Mathematics - 2016-04-11 (page 2 of 3)

GPhys

06:01

at least, I can't think of a counterexample

PVAL

06:33

@MikeMiller @BalarkaSen in dim 4 PL=Smooth

@MikeMiller Did you find the paper?

I haven't even read Smale's paper on the 2-dimensional case.

Mike Miller

@PVAL I was talking about triangulable, not PL... One needs to know that the link of a vertex is still a sphere. I guess it's a simply connected homology manifold which in dim 3 is probably now known to be a sphere, hence automatically PL.

I did find the paper. I can link it if you want.

There's a complex analysis proof. I didn't really understand Smale's paper.

Anthony

Ayooooooooooooo.

PVAL

@MikeMiller Can you link it.

I'll probably go over it with google translate or something at some point.

I think its only a page or two.

Mike Miller

mathnet.ru/links/de7ecc81e50a04ea03cf27c8e18faccb/rm2882.pdf

PVAL

Thanks

Mike Miller

06:44

Sure. What's in it?

PVAL

@MikeMiller Apparently he proved that the map from convex CR-structures to coorientable contact structures is a homotopy equivalence.

Mike Miller

hm

PVAL

I think theres probably only $C^\infty$ assumptions on the CR structures

and they might not necessairly come from a germ of an honest complex structure on M^3 \times I

(this is all only for compact-oriented 3-mflds)

I have seen notes of a talk which was obviously given long after the paper claiming that the map was a surjection (in the case where the CR structure comes from a complex str.) which would be implied by this already.

It's possible this person didn't know that this was well known, but its more likely the older paper proved something weaker.

I'd like to look at this proof and see if theres an obvious way of lifting a deformation of the CR structure to some deformation of the complex str. induced in a small neighborhood.

Mike Miller

Makes sense

PVAL

If I could do this, I'd have everything I wanted.

Well I can already guess who М. Л. Громов is.

Does this mean I can read Russian?

Mike Miller

06:55

Lol

hhh

07:12

How do you understand toric ideal?

How is toric ideal different from ideal?

Tobias Kildetoft

@hhh Hmm, so toric would mean that there is an action of the torus, so toric ideal probably meas one that is invariant under this action

hhh

Suppose an ideal I. Now is $I/\langle Ax \rangle$ a toric ideal? I have found toric ideals in the context of homogenous equations, kernel of a map such as here -- but I haven't found a specific definition.

Sturmfels introduced the topic here, perhaps I need to learn toric geometry to find the definition?

LifeOfPai

07:51

R composition S symmetric relation iff R composition S=S composition R?

Tobias Kildetoft

@LifeOfPai No

LifeOfPai

WHY?

Tobias Kildetoft

@LifeOfPai Because you can find examples where it fails

LifeOfPai

You have counterexample

Tobias Kildetoft

take an injective but not surjective function and its left inverse for example

LifeOfPai

07:54

I tried to find

ohhh sorry

skill patrol

@LifeOfPai please don't shout :P

LifeOfPai

R and S are symmetric relation

Tobias Kildetoft

@LifeOfPai Then the composition will always be symmetric. But they need not commute

@LifeOfPai Actually, disregard that last bit

I was thinking reflexive somehow

LifeOfPai

So if R and S are symmetric relation,
R composition S symmetric relation iff R composition S=S composition R?

Tobias Kildetoft

Right, because taking opposite relation switches the order of composition

LifeOfPai

07:59

ya!

thank you :)

skill patrol

08:47

user174558

10:38

@robjohn I am back, LOL.

Mary Star

11:01

Could someone of you take a look at my question:

0

Q: Do we have to show that there is an ideal that satifies the conditions?

Let $p$ be a prime. We define $$R=\{m/n\in \mathbb{Q}\mid m,n\in \mathbb{Z} \text{ and } p\not\mid n\}$$ I want to show that $R$ is a local subring of $\mathbb{Q}$. To show that, do we have to show that there is a $I\subseteq R$ which satisfies the following conditions? $I$ is the only m...

?

Tobias Kildetoft

@MaryStar Is that really the definition of local you have been given?

Danu

I was just told in the TeX - LaTeX chat that I shouldn't be using "math operator" styles to denote categories in tex. However, they have not suggested what I should be doing: Anyone have any idea what is standard?

Tobias Kildetoft

@Danu Did they say why?

Ohh, never mind, it is due to te way spacing works

Danu

@TobiasKildetoft Probably, yes.

Tobias Kildetoft

any font would be fine, just note that math operator is not a font, so it should not be used for objects (but, as the name suggests, for operators)

Mary Star

11:08

At an other exercise I had to prove these conditions and at the end of that exercise there is the note:
Rings that satisfy these conditions are called local.
I haven't found anything else about local rings in my notes.
Can we not show it in that way? @TobiasKildetoft

Tobias Kildetoft

@MaryStar What do you mean you had to prove these conditions? That you had to prove them for some specific ring?

Mary Star

I was given that $R$ is a ring and $I\subset R$ is the only maximum right ideal of $R$.
I had to prove the following:
- $I$ is an ideal
- each element $a\in R-I$ is invertible
- $I$ is the only maximum left ideal of $R$
@TobiasKildetoft

Tobias Kildetoft

@MaryStar So do you see why you do not need to prove all those things?

also, it is maximal, not maximum (as that means something stronger)

Mary Star

@TobiasKildetoft Not really... Why do we not have to prove all these conditions?

@TobiasKildetoft Ah ok...

Tobias Kildetoft

@MaryStar Because you just did an exercise telling you this

Mary Star

11:14

So, do I have to find the only maximal right ideal of $R$ ? @TobiasKildetoft

Tobias Kildetoft

@MaryStar Well, you have to show that it has a unique one (so finding it seems like a good idea)

Mary Star

So, to find the right ideal we have to find an additve subgroup of $R$, say $I$, that satisifes the condition $ar\in I, \forall r\in R, \forall a\in I$, right?
@TobiasKildetoft

Tobias Kildetoft

@MaryStar Yes, that is what it means to be a right ideal

Jester Tran

With x = (x_1, x_2), is the supremum with ||x||_2 = 1 of some expression f(x_1, x_2) just the expression alone since there is only one thing to consider?

Tobias Kildetoft

The third of the properties should help you see which elements to pick

@JesterTran No, that means the supremum over all values of the function on vectors of length 1

Mary Star

11:18

Do you mean that it must be also a left ideal? @TobiasKildetoft

Tobias Kildetoft

there are certainly more than one of those unless you make some more restrictions

@MaryStar I meant the one involving invertible elements

Jester Tran

@TobiasKildetoft I get it now lol

Mary Star

Ah ok... I will think about it... Thank you!! :-) @TobiasKildetoft

Mary Star

11:47

Aren't all elements of $R$ invertible besides of $0$ ? @TobiasKildetoft

Tobias Kildetoft

@MaryStar Only if $R$ is a field, which it was not in the linked question

Mary Star

The elements of $R$ are of the form $\frac{m}{n}$. So, aren't the inverses the elements of the form $\frac{n}{m}$, when $m,n\neq 0$ ? @TobiasKildetoft

Tobias Kildetoft

@MaryStar What is $n$ allowed to be?

Mary Star

$n\in\mathbb{Z}\setminus\{0\}$, right? @TobiasKildetoft

Tobias Kildetoft

@MaryStar no, read the question again

Mary Star

11:53

Ah and it is not divisible by a prime. @TobiasKildetoft

Tobias Kildetoft

@MaryStar Ok, so what does a non-invertible element look like?

Mary Star

It is of the form $\frac{m}{n}$ where $m,n\in \mathbb{Z}\setminus\{0\}$ and $m$ is divisible by a prime, or not? @TobiasKildetoft

Tobias Kildetoft

@MaryStar what do you mean by "or not?"?

Mary Star

If it is correct...

Tobias Kildetoft

rather than just immediately ask, try to verify it yourself

Mary Star

11:59

I believe that it is correct since then the inverse wouldn't be an element of the ring. @TobiasKildetoft

Tobias Kildetoft

Good. So what would be your guess as to what the maximal ideal $I$ should be?

Mary Star

$I=\{\frac{m}{n}\in \mathbb{Q} \mid m,n\in \mathbb{Z}\setminus\{0\} \text{ and } p\not\mid m,n\}$
@TobiasKildetoft

Do we have to prove know that it is maximal?

Tobias Kildetoft

@MaryStar That is the set of invertible elements.

Mary Star

Oh sorry...

$I=\{\frac{m}{n}\in \mathbb{Q} \mid m,n\in \mathbb{Z}\setminus\{0\} \text{ and } p\mid m, p\not\mid n\}$ @TobiasKildetoft

Tobias Kildetoft

@MaryStar Ok, now you need to show that this is indeed an ideal

Mary Star

12:15

$I=\{\frac{m}{n}\in \mathbb{Q} \mid m,n\in \mathbb{Z}\setminus\{0\} \text{ and } p\mid m, p\not\mid n\}$

If $a\in I$ then $a=\frac{m}{n}$ for $m,n\in \mathbb{Z}\setminus\{0\} \text{ and } p\mid m, p\not\mid n$.

Then $-a=-\frac{m}{n}$ for $m,n\in \mathbb{Z}\setminus\{0\} \text{ and } p\mid m, p\not\mid n$. That means that $-a\in I$.

If $a,b\in I$ then $a=\frac{m_1}{n_1}, b=\frac{m_2}{n_2}$. Then $a+b=\frac{m_1}{n_1}+\frac{m_2}{n_2}=\frac{m_1n_2+n_1m_2}{n_1n_2}$. We have that $m_1n_2+n_1m_2, n_1n_2\in \mathbb{Z}\setminus\{0\} \text{ and } p\mid m_1n_2+n_1m_2 (\text{ since } p\mid m_1,m_2),…

Tobias Kildetoft

@MaryStar Not quite. You should not be restricting the numerator to be non-zero (clearly). Other than that it looks ok.

well, also apart from the use of $\implies$

@MaryStar Both

Mary Star

So, now we have to show that this right ideal is maximal and unique, right? @TobiasKildetoft

Tobias Kildetoft

@MaryStar the unique maximal ideal you mean (unique does not make sense on its own)

i.e. you need to show that if $J$ is any proper ideal, then $J\subseteq I$.

Mary Star

12:34

Let $J$ is any proper ideal with $I\subset J\subset R$.

We have that every element of $R-I$ is invertible. So, some of the elements of $J$ are invertible and some are non-invertible.
We take an element of $R-J$, which is invertible, say $x$ and an element of $J$ that is non-invertible, say $y$.
Then $yx$ is non-invertible, so it is an element of $I$.
Is this correct? @TobiasKildetoft

user174558

@evinda Aha!

Andrew Thompson

@TobiasKildetoft Are there any well-known results on when a f.g projective module has a free factor in the noncommutative case? I know Serre gave a sufficient condition in terms of the dimension of the maximal spectrum of a commutative ring (in the case it is connected as a topological space so that the rank of the projectives are constant), however I have been unable to find anything useful when the ground ring is not commutative.

Evinda

Hi @JasonBourne

Hey @robjohn

@robjohn I want to find $\lim_{\epsilon \to 0}\int_{|x|=\epsilon} \left( \ln{|x|} \frac{\partial{\phi}}{\partial{\eta}}- \frac{\phi}{|x|}\right) dS$, where $\phi$ is a test function and $|x|=\sqrt{x_1^2+ x_2^2}$.

$\int_{|x|=\epsilon} \ln{|x|} \frac{\partial{\phi}}{\partial{\eta}} dS= \int_{|x|=\epsilon} \ln{\epsilon} \frac{\partial{\phi}}{\partial{\eta}} dS= \ln{\epsilon }\int_{|x|=\epsilon} \frac{\partial{\phi}}{\partial{\eta}} dS $

and

$\int_{|x|=\epsilon} \frac{\phi}{|x|} dS= \frac{1}{\epsilon} \int_{|x|=\epsilon} \phi dS$

@robjohn Have I done something wrong? Because the last integral does not depend on $\epsilon$ ...

robjohn

@Evinda Didn't I show that that was $-2\pi \phi(0)$?

Evinda

@robjohn You just told me that the result would be this and I try to prove it...

Tobias Kildetoft

12:49

@AndrewThompson No idea

@MaryStar It is not enough to show this for ideals $J$ containing $I$. To show it is the unique maximal, you need to show that an arbitrary proper ideal is contained in it.

robjohn

@Evinda No, I also asked you what $$\lim_{\epsilon\to0}\int_{|x|=\epsilon}\log|x|\,\mathrm{d}S$$ and $$\lim_{\epsilon\to0}\int_{|x|=\epsilon}\frac1{|x|}\,\mathrm{d}S$$ were

Tobias Kildetoft

And you really ought to know that you should not be needing to say something like "some of them are invertible", as none of them will be

user174558

My underwear is invertible.

robjohn

Apr 4 at 14:41, by robjohn

@Evinda what is $\int_{|x|=\epsilon}\log|x|\,\mathrm{d}S$? $\int_{|x|=\epsilon}\frac1{|x|}\,\mathrm{d}S$?

user174558

My new movie will be out in July this year. The title is Jason Bourne.

Evinda

12:55

$$\lim_{\epsilon\to0}\int_{|x|=\epsilon}\log|x|\,\mathrm{d}S=\lim_{\epsilon \to 0}\log{\epsilon} \int_{|x|=\epsilon} dS=0$$ and $$\lim_{\epsilon\to0}\int_{|x|=\epsilon}\frac1{|x|}\,\mathrm{d}S=\lim_{\epsilon \to 0} \frac{1}{\epsilon} \int_{|x|=\epsilon} dS \to \infty$$

@robjohn right?

robjohn

No. What is $\int_{|x|=\epsilon}\mathrm{d}S$?

Evinda

Does it hold that $\int_{|x|=\epsilon}\mathrm{d}S=\frac{4}{3} \pi \epsilon^3$? @robjohn

robjohn

@Evinda That is the formula for the volume of a sphere in $\mathbb{R}^3$. We are looking at the circumference of a circle in $\mathbb{R}^2$.

Evinda

So $\int_{|x|=\epsilon}\mathrm{d}S=2 \pi \epsilon$, right? @robjohn

robjohn

@Evinda yes. Why are you posing your answers as questions?

Krijn

13:05

Is there a name for the Dirichlet problem where we take away a point $z_0$ in the domain $\Omega$ and then add the condition that a solution $u$ should also have that $u(z) + \log |z - z_0|$ remains bounded as $z \to z_0$ ?

Evinda

I don't know... :) Then we have the following:

$$\lim_{\epsilon\to0}\int_{|x|=\epsilon}\log|x|\,\mathrm{d}S=\lim_{\epsilon \to 0}\log{\epsilon} \int_{|x|=\epsilon} dS=0$$ and $$\lim_{\epsilon\to0}\int_{|x|=\epsilon}\frac1{|x|}\,\mathrm{d}S=\lim_{\epsilon \to 0} \frac{1}{\epsilon} \int_{|x|=\epsilon} dS= 2 \pi$$
@robjohn

robjohn

yes

Evinda

But the one limit in our case is: $\lim_{\epsilon \to 0}\int_{|x|=\epsilon} \ln{|x|} \frac{\partial{\phi}}{\partial{\eta}} dS= \lim_{\epsilon \to 0}\ln{\epsilon} \int_{|x|=\epsilon} \frac{\partial{\phi}}{\partial{\eta}} dS$.

How can we find the value of $\int_{|x|=\epsilon} \frac{\partial{\phi}}{\partial{\eta}} dS$ ? @robjohn

Mary Star

So, we have to show that every proper ideal must contain only non-invertible elements, right? @TobiasKildetoft

Evinda

13:24

It holds that $\int_{|x|=\epsilon} \frac{\partial{\phi}}{\partial{\eta}} dS=\phi(x)$ for $x$ such that $|x|=\epsilon$.
But can we find something about the value of x? @robjohn

Mary Star

But how could we do that? @TobiasKildetoft

Tobias Kildetoft

@MaryStar This is a completely basic exercise in ring theory

if an ideal contains an invertible element, then it is the entire ring

Mary Star

Because then the ideal contains also the identity $1$, and then by the definition of a right/left ideal we have that $r\in I, \forall r\in R$, right? @TobiasKildetoft

Tobias Kildetoft

right

Mary Star

Do we have the unique maximal because every other ideal must be the entire ring? @TobiasKildetoft

Tobias Kildetoft

13:38

@MaryStar No, there are other ideals. They are just contained in $I$

Mary Star

Is it because when an ideal contains only non-invertible elements, then it must be a subset of $I$, since $I$ contains all the non-invertible elements? @TobiasKildetoft

Tobias Kildetoft

@MaryStar Yes

Ayan Biswas

Please what are we talking about?? I just came in now

Obliv

What is the point of saying $a,b \in \mathbb{Z} - \{0\}$? is there any point in subtracting a set by the null set?

Huy

@Obliv: that means that $a$ and $b$ can be any integer except for $0$

Obliv

13:51

nvm i thought 0 was the null set but it's just a set containing 0.

thanks :p

user174558

14:02

@Huy Or you can say any nonzero integer.

ramsay

$\dfrac{d i}{dx}=?$, where $i$ is $\sqrt{-1}$

Mike Miller

It's constant, so...

@Krijn Yes, this has a name. I forget what it is.

ramsay

Mike: then derivative is $0$ or not?

Evinda

Hey @UserX

UserX

hey @evinda

Evinda

14:13

Where are you from? @UserX

UserX

A small town in Greece.

Evinda

Ok.

user174558

I am from Antarctica.

Mary Star

@TobiasKildetoft Ah ok... Thank you very much for your help!! :-)

Evinda

From Ioannina where you study? @UserX

UserX

14:16

@Evinda nope. My hometown is about 300 km away. Small towns don't have universities in them anyway.

user174558

Greece seems a mysterious place to me.

Evinda

Aha @UserX

user174558

I don't hear anything about Greece in the news.

Evinda

Which is your field? @UserX

UserX

@JasonBourne The capital controls make it seem like north korea if you live here. It sucks.

user174558

14:20

@UserX Wait till you come to Antarctica.

UserX

@Evinda I don't have a field yet. I major in mathematics. We have the option to choose no field at all but I'm thinking of going into engineering math later because of the job opportunities.

user174558

There is engineering and there is math, but can you go into engineering math?

Evinda

In which semester are you? @UserX

UserX

I didn't phrase that correctly. We can choose one of the following fields; analysis, algebra/geometry, engineering, computer science. Or choose no field at all.

@Evinda 2nd.

Kinda like american minors.

user174558

I think they should not let undergrads choose a subfield of math.

UserX

14:25

Thing is, no matter what field you choose you end up with a math degree. Not a CS one, or an engineering one if you chose those ones. So it's just a way to be more picky on classes.

Mike Miller

@ramsay One subtlety is that there's a number of things you might mean by the derivative of a complex-valued function. But no matter what you're thinking about, the derivatice of a fonstant is 0.

robjohn

@Evinda why do you need to find it? $\phi$ is a test function, so $\frac{\partial\phi}{\partial\eta}$ is bounded. We are multiplying it by something that vanishes.

Huy

is a fonstant both function and constant?

robjohn

@Huy then a cariable is both a constant and a variable?

ramsay

hehe

user174558

14:27

A liger is born from a lion and a tiger.

robjohn

@JasonBourne I have seen one of those.

user174558

A tigon is born from a tiger and a lion.

user174558

@robjohn Sometimes, I think I am born of a god, not a human.

robjohn

Liger

The liger is a hybrid cross between a male lion (Panthera leo) and a female tiger, a tigress (Panthera tigris). The liger has parents in the same genus but of different species. The liger is distinct from the similar hybrid tigon. The liger is the largest of all known extant felines. Ligers enjoy swimming, which is a characteristic of tigers, and are very sociable like lions. Ligers exist only in captivity because the habitats of the parental species do not overlap in the wild. Historically, when the Asiatic lion was prolific, the territories of lions and tigers did overlap and there are legends...

UserX

That was fast

user174558

14:28

There has been speculation if a chimpanzee and a human can mate.

user174558

But any such experiment is bound to be banned by the politicians.

robjohn

@JasonBourne feel free.

user174558

Perhaps there is a chimp and human offspring in top secret underground bunkers.

user174558

Now supposedly cats and dogs cannot mate, but there are reports that they have produced offspring.

UserX

I think the weirdest one I've heard is a horse mating with a zebra.

Evinda

14:32

@UserX Have you passed all the subjects so far?

user174558

@robjohn It seems that very often these hybrids are not fertile though.

ramsay

how to integrate....

$user image$

UserX

@Evinda Nope, not a single one. I skipped the whole first semester's exam period because I was heavily sick. And the profs won't let me give the june exams for the 4+ year students :(

ramsay

@Huy baby in avatar is cute :-D

user174558

@UserX If you are sick, best to redo.

Huy

14:33

@ramsay: that's not a baby. Asians look younger.

UserX

@JasonBourne It's ok, I got the september exams to make up for the lost semester. I studied for the subjects for the january exams so a revision would be enough to pass them in september.

Huy

@ramsay: do you know any function that almost results in itself again after differentiating twice?

user174558

@UserX In Sep, not on Sep.

UserX

oh well

robjohn

@ramsay Do you mean how to solve?

ramsay

14:35

@Huy yes

Huy

@ramsay: name some

ramsay

@robjohn yes, i mean to solve:P

@Huy 0

4

Huy

@ramsay: maybe a more interesting one

robjohn

@ramsay Do you know about integrating factors?

ramsay

@robjohn yup!

@Huy no idea!

user174558

14:37

Can we integrate multiples?

user174558

Why do we have differentiating factors?

robjohn

@ramsay Can you solve $\frac{\mathrm{d}x}{\mathrm{d}t}+i\sqrt{\frac km}\,x=0$

UserX

@ramsay Make the Ansatz that $x(t)=e^{\lambda t}$. I think you'll be able to solve the rest by yourself.

robjohn

@UserX it can be done in a more systematic way without happening to know the solution ahead of time.

ramsay

@robjohn no it has a complex function, if you remove it i may be able to solve

Evinda

14:41

And is an integral of a bounded function always bounded?

Also $\lim_{\epsilon \to 0} \frac{\phi}{|x|} dx=-\lim_{\epsilon \to 0}\frac{1}{\epsilon} \int_{|x|=\epsilon} \phi dS$.

Do we use now the dominated convergence theorem?

robjohn

@Evinda no... $\int_Df(x)\,\mathrm{d}x\le\max\limits_{x\in D}f(x)\int_D1\,\mathrm{d}x$

@ramsay how about $\frac{\mathrm{d}x}{\mathrm{d}t}+\lambda x=0$

Mary Star

Hello @arctictern !!
I thought again about what we were talking about yesterday...
When $I$ is a right ideal of $R$, why is $xI, x\in R$ also a right ideal?
So that it holds it must be: $ar\in xI, \forall r\in R, \forall a\in xI$.
But why does this hold?

Huy

@MikeMiller: any idea what $K(\pi_1(S),1)$ stands for?

(S being a surface with $\chi(S) \leq 0$)

ramsay

well(*first derivative behaves like a fraction*) , so
$\dfrac{dx}{x}=-\lambda dt$
am i correct?

then i will integrate

robjohn

@ramsay yes. Have you not had any complex analysis?

Huy

14:50

spotted the physics student

ramsay

@robjohn hmm, not so far :(

@Huy are you pointing me?

Huy

;)

ramsay

:-D

robjohn

@ramsay Oh, so you have not seen that $e^{ix}=\cos(x)+i\sin(x)$. I guess then you might need to use the method that UserX was indicating. Looking for functions which are the same when you differentiate them twice.

user174558

Such as 0, lol.

Mike Miller

14:52

@Huy $K(G,1)$ means an Eilenberg-MacLane space, a space with all homotopy groups zero except $\pi_1$, which is $G$.

This is equivalent to saying that $\pi_1 = G$ and the universal cover is contractible.

Any two $K(G,1)$s are homotopy equivalent.

Evinda

$\frac{1}{\epsilon}\int_{|x|=\epsilon} \phi dS \leq \frac{1}{\epsilon} \max_{|x|=\epsilon} \phi(x) \int_{|x|=\epsilon} 1 dS=2 \pi \max_{|x|=\epsilon} \phi(x) $ @robjohn

Is it right? Do we have to find now also an lower bound? @robjohn

ramsay

@robjohn haha. but i have see that $e^{ix}$ thing and i have used but i don't know complex numbers + calculus

Huy

@Mike thanks, I'll look up these spaces

ramsay

@JasonBourne hhehe:)

Mike Miller

I assume they're about to use some long exact sequences in homotopy groups like Birman's to understand the space of diffeomorphisms of the surface, which has the mapping class group as $\pi_0$.

robjohn

14:54

@Evinda why are you changing problems in the middle. we were working on $\frac{\partial\phi}{\partial\eta}$ which is multiplied by $\log|x|$

Huy

@MikeMiller: these are preparations in Farb & Margalit to prove Dehn-Nielsen-Baer (the extended mapping class group of a surface is isomorphic to the outer automorphism group of its fundamental group)

Mike Miller

Makes sense.

Extended?

Huy

with orientation-reversing homeos

Mike Miller

Oh, sure. I would probably just call that the mapping class group.

ramsay

@UserX still i don't know how to proceed

robjohn

15:00

@ramsay Consider differentiating $\sin(\lambda t)$ and $\cos(\lambda t)$ twice

user174558

@robjohn I forgot how to do that.

robjohn

@JasonBourne was I talking to you? ;-p

user174558

=) no need to add the smiley

ramsay

Oh :-D
:(

user174558

Where is the great Chris?

Evinda

15:06

Oh sorry... I thought that you meant that we do not use the dominated onvergence theorem for the other limit...

So in this case $\int_{|x|=\epsilon} \frac{\partial{\phi}}{\partial{\eta}} dS$ is bounded above since it is $\leq 2 \pi \epsilon \max_{|x|=\epsilon} \frac{\partial{\phi}}{\partial{\eta}}$ . Is it also bounded below?

ramsay

thank you robjohn, Huy

Akiva Weinberger

With homology, homotopy, homogeneity, and homomorphism…

why are there so many homos in math?

(Wait. That sounds like something else.

The question there would be "Why are there so few homos in math?".)

user174558

@ramsay You are welcome, lol.

ramsay

15:21

@JasonBourne but i never mentioned you, ;)

user174558

@ramsay Hence the lol, lol.

ramsay

hahahaaaaa:P

user174558

LOL

Evinda

Or is the above inequality maybe true for the absolute value, i.e.
$\left|\frac{1}{\epsilon}\int_{|x|=\epsilon} \phi dS \right| \leq \frac{1}{\epsilon} \max_{|x|=\epsilon} \phi(x) \int_{|x|=\epsilon} 1 dS=2 \pi \max_{|x|=\epsilon} \phi(x) $ ? @robjohn

ramsay

LOLLLL$\to \infty$

Mike Miller

15:34

@Akiva studying various notions of sameness is a natural thing to do

robjohn

@Evinda But $\epsilon\to0$, no?

@Evinda no, use the fact that $\lim\limits_{x\to0}\phi(x)=\phi(0)$

Evinda

15:53

@robjohn Yes. So we get that $\lim_{\epsilon \to 0}\int_{|x|=\epsilon} \frac{\partial{\phi}}{\partial{\eta}} dS \leq 0$.

@robjohn At which point do we use this?

robjohn

@Evinda put absolute values on the integrand even though it is multiplied by $\log|x|$, it still goes to $0$

qwertymk

How would I go about creating a function that would create different permutations of a string, but some chars have many options. For example: 'hello', the h can be either h or H but the o can be one of ['o', 'O', '0'], my initial thinking is to compute how many bits each char is and then loop from 1 to the max bits and do something like a base2 convert algorithm with the dividend being the proper bit. Does that make sense?

0 === 'hello'
1 === 'hellO'
2 === 'hell0'
3 === 'helLo'
...

Akiva Weinberger

@qwertymk I might be wrong, but I think this belongs better on the programming site than here

Evinda

16:09

$\int_{|x|=\epsilon} \frac{\partial{\phi}}{\partial{\eta}} dS$ $\leq 2 \pi \epsilon \max_{|x|=\epsilon} \frac{\partial{\phi}}{\partial{\eta}} \Rightarrow \left| \int_{|x|=\epsilon} \frac{\partial{\phi}}{\partial{\eta}} dS\right| \leq 2 \pi \epsilon \max_{|x|=\epsilon} \frac{\partial{\phi}}{\partial{\eta}} \to 0$

That's what you mean? @robjohn

LifeOfPai

16:24

Hello :)

A={1,2,3,4}
B={0,1,2}
Find an equivalence relation on sets of functions $B^A$ that all the equivalence classes have the same size...
I am really dont understand the question...

robjohn

@Evinda yes, except there is a $\log|x|$ in there, but does not change the limit

@LifeOfPai If every function is only equivalent to itself, then all equivalence classes have size 1

LifeOfPai

But $B^A={f:A->B}$

Why they equivalent to itself?

Evinda

16:40

You mean that it is like that: $\left|\int_{|x|=\epsilon} \ln{|x|} \frac{\partial{\phi}}{\partial{\eta}} dS\right| \leq \left| 2 \pi \epsilon \ln{\epsilon} \max_{|x|=\epsilon}{\frac{\partial{\phi}}{\partial{\eta}}} \right| \to 0$ while $\epsilon \to 0$ ? @robjohn

robjohn

$2\pi\epsilon\log(\epsilon)\max\limits_{|x|=\epsilon} \left|\frac{\partial\phi}{\partial\eta}\right|$

Axoren

How do you invite a user to a chat?

I've had people do it to me.

I want to do it to another user to move a discussion out of a comment thread.

ramsay

16:56

@Axoren is he now in any chat room

Axoren

@ramsay Couldn't say. I have no way to determining that, as far as I'm aware.

ramsay

well.... then i don't know to? sorry!

Axoren

Side note, @ramsay, your response to Huy looks like it has 10 times as many stars as it actually has upon first glance.

Evinda

@robjohn Nice... As for the other limit, can we calculate it as follows?

$\lim_{\epsilon \to 0} \int_{|x|=\epsilon} \frac{\phi}{|x|} dS=\lim_{\epsilon \to 0} \frac{1}{\epsilon} \cdot \lim_{\epsilon \to 0}\int_{|x|=\epsilon} \phi dS=\lim_{\epsilon \to 0} \frac{1}{\epsilon} \int_{|x|=0} \phi dS=\lim_{\epsilon \to 0} \frac{1}{\epsilon} \phi(0) \int_{|x|=0} 1 dS $.

But $\int_{|x|=0} 1 dS$ is equal to 0, isn't it?

Akiva Weinberger

It's fairly surprising that the proof of the Jordan curve theorem has basically no $\epsilon-\delta$.

Mike Miller

17:08

"The" proof?

Akiva Weinberger

The homology one.

J. M.

Ever since my e-mail address was included in a published article, I've been getting more and more spam. Pfft...

Akiva Weinberger

Maybe the existence of a Lebesgue number counts

Axoren

@J.M. I know the feeling. I'm getting emails from people who think I'm already a doctor.

Mike Miller

It's not terribly surprising. We've built a big machine to get away from point-set problems.

Akiva Weinberger

17:08

which is used somewhere, I think

J. M.

@Axoren Exactly my problem! :D "Dear Dr. J.M. ..." - dummkopf, I haven't even considered getting a doctorate yet!

Akiva Weinberger

But it's, like, used once, and never brought up in the rest of homology theory again.

Axoren

I ended up 3rd author on a journal paper in TKDE during my undergraduate and people have been doing it ever since.

Mike Miller

@J.M. I think mine is somewhere, but it's an address I no longer use, so it all works out.

J. M.

@MikeMiller :D Ostensibly, you're expecting correspondence from people who are curious about your ideas, not the girth of what's between your thighs...

Mike Miller

17:17

Luckily, I have no ideas.

Axoren

@J.M. Isn't it everyone's desire to want correspondence about both?

Personally, I enjoy my conversations with the Nigerian prince.

"desire to want" Good lord, I need to sleep more.

J. M.

@Axoren well, not in the address reserved for professional correspondence! :P

But taking the piss out of the prince is occasionally fun.

masfenix

Hey guys, I am looking for some clarification on Lipschitz functions. I have two functions $f_1(x) = x*\cos(\frac{1}{x})$ and $f_2(x) = x^2 \cos(\frac{1}{x})$ and $f_i(x) = 0$ defined on $(-1, 1)$. Initially I thought that $f_1$ is Lipschitz, but after taking the derivatives, I see that $f_1$ derivative blows up at $|x|->0$.

Any thoughts?

Evinda

@robjohn Or could we use somehow the dominanated convergence theorem?

Krijn

17:39

@MikeMiller Welp, then I'm gonna keep on searching for it

Evinda

Hey @Alessandro :)

Alessandro

good evening @evinda!

Evinda

How are you? @Alessandro

Alessandro

it's been quite some time since we last talked, how are you?

Mike Miller

Try Conway's complex analysis book in his section on uniformization. This might be in the sequel, I don't remember.

It's going to be something in potential theory, I just forget the name.

Evinda

17:40

Yes... I am fine, thanks... And you? @Alessandro

Alessandro

good thanks! I had a few exams last week but now I'm free for a while

Evinda

Has your semester ended? @Alessandro

Alessandro

no, it will end in June, but we have some so called "partial exams" in the middle of it

Evinda

Ah I see...

Mike Miller

@Krijn Conway 1, page 277, definition 5.1 $G$ a region in the plane, $a \in G$, a Green's function with singularity at $a$ is a function $g_a: G \to \Bbb R$ that's harmonic away from $a$, has $g_a(z)+\log |z-a|$ harmonic near $a$, and $g_a(z) \to 0$ as $z \to \partial G$.

Usually you then assemble these Green's functions into one big Green's function $g: G \times G \to \Bbb R$.

Blame Conway for naming the region $G$, not me.

Semiclassical

18:13

symplectic question for you, @mike

are action-angle variables a thing you've run into much?

Tobias Kildetoft

Hmm, a symplectic question. I wonder what those look like :)

Mike Miller

Nope

Semiclassical

nuts.

i'm trying to remember how Hamilton-Jacobi theory works at a level which isn't bullshit :)

i suspect there's a natural description of it in terms of symplectic geometry, but that don't mean i know it

Huy

what's bullshit-level

hhh

ti.inf.ethz.ch/ew/lehre/extremal04/heer.pdf

Huy

18:26

ETHZ.ch

I know that one

hhh

I am trying to understand the antichains, {1,2,3}, {12,13,23} -- what is the third?

${\emptyset,123}$ or only {123}?

Suppose poset has the largest chain of size r. Then the poset can be partitioned to r antichains.

Tobias Kildetoft

@hhh why third? There are 4 antichains there

hhh

${\emptyset,1,12,123}$ is a chain of size 4 or of size 3?

@TobiasKildetoft Thank you, that explained!

...of size 4 :)

I am trying to understand this oeis.org/A000372

the number of antichains of subsets of n-set, what does it mean?

Elements vs subsets

0 --> 2: \emptyset, no set
1 --> 3: \emptyset, no set, 1
2 --> 6: ...

I find it interesting that emptyset has the Dedekind size 2, can someone explain this?

Ted Shifrin

18:42

@Semiclassic: Do you know Abraham-Marsden? Excellent reference for physics-y folks.

Danu

18:54

Man, category theory is so awesome---I'm learning the basic basics of it, and it's so supremely enlightening to me!

Tobias Kildetoft

@Danu Yeah, it is a great topic

Danu

I love how it allows one to e.g. precisely state the goals of alg. top. as "finding functors from a top. cat. to some alg. cat." :)

Tobias Kildetoft

It provides both a more general picture of why certain things should be considered "the same" (via universal objects for example)

And it also provides new interesting structures to study in their own right, which has lead to interesting results in other areas as well

Danu

I am looking forward to its use in (co)homology.

Semiclassical

@ted i should take a look at that

@ted the other i should borrow from the library is Arnold's book on classical mech

Evinda

19:15

@robjohn Ok... I got that it is equal to $-2\pi \delta$.
If we would have $G(\overline{x},\overline{y})=-\frac{1}{4 \pi} \frac{1}{||\overline{x}-\overline{y}||}$, I wanted to find in the same way $\Delta G$.

I got that $\langle \Delta G, \phi \rangle=\lim_{\epsilon \to 0} \int_{|x|=\epsilon} G(|x|) \frac{\partial{\phi}}{\partial{\eta}} dS-\lim_{\epsilon \to 0} \int_{|x|=\epsilon} \frac{\phi}{4 \pi |x|^2} dS$

Am I right? But the first limit isn't equal to 0. Is it?

IdiotfromPrinceton

19:30

hello

Tobias Kildetoft

@IdiotfromPrinceton Hi

IdiotfromPrinceton

@TobiasKildetoft is every group of order 4 abelian?

Tobias Kildetoft

@IdiotfromPrinceton Yes

IdiotfromPrinceton

oh alright

also, if the kernel of a homomorphism is the identity, then do we say it is one to one? it this correct?

Tobias Kildetoft

@IdiotfromPrinceton Yes, a homomorphism is injective iff the kernel is trivial

IdiotfromPrinceton

19:37

thanks a lot for your time, i see now. these are not in my textbook thats why i am struggling

Tobias Kildetoft

How odd. This is usually one of the first things covered about homomorphisms

IdiotfromPrinceton

when i think abt it, it makes sense, but 'tis hard kinda.i use gallian

robjohn

@Evinda You can't take the limit in one occurrence of a variable and not in another

Tobias Kildetoft

It has to be in Gallian. That is such a standard text

hhh

How can I use the Binomial package in Macaulay2?

Evinda

19:46

Yes, I see that it can't be right... @robjohn
Did you see my other question?

hhh

I solved it, sorry disturbance, solution: loadPackage "Binomials";

Pichi Wuana

Hello

Semiclassical

@TedShifrin fyi---i was over by the math library for other reasons, so i grabbed A&M

so we'll see if I like it :P

Mike Miller

clickhole.com/article/…

Transcript for

$Mathematics$ Mathematics