Mathematics - 2015-05-25 (page 2 of 4)

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user147690

11:00 AM

@SohamChowdhury I haven't done those chapters of aluffi, and he definitely approaches the subject from a unique perspective (e.g. category theory, which isn't taught at my uni)

Soham Chowdhury

DF doesn't cover them?
And I know around three pages about them.

@AlexClark Oh.

user147690

@SohamChowdhury I haven't consistently covered any algebra text except Cohn's first text(which I was only around 80 pages deep in at the time I left it)

Soham Chowdhury

You're doing DF, right?

user147690

@SohamChowdhury The ring theory section yep

Soham Chowdhury

Pinging @BalarkaSen for when he comes online: are free abelian groups vector spaces?

@AlexClark only that bit?

user147690

11:03 AM

@SohamChowdhury Well I will do all of it eventually I hope

user147690

@SohamChowdhury But I wasn't using D&F until recently

vector spaces over what?

what's the ring/field you have in mind?

Soham Chowdhury

Listen. I know nothing about rings or fields "properly" (aside from the first three pages of Axler). Look at this:

user image

how can I answer your question if you don't know what your question means?

Soham Chowdhury

wait.

I guess vector spaces over ${0,1}$?

11:06 AM

Vector spaces over sets are nonsense.

Soham Chowdhury

ah.

so there's no connection?

look at the screenshot. that thing looks like a basis to me.

Yes.

That's because $\Bbb Z^n$ is a $\Bbb Z$-module.

user147690

@BalarkaSen You can define a vector space over a ring?

Soham Chowdhury

so I'll wait till next chapter.

for modules.

Module theory is all about that, yes.

Soham Chowdhury

11:08 AM

mmm

@Soham However, the connection is not really interesting. Every abelian group can be given a structure of a $\Bbb Z$-module, for one.

Soham Chowdhury

11:19 AM

Sarat Books says they can get me Aluffi for $90.

Goodness!

That's nice.

They've got a lot of books with them.

Soham Chowdhury

That's a lot of money, though.
How much do your books usually run you, on average?

I doubt they always have them in stock. You do a "special order", right?

Sometimes you do, but most of the times the books I want are there.

*I do

@AlexClark You're around?

user147690

11:40 AM

@BalarkaSen Yes, was just taking a break

Did you figure out that $\Bbb R[x]/(x^3 + 1)$ thingy?

user147690

@BalarkaSen Not yet, I was going to fully go through the relevant section in D&F to get it

There are better, easier ways to do it than what my hints lead you to. Just sayin'.

You're awake early, @Ted.

user147690

@BalarkaSen Oh?

howdy @Balarka @AlexC

11:50 AM

Fields have no zero divisors, @AlexC.

user147690

Hey @TedS

user147690

@BalarkaSen ahhh indeed

Well, in two days I need to be on an airplane at this time, and the airport is 1 1/2 hours away.

user147690

@TedShifrin Fair enough. Do some natural prep

So you're practicing waking up early?

11:51 AM

Plus, I'm about to switch time zones to three hours earlier ... Yes.

But, actually, I'm always up at this time; just didn't go on line when I was going to work.

Switching time zone can be a mess. US is a big place.

So is India.

user147690

So is Straya'

Well, now that we're done with that ...

user147690

Are we though?

11:56 AM

australian pronunciation rolls eyes

<--- likes 'Strayan accents. :P

user147690

We don't actually pronounce it like that lmao

user147690

@BalarkaSen You'll hear my lack of accent when I make a video for tropical geometry

Damn, Balarka. You've also learned eye rolling from me.

It took some practice, @Ted. Last time I tried to do it, one of the eyes popped out.

11:57 AM

You really do need to eat more, Balarka.

Dunno how you managed to roll all the 6 of 'em.

Oh oh, bananas is here.

iwriteonbananas

TED!!

<3<3

how is retirement?

Such a flirt ... :D

I'll let you know in six months or so.

iwriteonbananas

good, make sure u keep frequenting this chat

user147690

12:00 PM

Where are you @ᴇʏᴇs? Weird you being gone for 4 days

iwriteonbananas

@TedShifrin they're vile

LOL.

What kind of accent do you have, bananas?

iwriteonbananas

i dont know, i never speak to people in real life

the kind of accent everyone has when they're chewing on bananas, of course

user147690

Can someone listen to some of this, these guys are a band near where I live, and their accents are pretty normal Australian youtube.com/watch?v=kUfCHA0hZwg

12:02 PM

no

You're mute, bananas?

iwriteonbananas

people often tell me i speak like a robot because i speak very montonously

user147690

@BalarkaSen Not a song, it's an interview

iwriteonbananas

@TedShifrin no, just dont have any friends

user147690

@iwriteonbananas Actually?

iwriteonbananas

12:04 PM

cries

well, bananas, work on changing that. Even math geeks need friends.

iwriteonbananas

im sort of joking, but i live in a new city and most of my friends live in another city

find another math geek.

user147690

@iwriteonbananas That's hard. You are a grad student though right?

But math geeks are a set of measure zero, @Balarka, except here.

iwriteonbananas

12:05 PM

no, undergrad

@TedShifrin human population is a set of (lebesgue) measure zero

we're working on changing that, @Ted

oh, I thought you meant here here, @AlexC

user147690

Nah from my classes

user147690

Here it seems quite random

My closest friends through my life have not been math people, @AlexC, but I've had plenty of good friends who are.

iwriteonbananas

12:07 PM

most people i know from uni i only talk about math with

user147690

@TedShifrin Yep I think that makes sense. You have a larger sample for generic friendship qualities from the general population with lower expectancy for competitivity

iwriteonbananas

@AlexClark yeah, it sounds terrible

:P

user147690

@iwriteonbananas Really??

iwriteonbananas

absofruitly

user147690

I can't really hear American or Australian accents

12:08 PM

It doesn't need to be competitive, @AlexC. My best math research has been collaborative with very good people.

give me a good algebraic topology question, @iwriteonbananas

user147690

@TedShifrin I personally am not competitive, since it seems unnecessary conflict - but some of my classmates seem to be

iwriteonbananas

@BalarkaSen im looking for exercises regarding mayer vietoris

user147690

Well I will grab a coffee and get back to work, talk later guys

iwriteonbananas

take care alex

12:09 PM

compute homology of two torii pasted via boundary with identity map

Good Mayer-Vietoris exercise: Take two solid tori and glue them together with a linear map $A\in M_{2\times 2}(\Bbb Z)$ on the boundary. Compute the homology of that identification space.

@BalarkaSen LOL, how funny. Mine's just a bit more general.

iwriteonbananas

loll

okay

@TedShifrin the whole point of M-V is, of course, computing homology of 3-manifolds.

so no surprise that we've come up with similar exercises

"The whole point"? Very myopic view.

Mayer-Vietoris for deRham cohomology is very snazzy. You see everything explicitly (well, as explicitly as you can compute partitions of unity).

ok, most of the point.

12:13 PM

No, not most of the point.

OK, bubye for now.

@TedShifrin that sounds cool. can't wait to know about that stuff

iwriteonbananas

bye ted

once you do my exercise, i'll teach you a way to cheat your way through

iwriteonbananas

cool

hint for cheating (not recommended for official use) : two torii pasted through the boundary by identity map is a known space.

iwriteonbananas

12:17 PM

yeah im thinking about that right now

u mean solid torii btw. right?

don't. do it the hard way.

yeah

iwriteonbananas

i cant recall what space that is anyway right now (although i think u told me like 1-2 days ago)

im trying to figure out how to do it the hard way

no, i don't think i did

iwriteonbananas

ok

once you do this, try computing homology of the space obtained from a solid torus with a tubular nbhd of a circle (also homeomorphic to a solid torii) chucked out, and the boundary of the removed nbhd identified with the boundary of the original solid torus using the identity map (non-official hint : this is also a known space)

iwriteonbananas

12:21 PM

okok, gimme a minute

i have to run for now, will check in on you later.

iwriteonbananas

ok, talk to u later

Soham Chowdhury

12:41 PM

Is it anything special if $G \times G \cong G$ for a group $G$?

user147690

@SohamChowdhury journals.cambridge.org/…

Soham Chowdhury

Those are some weird constructions.

I know that $\mathbb{Z}^{\oplus \mathbb{N}}$ works.

user147690

$\oplus$?

user147690

oplus

user147690

@SohamChowdhury I don't know what this means

user147690

12:49 PM

Still don't know what that is

Soham Chowdhury

It's a "direct product of groups". I have a sort of vague idea of what that is atm, really.

You know what $\mathbb{Z}\oplus\mathbb{Z}$ is?

user147690

It is $\Bbb Z\times \Bbb Z$

Soham Chowdhury

Yes, products and coproducts are isomorphic in $\sf{Ab}$.

user147690

$\Bbb Z \oplus \Bbb Z = \Bbb Z\times \Bbb Z$

Hello everyone. I need some help with linear algebra.

user147690

12:52 PM

@SohamChowdhury and $\bf{Vect}$

Soham Chowdhury

@AlexClark I'll be back in a while.

Pretty new to it so I'm struggling a bit.

user147690

@nTuply Shoot

Soham Chowdhury

@AlexClark It'll take me some time to know that.

user147690

@SohamChowdhury Have fun!

Soham Chowdhury

12:52 PM

Oh, I'm having loads.

user147690

@SohamChowdhury Haha probably not, you are sprinting through this stuff

@AlexClark I'm trying to figure out whether a set is a vector space or not/

user147690

@nTuply A set?

I understand the general idea of what a vector space is and I understand the axioms/

However I find it hard to solve such problems. Lemme give you an example.

user147690

Ok

12:55 PM

$V = \{p\in P_{4}[x] : p(0) + p(1) = 0\}$

user147690

So polynomials of degree 4 such that $p(0)+p(1)=0$?

yes

Soham Chowdhury

@AlexClark I just go over things until they click. Which is sort of the canonical way to learn math, I guess.

But I intend to finish this chapter this week.

user147690

@SohamChowdhury What do you eat?

Soham Chowdhury

@AlexClark ?

user147690

12:56 PM

@nTuply Taking values from where?

user147690

@SohamChowdhury Just wondering what you eat since what you eat says a heap about your energy levels and focus

What values? This is the only thing that's given and I have to determine whether is a vector space or not

user147690

@nTuply I mean what is your field

Soham Chowdhury

@AlexClark Rice, fish, chicken, the occasional veg if I must. Standard Bengali fare.

@AlexClark You're confusing him. (or maybe not)

@AlexClark the Real numbers

user147690

12:57 PM

@nTuply So it's a set with the field $\Bbb R$

Yes

Soham Chowdhury

@nTuply Try checking if it satisfies all the axioms.

How to render latex in the chat btw?

user147690

So already we don't just have a set, which is good. Now you know all of the axioms?

Soham Chowdhury

Do you have ChatJax? @nTuply

12:58 PM

@Soham nop

Soham Chowdhury

It's off the starboard, shit.

user147690

May 10 at 10:15, by Paul Plummer

Chat guidelines | $\LaTeX$ in chat

Okay I got it from Google was on a UCLA site

Soham Chowdhury

Yeah, do what it says.

Okay my Latex works now

Soham Chowdhury

12:59 PM

Cooleo

p(x) is a function, right?

user147690

> Chat guidelines | $\LaTeX$ in chat

Soham Chowdhury

@nTuply check if it satisfies all the axioms, one-by-one

@Soham I understand that but how? I mean what does $p(0) + p(1) = 0$ mean in that case?

user147690

@nTuply Important to note what $P(0)$ is and you will know

Soham Chowdhury

$p(x)$ is a function, @nTuply.

p(0) is its value at 0.

1:01 PM

$p(0)$ is a polynomial of degree at most 4, where we replace all the x by 0 so this one satisfies the equation

user147690

So what does $P(0)$ look like?

now for $p(1)$ it's going to be just the sum of the coefficients of the linear combination

how can the be equal to 0

Soham Chowdhury

Take $p(x) = x^4+x^3+x^2+x^1-4$

ok

user147690

$P(0)=a\in\Bbb R$

1:02 PM

Makes sense.

Soham Chowdhury

$p(1) = 4 -4 = ?$

But what about $x^4 + x^3 + x^2 + x^1 - 1$

In that case it fails.

Soham Chowdhury

So essentially the problem says that for $p(x) = ax^4 + bx^3 + cx^2 + dx + e$, $p(0) + p(1) = a + b + c + d + 2e$ (check why that's true, simple substitution).

Do all such polynomials form a vector space? That's what you have to show.

Makes sense.

Yes. Because all obey vector addition and scalar multiplication and 0 is in the set

Soham Chowdhury

And are all linear combinations also in the set?

1:05 PM

But what about cases where the values don't equal to 0?

Yes.

Soham Chowdhury

@AlexClark help him out, gotta run to get some stuff from the store

For example say a random tuple a to e (-1,2,3,4,1)

That' will not be equal to zero although it obeys all the other axioms/

Soham Chowdhury

Then . . . ?

Is zero in the set, as you say?

no

iwriteonbananas

@BalarkaSen is $H_2(X) \approx \Bbb{Z} \approx H_1(X)$ ?

1:14 PM

what is $X$?

iwriteonbananas

two solid tori glued along their boundary via identity

then yes

and higher homologies are?

iwriteonbananas

i get that $H_3(X) \approx \Bbb{Z}$ but i think it should be 0

it is 0

iwriteonbananas

what decomposition do you use for mayer vietoris?

1:16 PM

$U$ is the first torus plus bit of the other torus

and $V$ is the second torus plus bit of the first one

iwriteonbananas

yeah that's what i did

$U,V\approx S^1$

and their intersection is a torus

yes

the intersection is a torus

iwriteonbananas

i just said that :P

i am doing too many things at once, so i didn't see that

iwriteonbananas

np

1:19 PM

what's the sequence you get at $H_3$?

iwriteonbananas

the problem is figuring out what the maps do

right, do it.

iwriteonbananas

ok

also, is this space just $S^2 \times S^1$?

indeed : how did you find out? :P

iwriteonbananas

i read it somewhere sometime...i think mike once told me

when we were talking about exercise 2.1.7

1:22 PM

But not when you throw in my general $A\in M_{2\times 2}(\Bbb Z)$. Then you get all sorts of interesting things.

right, let me tell you how to visualize it :

iwriteonbananas

he remarked that the space u get from gluing two solid tori along their boundary depends a lot on the homeomorphism, it could be $S^2\times S^1$ or lens space

or many other things

take two solid torii. take two circles at the same position in the boundary of the two torii.

iwriteonbananas

ok

when you glue by the identity map, you're gluing these two circles on the boundary of the torii.

iwriteonbananas

1:24 PM

yea

but the interior is left unidentified, right? so think about two disks bounded by each of the circles, and you're gluing the boundary of the two disks.

@Balarka: Am I taking a circle that is nullhomologous or the one that generates $H_1$ of the solid torus?

this gives you a sphere at every point, and doing this globally brings back $S^2 \times S^1$

@TedShifrin the nullhomologous circle

You'd best have said that? :)

yeah, my bad.

1:26 PM

Yeah, you're doing $D^2\times S^1$ and talking about $\partial D^2$.

BTW, in preparation, goodnight @MikeM.

@iwriteonbanana but, as @Ted says, this gets gradually impossible to visualize when your gluing maps are wildly complicated.

Well, what if you glue by reversing orientation, @Balarka? Let bananas think about that one.

take a wacky 3-manifold and it's heegard decomposition, for example.

Yeah, 3-manifolds people think about Dehn twists and can visualize it fine.

good problem, @Ted

oh?

iwriteonbananas

1:29 PM

i dont really understand your visualization. so you take nullhomologoous circle on each solid torus, then do the gluing. that is, you're gluing two disks along their boundary which gives $S^2$

right. and you do this for each pair of circles at the same position.

so you're sticking an $S^2$ at each point of the circle : that gives you $S^2 \times S^1$

@TedShifrin i'll be interested to listen if you can elaborate on dehn twists.

iwriteonbananas

ok, i guess

what's $S^2$ with antipodal points on the equator identified?

you want to know the cell decomposition?

iwriteonbananas

yeah

@TedShifrin is that $S^3$?

no.

iwriteonbananas

1:35 PM

ok

@iwriteonbananas $e^0 \cup e^1 \cup e^2 \cup e^2$ with $e^2$s pasted to $e^1$ via degree 2 maps.

iwriteonbananas

ok

trying to figure out what we get if we glue by reversing orientation

$S^3$ is two solid torii identified to each other via gluing the nullhomologous circle of one to the non-nullhomologous circle of the other.

how is that map $x \mapsto -x$?

iwriteonbananas

yeah, my bad, that's true

@BalarkaSen can you help me understand how $H_1(X)$ is computed w/ mayer vietoris?

we have the sequence $H_1(S^1\times S^1) \to H_1(S^1) \oplus H_1(S^1) \to H_1(X) \to H_0(S^1\times S^1) \to...$

i.e. $...\to\Bbb{Z}^2 \to \Bbb{Z}^2 \to H_1(X) \to \Bbb{Z}\to...$

what's the map $\Bbb{Z}^2 \to H_1(X)$?

i guess it's $(x,y)\mapsto x$

the generators of each circle are homologous in $X$

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