Mathematics - 2015-02-05 (page 2 of 2)

bolbteppa

16:00

I mean, if you look in a number theory book, you'll see them talk about divisors, then the Euclidean algorithm, then Bezout's theorem (reverse Euclidean algorithm to find the GCD), then Unique factorization theorem

Then they'll do congruence and Fermat, Euler etc...

You can do that in a few pages if you want, well ring theory just expands every step of this process out

So a 'ring' is analogous to the natural numbers, and quotient rings are analogous to congruences on the natural numbers. In a ring theory book they do those both at the start, but in a number theory book that's come after all this divisor stuff, so its hard to see the similarities already due to the presentation

I don't know how to describe ideals in the nicest way but intuitively they are related to the partition of the natural numbers you make when constructing a congruence

The idea is to unify numbers over N or congruences of N and polynomials and matrices and shit, and talk about the structures being similar via homomorphisms

Studentmath

I can see what you mean, yeah - they do try to explain the similarities in my book quite often, also toe xplain where the names come from, and how they expand on number-theory concepts to get to the new theorems, etc.

bolbteppa

Once you have this basis down the rest of the theory is just about divisibility, as you'd do in a number theory book, only for numbers, polynomials etc... all at the same time :)

Yeah I seen all that too but I really had no idea wtf was going on haha

Studentmath

I get that feeling no matter what :P

bolbteppa

So a domain generalizes this notion of solving polynomials by factoring, x^2 - 1 = 0 ---> (x - 1)(x + 1) = 0, if ab = 0 then one of these equals zero, a = 0 or b = 0, but the same notion kind of applies in congruences, i.e. 3.2 = 6 = 0 where 3=/=0 or 2=/= 0 so we have to be careful about when it actually applies

An integral domain just makes sure that when 1*(x - 2) = 0 that 1 =/= 0, i.e. that x = 2

So integral domains are domains, ab = 0, where the integral element 1 is not zero

Then there are 3 common types of integral domain that arise just mimicking this basic number theory stuff

So a Euclidean domain is an integral domain where you can perform the Euclidean algorithm (notice that was the first number theory topic I mentioned), then a Principal Ideal Domain is the name given to a structure in which we can perform the Reverse Euclidean Algorithm to find the GCD of two numbers, it's just formalizing that basic simple example, that's the motivation for it :)

Finally a Unique Factorization Domain just formalizes the unique factorization theorem for the integers that commonly comes after all this theory in a basic number theory book, and they relate it all

That took me so long to see

Mathy Person

16:21

Oh wow ^

Can anyone help me with this? "Find the imaginary part of $(\cos12^\circ+i\sin12^\circ+\cos48^\circ+i\sin48^\circ)^6$."

I've noticed that whenever I enter this room......

....everyone mysteriously goes quiet O_O

bolbteppa

@MathyPerson maybe if you write it as (e^i12' + e^i48')^6 and expand it out

Daniel Fischer

Better, write it as $$\Bigl( e^{\pi i/6}\bigl(e^{\pi i/10} + e^{-\pi i/10}\bigr)\Bigr)^6.$$

Mathy Person

@bolbteppa: What is '?

bolbteppa

@MathyPerson youtube.com/watch?v=qjT3XvS7Qno

Can anyone give an intuitive explanation of modules like I did with ring theory? :)

Exterior

Not sure about intuitive, but a naive-yet-useful way to view modules is as "almost-vector-space", except now the structure of the ring has an enormous impact on how much linear algebra you can do.
You could also say modules are rings acting on abelian groups.

bolbteppa

16:32

Yeah I know those thanks :) You could say a similar thing about ring theory (Oh I did sadly haha) and miss the above beautiful explanation...

Exterior

@bolbteppa hehe It was pretty clear to me from the get-go I won't be teaching you anything new about algebra :p

bolbteppa

Haha well you definitely could, I have tons of gaps :D

Link

Hi, question. I need to use a chi-square test for one of my bio lab reports. But I'm not sure what the actual test is used for? Like what does the value it give represent?

bolbteppa

I've found something beautiful for group theory, for ring theory, for Galois theory, but not modules, even Category theory (related to modules in a lot of places) has a nice intuitive motivational concept stringing lots of it together (homotopy)

Mathy Person

@DanielFischer I reduced that down to $$(\cos (\frac{\pi}{6}) + i \sin (\frac{\pi}{6}))(2 \cos (\frac{\pi}{10})$$. What do I do next?

Exterior

16:35

@bolbteppa how does homotopy string together things like groups and rings?

Daniel Fischer

@MathyPerson Raise each factor to the sixth power.

bolbteppa

No it strings some aspects of category theory together sorry :)

Exterior

@bolbteppa can you elaborate more :)?

Mathy Person

@DanielFischer: So something like $(\cos^{6}(\frac{\pi}{6}) + i \sin^{6} (\frac{\pi}{6}))(64 \cos^{6} (\frac{\pi}{10})$??

bolbteppa

This math.stackexchange.com/a/438697/82615 & this math.stackexchange.com/a/399960/82615 (the whole page is good) are pretty cool, once I thought of a category as formalizing the notion of paths in the plane it made sense of everything for me

Daniel Fischer

16:38

@MathyPerson The first is wrong. $$\bigl(e^{\pi i/6}\bigr)^6 = \,?$$

bolbteppa

There are more concepts I have to go over and fit into this little idea but they seemed to fit when I went through them ages ago

Mathy Person

@DanielFischer: Would that be $e^{\pi*i}$?

Daniel Fischer

Yup.

Exterior

Ah, yes, the first link is essentially what inspired Eilenberg and Maclane

homotopies were the original 2-arrows :p

@bolbteppa there's a great book - From a Geometeric Point of View - A Study of the History and Philosophy of Category Theory, It's a very interesting read

bolbteppa

Yeah I'm so happy this stuff was not actually abstract nonsense but had nice pictorial motivation haha

Sweet, thanks man

Exterior

16:40

Thanks for the ring explanation!

bolbteppa

Woah, it links to Klein's Erlangden program at the very start!

Mathy Person

@DanielFischer: Does that mean, when simplified, it is: $-(2 \cos (\frac{\pi}{10}))^{6}$?

Exterior

@bolbteppa From a Geometric Point of View :p

Mathy Person

@DanielFischer: Does the answer factor out into a nice, whole integer?

Daniel Fischer

@MathyPerson It is.

@MathyPerson Yes.

Mathy Person

16:44

@DanielFischer: $\frac{\pi}{10}$ is equivalent to $18$ degrees. How would it factor out into a nice, whole integer?

Daniel Fischer

@MathyPerson Look what you really want to have. That is a nice integer.

Mathy Person

@DanielFischer: Are you referring to -(2)^{64}?

Daniel Fischer

@MathyPerson No. Check again what the asked for result is.

Mathy Person

@DanielFischer: It is to find the imaginary part.

@DanielFischer: Oh! Is it 0?

Karlo

does someone know the combinatorial proof of $ \binom{n}{m} = \binom{n-1}{m}+\binom{n-1}{m-1} $ . Something with team and coaches?

Ramanewbie

16:49

how to solve for $x$ $4x^2-4x-\sqrt{2x-1}-1=0$

Mathy Person

@DanielFischer I'm assuming it is 0 because the imaginary parts cancelled each other out.

Daniel Fischer

@Ramanewbie Substitute $t = \sqrt{2x-1}$.

@MathyPerson Don't assume, show. Which imaginary parts cancelled each other?

Ramanewbie

@daniel I'm trying to make a polinomial form, but the problem is that there's $x$ under the root...

Mathy Person

@Ramanewbie: Did you try squaring the entire equation?

Ramanewbie

@MathyPerson I can't solve 3rd degree polynomials...

Daniel Fischer

16:52

@Ramanewbie That's why I suggested $t = \sqrt{2x-1}$.

Not that it will give a pretty solution. If it were $4x^2 - 4x + \sqrt{2x-1} - 1$, that would be nicer.

Mathy Person

@DanielFischer: We are left with $-(e^{\frac{\pi}{10}i} + e^{-\frac{\pi}{10}i})^6$. In the parentheses, this simplifies down to $\cos (\frac{\pi}{10}) + i \sin (\frac{\pi}{10}) + \cos (\frac{\pi}{10}) - i \sin (\frac{\pi}{10})$. This is where the imaginary parts cancel out.

Daniel Fischer

@MathyPerson Yes.

Mathy Person

@DanielFischer Therefore...it is zero?

Daniel Fischer

@MathyPerson Therefore, the entire thing is real, and hence the imaginary part is $0$.

Mathy Person

@DanielFischer: Cool! Thanks for all of the help!

Ramanewbie

17:07

@daniel I get $4x^2-4x-t-1=0$ then? And ? That's not a polynomial...

user159870

Hello!

I have a question. The moderators of math.stackexchange are getting paid for being a moderator?

@ArthurFischer @DanielFischer

Daniel Fischer

@Ramanewbie You need to substitute more. Does $4x^2 - 4x$ remind you of something having to do with $2x-1$?

@user159870 We get exactly $0$ units of [relevant currency] for doing it. Next year, it will be doubled.

Ramanewbie

ri

user159870

So you are doing it just for fun? @DanielFischer

Ramanewbie

@daniel $(2x-1)^2=4x^2-4x+1$

Daniel Fischer

17:11

@user159870 No, the fun part isn't large. We do it because somebody has to do it.

@Ramanewbie Right. So if you write the entire thing in terms of $t$, you get?

user159870

Ok @DanielFischer

Sayan Chattopadhyay

@Ramanewbie take the square root to the other side an then square it

Do that I will get it

Ramanewbie

@sayan is I square both side I will get a third degree polynomial, and I don't like that...

Sayan Chattopadhyay

U

Ramanewbie

@daniel the problem is that there isn't $4x^2-4x+1$, but $4x^2-4x-1$ in my equation

Sayan Chattopadhyay

17:18

Take the square root to that side

Then square it up

Daniel Fischer

@Ramanewbie That's why you don't get something really nice. If it were $+1$ or $+\sqrt{2x-1} - 1$, you'd get something nicer.

$t^4-t-2$ is what you get. (And I'm off to dinner, see you later.)

Ramanewbie

@daniel see you

Arthur Fischer

@user159870 I got a T-shirt, a cap, and a sticker.

I didn't go for the optional business cards.

Ramanewbie

anyone else have an idea ? I'm trying to solve for $x$ : $4x^2-4x-\sqrt{2x-1}-1=0$

Sayan Chattopadhyay

I hav one raman

Ramanewbie

17:23

@sayan yes ?

bolbteppa

@Exterior @Studentmath I think ideals are literally the formalization of division/factoring math.stackexchange.com/a/63918/82615 !!!

Sayan Chattopadhyay

Take only $\sqrt of 2x-1$that side.. Then on the left hand side u will have $4x^2-4x-1$

Find the zero for the left hand side polynomial

U will get a value for x

Ramanewbie

@sayan "$\sqrt of 2x-1$" ?

Sayan Chattopadhyay

I can't write that

Its the square root part

Ramanewbie

show a pic !

Sayan Chattopadhyay

17:26

I m using a phone so cant

Ramanewbie

@sayan oh ok... where are you ?

Sayan Chattopadhyay

Sitting in front of my whiteboard

Ramanewbie

@sayah your whiteboard ? what's your job again ?

Sayan Chattopadhyay

Oh I bought one

I love whiteboards and chalkboards

A student

Ramanewbie

@SayanChattopadhyay are you @sayan's father ??

Sayan Chattopadhyay

17:29

Nope @Ramanewbie

He would have not let me do maths

Ramanewbie

@sayan I really don't understand what's going on...

there are both "sayan" and "SayanChattopadhyay"

Sayan Chattopadhyay

Nope there is only @sayan chattopadhyay

Chuck that topic @Ramanewbie

Ramanewbie

but why can I see both on the left side of the screen ?

Sayan Chattopadhyay

Let's do maths

Huy

What's up, @TedShifrin?

Ramanewbie

17:31

@ted I didn't even saw you were there...

Mike Miller

Morning, @Ted

Sayan Chattopadhyay

I don't know malfunction in your pc

Ramanewbie

@mike can you see both written "sayan" and "SayanChattopadhyay" at the left side of your screen ?

Sayan Chattopadhyay

@Ramanewbie he is ignoring me

Ramanewbie

@sayan oh bad luck... anyway !

Varun

17:34

Hi, I would like a recommedation for a book on Statistics. I have a fairly strong background in Mathematics.I am taking a course on Statistics and Probability but loooking at the content in books like Joel, Spencer the treatment is so simple, that it feels to boring intellectually

Ramanewbie

@sayan so what did you propose me to do for that equation ? I didin't really get it...

Sayan Chattopadhyay

Take the square root part to the right hand side

Now there is a quadratic equation on the lhs

Solve it for x

Ramanewbie

what's the right "hand" @sayan

Sayan Chattopadhyay

Oh okhay that side where zero is there

Famous Blue Raincoat

@Ramanewbie If a message is long, there is more room for the name, so more of it gets shown.

Ramanewbie

17:36

of course... @sayan

@famous what are talking about ?? Didn't you mistake destinary ?

Sayan Chattopadhyay

Now u get a polynomial in x on the other side where the square root is not there

Solve it for x

Then put this value for x in the square root part

U get the zeros of x

Ramanewbie

@sayan I haven't done anything yet, because taking the root of the right side is nonsense (it's $0$).

@sayan from the beggining, what must I do ?

Sayan Chattopadhyay

It is going to be $0-sqrt of 2x-1$

When u take it

Take the full value not only the sqrt

Ramanewbie

@sayan why don't you use $LaTeX$ ??

Sayan Chattopadhyay

How

Ramanewbie

17:41

you put the latex expression into "$"

Sayan Chattopadhyay

I'm but nothing is happening

Ramanewbie

??

do you have chatjew installed @sayan ?

Sayan Chattopadhyay

Nope

Ramanewbie

then go there : math.ucla.edu/~robjohn/math/mathjax.html @sayan

Sayan Chattopadhyay

From play store right

Ramanewbie

17:43

@sayan no

@sayan just go on that page and pin it to your bookmarks

Varun

Hi, I would like a recommedation for a book on Statistics. I have a fairly strong background in Mathematics.I am taking a course on Statistics and Probability but loooking at the content in books like Joel, Spencer the treatment is so simple, that it feels to boring intellectually

Ramanewbie

@sayan so what must I do ?

Mary Star

18:18

http://langacademy.net/vb/showthread.php/38930-Fit-furs-Goethe-Zertifikat-C2-Grosses-Deutsches-Sprachdiplom-2012
Can this be downloaded??

@AnnaLear I want to register but they require an invitation code.. How can I get one??

Ramanewbie

@hippa how can I find $x$ for $4x^2-4x-\sqrt{2x-1}-1=0$ ?
I'm trying to get a polynomial, or to factorise sth, but I can't...

Hippalectryon

@Ramanewbie Where is that from ?

Ramanewbie

18:34

@hippa my "shaken hands" problem

Sawarnik

@Parth Why are you nowhere to be seen?

Hippalectryon

@Ramanewbie easy

Ilya_Gazman

Hello

Can some please look in to my question and tell me if it is structured well?

Ramanewbie

@hippa any idea ?

Hippalectryon

@Ramanewbie $x=\frac{1}{3} + \frac{5 \sqrt[3]{2}}{3 \sqrt[3]{3 \sqrt{249} + 79}} + \frac{2^{\frac{2}{3}}}{12} \sqrt[3]{3 \sqrt{249} + 79}$

Anna Lear

18:41

@MaryStar What?

Ramanewbie

@hippa well thak you... But how did you find that ??

user159870

Helloooooo

Is it me you're looking fooooor???

I can seeee it in your eyesss. I can seee it in your smilee :)

@Ilya_Gazman

HAHAHHAHAHAHAHAHHAH

Sawarnik

@Hippa @Raman Why is one of you online if and only if the other one is?

Ramanewbie

@Sawarnik haha...

@hippa tox works again !!

Ilya_Gazman

@user159870 lol

Ramanewbie

18:48

so @hippa

Ilya_Gazman

@Sawarnik you did not defined who is the one who is the other. So it could be that the other online and the one is not.

Hippalectryon

@Ramanewbie It's just a 3rd degree equation

Mary Star

@AnnaLear How can I download it?? I need it...

Anna Lear

Download what?

evinda

@AnnaLear The book

Anna Lear

18:53

How would I know?

evinda

Aren't you a computer scientist? @AnnaLear

Mary Star

@AnnaLear Yes I want the book

Anna Lear

Uh. I have a CS degree, yeah. Why does that mean I know how to download some book of a website I've never seen before?

evinda

@AnnaLear I don't know.. @MaryStar may thought so..

@AnnaLear Which is your favorite domain from computer science?

Hi @Hippalectryon @Ramanewbie

Anna Lear

@evinda I don't know that I have a favorite, but I enjoyed graphics programming and AI the most out of university. Been a while since I've done either, though.

evinda

18:59

@AnnaLear Aha!

Arthur Fischer

@AnnaLear Maybe you could build some sort of AI to download books from website you've never seen before.

Just a thought.

Anna Lear

That's an excellent idea. And why limit it to the books that have already been written? I could have another AI that would write books on any subject.

Arthur Fischer

@AnnaLear You know you don't need any AI for that! Just an infinite number of monkeys, an infinite number of typewriters and an infinite amount of time.

user61230

@Arthur Producing infinitely long books...

Balarka Sen

And (uncountably?) infinite amount of garbage along the way

Arthur Fischer

19:06

@Emrakul Yikes! Won't be able to get all the infinitely long books that way.

Need more monkeys.

Balarka Sen

$\aleph_1$ monkeys should do it, @ArthurFischer

Arthur Fischer

@BalarkaSen Only if you assume CH.

Anna Lear

@ArthurFischer Yeah, but then I have to hire monkey keepers... That sounds very 20th century. Needs moar cloud.

Synergy.

user61230

@Anna Just ask Amazon for their Mechanical Infinite Collection of Monkeys.

Balarka Sen

@ArthurFischer I daren't imagine how fuzzy set of monkeys will look like.

Arthur Fischer

19:10

@BalarkaSen I thought most monkeys are fuzzy.

@AnnaLear So crowdsource the monkeys? Could work.

Balarka Sen

OK, way too much set theory. Let's introduce general nonsense in the play and start thinking about comonkeys.

Arthur Fischer

Every cobook will eventually cowrite every coinfinite set of comonkeys?

3

user61230

Ha. Neeeerds.

Balarka Sen

cofinite, yeah.

Anna Lear

This is how scientific breakthroughs are made.

Arthur Fischer

19:15

@Emrakul Calm it down there, Fred Palowaski.

Balarka Sen

[actually I was thinking about a bad joke going something along the lines of a trivial monkey bundle over books, but thought better of it]

Pyraminx

In every chat, the ratio of on topic to off topic is about 30:2

user61230

Ahh! A ratio of integers sharing common factors! It buuuurns!

Pyraminx

oops, 15:1

user61230

phew, close one

Pyraminx

19:17

sry

Anna Lear

Also, seems like it should be off-topic to on-topic. Unless the chat is super strict with the rules. :)

user61230

Well, this is my allotment of off-topic statements for the day. If I say one more thing, they might kill m--hhrk---...

Pyraminx

nooooooo

Arthur Fischer

@Emrakul No! You can't die! Think of how happy that'd make rand al'thor!

2

Pyraminx

nooooooo

user61230

19:20

all of my frustration and anxiety just came out as a giant fit of giggling and laughter

Balarka Sen

I... didn't get it.

Arthur Fischer

@BalarkaSen It's a puzzling thing, that....

Ramanewbie

hi @evinda

@hippa sorry, I had to go for dinner just after I posted my ast post

@hippa you said it's just a3rd degree equation... Well it's a shame I've no idea how to resolve it !

Hippalectryon

By substitution

Ramanewbie

@hippa I thought substitution was for systems

@hippa how can I do a substitution with only one equation ?

@hippa do I have to make myself another equation ?

Pyraminx

19:28

oo??

anyone on

Ramanewbie

@Pyraminx no -_-

@hippa are you gone ?

Pyraminx

No one is on anymore

bye

by

bye

hah i was counted as spam

byby

Ramanewbie

19:45

I love the baby camel sound when someone star a post...

Ilya_Gazman

20:37

Can you please explain to me what Quickbeam2k1 meant?

If $k_2\neq k_3$ your function is not well defined if $i=1$ is allowed. — Quickbeam2k1 55 mins ago

Daniel Fischer

@Ilya_Gazman On the one hand, you have the definition $f(3) = k_3$, and on the other $f(i+2) = k_2$. If you take $i = 1$ in the latter, you have two definitions of $f(3)$.

Ilya_Gazman

@DanielFischer tnx, I edited it

Don Larynx

21:42

Prove if $R, S$ are relations that $(R \cap S)^{-1} = R^{-1} \cap S^{-1}$

Well, $x(R \cap S)^{-1}y = xR^{-1}y \cap xS^{-1}y$ so then $R^{-1} \cap S^{-1}$

We are halfway done at this point aren't we?

user112495

22:38

I am trying to follow a proof that if$G$ is a group of order $6$, the G is isomorphic to either $D_6$ or $C_6$.

At one point, they say that let $N=\{1, a, a^2\}$ (a is an element of order 3) which is normal as its index in G is two. I understand and am fine with this. They then go on to say choose $b \in G-N$. The $G=N \cup Nb$.

I don't understand why this is the case. And why did we need to show that N was a normal subgroup? Was this necessary in establishing that $G=N \cup Nb$?

Ted Shifrin

@user112495: You're right, that is not yet using normal.

Ramanewbie

hi @ted !

user112495

@TedShifrin Hmm. I don't think I can see anywhere in this proof that requires N to be normal.

Ted Shifrin

Hi @Ramanewb. Tu devrais t'endormir :P

user112495

Can I send you a link to it for you to have a look at?

Ted Shifrin

22:45

@user112495: You don't need to use normality to do this proof.

Ramanewbie

@ted I'm going soon... but I don't have a lot of class tomorrow

Ted Shifrin

You could use $bab^{-1} \in N$, @user112495, but you don't need to.

user112495

@TedShifrin Oh... Maybe they just put it in there for the sake of it then...

Ted Shifrin

You can just do the argument by thinking about possible cases, @user112495.

user112495

@TedShifrin Yeah. I've got the proof in front of me. I was just confused as to why they felt the need to point out that it was a normal subgroup. Also, as for why we have $G=N \cup Nb$, is this because they are disjoint, and the order of their union is equal to $G$?

Ted Shifrin

22:49

The cosets of any subgroup partition the group, @user112495.

Mike Miller

Morning, @Ted

Ted Shifrin

good night, @Mike

Ramanewbie

@mike you keep saying good morning....

Mike Miller

Just got out of office hours

The top joint on one of my fingers hurts... I always scrape that against the blackboard :(

Ted Shifrin

Me too, @Mike, an hour ago ... was meeting with a student writing her Honors thesis. Not what I need to be doing in my last few months.

I knew you were a Neanderthal, @Mike.

Mike Miller

22:54

I was talking to my thinking student again.

Ted Shifrin

ah, so that made you more ape-like?

what did he ask today?

Mike Miller

Indeed.

We talked in discussion about the algorithm to find a basis for the kernel of a matrix

Ted Shifrin

based on reduced echelon form, of course

Mike Miller

(identify free variables; set first free variable equal to 1, rest equal to 0; make vector; rinse, repeat)

Ted Shifrin

yuppers ...

Mike Miller

22:55

He wanted to know why it worked so we proved together that it did

Ted Shifrin

not that I understood this stuff before I wrote a linear algebra book :D

Mike Miller

And then he wanted to know why 1,0,0...0

user112495

@TedShifrin Ah, okay. I've been given the problem of showing that if p is prime, then a group $G$ of order $p^2$ is isomorphic to either $C_p \times C_p$ or $C_{p^2}$.

I've said that for any $g \in G$, we can only have $|g| \in \{1, p, p^2\}$ by Lagranges theorem. If we have a g such that $|g|=p^2$, then $G \cong C_{p^2}$. So suppose not. If we had $|g|=1 \forall g$, then the order of $G$ would be one. But 1 isn't prime. So there must be an element of order p.

Mike Miller

it took some doing to convince him that what was going on was just that we were picking a basis for $\Bbb R^n$ and just using that as a basis for the values of the free variables :)

But then we did the same thing with (1,1,1), (1,0,0), (0,2,-1) instead.

Ted Shifrin

Huh? @Mike That maketh no senseth to me.

user112495

23:00

@TedShifrin I now need to show that this group is generated by two elements, $a, b$ such that $a^p=b^p=1$. As $G$ is abelian, we will then have the result that $G \cong C_p \times C_p$.

Ted Shifrin

Why must $G$ be abelian, @user112495? That's a big deal.

user112495

@TedShifrin We've been told in the question that G is abelian. (Well the condition was that $G$ be an abelian group of order $p^2$).

Ted Shifrin

Oh. :)

It turns out (later) that it must be, @user112495, but you don't have the tools yet.

user112495

@TedShifrin I'm having trouble showing that $a$ and $b$ generate $G$.

Ted Shifrin

Count, @user112495

Mike Miller

23:03

@TedShifrin You have three free variables. For your basis of the kernel, pick their values to be, say, (1,1,1) for the first vector, (1,0,0) for the second vector, and (0,2,-1) for the third vector. Because those define a basis for $\Bbb R^3$, the vectors you produced above define a basis for the kernel.

Ted Shifrin

I'm confused, @Mike, because you seem to have a $3$-dimensional kernel in $\Bbb R^3$, so that seems not to correspond to vectors you'd get from a reduced echelon form, at all.

user112495

@TedShifrin I'm confused? If I follow the same method used for a group of order 6, then I can only show that they generate a group of order $2p$, not $p^2$.

Semiclassical

it's always a little disappointing when you give a fancy answer to something and they prefer the simpler one. (not that they're wrong to prefer it, mind)

Mike Miller

@TedShifrin I have a 3-dimensional kernel. I'm identifying "values your free variables can take on" with "$\Bbb R^3$. Since, like, they're the same thing.

Ted Shifrin

Well, $6 = 2\times 3$ and $p^2= p\times p$, @user112495 :)

You're confusing me, @Mike, so I don't approve of your pedagogy :P

Mike Miller

23:06

And you're frustrating me, so we're both happy ending this conversation./

Ted Shifrin

Great. I'll say hi to @Pedro and then go make dinner, eat, and go to a concert.

Hi @Pedro

user112495

@TedShifrin I used a slightly different method for the order 6 one though. In that, my group essentially consisted of the powers of a, and those powers multiplied by b, which gave me double the three powers. Using that same method, I can only get to a group of order 2p.

Pedro Tamaroff

@TedShifrin Hello.

@TedShifrin I have a question.

Ted Shifrin

No, @user112495, if you take $a^ib^j$ with $0\le i,j\le p-1$, you should have ...

Mike Miller

Enjoy your dinner and concert, @Ted :)

Ted Shifrin

23:09

Que? @Pedro

user112495

@TedShifrin Ooooh, yep. That'll work :p. Thanks :).

Ted Shifrin

You're welcome, @user112495

Pedro Tamaroff

@TedShifrin Suppose you have the plane $\mathbf R^2$ minus a point $p$, and a form over this that is closed. Moreover, you know this form integrates over $0$ on a circle around $p$. Then this form is exact. More generally, this is true for the plane minus finitely many points provided the circles are small enough they just wind around single points.

Can one prove this using homology, inductively on the number of points?

@MikeMiller said yes. My question is actually "How?"

Ted Shifrin

Sure, @Pedro, but you should be able to prove it for forms without all that.

Pedro Tamaroff

@TedShifrin Yes, I know I have to take half planes that intersect at connected components and show the forms glue nicely.

But it's so bothersome.

Ted Shifrin

23:12

Well, just as there is Mayer-Vietoris for homology, there is that for cohomology (with forms), but I think you can do it more bare-hands here.

If $\int_{C_i} \omega = 0$ for all the curves around the points, you should be able to argue reasonably precisely that $\int_C \omega = 0$ for any closed curve.

Pedro Tamaroff

@TedShifrin No no, not all.

Just for one circle.

Ted Shifrin

You can deform any closed curve so that it becomes $\sum n_iC_i$ for some integers $n_i$. Officially, of course, you need winding numbers, which you know.

To make this totally precise, you can use rectangular approximations of all curves.

Pedro Tamaroff

Say the points are $p_1,\ldots,p_n$ and $r$ is smaller than the distances between all the points, and we know the form integrates over the circles $\partial B(p_i,r)$ to $0$.

Ted Shifrin

Yes, those are my $C_i$.

Pedro Tamaroff

Yes, I get your point.

Ted Shifrin

23:15

I will be going to a concert tonight, but you can email me if there are unresolved issues :P

Pedro Tamaroff

@TedShifrin Just one more thing.

Ted Shifrin

Yes?

Pedro Tamaroff

Suppose I have a circle that encloses two points. How do I handle that one?

I have to deform it to sums of the other circles.

Ted Shifrin

You pinch it and show what I said above. Right

Pedro Tamaroff

Ah, duh.

I shouldn't have asked. =)

Ted Shifrin

23:19

You can actually draw the picture of the region whose boundary is the original circle and the other two circles. Then use Stokes's (Green's) Theorem.

Pedro Tamaroff

Yes, of course.

This is nice. Thanks.

Ted Shifrin

You're welcome.

Pedro Tamaroff

Enjoy your concert.

Ted Shifrin

Thanks :)

user112495

@TedShifrin Why did we need that G was abelian there?

@TedShifrin Oh wait, it was a condition of the proposition. Don't worry

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$Mathematics$ Mathematics