Mathematics - 2018-03-11 (page 2 of 3)

Ted Shifrin

06:00

(I did everything with right actions instead of left, but you can change it to make it more what you like.)

Not quite, @Balarka.

Mike Miller

Cool

Balarka Sen

I have forgotten what $\text{Ad}$ does. $\text{Ad}_g \in \text{Aut}(\mathfrak{g})$ is derivative of $g(-)g^{-1}$ at the identity?

Ted Shifrin

Yes.

We're using the hypothesis that $[\mathfrak h,\mathfrak m]\subset\mathfrak m$ for this to make sense.

Balarka Sen

Ahh

Ted Shifrin

That's basically the whole point of a locally symmetric space.

Balarka Sen

06:05

Right, so $\text{Ad}_h$ is well-defined on $\mathfrak{m}$ for any $h \in H$

Ted Shifrin

yup.

Akiva Weinberger

What's the significance of the $[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0$ thing? It's a nice-looking equation, and it's not hard to prove (write $[X,Y]=XY-YX$ and expand), but it's not obvious why we would care about it

Then again, it's probably more important in Lie theory than Riemannian geometry

Balarka Sen

It's the closest thing to associativity in Lie algebras for one

Ted Shifrin

It's called the Jacobi identity and is part of what you need to have a Lie algebra.

Balarka Sen

Also, it looks like the Riemann curvature tensor if you write it in terms of Lie derivative :P

Ted Shifrin

06:08

It certainly shows up in proofs/calculations in Riemannian geometry.

Akiva Weinberger

@TedShifrin If it satisfies it, you can build a Lie space around it? Such that the algebra is it's Lie algebra?

Balarka Sen

$\mathcal{L}_X \mathcal{L}_Y Z - \mathcal{L}_Y \mathcal{L}_X Z - \mathcal{L}_{[X, Y]} Z = 0$

Akiva Weinberger

Hm

Ted Shifrin

Yeah, you can get a Lie group with that Lie algebra. That's one of the main theorems.

Balarka Sen

Riemann curvature tensor bruh

Akiva Weinberger

06:09

Strange. Not the most obvious thing to require

Ted Shifrin

It's an integrability condition, I suppose.

Akiva Weinberger

I guess it probably falls out of the equations; like, if you try to build the Lie group, something goes wrong if it doesn't equal 0

Balarka Sen

@AkivaWeinberger Well it means if you translate/flow around $Z$ on that truncated box picture consisting of the flowlines of $X$, $Y$, $Y$, $X$ and $[X, Y]$, you'll end up with no change

Ted Shifrin

Yup, start writing out $e^{tX}$, etc.

Silent

Suppose a function $f: X\to Y$ where $X,Y$ metric spaces and $E\subset X$, then if $f(X)\subset \overline{f(E)}$, then $f(E)$ is dense in $f(X)$. Is this statement true?

Akiva Weinberger

06:10

Do things not end up well-defined if it's not 0?

Would depend on the construction maybe

Daminark

@Akiva wait so now let's say we're in the plane, what are you trying to measure with the area of a loop? The area of the set it encloses?

Akiva Weinberger

Yes

Ted Shifrin

Yes @Silent.

Akiva Weinberger

And some weirdness if it intersects itself or goes the wrong way

but yeah

Silent

@TedShifrin how do we get other inclusion?

Ted Shifrin

06:13

I'm not sure you automatically do. You'd need $f(X)$ closed in $Y$.

Daminark

Hmm, so now if you have a non-plane curve it's one of those things where there isn't really a nice volume it necessarily encloses but somehow I feel the generalization you want would be...

MatheinBoulomenos

One more thing, I'm assuming that the correspondence between representations of $\pi_1(X)$ and flat bundles is a category equivalence. There's another category which is equivalent to real representations of $\pi_1(X)$, namely locally constant sheaves with values in $\Bbb R$-vector spaces. So we have some bundles, i.e. modules over the locally ringed space $(X, C^\infty)$ on the one side and locally constant sheaves, so modules over $(X,\Bbb R_X)$ on the other side.

Daminark

Let's say it's some smooth curve, try to see if it's the boundary of some surface

Now there isn't a unique way to do this even if it exists

So in principle we'd want to minimize the area

Balarka Sen

I asked Ted above ^^^ and he said there's a minimal area surface which encloses it

Akiva Weinberger

@Daminark My generalization isn't a number, it's a vector (aka a triple of numbers)

You project it onto the three coordinate planes, measure its area on those, and make a vector out of them

Daminark

06:15

What information does that encode?

Also, is this preserved by isometries of the space?

MatheinBoulomenos

Since $(X,C^\infty)$ is a sheaf of $\Bbb R$-algberas, there's also an obvious way to go from modules over $(X, C^\infty)$ to modules over $(X, \Bbb R_X)$

Balarka Sen

Rotations definitely fuck the vector up

Akiva Weinberger

Isometries of the space where the loops live in correspond to isometries of the vector space where the result lives it @Daminark

Ted Shifrin

Actually, DogAteMy probably does want this to live in $\mathfrak{so}(3)$ rather than $\Bbb R^3$.

Mike Miller

Are we saying anything interesting I should weigh in on

Balarka Sen

06:16

Akiva is onto something

Ted Shifrin

LOL, weighing is too expensive, Mike.

Balarka Sen

So maybe look at what he says

Silent

@TedShifrin So, if $f$ continuous, then $f(X)$ closed in $Y$, hence we get the result, i think.

Balarka Sen

The rest of us are just shitposting

Ted Shifrin

@Silent: Why is that true?

Mike Miller

06:17

Too much work

Akiva Weinberger

@TedShifrin Except the Lie bracket structure doesn't seem to be relevant, and as a vector space it's just $\Bbb R^3$

Ted Shifrin

I'm not convinced, DogAteMy, since you really are trying to relate this to $SO(3)$ in the end.

Silent

@TedShifrin Because $X$ is closed, and so $f(X)$ is closed.

Ted Shifrin

@Silent: Are you telling me continuous maps send closed set to closed sets?

Akiva Weinberger

You still need a basis to describe the result in $\mathfrak{so}(3)$, though, right?

Silent

06:18

@TedShifrin OMG! so sorry.

Ted Shifrin

LOL

@Silent: Think of $f$ as the inclusion of $\Bbb Q$ into $\Bbb R$.

Akiva Weinberger

Isometries of the loop in $\Bbb R^3$ will make isometries in $\frak{so}(3)$.

Balarka Sen

@MikeMiller I pointed out that the Jacobi identity is the origin of the Riemann curvature tensor, a rEaLiZaTioN that I am proud of

so maybe weigh in on that :P

Akiva Weinberger

I guess I could measure the length of the vector in $\mathfrak{so}(3)$, though.

Pros: agrees with the usual area whenever the loop lies in a plane. Cons: not additive anymore

(Pros: is invariant under isometries)

Ted Shifrin

I don't think doCarmo ever does things like Ambrose-Singer, but eventually you'll want to know that, DogAteMy.

Akiva Weinberger

06:21

(And that has to do with the Pythagoras-y thing you mentioned earlier @BalarkaSen)

MatheinBoulomenos

So hmm, everything I said is just that the holonomy representation of a flat bundle is also the monodromy representation of some locally constant sheaf, which I guess is not that interesting. Still it seems strange that one can pass from certain sheaves over $(X, \Bbb R)$ to certain sheaves over $(X, C^\infty)$ and vice versa

Ted Shifrin

I'm not paying too much attention, @Mathein, but the whole point is that a flat bundle has locally constant transition functions.

Daminark

Wait so you said it's invariant under isometry but... Let's call this $A_3(\gamma)$. Then are you saying that $A_3(T(\gamma)) = T(A_3(\gamma))$ for an isometry $T$?

Ted Shifrin

That makes no sense, Demonark.

Akiva Weinberger

Should be @Daminark

Ted Shifrin

06:22

Huh?

Balarka Sen

Structure groups of flat bundles are totally disconnected I think

Akiva Weinberger

Or maybe not, I don't know

Ted Shifrin

Is $A_3$ an area or a region in some plane? Either way, when you apply $T$ the plane changes.

Balarka Sen

Just follows from thinking about what happens to foliated bundles

Ted Shifrin

You don't need to be that fancy, Balarka.

Akiva Weinberger

06:23

His $A_3$ is an element of the three-dimensional vector space

Ted Shifrin

Follows from the $\pi$ representation, since it's a discrete group.

Balarka Sen

@TedShifrin Hah, fair

Thanks, I'm too used to foliations

Daminark

$A_3$ is the "area" of the loop, by which we mean each coordinate is the area of its projection into one of the coordinate planes

Balarka Sen

They are my friendos

Ted Shifrin

Oh, sorry, DogAteMy, @Demonark. I didn't know the notation.

Daminark

06:24

So I'm slightly suspicious that this is true

Akiva Weinberger

@Daminark I believe it is true, then, yeah

Ted Shifrin

I don't think that's quite right, but maybe it works with $T^{-1}$ on one side. I'd have to think about it.

Daminark

The reason why is that you have the issue of a projection being self-intersecting

But then you rotate a tiny bit and now it's not

Ted Shifrin

You have to coordinatize $A_3$ by normal vectors to planes, so something's screwy, guys.

Daminark

If that isn't an issue then yeah I buy something to this effect

Ted Shifrin

06:26

You have to use $\bigwedge^2 T$, I think.

Akiva Weinberger

This is essentially the integral of $(xdy-ydx,ydz-zdy,zdx-xdz)$ around the loop

or maybe half that

So you just need to look at what the isometry does to the form

Silent

@TedShifrin Rudin says that if $f$ continuous and a subset $E$ of domain $X$ is dense in domain then $f(E)$ is dense in $f(X)$. Then why $\overline{f(\Bbb Q)}=f(X)$ does not hold if $X=\Bbb Q$ as in your example?

MatheinBoulomenos

@TedShifrin this is probably very basic if one knows more about bundles, but I'm still a bit confused. Does this mean there's some simpler description to go from a locally constant sheaf of $\Bbb R$-modules on $X$ to a flat bundle than taking the monodromy representation and then the associated bundle?

Akiva Weinberger

Does isometry mean $T^{-1}=T^\top$? (Just double-checking)

Ted Shifrin

Because $\overline{f(\Bbb Q)} = \Bbb R$, which includes $\Bbb Q$, but isn't contained in $\Bbb Q$.

Akiva Weinberger

06:28

Yeah

Ted Shifrin

Yes, DogAteMy.

Daminark

Well, not necessarily a linear isometry but yeah

Akiva Weinberger

So I don't know what the general map on $A_3$ would be if you do an arbitrary map on the loop

@TedShifrin That's a map on $\bigwedge^2\Bbb R^3$?

Ted Shifrin

Yup.

Silent

@TedShifrin yes, that's my point. Doesn't $f(E)$ dense in $f(X)$ mean that $\overline{f(E)}=f(X)$?

Ted Shifrin

06:31

No, @Silent, only if $f(X)$ is closed in $Y$.

Akiva Weinberger

So maps on $\Bbb R^3$ determine maps on $\bigwedge^2\Bbb R^3$. Then, yeah, the latter (which is isomorphic to $\Bbb R^3$) looks like the natural place for $A_3$ to live in, and if $T$ acts on $\Bbb R^3$, then I think $\bigwedge^2 T$ acts on $A_3$.

MatheinBoulomenos

@Silent you have to be careful if you take the closure in the image (with the subspace topology) or the codomain

Akiva Weinberger

But in the special case of $T$ being an isometry and the loop being in 3D space, I think $T$ equals $\bigwedge^2 T$ under the natural(?) isomorphism.

Silent

Oh! so, generally, $E$ dense in metric space $X$ iff $X\subset \overline E$?

Akiva Weinberger

(But don't quote me on that)

Ted Shifrin

06:33

Because of the cofactor formula for inverses, DogAteMy.

@Silent: The problem here is that you have things that live in $Y$. If $E\subset X$ then $E$ is dense in $X$ if $\bar E = X$.

Akiva Weinberger

That's the formula that computes inverses one element at a time, right?

Ted Shifrin

Yes, DogAteMy.

Akiva Weinberger

I learned it and never used it again, essentially, 'cause of its inefficiency. What's the connection here?

Ted Shifrin

It's important for theory, not computation.

So $A^{-1}$ is $1/\det A$ (which is $1$ here) times the transpose of the matrix of cofactors. So the matrix of cofactors equals $A$ for someone in $SO(n)$.

Balarka Sen

I love that formula

Ted Shifrin

06:37

I actually had to use that somewhere in some integral geometry proof years ago.

Akiva Weinberger

And the matrix of cofactors of $T$ is $\bigwedge^{n-1}T\in\bigwedge^{n-1}\Bbb R^n\simeq\Bbb R^n$?

Ted Shifrin

So I think it does work out the way you and Demonark wanted, DogAteMy, but you should check.

Nooo ...

MatheinBoulomenos

I have used the cofactor formula for a difficult proof on integral bases in algebraic number theory, just saying

Ted Shifrin

It's a linear map on $\bigwedge^{n-1}\Bbb R^n$.

It's an important result, @Mathein.

Akiva Weinberger

Oh, I didn't mean $\in$

Balarka Sen

06:38

$\to \!\!\!\!\!\!\ \in$

This?

Akiva Weinberger

…Pitchfork?

Ted Shifrin

So, modulo all your identifications, DogAteMy, it might work out.

Akiva Weinberger

What do you call the set of square matrices on a vector space?

I wanna write ${\sf Sq}(V)$ but I'm sure that's wrong.

Ted Shifrin

No one does that.

$M_{n\times n}(\Bbb R)$

MatheinBoulomenos

$\operatorname{End}(V)$

Balarka Sen

06:40

$M_{\bullet\times \bullet}(\Bbb R)$

Akiva Weinberger

Ah

Ted Shifrin

But he means matrices, not linear maps.

MatheinBoulomenos

if you have a vector space without a basis, then matrix makes no sense

Ted Shifrin

He's working with a specific basis.

Akiva Weinberger

$\bigwedge^{n-1}T\in{\rm End}(\bigwedge^{n-1}\Bbb R^n)\simeq{\rm End}(\Bbb R^n)$

Ted Shifrin

06:41

^^

Akiva Weinberger

And that would be the matrix of cofactors, I hope

Ted Shifrin

You should figure out why :)

Akiva Weinberger

I'm just confused by the minus signs

Ted Shifrin

Well, you have to discuss orientations on the hyperplanes. Think about the minus sign in cross products.

This is why I use $dy\wedge dz$, $dz\wedge dx$, and $dx\wedge dy$ as the basis for $2$-forms ...

Akiva Weinberger

Right, that makes sense

Ted Shifrin

06:46

Anyhow ... Don't forget to change your clocks, everyone in the US.

Akiva Weinberger

And so swapping $x$ and $y$, for example, gives you two minus signs

or swapping any pair

Ted Shifrin

Huh?

Balarka Sen

@TedShifrin DST?

Ted Shifrin

Hell if I know, Balarka.

Akiva Weinberger

@TedShifrin If I swap $x$ and $y$ one these, I get $dx\wedge dz=-dz\wedge dx$, $dz\wedge dy=-dy\wedge dz$, and $dy\wedge dx=-dx\wedge dy$. So three minus signs, actually

@BalarkaSen T.

MatheinBoulomenos

06:48

@TedShifrin I'm thinking about taking the second part of the diff top course next semester. It looks like it's going to be a bit algebraic, he lists Bott&Tu and a book on fibre bundles as references (among others)

Ted Shifrin

If you swap $x$ and $y$, $dx\wedge dz$ becomes $dy\wedge dz$. I don't follow you.

MatheinBoulomenos

I would need to self-study the contents of the first part, of course

Ted Shifrin

That's cool, Mathein.

Akiva Weinberger

@TedShifrin Yeah, but $dx\wedge dz$ wasn't one of your basis vectors

Ted Shifrin

Huh? I'm totally not following you. But I'm quitting for tonight.

Akiva Weinberger

06:50

$dy\wedge dz$ was one of your basis vectors that you listed, and swapping $x$ and $y$ gives you $dx\wedge dz=-dz\wedge dx$.

MatheinBoulomenos

@TedShifrin Here's the announcement, if you're interested: mathi.uni-heidelberg.de/~banagl/pdfdocs/infodifftop2.pdf

Akiva Weinberger

(Rewriting that last one in terms of basis vectors.)

So if $T=\begin{bmatrix}0&1&0\\1&0&0\\0&0&1\end{bmatrix}$ then $\bigwedge^2T=\begin{bmatrix}0&-1& 0\\-1&0&0\\0&0&-1\end{bmatrix}$, the cofactor matrix. And that explains where the minus signs come from.

@TedShifrin

Daminark

07:29

@Mathein huh, that seems fun

Secret

09:48

[The Cult of Infinity]

(Because I am bored when working)

0. We start with some logic, which give us the symbols and the inference rules to construct objects

1.Our first axiom is the existence of the universe $\text{Uni}$ which contains every object we are going to construct. It should be noted the universe itself does not contain itself, by definition

2. The logic system we are using consists of the following logical symbols, with their associated inference rules:

Secret

10:05

But before we begin, we need to introduce a systematic way to notate propositions, truth values and maps:

Secret

10:15

2a. Truth values: There are 3 truth values T, ⊥,🛇 which corresponds to True, False and Null.

2b. Boolean map bool(P), which returns the truth value of P. The details of bool are programs that can verify the truth value of given inputs P, which as we shall elaborate later, if all programs does not halt for some given P, then return 🛇

2c. Propositions P, which are objects with truth values and may have parameters or variables

Now to introduce the inference rules:

0. Membership ∈

∈ intro:
bool(Q(P)) = T
--------
P ∈ Q

∈ elim:
P ∈ Q
--------
bool(Q(P)) = T

1. And ∧

∧ intro:
P ∈ Q
R ∈ Q
--------
P ∧ Q

∧ elim:
P ∧ Q
--------
P ∈ Q
R ∈ Q

2. Negation ¬

¬ intro:
bool(Q(P)) = F
--------
¬(P ∈ Q)

Note we have the following shorthand: ¬(P ∈ Q) is the same as P ∉ Q

¬ elim:
¬(P ∈ Q)
--------
bool(Q(P)) = F

Secret

10:54

3. Implication ⟹

Here will be a good idea to introduce the notion of a subcontext. A subcontext is a statement containing some proposition P such that P is true within it. Here's an example:

⟹ intro:
if P:
    Q
--------
P ⟹ Q

⟹ elim:
P ⟹ Q
P
--------
if P:
    Q

(btw all these rules are inspired from here though I cannot guarentee whether I have used them correctly)

typo

∧ intro:
P ∈ Q
R ∈ Q
--------
(P ∈ Q) ∧ (R ∈ Q)
∧ elim:
(P ∈ Q) ∧ (R ∈ Q)
--------
P ∈ Q
R ∈ Q

Silent

Where do we use the fact that any open set of reals is countable collection of disjoint intervals here, exercise 5? I mean, what role 'countable' plays there?

Secret

11:11

Exercise 4.5?

there is no exercise 5. in that link

Secret

Restart

[The Cult of Infinity]
0. We start with some logic, which give us the symbols and the inference rules to construct objects
1.Our first axiom is the existence of the universe $\text{Uni}$ which contains every object we are going to construct. It should be noted the universe does not contain itself, by definition
2. The logic system we are using consists of the following logical symbols, with their associated inference rules
But before we begin, we need to introduce a systematic way to notate propositions, truth values and maps:

2a. Truth values: There are 3 truth values T, ⊥, 🛇 which corresponds to True, False and Null.
2b. Boolean map bool(P), which returns the truth value of P. The details of bool are programs that can verify the truth value of given inputs P, which as we shall elaborate later, if all programs does not halt for some given P, then return 🛇
2c. Propositions P, which are objects with truth values and may depend on parameters or variables

We first start with inference rules without reference to the existence of the objects, for the bounded version depends on these rules:

The rules are based on this MSE

just want to get better

is $\|f_{n}g\|_{L^{1}}\leq\|g\|_{L^{1}}\|f_{n}\|_{L^{\infty}}$ true in general

Secret

11:34

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General chat for Physics SE (physics.stackexchange.com). For M...

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41

Busi

12:24

hey

I'm having trouble with an age problem.

A mother's age is 8 times older than her son. When her son's age is half (1/2) of mother's age, the sum of their ages becomes 60. What was the mother's age when her son was born?

First I've tried writting the correct ewuations

equations*

x = 8y

t = passed time

John

Hello, don't know if asking here is appropriate: if not, I apologize in advance. I'd like to contact via chat a moderator for a formal question about stackexchange math, how shall I do it?

Secret

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For informal chat with the site moderators about moderation, s...

2

John

thanks a lot

Silent

@Secret you are right ! that one only

Busi

A mother's age is 8 times older than her son. When her son's age is half (1/2) of mother's age, the sum of their ages becomes 60. What was the mother's age when her son was born?

Can someone help me solving this question?

I only wrote x = 8y

Secret

12:33

Let son age be y

mother is 8y

(still typing)

So I think... solve for the time when the son's age is half of the mother's and then find the mother's age when son's age is 0

ah forgot another equation

(8y+x)/2 = y+x

Secret

12:50

Silent: I suspect countablility is not really used in that question, rather it is a theorem that any open set of on the reals (in the usual topology) are countable disjoint union of open intervals (as otherwise you get a contradiction in terms of the cardinality of the rationals)

The reason why I don't suspect the countability is actually used in the proof, is because there is almost no conditions on the index k

yh016

Hello all, I'm having difficulty in coming up with the null hypothesis and alternative hypothesis for this word problem:

A previous health study had found that 5% of the population suffered from a blood disease. 23 people randomly selected from an area near a mobile phone transmitter have a medical exam recently and 3 of them are found to have that disease. The locals believe that the transmitter increases the likelihood of having the disease. A researcher wants to perform a formal hypothesis test on whether the mobile phone transmitter increases the incidence of the disease.

Kasmir Khaan

13:08

@MatheinBoulomenos mathein :D

here? :D

Long time no see =p

Secret

Try: Select 23 people far away from the transmitter and check if the instance of having disease is equal or higher than those of the transmitter group.

The null hypothesis is obeyed if the number of people sampled and those of which have the disease are not positively correlated

Mats Granvik: Trying to divide by zero aren't you :D

Mats Granvik

It was one of August Strindberg's jokes:
Since
0*0=0
and
1*1=1
then
2*2=2

Secret

lol I see

The + version of this joke is actually true though if you have a distributive algebra and then you try to divide by zero. This is because we always have:

0+0=0

thus acting the zero inverse on both sides and using the distributive law, you get:

1+1=1

which then all hell broke lose because you can then multiply both sides by x to get

x+x=x

thus you end up with an indempotent semigroup

It gets even worse, recall that we have:

0+x=x

Suppose a zero inverse q exists, then multiply both sides by q we get

1+qx=qx

and then you can multiply this by any y

y+y(qx)=y(qx)

and thus what you end up is a lot of elements will eat up elements on the left side of the +, and thus giving a rather trivial structure

(Note how this holds even if associativity is broken)

MatheinBoulomenos

13:29

@KasmirKhaan I'm here now

Kasmir Khaan

@MatheinBoulomenos yeeey :D

I kinda need help for a special Q -.-

MatheinBoulomenos

okay

Kasmir Khaan

I got stuck on it for 1 day now

so we have a group G

and a partition P of the group, such that for all two partitions A, B

AB is contained in P

we have 1 subset N that containes the identity of G

MatheinBoulomenos

What's AB?

Kasmir Khaan

we have to show that N is normal, and that P is the set of its cosets

AB= {ab : a inA, b in B }

Alessandro Codenotti

13:37

Hi @Mathei

Kasmir Khaan

I think i should write the question varbatim

MatheinBoulomenos

No, that doesn't make sense. A and B are partitions

Kasmir Khaan

I ll get my ipad :D

MatheinBoulomenos

so they set of subsets

Hi @AlessandroCodenotti

Kasmir Khaan

Yes sorry

P is the partition

A , B are parts of it

ill type the question now

Let P be a partition of the group G with the property that for any pair of elements A,B of the partition, the product set AB is contained entirely in an another element C of the partition

MatheinBoulomenos

13:39

Oh okay

Kasmir Khaan

Let N be the element of P that contains the identity of G

prove that Nn is normal subgroup of G and that P is the set of its cosets

@MatheinBoulomenos pls dont answer yet :D

MatheinBoulomenos

okay

Kasmir Khaan

Ill show you my work in 5 mins :D

brb 5 ! :)

okay :D

@MatheinBoulomenos first we have to show that N is a subgroup

and I showed that

A in contained in AN but AN must be some part of the partion say B

since partitions are either equal or have no intersection we conclude that A =B

now i want to show that those elements of the partition are cosets of N in G

AN =A

ie aN is contained in A

where i was stuck is at showing that A is contained in aN

i want to conclude that A= aN, hence it is a coset of N

am not sure if this is the right approch but ><

MatheinBoulomenos

There's also an element in the partition that contains the inverse of a

Kasmir Khaan

hmm

MatheinBoulomenos

13:54

So call B the element in the partition that contains the inverse of a. What can you say about BA?

Silent

3 hours ago, by Silent

Where do we use the fact that any open set of reals is countable collection of disjoint intervals here, exercise 5? I mean, what role 'countable' plays there?

@BalarkaSen Please look at problem 4.5

Kasmir Khaan

@MatheinBoulomenos wont that make BA =N ?

since the idenity will be there

MatheinBoulomenos

yes

Kasmir Khaan

are we trying to make a group structure on the elements of P ?

MatheinBoulomenos

that could work as well

but what does B look like?

you have already done this for A

You know that a'N is contained in B

Since BN=B

And also NA=A (by the same argument as AN=A)

so now try to put that together

Kasmir Khaan

13:59

Okay thinking :D

><

MatheinBoulomenos

$a^{-1}A=a^{-1}NA \subset BA =N$

Kasmir Khaan

i want to show what B look like?

MatheinBoulomenos

I answered that question myself with "a'N is contained in B"

that's what I meant, sorry

Kasmir Khaan

:D

no problems =p

i have hard time with such questions -.-'

so we have that a'N is in N

i dont see the full picture

MatheinBoulomenos

No, a'A is in N

what happens if you multiply with a from the left?

Kasmir Khaan

14:07

N is in aN

MatheinBoulomenos

No

Kasmir Khaan

but does the aritmatic rules continue to be valid in such expressions?

MatheinBoulomenos

A is in aN

Kasmir Khaan

like contained in and stuff

oups

MatheinBoulomenos

you can just check that from the definitions

Kasmir Khaan

14:08

that is what i meant , A is in aN

okay so now arguing in the same way, we show that each element of P

is a coset of N

using lagrange we can aruge that all elements of P has the same cardinality

we are close but not there yet ><

MatheinBoulomenos

does the group have to be finite?

I don't see how cardinality helps

Kasmir Khaan

there was no mention of that

just stated a group

MatheinBoulomenos

To show it is normal, note that aNa' is contained in (aN)N(a'N) which is an element of the partition which contains 1

Kasmir Khaan

aha

that does it then ><

thanks alot mathein :D

we have showed now N is normal and the partition is the set of its cosets

very weird question -_-'

MatheinBoulomenos

yeah, I don't really see the point of it

Kasmir Khaan

14:20

><

from artin book

chapter 2 setion 10 Q 3

tazdingo

hello! how are you all doing?

can someone tell me how this expresion D -> 0D + 1C + λ can be transformed into this D -> (0)*1C + (0)*λ using the kleene star?

i mean what it want basically its get rid of the recursion

Secret

Nehal Samee

15:32

Can anyone give me HINT regarding integration of of $\frac{1}{1+v^{4}}$...?

Astyx

@NehalSamee What are the roots of $1+v^4$ ?

@tazdingo What's $D$ ?

Nehal Samee

@Astyx...Could factorize up to 3rd power...What about the 4th power...?

Astyx

What ?

Nehal Samee

Is it (1+√2v+v)(1-√2v+v)

tazdingo

I'm working with regular expressions, and I'm trying to solve system of regular expressions equations

Astyx

15:44

Missing square but yeah

You can then factorize again

Nehal Samee

@Astyx...After this , what should I do...?

Astyx

And get the product of 4 monomials

Then there's a theorem that states that you can write your function as a linear combination of the inverse of those monomials

Wait

Ignore what I just said

You have ${1\over a+v^4} = {a\over 1+\sqrt{2}v+v^2} + {b\over 1-\sqrt{2}v+v^2}$ for some constants $a$ and $b$

Find those constants and it's the sum of usual integrals

Nehal Samee

@Astyx...After partializing , I tried to equate the coefficients of v² ... But to no purpose...

There , after equating , I get , a+b=0(for constant) and a+b= 1(for v)...???

Astyx

Wait

It's not $a$ but $a_0+va_1$

And same goes for b

Sorry I haven't done this in a long time

Nehal Samee

How can I post my working here...?

@Astyx...They still produce contradictory conditions...

Astyx

16:57

What do you mean ?

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