sure.

Huy

@Krijn: got any recordings?

Krijn

9:15 PM

No proper ones, I'm not in it for that

Huy

you don't have to "be in it for that" to record yourself :P

Krijn

Oh, I thought you meant professionally.

Huy

I have hundreds of recordings of myself just so I can listen to my amateur mistakes.

Krijn

I did record myself sometimes, I think. I wonder where those went though

user 1618033

@Huy You're perfectly right. In general I suspect that some I had tougher discussion with are somewhat older than the usual students that come here, like 50-60. For instance, I suspect Balarka is over 40-50. If I knew 100% he's just 14 as he claims, then most probably I would have never argued with him. Just an example. Once again, I agree with what you said!

Balarka Sen

9:18 PM

lol

user 1618033

Then my style of doing mathematics is very different, I'm very enthusiastic about all my work and natural in my reactions. I express joy fully naturally when talking about mathematics.

Huy

maybe over 40-50kg, if he starts to eat properly

Balarka Sen

somewhat near 40, actually

Huy

:P

what's your height?

Balarka Sen

eh

i dunno, never measured my height.

Krijn

9:20 PM

You should now approximately, right?

Balarka Sen

i don't have my centimeter scale with me right now. i lost it.

Huy

@BalarkaSen: ok, if you drop a pen from the height of your neck, how long does it take to reach the floor?

Balarka Sen

so i can't.

my guess without anything to compare with would be horrible

Krijn

Give us a 10 cm interval that you'd surely fit

Balarka Sen

@Huy pretty quick, but i don't think that's because of my height

user 1618033

9:22 PM

For instance, @robjohn showed his age some years ago on the status, and that was very surprising to me, he's so young (in the sense of being energic and involved in the stuff) that I thought he's just 25-30 at the beginning. It's really cool to be so, and I hope to be like him when I'll be his age.

Balarka Sen

i think i am moderately tall

user 1618033

Some are just cool, apart from the side that I talked a lot with robjohn. robjohn is cool!

0celo7

@BalarkaSen what

user 1618033

Anyway.

0celo7

Don't they tell you at the doctor's

user 1618033

9:25 PM

BBL (to write some more on my stuff)

Huy

@BalarkaSen: if you're moderately tall, being somewhat near 40kg seems rather underweight

user 1618033

@Huy Just one question before returning to my stuff. Have you ever seen Balarka? Would you bet that he's not 50?

Balarka Sen

@Krijn I am probably over 5ft, but less than 6ft. Can't tell you exactly were, probably a bit over 5+1/2ft.

0celo7

What's that in Freedom Units

Huy

@user1618033: not in person.

0celo7

9:26 PM

80 pounds? 90?

100?

Balarka Sen

@0celo7 Nope, only weight.

user 1618033

@Huy OK

Alessandro

40Kg is 88 pounds

now, if only I knew how many centimeters 5ft are...

Huy

@Alessandro: don't you ever measure your feet?

Krijn

@Balarka, you are definitely underweight then...

0celo7

9:27 PM

88 is underweight unless you're 5 feet, probably.

Huy

@Krijn: unless he's pretty much 5'

Balarka Sen

I know.

That's not a news.

0celo7

...eat more?

Alessandro

5.5ft is 167cm, I wouldn't describe that at moderately tall here, but average heights varies around the globe

@Huy I ony know they are big and a hassle when I need shoes

Huy

yeah, 5.5 for a guy should be roughly below average

Krijn

9:28 PM

He can still grow a lot though

Huy

@Alessandro: haha, what's your shoe size? :D

0celo7

He's like 7 though

Alessandro

@Huy 45, sometimes even 46 depending on the manufacturer, in European sizes, which is 12-13 in US sizes according to the first chart i pulled on google

Huy

ok, that's quite large

Balarka Sen

speaking of, I need new shoes.

Huy

9:30 PM

I'm like 39-41 depending on the manufacturer

but I'm also only ~165cm

Krijn

45 here also, you should go to the Netherlands, they are prepared for that

Huy

@BalarkaSen: there's a maths student frequenting the maths library at my university who doesn't wear shoes. not even in winter.

Balarka Sen

hah

Alessandro

uhm I haven't measured myself in a while, but I should be almost 180cm, like 178 or something like that

Balarka Sen

i don't actually wear shoes tho. just some sort of slippers.

if that's what you call 'em

Alessandro

9:33 PM

@Huy there was a guy like that in another high school in my city too! He was actually great at math competitions, I wonder what he's doing now

Huy

it's always the math guys.

Balarka Sen

i mean, what do they wear, then?

Krijn

Nothing, I'd say?

Just bare feet

Alessandro

calzoleriaromagnola.it/4996-large_default/pantille-paglia.jpg that

Balarka Sen

that's weird, even i wouldn't do that

Alessandro

9:36 PM

even in the winter tho

Balarka Sen

@Alessandro ah, my kind of guy

Huy

@BalarkaSen you should try this

Krijn

I wore those to uni yesterday!

Huy

and yes, they're bare feet

Alessandro

he often went barefoot inside buildings

Balarka Sen

9:37 PM

i should try barefoot.

seems appealing

Huy

please don't.

barefoot and not showering on a regular basis aren't a good combination

0celo7

@BalarkaSen you don't shower?

Balarka Sen

i do, but not as frequently

0celo7

...not daily?

Huy

@0celo7: did you put me off ignore? :o

0celo7

9:39 PM

@Huy Yes

Huy

why?

0celo7

Don't remember why I ignored you.

Huy

I do. :P

Balarka Sen

I do too

0celo7

Good for you

I know you bullied me

But I couldn't find it in the chat log

Huy

9:39 PM

@0celo7: I'm not a bully. I used to be one, in high school.

0celo7

You said something to piss me off. But it couldn't have been that bad because I've forgotten what it was.

Huy

ok

so how are you enjoying Apple Music?

0celo7

Amazing

Huy

glad you like it

0celo7

I need a new phone

128GB

Huy

9:40 PM

why, which do you have?

0celo7

64 ain't shit

Huy

I have 64

what do you store that takes up so much space

0celo7

I have like 70 gigs of music

Krijn

That's quite a lot

Huy

why not just stream it via Apple music?

Balarka Sen

9:42 PM

whenever i hear "bully", I think of the following.

0celo7

I don't have unlimited data and wifi coverage isn't 100%

Huy

ah, ok

0celo7

But I have a 5S, it's time for an upgrade anyway

Huy

true.

7's just around the corner

0celo7

Yeah, I'll get that

Huy

9:43 PM

same here

Balarka Sen

(whoa, that's way too long, i only intended to refer to the first 2 or 3 minutes of the video)

0celo7

Going to attempt some Hirsch exercises.

Balarka Sen

i wish you luck

0celo7

Thanks.

Balarka Sen

which exercises, out of curiosity?

0celo7

9:45 PM

My advisor said we'll "discuss" them, I don't think he expects me to actually solve any

@BalarkaSen I'll tell you in a bit.

Have to pack my bag and make my way to the library

Right next to the geometry and topology books

Huy

@BalarkaSen: any idea who I could ask about the residual finiteness of rings?

Balarka Sen

@Huy Tobias, probably.

He'd know.

Huy

good idea.

@BalarkaSen: BTW, why aren't you sleeping already?

Balarka Sen

Dunno. Maybe I'll bunk school tomorrow.

Huy

I'll report you.

Krijn

9:51 PM

He is a bully.

Huy

no, I'm a teacher. that's my duty.

Balarka Sen

For what? Students are allowed to not attend school.

Civil rights.

Huy

@BalarkaSen: but not to not sleep.

Balarka Sen

@Krijn Being a bully and a teacher at the same time is not a helpful combination

Not at all.

Huy

@BalarkaSen then why did I get such a good evaluation from the headmaster :>

Balarka Sen

9:55 PM

some bullies are skillful about bullying

Huy

:o

0celo7

Ok, let's see which exercises

Chapter 6

Sect. 1: 2,4
Sect 2: 5, 9
Sect 3: 6

ah, yes, 1.4 is a good problem

Balarka Sen

Oh Morse theory. OK, not interested right now.

0celo7

@BalarkaSen I'm doing Morse theory in my reading course, so yeah

Balarka Sen

Mike thinks an interested student can learn Morse theory later on, so better not to include in a differential topology course that he'd prefer, so I'm following his advice.

0celo7

10:03 PM

ok

Time to get to work

Balarka Sen

Well, not a "differential topology" course, but a "smooth manifolds" course.

0celo7

@BalarkaSen What's the difference?

I guess Hirsch doesn't assume $C^\infty$

Balarka Sen

nah, nah, not those horrible details about $C^{1 + \epsilon + \epsilon^2}$ etc. There's more geometry to smooth manifolds.

But I guess you're already familiar with a large chunk of it

0celo7

C^k is pretty horrible.

@BalarkaSen I don't think it's "geometry" until you have something besides the smooth structure.

i.e. a Lie group structure, Riemannian metric, symplectic form, etc.

Balarka Sen

Yes.

0celo7

10:08 PM

Lee is a mixed bag of topology and geometry in my mind

Balarka Sen

@0celo7 C^k is equivalent to C^infty for k > 0, I think

0celo7

@BalarkaSen I guarantee the proof of that is revolting :P

Balarka Sen

Me too.

0celo7

One day I want to read Chapter 2 of Hirsch.

Not now.

aras

Why is $\Bbb Z_4 + \Bbb Z_2$ not a free $\Bbb Z$-module?

I'm having a hard time understanding why

Balarka Sen

10:18 PM

$+$ means direct sum? $(1, 0)$ is a torsion element, e.g.

aras

Yes, + is direct sum.

What is a torsion element?

Is that a zero divisor, i.e. (1,0) * (0,1) = (0,0)?

Krijn

What is your definition of a free module

Balarka Sen

An elt $g$ in your Z-module is torsion if there is some nonzero $n$ such that $ng = 0$.

@aras Yes, but that's not relevant.

The important point is free modules cannot have torsion elements (do you know why?), whereas this one does.

aras

ah

The definition given of free module was a module that has a basis.

Krijn

Yes, but a basis needs to have linearly independent elements

aras

10:23 PM

A free module cannot have a torsion element because if it would, then there would be multiple distinct linear combinations for the same element, right? i.e. if g is torsion s.t. $ng = 0$, then $g = (n+1)g = ...$?

Krijn

So take any element $(a,b)$ in your supposed basis, then $4\cdot (a,b) = 0$

And $4 \neq 0$ in $\mathbb{Z}$

Balarka Sen

@aras Right. Also, listen to @Krijn, he's got a shorter proof.

Essentially the same, but said in a neat way.

aras

OK, but does this mean that Zp is also not a free Z-module because $p \cdot 1 = 0$?

Balarka Sen

Yep.

No cyclic group ever is a free module over Z.

Side-question: Does that tell you what the finitely generated free modules over Z are?

aras

Hm, well Z(a) for an algebraic element a is a free module I believe.

Krijn

10:27 PM

Or more in general, try to understand the structure of all $\mathbb{Z}$ modules!

@aras That's just isomorphic to $\mathbb{Z}$, right?

Balarka Sen

^

We're asking you to understand what they are upto isomorphism.

You need to know a more or less nontrivial theorem, which I am not sure if you're aware of.

Krijn

Two theorems, actually

aras

Are you referring to decomposition of abelian groups?

Balarka Sen

@aras Yes!!

aras

ok so then perhaps every Z-module would be only isomorphic to Z^n, because multiplying by any Z/pZ would add torsion elements to the group?

Krijn

10:30 PM

Every free Z-module

aras

Oh yeah sorry every free Z-module.

Balarka Sen

You need to say finitely generated there for the fundamental theorem of f.g. abelian groups to apply though.

What about non-finitely generated ones?

aras

OK.

0celo7

@BalarkaSen Good lord, the proof of this one thing involves submanifold geometry...

Principal curvatures, etc.

Hirsch is insane

Balarka Sen

@0celo7 :P

aras

10:32 PM

Would that be an infinite product $\Bbb Z \times \Bbb Z \times ...$?

I don't know much about infinitely generated groups

Balarka Sen

@aras Yes.

aras

Great, thank you for the help!!

Balarka Sen

Well, infinite direct sum, not product actually.

aras

Oh ok

$\Bbb Z \oplus \Bbb Z \oplus ...$

Balarka Sen

It's probably worth trying to come up with a torsion-free Z module which is not free.

You saw this cannot happen in finite dimensions.

So any example would be infinite dimensional.

Krijn

10:37 PM

Also, @Balarka, have you seen this one? en.wikipedia.org/wiki/…

Balarka Sen

Mhm.

It's primary decomposition, isn't it?

aras

@BalarkaSen I have to go now but I'll think about that question

Balarka Sen

Sure, sure.

aras

also thanks @Krijn

Krijn

Np.

There's so much I'd like to read if only I had the money :(

Balarka Sen

10:41 PM

For me it's time more than money: I can't even get the time to read the books my money can afford.

Let alone get more books, for which I obviously don't have money.

Krijn

I should really be reading right now

Remind me to start reading before the full hour

Balarka Sen

@0celo7 Given two oriented submanifolds intersecting transversely inside a big oriented manifold, how can I tell which point has intersection number +1 and which -1 immediately?

Just from seeing the picture?

@Krijn I'll have to run to bed soon though. Get an alarm clock.

@0celo7 Nevermind, pretty easy to do that.

0celo7

@BalarkaSen I don't have a good intuition for that. I think you take a basis of the tangent space of the first one, then start adding vectors from the second one until you get a basis of the whole manifold. Then you look at the orientation of that

The order is important

Balarka Sen

Right, just look at basis of $T_p(X)\oplus T_p(Z)$.

And what the sign of that is, relative to the orientation on the big manifold.

0celo7

Wow, you have to relate the critical points of the Morse function to the curvature of the submanifold

Hirsch, you madman

How is anyone supposed to solve this

Balarka Sen

10:49 PM

with a pen and a paper, most likely

Pushing forward orientations by path is a useful idea. I have used that in the past but now it's much clear what happens.

Should keep it in mind

Krijn

Woow I made it. I am reading.

Balarka Sen

:P

Ashwin Gupta

hey people. quick question

does cross product return the cosine of the angle between two vectors or the sin of the angle between two vectors?

0celo7

use google

Ashwin Gupta

I did, I'm still confused.

0celo7

10:59 PM

explain your confusion

Ashwin Gupta

well, It says that it is calculated with sin.

However, my dad once told me it was cosine of the angle

and it doesn't say anything about what it returns

Balarka Sen

Length of the cross product of two unit vectors is sine of the angle between them.

Ashwin Gupta

ok, thanks

0celo7

@BalarkaSen Length of the cross product.

Ashwin Gupta

that is all I wanted to know

Krijn

11:00 PM

Just remember $|| a \times b || = || a || \cdot || b || \cdot \sin \theta$

Balarka Sen

@Ashwin It tells you about area spanned by the two vector.

0celo7

Does anyone actually prove that?

Balarka Sen

Area of a parallelogram with sides $a,b$ is $ab\sin(\theta)$.

Ashwin Gupta

okay

Balarka Sen

@0celo7 Huh?

0celo7

11:01 PM

@BalarkaSen I don't think I've ever seen a proof of @Krijn 's formula, but I haven't looked.

Ashwin Gupta

well TBH I don't know much Trig. I've taken Geometry, haven't taken Algb2/Trig yet. I know basic stuff like law of sins, law of cos, sin area formula, etc.

Krijn

I did once, in a homework exercise @0celo7

Ashwin Gupta

I just needed to know because I'm programming lighting.

Balarka Sen

@0celo7 It's the definition, to me. :)

0celo7

@Krijn Is there a trick or is it a boring computation?

@BalarkaSen Of what?

Balarka Sen

11:02 PM

Cross product.

0celo7

The cross product is the wedge composed with the Hodge star.

Krijn

@0celo7 There's a small trick, I believe

Balarka Sen

Orthogonal to $a, b$, with length the signed area of the parallelogram they span.

Krijn

It was almost 4 years ago so I wouldn't know

0celo7

@BalarkaSen + right hand rule for the sign.

Balarka Sen

11:03 PM

You can translate that into determinant thingy

Note that area of the parallelogram $a, b$ span is the same as volume of the parallelpiped $a, b, n$ span where $n$ is the unit vector orthogonal to $a, b$.

Then that's $\text{det}(n, a, b)$.

That is all there is to it.

0celo7

@BalarkaSen Yeah, I imagine that's how the proof would go.

Balarka Sen

OK, gotta sleep.

0celo7

bye

aste123

11:19 PM

Hello, I'm facing difficulty in formally proving that a function is onto function. If it is not onto function, then the proof is easier because I can show a range of output for which there is no possible input. But if it is a onto function, then how to prove it?
Take a look at this please: http://math.stackexchange.com/questions/1872214/how-to-formally-prove-whether-this-function-is-onto-or-not-kx-x-2-where

aste123

11:31 PM

@0celo7 hey, can you help me with this question? I assume that it boils down to basically proving that every non negative real number has a square root that is non-negative in itself.

@Krijn

@Semiclassical can you help me with this ? :)

0celo7

Existence of square roots is a nontrivial theorem in analysis

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$Mathematics$ Mathematics