Mathematics - 2016-08-01 (page 1 of 2)

0celo7

00:50

I'm back

How did you poor souls survive without me

CRAZY GAY SHERIFF

01:02

uni tennesse lulz

0celo7

whut

Akiva Weinberger

02:21

@BalarkaSen I haven't read beyond the title and first sentence of this yet, but it seems to be related to what we were talking about a few days ago

mi.sanu.ac.rs/vismath/carter2011mart/Carter.pdf

Forever Mozart

02:38

for every integer $n \geq 2$, there exist positive integers $x, y$, and $z$ such that

${\displaystyle {\frac {4}{n}}={\frac {1}{x}}+{\frac {1}{y}}+{\frac {1}{z}}.}$

how to prove

tell me how and I will give $100

@PVAL you are topologist

Forever Mozart

03:59

Is it just me, or has the topology section become amateur hour? Are these people just posting homework problems?

0celo7

04:10

link?

I have a hard-ish topology problem

Balarka Sen

04:30

@AkivaWeinberger Indeed.

@0celo7 What's that?

@Huy I woke up at 10 in the morning, contrary to what you predicted. I guess I'll be fine :)

Forever Mozart

04:46

@0celo7 what is the problem?

the list of problems under the topology tag is depressing

Balarka Sen

05:19

Can I prove max mod principle for entire functions like this? Suppose $f$ is entire and $|f|$ attains global maximum at $p$. Then $f(p) = \frac{1}{2\pi} \int f(z)dz/(z-p)$ (integrate on a circle), which is the same as $\frac{1}{2\pi} \int_0^{2\pi} f(p + re^{i\theta}) d\theta$. Take absolute values to get $|f(p)| \leq \frac{1}{2\pi} \int_0^{2\pi} |f(p + re^{i\theta})| d\theta \leq \frac1{2\pi} \int_0^{2\pi} |f(p)| d\theta = |f(p)|$.

Hence some inequality there must be an equality: evidently the second inequality. Thus, $|f(p + re^{i\theta})| = |f(p)|$ for all $\theta$ and $r$. That means $|f|$ is constant on $\Bbb C$ - Liouville's theorem kicks in to show $f$ is constant.

I suspect a version of this would work for local maximums of holomorphic functions too. I don't think $|f|$ can be constant on a disk.

@PedroTamaroff You asked me to prove that proper entire functions are polynomial. I just read a proof in the book I am studying from of the fact that meromorphic endomorphisms of $\Bbb P^1$ are rational functions. That immediately proves it, because proper means extends to an endomorphism of one-point compactifications, and the only rational functions with no poles other than that at the infinity are polynomials.

It doesn't seem to use any significant theorem, though, but just a little trickery of local structure of functions at poles.

Tobias Kildetoft

05:35

Does anyone know how to see what was updated when orcid informs you that cross-ref has updated something in your profile? I can only see that it is in the "works" section, but it has updated like 5 times now without adding anything there that I can tell.

Forever Mozart

i am not an animal

Balarka Sen

huh?

Forever Mozart

elephant man

the movie

Balarka Sen

oh

Parcly Taxel

06:07

Just a question: how do I unsubscribe from all stack exchange emails?

Tobias Kildetoft

@ParclyTaxel The emails should contain whatever is needed to unsubscribe from them

Parcly Taxel

yes, I'm on preferences, now what do I do?

Which box do I edit to unsubscribe?

None seems obvious.

Tobias Kildetoft

Presumably whichever one you used to subscribe (I have never subscribed to those emails so no idea)

Parcly Taxel

Well I got an email that said "share your knowledge..."

It was basically a promo

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Is there not an "unsubscribe" link on the bottom?

Btw those^ should be required by law :P

Tobias Kildetoft

06:18

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ And they are in many places

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Really? Thanx for the info :-)

Parcly Taxel

There is one, but it links to the preferences page of my profil.

Tobias Kildetoft

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ Or at least a link that can be used for unsubscribing (no idea how it is formulated exactly, and this might depends on the location(

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Ok, let check my profile too @ParclyTaxel

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

06:30

Try looking in "edit your settings" ----> "web presence"

Parcly Taxel

yes?

I see a website link, a Tumblr link and a GitHub link, all of which are filled

And then below I see "private information" and in there is my email address

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Unfill them

Mine are blank

Parcly Taxel

Unfill what?

The email address?

Tobias Kildetoft

Still wondering whether any examples of whatever this paper is about actually exist arxiv.org/abs/1607.07345 (that is, examples which are not actually groups)

4

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Not the email address, just the others.

Parcly Taxel

06:38

The Tumblr and GitHub things?

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Yup

Parcly Taxel

Are you sure that will unsubscribe me?

(I should unfill the website link too?)

Actually, never mind.

That's a one-shot.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Mine are blank and I don't get any emails.

If that doesn't work send an email to the administration team.

Huy

07:34

@BalarkaSen: that's not so bad, is it?

Tobias Kildetoft

07:52

@Huy Up for trying to find an example of what I mentioned above?

Huy

@TobiasKildetoft: might be a question for main if you really want to find an example

don't really feel up for trying because I don't think they will be very fruitful if there are examples

Tobias Kildetoft

@Huy If I asked on main, I would feel like I was "calling out" the author for what is probably a poor paper

Huy

well I think any paper that introduces a supposedly "new structure" without providing a single example is definitely a poor paper

or at least shows they exist and are different from the other structures

reminds me of this

Tobias Kildetoft

Yeah, I am not even sure if examples exist (also, the definition given is slightly ambiguous, as it is not clear whether the identities and inverses are part of the data of the structure, since the identities need not be uniqye but seem to be treated as unique when defining inverses)

@Huy Yeah, would be sortof funny if these turned out to really just be groups "disguised".

Especially since it seems that the one place in which these showed up for the author is in a context where it turned out to be a group after all

Huy

@TobiasKildetoft: if you want I can ask on main, I don't intend to pursue a career in academy so I don't mind if people think I called someone out

you can always just email the author otherwise

Tobias Kildetoft

08:03

@Huy Nah, I don't really care enough. I was just curious

Huy

yeah, neither do I to be honest. :P

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

me three :-)

Mithlesh Upadhyay

09:49

265 answers with negative score math.stackexchange.com/users/175066/…

Martin Sleziak

10:13

SEDE: Users by percent of answers that have negative score

I suppose the query could be modified to show the users with largest number of negatively scored answers.

@TobiasKildetoft I do not think that is a problem. There was a recent question I and Brian M. Scott answered. Somebody studying some paper asked how some steps are derived. Both mine and Brian's answer said that the proof in the paper is not correct. I believe we both have written it in rather polite way.

On the other hand, it was an already published paper. I see that you are talking about arxiv preprint.

Tobias Kildetoft

@MartinSleziak Sure, but here I would feel like I was calling out a poor paper by someone who seems to be earlier than myself in their career, which just does not seem quite right

Martin Sleziak

Of course, it's your decision to make. (And it all depends on how much you are interested in that topic.)

I just wanted to mention that something similar (to some extend) has been asked before. And it did not come from an effort to "expose" bad paper, rather than from an effort to understand results published somewhere.

The name disguised-groups gives the title of the paper a bit esoteric feel.

Tobias Kildetoft

@MartinSleziak The paper generally feels like it was written by someone not particularly used to math research

Danu

11:16

@TobiasKildetoft I don't think it's impolite or anything to ask a question like that.

Emilio Pisanty

12:31

@MartinSleziak Sure. Take the latest version, data.stackexchange.com/math/query/396403/…, and change the ORDER BY to 'no. neg.'

Or more simply, just click on the appropriate column header in the results table ;-)

0celo7

13:23

@BalarkaSen @ForeverMozart The fundamental group of a complete, nonpositively curved Riemannian manifold contains no torsion elements.

All I have is a result about periodic isometries in the universal cover, but no seems to know what torsion elements imply so I'm stuck

Mithlesh Upadhyay

13:51

@EmilioPisanty , I were counted manually and that 265 is correct too.

Emilio Pisanty

@MithleshUpadhyay Fair enough, to each the technology they prefer. Maybe an abacus would help? =P

Mithlesh Upadhyay

@EmilioPisanty , off course.

Semiclassical

14:36

morning chat

Balarka Sen

@Huy Nope, perfectly fine.

@0celo7 Do those not have universal cover $\Bbb H^n$? Aka, such manifolds are aspherical (K(G, 1)), so if $G$ has torsion the manifold can never be a finite dimensional CW complex - see eg page 149 of Hatcher.

That's clearly garbage.

skill patrol

morning chatters

:-)

0celo7

I think you need to have constant curvature for the cover to be $H^n$, @BalarkaSen.

Balarka Sen

Hmm, I see.

0celo7

This result is cleverly titled "Cartan's theorem"

Balarka Sen

14:47

If universal cover is contractible, at least, then you're done.

0celo7

Which narrows the literature search to a few thousand papers

Especially because he as multiple Riem. geo. theorems

Balarka Sen

Seems like it's this.

Which says the universal cover is diffeomorphic to $\Bbb R^n$ - good enough for me.

0celo7

No, it's a corollary to: "if $(M,g)$ is complete, simply connected, and nonpositively curved, then any periodic isometry has a fixed point"

It might follow from Cartan-Hadamard

But I think that is proved after this

Balarka Sen

Huh? What is a corollary to what?

0celo7

@BalarkaSen The torsion thing is a corollary of the thing in quotes.

Balarka Sen

14:50

Could be, but I just proved it from Cartan-Hadamard.

Now I don't know if Cartan-Hadamard has anything to do with this quoted thing.

That's your job to figure out.

0celo7

I'm fairly sure Cartan-Hadamard is proved after this

But what does $\Bbb R^n$ have to do with torsion elements

Balarka Sen

Did you read what I wrote?

0celo7

The contractible thing?

I don't see that either

Balarka Sen

$\Bbb R^n$ is contractible - that means the space is a K(G, 1) (only homotopy group at $\pi_1$). K(G, 1) is finite dimensional only if G is free.

0celo7

Ayy that's too much algebraic topology

Balarka Sen

14:53

True. This is not so trivial.

0celo7

Is there maybe an isometry that shifts between the leaves of the covering?

Like on the universal cover of $S^1$, rotates everything up

Balarka Sen

I don't know what that means, but I don't know crap about Riemannian geometry.

if there is a torsion, there is a deck transformation which has finite order, yes

0celo7

Aha!

I bet deck transformations are isometries

(that's the kind of result I've been asking for)

Deck transformations are isometries.

It's on page 30 of this book

@BalarkaSen Do deck transformations ever have fixed points?

Balarka Sen

Only if it's identity.

0celo7

One that's not the identity

I didn't learn any math this weekend

Not a single thing

Balarka Sen

15:01

I learnt a bit about harmonic functions.

0celo7

Yeah?

CRAZY GAY SHERIFF

@ForeverMozart how would distinguish those questions from your questions, though?

Balarka Sen

@0celo7 Nothing terribly exciting.

I did learn a harmonic function on $\Bbb C$ is always real part of a holomorphic function - which makes a lot of results for holomorphic functions push through for harmonic ones too.

0celo7

Huh

skill patrol

How about a semi-amateur half-hour? @CRAZYGAYSHERIFF :-)

0celo7

15:09

That's probably proved in Jost RGGA or Jost Riem. Surfaces

Balarka Sen

It's funny how you always refer to books whenever mentioned a fact or a theorem :P

0celo7

I don't know the proof, but I know where it probably is

Why is that funny?

My advisor does it too

He just has way more books

Balarka Sen

Normal people probably care about the mathematical content of the fact or the theorem and comment on that, rather than mentioning they have seen the proof in [XYZ book].

0celo7

Normal people do not care about mathematics, @BalarkaSen

Balarka Sen

It is clear I meant normal people who does mathematics.

(yes, I know I shouldn't use the word "clear"...)

0celo7

15:15

@BalarkaSen This is trivially clear and obvious.

Balarka Sen

More annoying would be "It is clearly and obviously trivial"

0celo7

Clearly.

@BalarkaSen Can you give me a summary of the "argument principle" please

From the complex analytic PoV

Balarka Sen

Take a meromorphic function $f$ on $U \subset \Bbb C$.

Take a closed disk $D$ in $U$.

Then $\int_{\partial D} f'(z)/f(z) dz = n - p$ where $n$ is the number of zeroes of $f$ in $D$ and $p$ is the number of poles in $D$.

0celo7

Why does GP write that as $\oint d(\mathrm{arg}\,f(z))$

Balarka Sen

Intuition is that $f'(z)/f(z)$ is the derivative of $\log f(z)$. Then if $f$ transverses the circle $\partial D$, something happens up in the cover.

@0celo7 Well, $\log(f(z)) = \log|z| + i\text{arg}(f(z))$, not? I guess he's just writing that out like that.

BTW, there should have been a $1/{2\pi i}$ outside of my integral.

0celo7

16:00

@BalarkaSen Do you mean $\log |f(z)|$?

Forever Mozart

19:45

@CRAZYGAYSHERIFF @skillpatrol the difference is that my questions cannot be found in Munkres

0celo7

@ForeverMozart If your point set topology is not in Munkres, I don't know what to say

You're probably thinking too hard about point set topology

@BalarkaSen Are you aroundeth?

0celo7

20:03

@BalarkaSen Homological orientation of manifolds is discussing in Dold.

Forever Mozart

20:48

@0celo7 Topology fundamentals are in Munkres, yes, but these are not the things I have questions about. Munkres is not the sum total of point set topology. There are tons of open problems and new developments.

0celo7

but are they worthy of human effort ;)

Semiclassical

considering that's his research you're referring to, I would presume so.

Forever Mozart

depends on your taste. Physics folks probably say that very little of pure mathematics is worth anything

Semiclassical

meh. I don't really like that logic. To me there's not 'worthwhile math' and 'not worthwhile math' but 'math I find interesting' and 'math I don't find interesting.'

0celo7

Damn Poincare recurrence

What does "recurrence" even mean

Forever Mozart

20:52

happens again

returns to initial state

0celo7

You think you're some kind of comedian

Semiclassical

you're the one who asked what recurrence meant

0celo7

I need a precise definition

"almost all points are recurrent"

I've had to construct a Borel measure to make "almost all" precise

Now what the hell does "recurrent" mean

Forever Mozart

something like if you keep applying $f$ to a point, you get a certain point

but I cant remember the theorem

Semiclassical

In continuous systems, it's usually not phrased as 'you get back to where you started' but 'you get arbitrarily close to where you started.'

0celo7

20:54

I think Poincare recurrence is not the same as the existence of closed curves in phase space

@Semiclassical Yes

So it should be that there's an open set that you return to

I'm thinking it means that given open $O$ and $t>0$, we have an integer $n$ such that $(\varphi_t)^n(p)\in O$

where $\varphi_t$ is the Hamiltonian flow

maybe one can define it in terms of a limit

like, as $t\to\infty$ one tends closer to $p$ around each trip...

Semiclassical

is $p$ a point in configuration space or in phase space?

0celo7

@Semiclassical phase space

I have a proof in mind, actually

But I need to define recurrent first...

Because if I can make this definition of recurrent be the same as this other definition...I'm in the money

Forever Mozart

ocelot

0celo7

yes

Forever Mozart

Ocelot

The ocelot (/ˈɒsəlɒt/; Leopardus pardalis), also known as the dwarf leopard, is a wild cat distributed extensively within South America including the islands of Trinidad and Margarita, Central America, and Mexico. It has been reported as far north as Texas. North of Mexico, it is found regularly only in the extreme southern part of Texas, although there are rare sightings in southern Arizona. The ocelot is similar in appearance to a domestic cat. Its fur resembles that of a clouded leopard or jaguar and was once regarded as particularly valuable. As a result, hundreds of thousands of ocelots were...

0celo7

21:00

God tier animal right there

I'm stupoid

Forever Mozart

you should make that your picture

0celo7

I like my current picture

Forever Mozart

of what?

0celo7

how old are you

Forever Mozart

10

0celo7

21:06

13 is the min age, that's a bannable statement

Forever Mozart

there is a min. age for this room?

0celo7

13

for the site in general

Forever Mozart

ok well I am beetween 13 and 70

0celo7

I'll guess 25

Forever Mozart

close

0celo7

21:08

You're probably not hip since you study topology

Forever Mozart

the graph theory kids are much cooler

0celo7

Geometers get the most tail

Forever Mozart

incidentally, most geometers I know are least like other mathematicians

very different

0celo7

what do you mean

Huy

WHAT DO YOU MEAN @FOREVERMOZART

0celo7

21:18

^

what is a limit, anyway

@ForeverMozart I need some point set topology!

What could the limit of a sequence mean in a general topological space

Forever Mozart

just personality-wise

$p$ is the limit of $(x_n)_{n\in\omega}$ if every open set containing $p$ contains a tail of the sequence

if $U$ is open and $p\in U$ then there exists $N$ such that $x_n\in U$ for all $n\geq N$.

Carlitos_30

21:34

Hello.

0celo7

@ForeverMozart I have a neighborhood basis of my point, and a sequence such that every point is contained in each set of the neighborhood basis at least once.

Does the sequence converge to the point?

Forever Mozart

not necessarily

0celo7

what else do I need

Forever Mozart

like the sequence $1,0,1,0,...$ does not converge to $0$, even though you can find a basis of $0$ that contains each point 1 and 0 at least once

0celo7

what if things are also compact

all I need is a convergent subsequence, really

Forever Mozart

21:37

you at least need every open set to contain infinitely many points of the sequence

Carlitos_30

Can I ask something?

Forever Mozart

yes

Carlitos_30

May I?

0celo7

no

Carlitos_30

ok

From Apostol book

about Supremum

every nonempty set and bounded above

Owatch

21:39

Evening.

Carlitos_30

has Supremum

The set can be {1,2}

it has supremum, right?

It would be 2

0celo7

sure

Carlitos_30

0celo7

the sup of a finite set is the max

Carlitos_30

x > sup S -h

h > 0

He says

0celo7

21:42

what

Carlitos_30

that this property that every set with supremum

Danu

This guy is basically posting full homework sets.

Carlitos_30

contains numbers near the supremum

0celo7

@Danu So?

Carlitos_30

but this set only contains two numbers

Forever Mozart

21:43

it is so fucking hot outside that I might have a stroke if I go running

0celo7

@Carlitos_30 Please stop typing fragments

Ask your whole question in complete sentences in one post, and use MathJax.

Forever Mozart

wish me luck

Carlitos_30

ok

Danu

@0celo7 So I downvoted all of his posts :)

Carlitos_30

ok im going to construct the thing first

Forever Mozart

21:44

if I'm not back in 30 min., call the police

0celo7

@Danu Savage!

Danu

Need to discourage that BS

Forever Mozart

mozart out!

Danu

You're not Mozart brah

0celo7

And you're not Danu

Danu

21:45

You must be a comedian

0celo7

No, but you know who is? Antoni Kosinski.

He defines orientation via relative homology, what the hell is this

Danu

Standard definition.

0celo7

Aren't you a smart one

Danu

I'm in fact typing notes on stuff related to that right now.

But it really is the standard definition

0celo7

I distinctly remember ACM telling you that in the chat and you being amazed

So don't give me that shit

Danu

21:48

Yes, indeed

I didn't say it's not great, or hard

I'm just telling you it's the standard definition.

(at least when you read topology)

0celo7

Not in Hirsch, Milnor, GP, Bott & Tu, etc.

Danu

That's all in the smooth setting, isn't it?

0celo7

The book I'm reading is about smooth manifolds.

@Danu Hirsch is not smooth.

Danu

differential topology

Sounds smooth to me

0celo7

Sure

You always know better

@Danu Regardless, I don't see why Kosinski defines orientation via homology, then states the Jacobian thing without proof.

Danu

21:51

Why are you trying to create drama? Take a chill pill

0celo7

Why not define it via Jacobians from the beginning?

What intuition is one supposed to glean from relative homology that the Jacobian definition does not give

@Danu I'm not. Hirsch is famous for not restricting to the smooth category, but I guess the title tells you otherwise.

Transcript for

$Mathematics$ Mathematics