Mathematics - 2020-05-12 (page 1 of 2)

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Shaun

00:24

I'm offering a bounty of 50 points for the following, should-be-simple question.

1

Q: Bijection from $A \rightarrow \varnothing$

My thoughts. We need to prove that: 1 $\forall x,y \in A, \text{ if } f(x) = f(y) \rightarrow x = y$ 2 $\forall y \in \varnothing, \exists x \in A, f(x) = y$. In (1), $f(x) = f(y)$ is false, since neither $f(x)$ nor $f(y)$ have a value, so (1) is vacuously true. Also, $\forall y \in \var...

elementary-set-theory

anakhro

00:54

@Shaun what do you not like about the selected answer?

EnjoysMath

01:14

0

Q: What functional equation does the "first in a twin prime pair" indicator function satisfy?

Basis for the $\Bbb{Z}$-module of eventually vanishing functions on $\Bbb{N}$. Let $M$ be the module of all eventually vanishing functions and for $f \in M$ define $N_f$ to be the vanishing point, i.e. it's the minimal $N_f \in \Bbb{N}$ such that for all $n \gt N_f$ we have $f(n) = 0$. Certain ...

elementary-number-theory prime-numbers natural-numbers open-problem prime-twins

K, I've done it now

Proof completed nearly

1 hour later…

Trey

02:25

can some help me with this question math.stackexchange.com/questions/219941/…

2 hours later…

user10478

03:58

When I compute $e^{tA}$ here (i.imgur.com/amXWu0Y.png) I get a different answer than when I compute $YY^{-1}(0)$ here (i.imgur.com/K9ieAdO.png), where the fundamental solutions for $Y$ are shown here (i.imgur.com/gnZhJeN.png) and $Y$ is inverted here (i.imgur.com/75QeXTk.png). What am I doing wrong?

abhas_RewCie

04:21

Morning

Pig

04:54

@BalarkaSen Hey - sorry I haven't read your previous messages yet

1 hour later…

robjohn

06:18

@Trey where are you having trouble?

skullpatrol

06:34

Did you read about the first black valedictorian at Princeton? @robjohn

robjohn

@skullpatrol Nicolas Johnson? No I haven't ;-P

skullpatrol

:D @robjohn

user434058

Is there any general method/procedure to find the maximum and minimum values of $|f(z)|$, where $f(z):\mathbb C\rightarrow \mathbb C$ is a complex function analytic everywhere?

Leaky Nun

06:52

@FakeMod-Inactiveaccount by Liouville's theorem the maximum (you mean supremum) must be infinity and the infimum must be 0

unless the function is constant

user434058

@LeakyNun Thanksl Now what about constrained optimisation. In the sense that $g(z)=0$ is the constraint and now we are supposed to find the maximum and the minimum values of $f(z)$?

Leaky Nun

$g(z) = 0$ defines a discrete set of points, unlike what you might expect

Thorgott

(if $g$ is analytic)

Leaky Nun

so you can't continuously trace through the zeroes of $g$ to maximize/minimize $|f(z)|$

so I don't know how you would do it

user434058

Mostly, I encounter questions where $g(z)=|z|-a$ (where $a$ is a constant). A typical example: Maximize $f(z)=|z^3-z+2|$ under the constraint that $|z|=1$.

user434058

07:02

@LeakyNun you can't. But could you use any complex equivalent of Lagrange Multipliers?

Leaky Nun

aha, then $g$ isn't analytic

cool fact for you: by Maximum Modulus Principle (MMP), the maximum in $|z| \le 1$ occurs on $|z| = 1$

user434058

@LeakyNun nope :P

user434058

@LeakyNun um... But that won't give the correct answer in my "typical example", which is $\sqrt {13}$...

Leaky Nun

why not?

are you sure it's $\sqrt{13}$?

user434058

@LeakyNun Oh sorry, it would :P I misread your message...

user434058

07:05

@LeakyNun definitely sure.

user434058

Though I know the long method of using polar form and then applying single variable calculus, however it's too cumbersome and boring :(

Thorgott

$|z|=1$ is a nice submanifold, so Lagrange multipliers should work

user434058

07:24

@Thorgott Could you please explain how (if convenient)?

Rafael Nadal

How do i find range of (x+1) + 4/(x+1) ? without using differentiation @LeakyNun

Leaky Nun

use differentiation

use AM-GM to find lower bound lol

Rafael Nadal

It's mentioned not to use differentiation

Oops!!

Thanks!! Just a glitch

Shaun

@anakhro It says, on one hand, that $A$ must be zero, yet, on the other, that $A$ can be arbitrary.

Leaky Nun

what glitch

Rafael Nadal

07:30

Glitch in my mind

Leaky Nun

@Shaun if a function $A \to \varnothing$ exists then $A$ must be empty

Rafael Nadal

What about this one (x+1) + 4/(x+9) ?@LeakyNun

Leaky Nun

-8 + (x+9) + 4/(x+9)

Rafael Nadal

I'm really confused in finding range when fn is of type like f(x) + 1/ g(x) ??

@LeakyNun thanks once again

Thorgott

@FakeMod I never thought this through, but you should just be able to translate it back to $\mathbb{R}^2$, apply the usual Lagrange multipliers there and translate it back

user434058

07:37

@Thorgott Oh! That's what I did for the first time. But again, it's cumbersome :) I suppose it's the only method...

Thorgott

I don't see any other method. The behavior of an analytic function on the boundary of a circle can still be rather arbitrary, I think, so holomorphicity may not make that much of a difference

I don't know anything about optimization except Lagrange though, so don't take that as a definite no

Balarka Sen

@Thorgott You proved smoothness of $\Phi$ on charts, right

user434058

@Thorgott alright Thanks :)

Balarka Sen

@Pig No worries! I talked to Ted about it extensively, I think I understand the issue now. You raised a good point

Thanks for the discussion!

Thorgott

@Balarka not yet, should I?

I guess I should

wait, the $U_i$ constitute an atlas already

Balarka Sen

07:53

@Thorgott Given any continuous function $g : S^1 \to \Bbb R$ there exists a continuous function $f : \Bbb D \to \Bbb C$ such that $f$ is holomorphic on the interior of the disk and $\Re(f)|\partial \Bbb D = g$.

In fact you can just demand $g$ to be $L^2$

Then $f$ will similarly be an $L^2$ functions such that ...

Thorgott

wha-

Balarka Sen

Wild, right?

Thorgott

damn, I knew there was something like this

but I didn't picture it to be that bad

Balarka Sen

@Thorgott yeah

and on each $U_i$ it's smooth

it agrees on the overlaps $U_i \cap U_j$ by uniqueness

Thorgott

right, so they essentially glue together by definition?

Balarka Sen

08:00

yup

Thorgott

so gluing together smooth functions on manifolds is for some reason easier than gluing them together on $\mathbb{R}^n$

well, cause $\mathbb{R}^n$ has an atlas only given by one chart, I suppose

Balarka Sen

haha yeah

the curves $\gamma_x = \Phi(x, \cdot)$ are called the "flowlines" of $X$ (starting at $x$), the map $\Phi$ is just following this flowline for time in $(-\varepsilon, \varepsilon)$. The flowline is a canonical object on the manifold $M$, but you have proved that on $U_i$ this "following the flowline" map is smooth in $x$ parameter - so it must be that it is globally smooth in $x$ parameter

So you do get a smooth map $\Phi : M \times (-\varepsilon, \varepsilon) \to M$. Can you tell me why $\Phi(\cdot, s + t) = \Phi(\cdot, s) \circ \Phi(\cdot, t) : M \to M$, for all $s, t, s + t \in (-\varepsilon, \varepsilon)$?

Thorgott

$\gamma_x$ is basically tracing the path the point $x$ traverses on the manifold if it moves in the direction given by the vector field at each point with time?

Balarka Sen

Yup

Doing this globally on $x$ paints the picture of a "flow", every point moves time $t$ in the direction given by $X$ at each point of time

Thorgott

I'll think about the homomorphy in a couple minutes (currently in a lecture, but there's a break soon)

Balarka Sen

08:08

Think of the vector field on $S^1$ given by the unit tangent counterclockwise, which I like to call $\partial/\partial \theta$. The flow along this vector field for time $t$ is given by rotating $S^1$ counterclockwise by angle $t$, so this particular flow keeps rotating $S^1$ counterclockwise forever and ever with unit angular speed

Ya sure

More pictures: Consider the vector field $X = p \, \partial/\partial x + q \,\partial/\partial y$ on $\Bbb R^2$ and scale it to be unit by $T = X/\|X\| = X/\sqrt{p^2 + q^2}$, just for convenience. Note that $X$, hence $T$, is invariant under the action of $\Bbb Z^2$ on $\Bbb R^2$ by vertical/horizontal translations.

Therefore $T$ descends down to a vector field on the torus $T^2 = \Bbb R^2/\Bbb Z^2$.

What does the flow look like?

Mike Miller

@BalarkaSen $f$ will be smooth on the interior and $L^2_{1/2}$ up to the boundary :P

Sha Vuklia

08:42

@MikeMiller I was hoping a night of sleep might give me some new insights, but alas. I am trying to figure out your four-line-one-line proof. All I could think of was to take for each $(g,g)$ on the diagonal a small neighbourhood $V_g\times V_g'$, such that for $(x,x')\in V_g\times V_g'$ we have $\vert h(x)-h(x')\vert<\epsilon$.

Ted suggested to find a $U$ such that $(g,g')\in \tilde{U}\iff gg'^{-1}\in U$. For a fixed $g\in G$, we have $(g,g')\in V_g\times V_g'$ iff $g'g^{-1}\in (V_g')g^{-1}$, which is indeed a neighbourhood around zero. However, the problem is that I fixed $g$, so this is useless I think. I don't see any other way tho.

Mike Miller

Why are you fixing $g$

Consider the map $\varphi: G \times G \to G \times G$ sending $(g,h) \to (g, hg^{-1})$

What is $\varphi(V)$ a neighborhood of?

Or rather, what is $\varphi$ of the diagonal?

Balarka Sen

@MikeMiller Ah

Mike Miller

Sobolev trace lemma says that restricting L^2_k to boundary loses 1/2 of a derivative

For L^p should be 1/p but I always forget some details (does it hold for all k? can I do it for boundary hypersurfaces? I think these are true but I forget)

Balarka Sen

I should learn this stuff eventually

Next semester I will probably, when I take complex analysis and functional analysis afterwards. Doubtful they'll do proper functional analysis but can't hurt to learn

Mike Miller

I think I wrote how to get the result you mentioned from APS somewhere

Balarka Sen

08:55

Oh possibly

Btw, what are integrable functions which simply satisfy the mean value property? In what sense do they solve $\Delta f = 0$?

Sha Vuklia

@MikeMiller Ah, so, $\phi$ is a homeomorphism, so we know $\phi(V_g\times V_g')$ is open. And since $(g,g)\in V_g\times V_g'$, we know that $\phi(V_g\times V_g')=V_g\times U$, where $U$ is a neighbourhood of the identity. Now, we have $(x,y)\in V_g\times V_g'$ iff $(x,xy^{-1})\in V_g\times U$. However, and this was my problem which is why I fixed $g$, we still don't have $xy^{-1}\in U\implies (x,y)\in V_g\times V_g'$.

Mike Miller

What is $V_g \times V_g'$ wtf

Who introduced that

Sha Vuklia

oh, I did before, sorry should have clarified

so $V_g\times V_g'$ is the neighbourhood around $(g,g)$

Mike Miller

Yeah I didn't say pick a small neighborhood of one point, that's not going to help you

Sha Vuklia

such that for $(x,y)\in V_g\times V_g'$ we have

o..

right

Mike Miller

08:59

You're trying to do this over all of $G$. Why would you restrict to a neighborhood of one point, right?

to START with you have $(g,g') \mapsto |h(g)-h(g')|$, which is a map $H: G \times G \to \Bbb R$

And then $V = H^{-1}(-\epsilon, \epsilon)$ is an open set which contains the diagonal, because $H(g,g) = 0$ for all $(g,g)$ on the diagonal

You're trying to show that there is a neighborhood of the origin $U$ so that if $g' g^{-1} \in U$, then $(g,g') \in V$

It will not help to do this any less than globally, over all of $G$ at once. Anyway, this is the $V$ I was talking about

Try the hint for that

Sha Vuklia

Right, I will think about it, thanks

Rudi_Birnbaum

Hi there, anyone here interested in both surreal numbers and quasicrystals?

Sha Vuklia

09:39

@MikeMiller Right, I finally see it x) So, $\phi$ is a homeomorphism, whose inverse is given by $(x,y)\mapsto (x,y^{-1}x)$. We have $\phi(V)=G\times U$, for some open $U\ni e$. Now, if $gh^{-1}\in U$, then $(g,gh^{-1})\in G\times U$, and hence $(g,h)\in V$.

or actually... I haven't used compactness now, I think

Mike Miller

No we don't, @ShaVuklia

We have that $\phi(V)$ contains $G \times \{e\}$

There's no reason to believe that $\phi(V)$ is a product itself though

And indeed you haven't used compactness of $G$ yet. From what I said above, what can you get about $\phi(V)$?

Sha Vuklia

Ah, man x) Right so, for each $g\in G$, we can find a product, $V_g\times U_g$, where $V_g$ is a neighbourhood about $g$, and $U_g$ a neighbourhood about $e$. By compactness of $G$, we only need finitely many to cover $G$. Taking this finite intersection of $U_g$, should give us $U$ then.

Mike Miller

don't you know a lemma that means you don't have to do more work?

there should be a lemma that immediately applies to the situation above

Sha Vuklia

oh, tube lemma?

Rudi_Birnbaum

Anyway, lets interprete a 1D quasicrystal as a surreal number. The surreal numbers of day $\omega$ e generated from a sequence of right, left operations. If we take for example the Fibonacci chain which is encoded as a series of such two different symbols, we would obtain a day $\omega$ surreal number.

As the Fibonacci chain is not a uniquely defined sequence of symbols (the paricular sequence depends on how it was generated exactly, e.g. substition, multigrid, cur and project, ...) I ask myself how the corresponding set of surreal numbers is composed.

Sha Vuklia

09:49

ye okay, that is nicer x)

Mike Miller

Yes, the whole point of this setup is that you wanted a sort of tube around the diagonal, right?

But there isn't a "diagonal tube lemma"

So we use the group structure to turn the diagonal into a slice

At which point we just have the tube lemma

Leaky Nun

@Rudi_Birnbaum how high are you

CosmicChalk

I haven't read the chat guidelines .I am new here. Hope it is not very strict chat which kicks you out for asking question.

Mike Miller

Nobody here has the power to kick you

Leaky Nun

if you read the chat description you'll see "Just ask; don't ask to ask."

Mike Miller

09:54

The worst case scenario is that nobody can / wants to answer your question, which can happen

Sha Vuklia

@MikeMiller Oh, that's what that $\phi$ did. Right:D thanks, that makes it more intuitive.

Rudi_Birnbaum

@LeakyNun Its not easy to quantify ...

Mike Miller

That was why I did it in the first place :)

Rudi_Birnbaum

It possibly has a nice answer: I conjecture, that the difference between two such Fibonacci-chain surreal numbers is a finite day generated number.

The idea is that the different Fibonacci chains are identical up to a finitely long sequence of elements.

@LeakyNun So what I am aiming at would be Freeman Dysons proposal of a classification of 1D-quasicrytals.

(Of his "birds and frogs" talk)

Edward Evans

10:21

Jo zeeeas @Rudi

robjohn

@MikeMiller ?

Mike Miller

@robjohn At that moment... ;)

robjohn

@MikeMiller Ah, I thought I was still in the room. I must have floated off...

Rudi_Birnbaum

10:57

@EdwardEvans Zeeeaas @Edward

wia gäds?

bist in DE oder GB?

Leaky Nun

@Rudi_Birnbaum how dare you call Britain gross! :P

Rudi_Birnbaum

@LeakyNun right, UK is anyway more accurate ..

Gross Britain

Thorgott

@Balarka Ok, I claim that for a function $\Phi\colon M\times(-\varepsilon,\varepsilon)\rightarrow M$ satisfying $\Phi(p,0)=p$ for all $p\in M$, the following two properties are equivalent.
1. $\Phi(p,s+t)=\Phi(\Phi(p,s),t)$ for all $p\in M$ and $s,t$ such that $|s|,|t|,|s+t|<\varepsilon$ and $d\Phi(p,-)\vert_0=X(p)$ for all $p\in M$.
2. $d\Phi(p,-)\vert_t=X(\Phi(p,t))$ for all $p\in M,|t|<\varepsilon$.
Suppose $\Phi$ satisfies 1., then $d\Phi(p,-)\vert_t=d\Phi(p,t+\cdot)\vert_0=d\Phi(\Phi(p,t),-)\vert_0=X(\Phi(p,t))$ (hard abuse of notation), so $\Phi$ satisfies 2.

Edward Evans

@Rudi bin noch in DE hehe

Rudi_Birnbaum

Hast den lockdown gut überstanden?

Edward Evans

11:10

Jaaa für mich war's eh business as usual

Rudi_Birnbaum

Was meinst wie das weitergeht?

Edward Evans

irgendwie hab ich nichtmal was gemerkt hahaha

Rudi_Birnbaum

lol

Edward Evans

Ich hab ehrlich gesagt keine Ahnung, aber mich freut's wenn ich von daheim aus arbeiten kann hahaha

Rudi_Birnbaum

(ehrlich gesagt: mich auch ...) lol

nächste Woche muss ich mal in die uni und einen Lehrfilm aufnehmen

das erste mal seit 2 Monaten...

wird hart, aber was tut man nicht für die Studis

Edward Evans

11:12

ooof, ich halte einen Seminarvortrag in einem Monat ungefähr, den ich auch von daheim halten werde lol

Rudi_Birnbaum

Sind deine Leute daheim ok?

Edward Evans

Ja denen geht's gut, aber die englische Regierung ist echt kacke

Rudi_Birnbaum

BoJo ...lol

aber eigentlich ned lustig

Edward Evans

ja überhaupt nicht, aber man kann nur halb lachen lol

Absoluter Trottel

Rudi_Birnbaum

und dann der krasse Typ da sein Beraterkumpel wie heisst der?

Edward Evans

11:15

Dominic Raab?

Rudi_Birnbaum

Dominic Cummings

?

Thorgott

Humor ist, wenn man trotzdem lacht

Edward Evans

ahhhh kann sein

Dominic Raab ist der Außenminister lol

irgendwie hab ich seit der Parlamentwahl angefangen die britische Politik zu ignorieren

und die Tagen zu zählen bis ich die deutsche Staatsbürgerschaft beantragen kann hahahaha

Rudi_Birnbaum

@Thorgott ja

Alessandro Codenotti

The Germans have taken over the chat

Edward Evans

11:27

d'Deeeeitschn

Rudi_Birnbaum

;-)

Germs?

Edward Evans

Germknödel

Rudi_Birnbaum

lol

T_01

11:48

I have to "find a condition" such that the image of an ideal under a ring-homomorphism $f$ is again an ideal.
I think the task is strange, since I could say $f$ is surjective and I'm obviously done - but what is the minimal condition for this?

Or, wait, is this condition minimal? If $f$ is not surjective I can always choose a field for the target Ring, and then $f$ yields a non-ideal as image as long it does not map everything to zero, right?

Edward Evans

@Alessandro I feel like functional analysis is just the triangle inequality with extra steps

Alessandro Codenotti

There's an extra "functional" in your sentence

Edward Evans

rofl

just about to have a seminar talk on semisimple rings and Artin-Wedderburn

and then directly after a talk on local fields lol

AND my copy of Lang's algebra is arriving today; I have a wheelbarrow ready to transport it up to my room once it arrives

Alessandro Codenotti

Nice, good luck for your talks!

Rudi_Birnbaum

12:05

@EdwardEvans lol

this one?

Balarka Sen

@Thorgott Excellent

I claim you can just extend that to an actual homomorphism $\Bbb R \to \text{Diff}(M)$ by going $\varepsilon/2$-much everytime

Knight

12:28

Can I find this limit $$ \lim_{n\to \infty} \frac{q^n \pi}{n !} \left(\frac{\pi}{4}\right)^n $$

Thorgott

why not $2\varepsilon$?

Knight

Please guide me

The most I can think is that the denominator grows faster than numerator and hence the limit may be zero.

Thorgott

ratio test will do the job

abhas_RewCie

@Knight Use Stiriling's approximatio

Thorgott

that's way overkill

abhas_RewCie

12:37

$n! \approx \sqrt{ 2 \pi n} (n/e)^n$

boom

Solved now?

Thorgott

the limit is trivial, Stirling's approximation is not, there's no reason to use it

abhas_RewCie

why?

Thorgott

I just stated why

Simply Beautiful Art

13:17

Out of curiosity, is there any way to swap the antidiagonal elements of a 2x2 matrix using matrix multiplication?

That is: $$\begin{bmatrix}a&b \\c&d \end{bmatrix} =\begin{bmatrix}? &?\\?&?\end{bmatrix} \begin{bmatrix}a&c \\b&d\end{bmatrix} \begin{bmatrix}?&? \\?&?\end{bmatrix}$$

Knight

@Thorgott What is ratio test? (if that reply was for me)

Balarka Sen

@Thorgott I mean, you don't know if the flow exists at time $\varepsilon$. You know it exists at time $\varepsilon/2$, so you extend it to exist for $(-\varepsilon, \varepsilon) \cup (\varepsilon/2 - \varepsilon, \varepsilon/2 + \varepsilon) = (-\varepsilon, 3\varepsilon/2)$

And so on

@EdwardEvans Listening to Autumn Eternal by Panopticon

Seems good

Thorgott

@Knight if $\limsup\Big\lvert\frac{a_{n+1}}{a_n}\Big\rvert<1$, then $a_n\rightarrow0$

Balarka Sen

@Thorgott Basically what I am saying is for any $t \in \Bbb R$, choose $n \in \Bbb Z$ such that $n\varepsilon/2 < t < (n+1)\varepsilon/2$. Then define the flow as follows: For notational ease I will denote $\Phi_t(x):=\Phi(x, t)$. Define $\Phi_t := \Phi_{\varepsilon/2}^{\circ n} \circ \Phi_{t - n\varepsilon/2}$

Knight

@Thorgott Thank you.

Balarka Sen

13:28

Where $\Phi^{\circ n}_{\varepsilon/2} = \Phi_{\varepsilon/2} \circ \Phi_{\varepsilon/2} \circ \cdots \circ \Phi_{\varepsilon/2}$, $n$ times.

Thorgott

What I was thinking of is each real having a unique representative in $(-\varepsilon,\varepsilon)$ $\mod2\varepsilon$

Balarka Sen

These are well defined, because a-priori you knew $\Phi_s$ makes sense for $s \in (-\varepsilon, \varepsilon)$

Thorgott

but I actually can't take steps that large, so that doesn't make sense

Balarka Sen

Yeah what you said just needed to be modified a little I think

Great, so we have proved flowing along a vector field on a compact manifold always makes sense

Thorgott

compactness doesn't play a role in extending $t$ to all of $\mathbb{R}$, I take it?

geocalc33

13:32

given these transformations how do you find the "net" transformation that is some combination of the two? $x\mapsto x^2$ and $(1-x) \mapsto (1-x)^2$

is there such a thing?

Balarka Sen

Yeah, it doesn't, because what you proved gives us a uniform $\varepsilon > 0$ irregardless of $x \in M$ for which $\Phi_t(x)$ is defined for all $|t| < \varepsilon$.

In general, given a vector field on a noncompact manifold, this $\varepsilon$ will vary point-to-point. Indeed, take the example of the vector field on $\Bbb R$ I gave by pushing $d/dt$ forward along a diffeomorphism $\tanh^{-1} : (0, 1) \to \Bbb R$ so that the norm of the vector in the resulting vector field on $\Bbb R$ gets very huge as you go towards infinity. There $\varepsilon$ varies from point-to-point so much that you cannot obtain a uniform $\varepsilon > 0$ that works.

Simply Beautiful Art

@SimplyBeautifulArt nvm lol

Thorgott

right, but if we have one $\varepsilon$ for a range of $x$, then we can always extend to all of $\mathbb{R}$ (but not necessarily to all of $M$)

Balarka Sen

I think if you inspect the proof closely you will be able to obtain a smooth function $\varepsilon : M \to \Bbb R_{+}$ such that $\Phi_{t}(x)$ is well-defined for all $|t| < \varepsilon(x)$.

Right

Thorgott

that's saying the solution interval of the IVP depends smoothly on the parameter

sounds believable

Balarka Sen

13:36

Yeah

Tapi

Can some help me with this?
$f(x)=\frac {2020^{2x}}{2020+2020^{2x}}$. Find $\sum f(\frac {k}{2019})$ with k from 1 to 2019.

Balarka Sen

@Thorgott You remember the notion of tangentspacewise dot product I defined earlier, before the flow story?

Leaky Nun

@BalarkaSen what is this, German?

Balarka Sen

Lol

Thorgott

I'm still in the process of convincing myself this $\Phi_t$ is still additive

Balarka Sen

13:47

Fair enough. It's instructive to understand this in terms of the flowlines

EnlightenedFunky

Does this site do migrations any more?

Leaky Nun

@EnlightenedFunky no. unfortunately, for migrations, one should seek their closest embassy

EnlightenedFunky

like I have a question a meta, and I want it migrated and I remember when I opened my account they can migrate their question to the main stack

Like it was a whole a thing the question title would or some where around there would say migrated from MathOverFlow.SE

feynhat

@Thorgott What does additivity mean in this context?

Thorgott

$\Phi_{s+t}=\Phi_s\circ\Phi_t$

Balarka Sen

13:54

Additive in time parameter

feynhat

okay.

Thorgott

Ok, say we have $m\varepsilon/2\le s<(m+1)\varepsilon/2$ and $n\varepsilon/2\le t<(n+1)\varepsilon/2$. Then $\Phi_s\circ\Phi_t=\Phi_{\varepsilon/2}^{\circ m}\circ\Phi_{s-m\varepsilon/2}\circ\Phi_{\varepsilon/2}^{\circ n}\circ\Phi_{t-n\varepsilon/2}$. But $s-m\varepsilon/2<\varepsilon/2$, so these commute (and the choice of $\varepsilon/2$ really was the right one), hence $\Phi_s\circ\Phi=\Phi_{\varepsilon/2}^{m+n}\circ\Phi_{s-m\varepsilon/2}\circ\Phi_{t-n\varepsilon/2}$.
Now we are guaranteed to have $(m+n)\varepsilon/2\le s+t<(m+n+2)\varepsilon/2$. If, in fact, $s+t-(m+n)\varepsilon/2<\var…

fuck

ok was just missing a dollar

Balarka Sen

Ya we extended something additive in (-eps, eps) additively to all of R so you expect that to be additive

Good to write down the details though

Thorgott

yeah, it's not surprising

just had to figure out that I can commute them through to collect exponents and then the case distinction

Balarka Sen

yup

Thorgott

13:58

I should also convince myself this is smooth in $t$ still

wait, that's obvious

Balarka Sen

u should join leagues with feynhat

Thorgott

but is it jointly smooth still

no wait, I'm being nonsensical

for each $t\in\mathbb{R}$, $\Phi_t\colon M\rightarrow M$ is smooth as composition of smooth functions (in fact, all diffeomorphisms, so itself a diffeomorphism)

but, if I fix $p\in M$, is $t\mapsto\Phi_t(x)$ smooth?

Balarka Sen

right thats what i meant when i said think in terms of flowlines

this follows from uniqueness, and the fact that smooth functions on open things glue to smooth functions when they agree on overlap

Anonymous

Hi. If $A$ is a non-abelian normal simple subgroup of a group $G$ and $C$ is the centralizer of $A$ in $G$, could someone explain why $A\cap C = 1$?

Thorgott

for some reason I have a mental barrier in my head that's telling me smooth functions on open sets don't glue together that easily

Balarka Sen

14:04

the sticky point ur thinking about is smooth functions on closed intervals agreeing on endpoints of the closed intervals dont glue to a smooth function

Thorgott

yeah lol

Balarka Sen

but since we're doing (-eps, eps) U (-eps + eps/2, eps + eps/2) and you have smooth curves defined on each of those intervals which agree on the intersection by uniqueness, the whole thing glues to a smooth curve

Thorgott

I was just trying to construct a counter-example using bump functions and realized I'm being an absolute buffoon

Balarka Sen

no this is a fine point

it tells you that the gluing lemma is not really the sheaf axiom

Thorgott

the fact that they glue together here is just stating that smoothness is a local property, essentially

Balarka Sen

14:06

cuz the most useful form of the gluing lemma is gluing continuous functions over closed guys which is stronger than gluing them over open guys

ya that smooth functions form a sheaf, to be precise :)

Thorgott

actually, they agree on the overlap, because of additivity, no?

but well, additivity follows from uniqueness, so same thing, I guess

Balarka Sen

@Triskelion if a nontrivial element of $A$ centralized $A$ then that would be in $Z(A)$, but centers are normal subgroups...

@Thorgott ye thats true

Thorgott

ok, after two days, I think I'm finally ready to believe you that flows are a thing

Balarka Sen

next thing i wanted to tell you that if you have a vector field such that for a long stretch of time nothing happens on a region in the manifold then the topology of the manifold is trivial on that region

Mike Miller

Your life would have been easier if you didn't believe

That's too vague bro I don't know what that means

Balarka Sen

14:13

i will tell him what it means

i have to clickbait first

Anonymous

@BalarkaSen Oh, and centers are also supposed to be abelian subgroups (which isn't admissible here as $A$ is non-abelian). So contradiction, I see. Thanks!

Mike Miller

I need to explain to you this theorem of Fenichel once I understand if

Balarka Sen

no thats the wrong contradiction

Thorgott

Social Experiment: having a vector field such that for a long stretch of time nothing happens on a region (gone wrong, nearly trivialized the topology)

2

Balarka Sen

@MikeMiller see, this is scary

i misread fenichel as fenchel even though u gave me a heads up

awful

Anonymous

14:16

@BalarkaSen Non-abelian simple groups are also centerless I think

Balarka Sen

correct, you need simplicity

abelianity is just to avoid Z/p

feynhat

Why are you guys splitting the domain into pieces? Is it to avoid saying things like 'whenever it is defined'? Wouldn't additivity follow trivially from uniqueness: For all $s$, $t \mapsto \Phi_{s+t}(p)$ defines a curve starting at $\Phi_s(p)$, which is by ~~definition~~ uniqueness $t \mapsto \Phi_{t}(\Phi_s(p))$? (Of course I have completely overlooked the fact that $s+t$ may not even be in the domain, but whenever it is, additivity holds).

Balarka Sen

it follows from uniqueness sure

Thorgott

we are splitting it into pieces precisely to define it everywhere

additivity on the initial interval (whenever defined) followed from uniqueness

feynhat

Ah okay. Global flows? Have you assumed something about the field? (Sorry, I haven't been following this discussion).

Anonymous

14:21

@BalarkaSen Heh, I momentarily thought that non-abelian groups only have non-abelian subgroups. Which is, of course, terribly false.

Balarka Sen

pedants joining leagues

time to scurry

feynhat

:|

Thorgott

this discussion started two days ago, I'd be worried had you been following

the manifold was assumed to be compact

feynhat

oh... okay. So, on compact manifold all fields are global?

Mike Miller

Yes, Thorgott is writing down a proof, though you can too

feynhat

14:24

Can? Probably. Will? Strong No.

Thorgott

I'm not sure what that means, what we've been discussing is that a smooth vector field on a compact manifold generates a unique, global flow

feynhat

Yes, I was asking if you can always extend local flow (which always exists) to $\Bbb R \times M$.

Balarka Sen

thats indeed what thorgott means by a global flow

feynhat

Cool.

You chatted up with the monk yet?

Balarka Sen

sorta

Thorgott

14:28

if you have a flow on $(-\varepsilon,\varepsilon)\times M$, you can extend it to $\mathbb{R}\times M$, this is what we did above

Balarka Sen

he told me probabilistic geometry is just a geometry bundle over a measure space

Thorgott

but you can't necessarily extend on $I\times U$ to $I\times M$ if $M$ is non-compact, it seems

feynhat

@BalarkaSen shivers

Thorgott

also, if we have that it's smooth in $x$ and smooth in $t$ if we fix the other each, it's also smooth in $(x,t)$, right? cause an atlas on the product manifold is given by the products of charts

Balarka Sen

oof, no. not even true for a function $f : \Bbb R^2 \to \Bbb R$

ah C^1

ok

continuous partials imply multivariable differentiability, thats what you need

feynhat

14:33

@BalarkaSen Were you thinking of continuous functions?

Thorgott

yeah

Balarka Sen

its not true for differentiable functions

Thorgott

but this is clearly given since everything's smooth

Balarka Sen

right

Thorgott

but actually, we don't need that

smoothness in Euclidean spaces is defined by the existence of partials of all orders

and you get partials by fixing all but one variable and differentiating wrt to the one that's left

(and you can do that all orders if fixing all but one argument gives you a smooth function each)

Balarka Sen

14:35

thats the right definition because that agrees with f is differetiable, Df is differentiable, DDf is differentiable, ...

Thorgott

I have never thought about DDf and I don't think I want to

feynhat

Why man? Hessians are nice. Symmetric for C^2. And also help generalize the second-derivative test for extremas.

Thorgott

yeah, I like Hessians

Hessian = mixed partials, DDf = no clue tbh

Mike Miller

lol

Thorgott

but bordered Hessians are the real shit

feynhat

14:41

no clue tbh

Balarka Sen

useful for lagrange multipliers iirc

Thorgott

yeah, that's why I know them

you formulate sufficient second-order conditions for maximization on a smooth submanifold using the bordered Hessian

what you actually want is for the Hessian of the Lagrange function to be positive- or negative-definite when restricted to the tangent space

and you can check that by calculating minors of the bordered Hessian

feynhat

@BalarkaSen Roushon asked me to read an intro paper on Lie Groupoids... And Thurston's notes on orbifolds.

Balarka Sen

Cool!

Teach that shit to me, I need to know Haefliger structures

feynhat

w h a t

Mike Miller

14:45

no way that's in the lie groupoid stuff

feynhat

No clue. But that H-word appears thrice in this paper.

Mike Miller

It's a guy's name

Balarka Sen

oh i misread orbifold as foliations

feynhat

Thanks Mike.

Balarka Sen

idk what lie groupoids are

Mike Miller

14:48

Yeah any time

I guess Haefliger structures were one of the first places that topological groupoids mattered to people

Balarka Sen

it makes sense that a k-dim foliation is classified by maps to something its not super clear to me what

germs of homeomorphisms R^n -> R^n preserving the k-dim plaques of the std foliation?

Mike Miller

The Haefliger groupoid on $X$ is the groupoid of germs of maps $U \to V$ where $U, V \subset X$

Balarka Sen

and that germ stuff is not going to be a group bcuz youre not fixing anything

its flowing along the leaves

so groupoid, i guess

Mike Miller

You bake in both the topology of $X$ and the germs of maps

And then you define "Gamma-structure" like you would for "G-bundle" in terms of cocycles, but apparently the theory for groupoids is much richer

Balarka Sen

sounds annoying to set up

Mike Miller

14:52

paper looks short

Balarka Sen

oh this should be simple enough

Let $G$ be the groupoid of germs of diffeomorphisms of $\Bbb R^k$, then on a manifold with a $k$-dimensional foliation you should naturally have a $G$-valued cocycle

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