Mathematics - 2014-12-03 (page 2 of 6)

Balarka Sen

12:05

@UserX

UserX

Yea?

I just learnt intro to biotechnology. We have sheep that produce milk full of insulin in gene farms. That's so crazy.

Balarka Sen

Did you successfully eject that BS cayley tables out of your head and classified groups of order p^2, @UserX?

UserX

@BalarkaSen no, I didn't eject the cayley table method. I did classify(with Alizter's help) p^2 groups though

Balarka Sen

OK? How did you do it?

UserX

How can I classify the groups of order n for any n? Any algorithm/method?

Balarka Sen

12:09

@UserX Patience daniel-san.

Committing to a challenge

Balarka has his 'focus only on math topics here' mode active

Balarka Sen

That's where the hard stuff comes.

I am slowly leading you there, @UserX

Integrator

@beginner Are you interested in accepting answer to this

Committing to a challenge

He has a 62% accept rate, not great, I guess he forgets or doesn't accept when there are competing answers

UserX

@BalarkaSen I showed that there are only two groups up to isomorphism, Z_p × Z_p and Z/p^2Z, then used the fact that only one is cyclic and Z_p×Z_p isn't to show they're not isomorphic, then that they got elements of order 1,p,p^2 , to show(with lagrange's theorem) that every group is either cyclic of isomorphic to Z_p×Z_p (not including the trivial group)

Committing to a challenge

12:16

I have a 91% accept rate :)

UserX

But my line of thought was "guess the number of groups, then prove that there are no more" which can't be done for any order

Balarka Sen

@UserX sure that works

but maybe you could just do it by looking at the order of the elements, @UserX

UserX

By the way, @MikeMiller talked about a Z/<a> trick

Or something like that.

Balarka Sen

G/Z(G) trick, yes

UserX

What's this?

Balarka Sen

12:18

@UserX you know what center of groups are?

UserX

No

Venus

@Chris'ssis @robjohn Do you know how to evaluate this integral $$\int_{0}^{\pi/2} x^2 \tan x\:\mathrm{d}x$$
I have a feeling that this integral has been discussed on MSE.

Balarka Sen

@UserX center of a group G is a subset of G consisting of elements in G which commute with every other element in G. rigorously, Z(G) = {z \in G s.t. zg = gz for all g \in G}

exercise : prove that Z(G) is a group too

this shouldn't be too hard but it's good to nail down the details, @UserX

Chris's sis

@Venus By the way, did you see my last question here? I was preparing to ask it again ...

Evaluate in one line $$\int_0^{\pi/2} \arctan(\sin(x)) \ dx$$

3

A: Finding $\int_0^{\frac{\pi}{2}}\arctan\left(\sin x\right)dx$

Using the integral definition of the arctangent function, we may write $$\arctan{\left(\sin{x}\right)}=\int_{0}^{1}\mathrm{d}y\,\frac{\sin{x}}{1+y^2\sin^2{x}},$$ thus, transforming the integral into a double integral. Changing the order of integration, we find: $$\begin{align} \mathcal{I} &=\in...

Venus

@Chris'ssis What question? I use my tab to online now

Chris's sis

12:21

@Venus Finish the integral in one line $$\int_0^{\pi/2} \arctan(\sin(x)) \ dx$$

Venus

@Chris'ssis In on line?? You're kidding, aren't you?

Chris's sis

@Venus No, I'm not kiding at all.

Venus

Then how?

UserX

@BalarkaSen this follows right away from the axioms. By this definition it also has to be abelian right?

Balarka Sen

indeed

Chris's sis

12:26

@Venus Did you hear of Legendre Chi function? mathworld.wolfram.com/LegendresChi-Function.html

Balarka Sen

Z(G) is the "greatest abelian part" of any group G, @UserX

Venus

@Chris'ssis BTW, do you know how to evaluate my question? I think it has been asked here on MSE

UserX

@BalarkaSen that's obvious from that definition yea. I have to eat, write down the G/Z(G) trick, I'll see it in 5 mins

Balarka Sen

@UserX it's a sophisticated trick. i'll only write it up if you know what normal subgroups are.

Chris's sis

@Venus Wait, something is wrong there ...

Venus

12:27

@Chris'ssis I think I have heard it in sos440's answer, but I forget which answer

Chris's sis

@Venus Is this integral correctly written? $$\int_{0}^{\pi/2} x^2 \tan x\:\mathrm{d}x$$

Venus

I mean I've seen it but I forget where it is

@Chris'ssis Yes

Chris's sis

@Venus The integral diverges.

Venus

But W|A says, it converges

UserX

@BalarkaSen I don't :P

Chris's sis

12:33

@UserX Really?

Balarka Sen

Then forget about it, @UserX

Study cosets and normal subgroups, then we're gonna talk.

Chris's sis

@Venus Well, near $\pi/2$, the integrand behaves like $$\frac{x^2}{\pi/2-x}$$ whence Q.E.D.

Balarka Sen

@Venus. I'm sorry if I acted rudely yesterday. I apologize.

beginner

@Integrator i dont know who to accept sorry

Venus

@BalarkaSen It's OK. No worries ^^

Balarka Sen

12:41

=)

beginner

=-)

Chris's sis

@Venus Now, look at that

$$\int_0^{\pi/2} \arctan(\sin(x)) \ dx=2\chi_2({\sqrt{2}-1})$$

Balarka Sen

Now, @UserX. Classify all groups of order 6.

beginner

@BalarkaSen what are groups of order 6?

Venus

@Chris'ssis Look what?? I'm online on my tab. The feature of the chat is really poor

Balarka Sen

12:43

@beginner are you familiar with groups?

beginner

@BalarkaSen are they like sets?

Venus

@Chris'ssis I don't know how to use that special function

Balarka Sen

more or less. i'd rather not start surveying about groups.

beginner

ill read them

Balarka Sen

@Chris'ssis Dirichlet charecter?

Chris's sis

12:44

@Venus I just wrote the integral I asked you is 2 mutiplied the legendre chi function of order $2$ of $\sqrt{2}-1$

Balarka Sen

Ok thank god it's only legender chi.

:P

beginner

so $\mathbb{Z}$ is a group

so it is a set with a bunch of rules(axioms)

Venus

@Chris'ssis OK, I'll note that. Legendre Chi function

beginner

and + or * work

Balarka Sen

@beginner no, it's just a set. a group is a pair, with a set and a binary operation.

beginner

12:46

ok what does order 6 mean?

Venus

@Chris'ssis How about this one $$\int_{0}^{\pi/2} x^2 \cot x\:\mathrm{d}x$$

UserX

6 elements

Balarka Sen

that the underlying set has cardinality 6, @beginner

@UserX can you guess how many are there of order 6?

Chris's sis

@Venus Before that, did you like my half-line proof?

Wait, let me write more ...

UserX

First guess 2

beginner

12:47

i guess i cant do this because having 6 elements makes no sense with the closure axiom

Venus

@Chris'ssis No! It's too high-level to me :D

UserX

We have 2n and n! both being able to equal 6

So a symmetric and a cyclic

Chris's sis

@Venus $$\int_0^{\pi/2} \arctan(\sin(x)) \ dx=2\chi_2({\sqrt{2}-1})=\frac{\pi^2} {8}-\frac{1}{2}\log^2(\sqrt{2}+1)$$

Balarka Sen

@beginner it does. you can't expect to do (somewhat hard) problems in group theory without knowing what a group is, you know =)

try reading up something on it

a textbook, say

@UserX Z_6 and S_3, right

UserX

No

Z_3

Balarka Sen

12:49

that's not of order 6, @UserX

UserX

Oh wait yes

I was thinking of D_3

Venus

@Chris'ssis Why don't you post your one-line answer there? I'll upvote it

beginner

but if you have 6 elements adding or multiplying the elements with eachother will always be bigger than 6?

Balarka Sen

yeah well, @UserX. D_3 is isomorphic to S_3.

UserX

Now I only gotta prove that they are the only ones up to isomorphism

Balarka Sen

12:50

@beginner your question makes no sense. define "bigger".

beginner

i mean that if i have $\{0,1,2,3,4,5\}$ and add or multiply $4,5$ i will be outside of the set

UserX

Well, there is only one group of order 6 that's cyclic, and we already got that, so the other one has to be non-cyclic so they're not isomorphic

@beginner keyword; modular arithmetic

Venus

@Chris'ssis My question please... ^^

Balarka Sen

what @UserX said, @beginner

Chris's sis

@Venus OK, let me look at it now

Balarka Sen

12:52

you need to define an operation. you can always do that. a canonical choice is addition modulo 6, @beginner

Venus

@Chris'ssis Thanks.

UserX

@beginner also, the binary operation doesn't have to be addition,multiplication etc. Groups are abstract, you can define an operation between elements however you want.

beginner

can the binary operation be a function?

Balarka Sen

@beginner define "function"

beginner

like the identity function?

Balarka Sen

12:54

i understand a binary operation as a map.

UserX

Saying it can be a map yea

beginner

$f:A\to A$, $x \mapsto x$ $f(x)=x$

Balarka Sen

that's not a binary operation @beginner

Chris's sis

@Venus Simple integration by parts combined with the Fourier series of $\log(\sin(x))$

Venus

@BalarkaSen Have you already got along with Huy?

Balarka Sen

12:54

a binary operation picks two elements from A and maps to A

UserX

@Venus good luck

Balarka Sen

so it'd by f : A \times A \to A

Will Hunting

@venus Did you get any answer to your meta question before it was deleted? =)

Balarka Sen

@Venus no, but i've never been rude with him

Venus

@Chris'ssis I know that why. Can we have other ways without using Fourier series?

*way

Chris's sis

12:55

@Venus Sure

beginner

ohhhh okay balarka thank you i think i have seen vector problems like that

Balarka Sen

not sure what you're referring to but your welcome.

beginner

what field of math is this?

Venus

@UserX Good luck for hat?

Balarka Sen

it's called group theory, @beginner?

Venus

12:57

*what

beginner

is that in algebra?

Balarka Sen

er, yeah

beginner

oh cool so vectors are algebra and group theory is in algebra

Venus

@WillHunting No, I didn't :D

Will Hunting

@Venus Hehe, I told you it would be deleted.

Balarka Sen

12:57

they're related, @beginner. vector spaces can be rigorously defined in the setting of algebra.

UserX

@Venus talking non-math with Balarka

Balarka Sen

it's something called a module.

Will Hunting

Why bring in modules? A vector space is a vector space, full stop.

beginner

@BalarkaSen when you were 11 were you at my level? i want to be at your level

Balarka Sen

@WillHunting It's because I feel natural to think about modules as vector space over fields.

Venus

12:59

@BalarkaSen So why were you only rude to a nice girl like me? ^^

Chris's sis

@Venus You can write $2\log(\sin(x))=\log(\sin^2(x))=\log(1-\cos^2(x))$ and then develop the last expression as a Taylor series, and then use the variable change, $\cos(x)=t$.

beginner

'In abstract algebra, the concept of a module over a ring is a generalization of the notion of vector space over a field, wherein the corresponding scalars are the elements of an arbitrary ring.' yeah i think i will read this later it makes me feel stupid

Balarka Sen

@venus because of that stupid suggestion starred by 4 people (yeah, yeah, 11 wants me to get smacked so this is nothing compared to it).

i'd rather quote Daniel Fischer "in internet, boys are girls and girls are FBI agents"

Venus

@WillHunting Men (refer to most of users here) are over sensitive :P

Will Hunting

@Venus Most people in here are too serious, lol. They should be like me, lol.

Chris's sis

13:01

@BalarkaSen What is the meaning of that? "in internet, boys are girls and girls are FBI agents"

Am I an FBI agent? :-)

Venus

@Chris'ssis I'll check it first

UserX

@BalarkaSen okay. Let G be a group with 6 elements. If it has an element of order 6, it's isomorphic to Z_6. If it has no element of order 6, we got 1,2,3(the divisors). We have an identity element, so there's our order 1 element. If they're all order 3, it becomes impossible. If it has elts all of order 2 it's impossible too. Then if we got an element of order 2 and one with order 3 we got a group isomorphic to S_3. Did I miss any cases?

Venus

@BalarkaSen but I'm a CIA agent :D

Balarka Sen

@WillHunting we are less serious than the guys in the homotopy chat room though

beginner

@BalarkaSen what that 11 thing to my question i dont know

Venus

13:02

@WillHunting I couldn't agree more with you :)

Chris's sis

@Venus I have a 3rd way for you: begin with the variable change $x\mapsto x-\pi/2$, and then use $\tan(x)$ as a series ...

beginner

@Chris'ssis what is a problem that i can do just with a big sum $\sum$?

@Chris'ssis can you give me one

Balarka Sen

@UserX The last statement is yet to be verified. Why is it true that having no group with elements entirely of order 2 or entirely of order 3 should mean that one has order 2 and one has order 3? it might be that two has order 3 and one has order 2. and there can be more than one non-cyclic groups of order 6 in this way.

UserX

@BalarkaSen actually the last statement is verified by the table, but I wanted to avoid stating this part :D

Balarka Sen

not with the tables again

throws cayley's tables at himself

Chris's sis

13:06

@beginner $$\sum_{n\ge 1}\frac{1}{n(n+1)}$$

UserX

Man they're just too useful

Venus

@Chris'ssis Yes, please. The second one looks not easy. How on earth deal with $$\int x\cos^{2n}x\,dx$$

Balarka Sen

@UserX not when you get to groups of order 657

beginner

does that mean $\sum \limits_{n=1}^\infty \frac{1}{n(n+1)}$?

UserX

@BalarkaSen I'm hoping I learn a general method for any order n by then :P

@beginner $\sum_{n=0}^m x^n$

Balarka Sen

13:08

there is no general method actually. but there are procedures to handle on a case-by-case basis

Chris's sis

@Venus It's an indefinite integral?

@beginner yeap.

UserX

@BalarkaSen you just crushed my dreams

beginner

@Chris'ssis thank you! i will try to do it, noone tell me please

UserX

@BalarkaSen how can I do it without inspection?(classifying order 6 groups)

Venus

@Chris'ssis No, for the second way the bounds will be from 0 to pi/2

Balarka Sen

13:10

it's really nonobvious @UserX. I gave it to you only to make you believe that it's nonobvious. you'll learn how to do it when you get to Sylow theory.

i.e., real group theory stuff.

Venus

I can't edit my chat @Chris'ssis

Sorry

UserX

@BalarkaSen I have to learn the fake group theory first(lol)

Balarka Sen

where are you at the moment, @UserX?

UserX

@BalarkaSen under my AC in my house

beginner

can i have a tiny clue that isn't big please

Balarka Sen

13:12

facepalm i mean how much have you studied in groups

UserX

@beginner its 1

beginner

that is a massive clue if it is true

UserX

@BalarkaSen i'm at cosets

Balarka Sen

oh you're close to normal subgroups and largange

UserX

@beginner actually the answer itself doesn't really give a clue.

Balarka Sen

13:13

that is almost real group theory's realm, @UserX

;)

UserX

@BalarkaSen cosets are not intuitive at all though

beginner

it is a big clue cause i wanted to work it out

UserX

I mean, I can't think of any use of them

@beginner you can still work it out. Pretend I answered $5.45$ and prove me wrong

beginner

ok

Balarka Sen

@userX ok... why do you think cosets are not intuitive?

beginner

13:15

ok that clue didnt help because i dont know how to do it still you were right userx

can i have another tiny clue

wait dont i will try drawing it

Venus

@Chris'ssis I gotta go, we'll discuss this later. Thanks for your help

Chris's sis

@Venus Welcome. :-) OK

Integrator

@beginner I didn't get it

UserX

@beginner telescopes

Committing to a challenge

@Integrator What don't you get?

beginner

13:20

is telescopes a hint or just a random word hehe

UserX

@beginner hint

Integrator

@Committingtoachallenge I don't get what he said in last response to me.

Balarka Sen

.

beginner

what did i say?

Balarka Sen

@UserX think of that as a group

UserX

13:21

@beginner whatever, here's the hint that will get it done, 1/(n(n+1))=1/n-1/(n+1)

Balarka Sen

and think of the subgroup $H$ as the dots in a row

beginner

oh i said that i dont know whose answer to accept

Balarka Sen

the coset $G/H$ then looks like $\bullet \;\;\; \bullet \;\;\; \bullet$

UserX

@BalarkaSen that's hard, considering I can't render latex

Integrator

@beginner You said

@beginner why is that so?

beginner

13:22

i didnt know whose answer was best

Balarka Sen

it was just 9 equidistant dots arrayed in a box, @UserX

Integrator

@beginner accept any one of those

UserX

I took it to mathb.in dunno what I should be seeing

@BalarkaSen G/H and H look the same to me...

beginner

balarka what textbook teaches your group theory stuff?

UserX

Also, don't we have to specify left or right cosets?

@Chris'ssis what did you mean by "Really?"

Chris's sis

13:27

@UserX where?

beginner

when did she say that??

Balarka Sen

@UserX nooo. consider Z. G be the subgroup of Z which consists of integer multiples of 4. Z/G (the left cosets) is then integers modulo 4. i.e., you are contracting all the multiples of 4 in Z to a single point in Z/G, contracting all the integer multiples of 4 plus 1 to a single point in Z/G, etc

@beginner i learned group theory from dummit-foote

beginner

i found it on google yay

oh it has your modules at p337

Balarka Sen

sure.

it's an undegrad survey on algebra

beginner

why you say survey?

Balarka Sen

13:30

because it is a survey :P

beginner

i have never heard a book called a survey

@UserX when did she say that

UserX

1 hour ago

beginner

@WillHunting can you tell me if 'it's an undergrad survey on algebra' makes sense i never heard that

Will Hunting

@beginner Why do you ask me?

beginner

@WillHunting because you know good english

you have 2000 rep on the english stack exchange

Will Hunting

13:32

@beginner Hmm, I think you can try to look up 'survey' in a good dictionary. I think it does make sense in this case.

Integrator

@beginner I have 5000 rep on Math SE and I suck at math!

Balarka Sen

ok i guess i have ranted enough about groups. i leave you on your own @UserX. make sure to finish normal subgroups

Integrator

@venus is English your -1 th language?

user2179021

I would like to understand Did's answer here math.stackexchange.com/questions/734267/…

beginner

@BalarkaSen are you leaving

user2179021

13:34

what is a good resource to learn discrete Fourier theory of this sort?

beginner

@Integrator you dont suck at math!

@BalarkaSen thank you for being a good teacher

Integrator

@beginner really?

UserX

@beginner apart from falling into the calculus trap, @Integrator is pretty good

Integrator

@UserX What do you mean? trap?

Committing to a challenge

Rushing Calculus instead of focusing on real Math is the premise of the Calculus trap

Integrator

13:38

Hi @robjohn

@Committingtoachallenge what is Real math Set theory? Number theory? Yuckk!

UserX

@Integrator rigorous, proof based subjects. If you like calculus, try real analysis.

Integrator

@UserX Ohh..

@UserX Should I start with nptel.ac.in/courses/111105069 and nptel.ac.in/courses/111106053 ?

Chris's sis

13:56

Can we finish this one in one line (in the spirit of the art as Ramanujan used to do it)? $$\int_0^{\pi/2} \arctan(2\tan^2(x)) \ dx$$

(without the differentiation under the integral sign)

Sven

hey guys

suppose we have n number of persons, all with different heights. We need to stack them such that only x are seen from the front and y are seen from the back.

we need to calculate the number of configurations we can do with those n-persons that satisfy the x,y requirement

let's take the biggest x persons and put them in the front, the y-1 persons and put them in the back, the rest goes between

(n-(x+y-1))! was my first try, but I forgot that we can have different combinations of putting the (x+y-1) guys either in the front or the back, now I got (n-(x+y-1))! + (x+y)!

Chris's sis

$$\int_0^{\pi/2} \arctan(2\tan^2(x)) \ dx=\pi \arctan\left(\frac{1}{2}\right)$$

Sven

this is correct for 3/5 test cases, it seems that I've forgot some additional cases

do you have some ideas?

Chris's sis

14:21

Waittttttttt.

I was about to miss this one (in one line) $$\int_0^{\pi/2} \arctan^2(\sin(x)) \ dx$$

DanZimm

@Chris'ssis I've noticed you're quite the integral guru ;D

Chris's sis

@DanZimm Thank you, but I'm not yet like Ramanujan, this is the only thing that matters to me as a comparison ... :-)

DanZimm

;P

Committing to a challenge

14:36

I must go to sleep now, clearly my sleeping patterns were ruined staying up for mark release, it is 12:35AM :( (I normally get up at 4-5am...)

Don't have too much fun @Chris'ssis ;)

Chris's sis

@Committingtoachallenge :D

hmmm. but how about this one ... $$\int_0^{\pi/2} \arctan^3(\sin(x)) \ dx$$?

BBL

Wait, this was already established here

33

Q: Integral $\int_0^1\frac{\arctan^2x}{\sqrt{1-x^2}}\mathrm dx$

Is it possible to evaluate this integral in a closed form? $$I=\int_0^1\frac{\arctan^2x}{\sqrt{1-x^2}}\mathrm dx$$ It also can be represented as $$I=\int_0^{\pi/4}\frac{\phi^2}{\cos \phi\,\sqrt{\cos 2\phi}}\mathrm d\phi$$

BBL (let's see then the cubic variant)

Anonymous

15:16

@Committingtoachallenge You need to get a lot of sleep

Anonymous

@Chris'ssis Why do you people love integrals so very much?:D

teadawg1337

@Ashwin because they're utterly fascinating

Anonymous

@teadawg1337 for people who are good at it :D

Anonymous

@teadawg1337 I can't even start solving one of those

Daniel Fischer

@AshwinGokhale For people who are fascinated by them.

Anonymous

15:23

I am not fascinated by such things due to several reasons

Anonymous

@DanielFischer Are those of any importance when solving real world prolems?

teadawg1337

@Ashwin Volume, surface area, electrostatics, etc.

I almost forgot center of mass

Daniel Fischer

@AshwinGokhale I don't know. It's hard to imagine a situation where having a closed form expression for such an integral is important.

Anonymous

I do consider your comments @teadawg1337 @DanielFischer

UserX

Damn a pcr machine is expensive

Anonymous

15:30

b.b....but I am not good at it

teadawg1337

Mathematics, like most anything else, takes lots of practice

Daniel Fischer

@UserX Why do you care? (What is a pcr machine anyway?)

UserX

@AshwinGokhale no. Everything is approximated numerically if you need it in real life. @DanielFischer a machine that exponentiatally grows genes

Anonymous

I agree with you @UserX

Daniel Fischer

@UserX Ah, those. Well, first question stands, why do you care?

Anonymous

15:34

He wants one

teadawg1337

It's been shown that you only need 39 digits of pi to measure the circumference of the universe within the width of one hydrogen atom

So decimal approximations can be accurate to a fault

Anonymous

@teadawg1337 Do you support Copenhagen interpretation or the many world one?

Mike Miller

long time no see, @DanielFischer

Daniel Fischer

@MikeMiller see <- There.

Mike Miller

Short time see, @DanielFischer

user2179021

15:41

anyone know where I can learn the material for math.stackexchange.com/questions/1049886/… ?

Balarka Sen

@MikeMiller being a galois cover implies that fiber over a point in the basespace has group structure, am i right?

Mike Miller

I don't remember what a galois cover is.

Balarka Sen

regular cover

Daniel Fischer

@MikeMiller Has Pedro had any nice complex analysis questions for you to answer in the last days?

Mike Miller

Nope, @DanielFischer

@BalarkaSen Any set has a group structure. What group structure are you thinking of?

Daniel Fischer

15:43

@MikeMiller Un-nice ones?

Balarka Sen

ah i knew you remember what a regular cover is, @Mike :P

Mike Miller

Nope, @DanielFischer. Sorry :(

robjohn

@Venus did you get this, or do you still need help?

Mike Miller

I've been desparate for questions lately, too.

robjohn

@Integrator hey there

Balarka Sen

15:45

@MikeMiller i mean the fiber acts on itself, right?

Anonymous

@BalarkaSen You at TIFR?

Mike Miller

In what manner?

Balarka Sen

no, @Ashwin

teadawg1337

@Mike I dunno if you saw @TedShifrin and I converse last night, but he resolved my question and I ended up not posting on MO

Mike Miller

OK.

Anonymous

15:46

@BalarkaSen Are you in Grad School?

user2179021

does anyone know any good resources to learn about fourier theory over $\mathbb{Z}_n$ ?

Balarka Sen

way too much crowd here

Mike Miller

@user2179021 Is there an interesting fourier theory over $\Bbb{Z}_n$? I think Rudin's book is the canonical place to learn about fourier theory on locally compact abelian groups.

user2179021

@MikeMiller I think so.. discrete fourier theory is big in theoretical computer science for example

Mike Miller

OK, well I should avoid talking about things I don't know.

user2179021

15:48

I can't avoid it :)

Balarka Sen

@MikeMiller ok lets see. f : E \to B be a cover, then the deck transformation group Aut(f) acts on the fibers over B.

hmm. we want a natural action of the fiber f^{-1}(b) on itself.

user2179021

@MikeMiller on a related note... I am trying to understand the comments of Emil Jeřábek at mathoverflow.net/questions/168474/…

but I am not even sure what you would call that area of math.. ?

I basically need to read some lecture notes but I don't even know which ones to look up

Mike Miller

I don't know anything about it, sorry.

user2179021

no problem

Balarka Sen

i am actually not sure about this @Mike. someone told me that the fiber has group structure. i don't see why.

Anonymous

15:57

@BalarkaSen Have you gone through this book:amazon.com/Visual-Classroom-Resource-Materials-Problem/dp/…

Mike Miller

Well, you know the fiber is (as a set) the deck transformation group. This is not a very natural thing to say. It's more interesting in terms of the representation of the deck transformation group on the permutations of the fiber. I'll let you think about that.

Balarka Sen

@MikeMiller that's just monodromy, @Mike

Mike Miller

Yes.

"Just".

Balarka Sen

take an elt in \pi_1(X, x_0), lift up and the endpoint is still in the fiber

Anonymous

@Committingtoachallenge Hey

Mike Miller

15:58

Yes.

Balarka Sen

this is an action of \pi_1(X, x_0) on Aut(f^{-1}(x_0)), and as it's a galois cover, it's an action of Aut(f) on Aut(f^{-1}(x_0)), @Mike

Mike Miller

Sure.

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$Mathematics$ Mathematics