Mathematics - 2019-02-06 (page 1 of 3)

Mike Miller

00:00

Though I prefer the interval factor first.

Ted Shifrin

Oh, gotcha. You're starting with an arbitrary one. Sorry.

Yeah, yeah, I'm rusty.

OK, I'm caught up. Now you can finish after the dentist. :P

Mike Miller

No need to be sorry.

Ted Shifrin

LOL, no need to be hyper-solicitous, either :P

Mike Miller

Yeah, seems right.

Anyway, you cap off the boundary components and have a simply connected closed 3-manifold.

In other words (Perelman) you have S^3.

Now you just need to delete the balls.

Ted Shifrin

Oh, I see.

Mike Miller

00:04

Ok, I'm gone. T^2 is harder.

Ted Shifrin

Thanks :P

Mike Miller

It's a surgery argument (like one I have in a post about 3manifolds with fundamental group Z).

Ultradark

I got a textbook

Daminark

Same

Eh I guess "I got" suggests recent so I can't make that joke

:/

Dair

what textbook? lol

lol what joke?

i really need to stop saying lol. I do it too much irl...

Ultradark

00:19

you say lol in spoken language?

Dair

yeah lol.

Ted Shifrin

heya Dair.

Dair

i have problems.

@TedShifrin Hey! I believe you can confirm is say lol in real life?

Ted Shifrin

I don't usually, no.

Dair

ok. nvm. haha.

Érico Melo Silva

00:22

hlo

Dair

@Ted The weather has been crazy today.

Ultradark

hi

Daminark

@Dair it's a joke I make sometimes. If someone starts a story saying "So the other day I was taking a walk" I immediately interject and say "same".

Ted Shifrin

Yeah, Dair.

Daminark

And yeah sometimes I say "lol" in real life but it's rare

Dair

00:25

it just rolls of the tongue so well. And it's less cacophonous than haha.

Érico Melo Silva

if u say lol in real life ur a nerd don’t at me

Ted Shifrin

Do you say "lol" or L-O-L ?

Dair

@Ted "lol"

I don't actually spell it out.

Ted Shifrin

I know no one (else) who does that.

Dair

I was about to say... Until you corrected your message

Daminark

00:27

I also say it as "lol"

Ted Shifrin

sigh

Daminark

I also do that with "lmao"

luh-mao

Dair

L'Mao

Yeah definitely I feel it.

Érico Melo Silva

that one is way worse

Ted Shifrin

next you'll say Lao-Tzu.

Dair

00:28

next you'll be saying raful.

too bad there isn't really a good way to wtf except for spelling it out...

Rithaniel

It's like a french cat: "Le Mow."

Dair

@ÉricoMeloSilva You seem to be greatly troubled by this.

Ted Shifrin

En français on dit "miauler" ...

Érico Melo Silva

im not troubled by anything lol

Dair

well, except for your GR pset...

:P

Érico Melo Silva

00:32

insulted is a better word tbh

Rithaniel

Gotcha, Ted

Dair

Miauler sounds like some sort of bastardization of Mahler

Ted Shifrin

not if you pronounce it in French :)

Dair

Miauler's Symphonie Fantastique.

Rithaniel

It looks like it'd be closer to "moo-lay"

Ted Shifrin

00:36

that's Berlioz, Dair :P

Dair

ohhh yeah.

didn't Mahler make like that depressed version of Frera Jacque?

i forget how to spell it lol.

wait lmao i should know this.

Ted Shifrin

Yes, Mahler 1st symphony has that.

Frère Jacques

Dair

except creepy af

and Beriloz has the massive Eb clarinet part.

Ted Shifrin

Berlioz has serious brass stuff

Dair

Eb clarinet part is super rare. nobody likes the Eb clarinet.

Leaky Nun

00:40

@TedShifrin what is the physical significance of the laplacian?

Ted Shifrin

divergence of the gradient

It measures how much the value at the center of a sphere differs from the average values on shrinking spheres centered at that point.

Leaky Nun

is there like an "integral" formulation?

khan calls them "formal definitions" which I disagree with

Ultradark

what do you call a derivative that is not truthful

Mike Miller

@TedShifrin and hence governs heat flow :)

Ted Shifrin

Green's identities are the answer to that, Leaky.

Did the dentist pull any teeth, @MikeM?

Mike Miller

00:50

No, but I have to come back. :(

Ted Shifrin

Story of my life ....

Ultradark

dentist

I know one of those

N. Maneesh

0

Q: How do I prove the $f:X\to I$ by $f(x)=\min\{f_i(x_{\beta}):i=1,2,...,n\}$ is continuous.

How do I prove that $f:X\to I$ by $f(x)=\min\{f_i(x_{\beta}):i=1,2,...,n\}$ is continuous. My attempt:- Minimum of two continuous real-valued functions are continuous. $$f_i \circ \pi_{\beta_i}: X\to I$$ be continuous function.So, $f(x)$ can be re-defined as $f(x)=\min\{f_i(x_{\beta_i}):...

Dair

@Ultradark It's probably a good idea that everyone in chat knows at least one dentist.

N. Maneesh

So, $f(x)$ can be re-defined as $f(x)=\min\{f_i(x_{\beta_i}):i=1,2,...,n\}=\min\{f_i\circ \pi_{\beta_i}(x):i=1,2,...,n\}$.

Is it necessory to redefine the function like this?

Érico Melo Silva

00:55

@MikeMiller heat flow best flow lets go

N. Maneesh

$f(x)=\min\{f_i(x_{\beta_i}):i=1,2,...,n\}$ Can I appy the result 'Minimum of continuous real valued function is continuous directly here?

Mike Miller

Hey want to read that Ricci flow on surfaces paper together later? Month from now maybe?

Ted Shifrin

@N.Maneesh: Note that your title makes no sense. You need the subscript on the $\beta$.

N. Maneesh

okay. I have corrected it.

Ted Shifrin

The tricky part here is that you have a function of $x$, not just of the finitely many $\beta_i$. But it's not very tricky.

N. Maneesh

00:58

@TedShifrin What about my argument? Is there any mistake?

Ted Shifrin

Think about the product topology. If you have $X = \prod X_\alpha$ and you fix one $\alpha$ and set $f(x) = g(x_\alpha)$, if $g$ is continuous, does it follow that $f$ is?

I haven't read it.

Mike Miller

I think I'm going to try to read Kirby-Siebenmann.

Ted Shifrin

Yeah, your argument is right, @N.Maneesh. Once you know the min of two continuous functions is continuous, it's just induction to get $n$.

N. Maneesh

@TedShifrin Thank you very much. :)

Ultradark

heat check, get wrecked, ricci flow cruisin on the manifold. top down, windows low, can't stop, modular flow over the space of lattices, you already know

6

Secret

01:52

The mathematics of emptiness:

$\varnothing = \text{class}(\varnothing) = \text{object}(\varnothing) = 0 < \text{null}$

But for some reason, I felt zero is less empty than the emptyset

Meanwhile, null is much less empty than zero because zero is a subset of null objects, which includes any object not necessary zero such that given some relation, it evaluates to zero

A common example are nilpotent elements in algebra where there exists some nonzero $\epsilon$ such that $\epsilon^2=0$, another example are null vectors in inner product spaces

Ultradark

interesting

Secret

02:12

For most who are untrained in the perception of emptiness, all notions of emptiness, such as nonexistence, nothingness, phantom, ghost, vacancy, hollow, emptiness, blank, conceal, unseen, hidden, subtle, counterfactual (ok this one is not), etc. look the same and considered to be the same equivalence class of nothingness

But the reality is, there are different shades of nothing, and they are all different

Back to maths, no matter how many times an empty set is partitioned, (can be rigorusly done by letting a partition function $\pi_i$ and $i \in I$ where $I$ is some index set, and then put elements under $\pi_i$ into the ith equivalence class) the resulting sets are always empty

Newbie

02:49

@TedShifrin very nice video youtube.com/watch?v=ez1rWBPznEc

Secret

Optimistic Nihilism

►

Silent

Let $f:\Bbb R\to\Bbb R$ be such that $f''$ is continuous on $\Bbb R$ and $f(0)=1, f'(0)=0, f''(0)=-1$. The $\displaystyle{\lim_{x\to\infty}\left(f\left(\frac{\sqrt2}{x}\right)\right)^x}$ is .....

I did this using particular function $1-\frac{x^2}2$, but how to do generally?

Mary Star

05:52

Hello!! How can we calculate the supremum of $-\frac{2}{9x^{\frac{4}{3}}}$ ? I got stuck right now.

Akiva Weinberger

Apparently xkcd wrote another book

Secret

hmm?

Marystar: Isn't the supremum of that just the asymptotic value?

Silent

06:08

@AkivaWeinberger, hi. Will you please let me know why does this happen: This here says that line integral over scalar field is a special case of line integral over vector field, but while line integral over vector field changes sign when path reversed, this says that line integral over scalar field is independent of path direction!!

Akiva Weinberger

If we have a line integral over a vector field, but the vectors are all tangent to the path of integration, then it's the same as the line integral over the scalar field of their magnitudes

If you reverse the direction of the path then the vectors are no longer tangent to the path of integration

Er, I guess we need "tangent and pointing in the same direction"

and so if it points oppositely it counts as negative

@Silent

Silent

ok! thank you very much for that modification

@AkivaWeinberger, so modification is needed in math.se answer, right?

Akiva Weinberger

The answer in math.se says exactly what I just said

(Note that $r$ is the path)

Silent

06:36

@AkivaWeinberger I am sorry, but i can't figure out how $\cos \theta=-1$ possibility says "tangent and pointing in the same direction"

Akiva Weinberger

If it's $-1$ then it's pointing in the opposite direction and counts negative

(and $\theta=180^\circ$)

Silent

So, if we reverse path, then line integral on scalar field be negative, right?

Zee

07:05

Who knows

Silent

@Zee I still think that reversing path reverses sign even in scalar integral, i took path parallel to x axis and tried both directions, and sign seems to reverse!

Akiva Weinberger

@Silent No, it should stay the same

for a scalar field

Silent

07:21

Here blue arrows represent f, while red arrows are path's direction. Please explain how does line integral does not change.

@AkivaWeinberger

(Maybe i need to provide more info, but f(x,y)=(1,0) is what in my mind, and (t,0) is original-first path.)

Lucas Henrique

Hi chat.

Akiva Weinberger

08:29

@Silent You've drawn a line integral of a vector field though

Note that in our recipe for converting a line integral of a scalar field into a line integral of a vector field, the vector field depends both on the original scalar field and the path

Silent

omg! let me think about that. thanks for pin pointing that.

Leaky Nun

Apparently you can't loop a song in Spotify even if you paid?

ÍgjøgnumMeg

Yes you can

Fargle

@LeakyNun Put the song into its own playlist, then turn loop on

ÍgjøgnumMeg

or just double click the repeat button

Fargle

08:41

Oh nice, I didn't know about that

ÍgjøgnumMeg

A little $1$ appears

Leaky Nun

@ÍgjøgnumMeg no I can't do that with the new update

ÍgjøgnumMeg

Hmm that's weird

on phone version?

Leaky Nun

ok I found the button it's hidden in the 3 dots menu

yes phone

ÍgjøgnumMeg

Yeah

I've just seen that lol, weird

wonder why they'd bother moving that

Leaky Nun

08:44

The Spotify team isn't the brightest

Akiva Weinberger

09:32

The Simons Center at Stony Brook

user616128

@Secret What are these different shades of nothingness?

Do you mean in reality, or in logic, or what?

Akiva Weinberger

yesterday, by Rudi_Birnbaum

@AkivaWeinberger Fourier-Transformation

@Rudi_Birnbaum Found this

jcrystal.com/steffenweber/dos/fourier.html

Akiva Weinberger

10:03

This is also a neat image

@Rudi_Birnbaum Go here jcrystal.com/steffenweber and click on "Fourier Tr." on the left

Heh:

Markings on the Penrose rhombs that enforce the matching rules

(with the result that these tiles can tile the plan, but they cannot do so periodically)

(and they can tile the plane in a locally unique way, in the sense of, if you have two infinite tilings with these shapes, any finite piece of one can be found infinitely many times in the other)

Secret

10:44

@user616128 nah forget that, its metaphysics and possibly not even wrong

Akiva Weinberger

11:08

Whoa:

Secret

how come this is a fractal. I thought quasicrystals have fixed sizes?

Akiva Weinberger

Oh my god there's a whole gallery of these GIFs

imgur.com/a/IZM3J

and here

imgur.com/a/MvLNc

Rithaniel

11:27

Whoa indeed

Akiva Weinberger

These are similar

biorobots.case.edu/personnel/adh/quasicrystal/examples

ÍgjøgnumMeg

@Akiva I like the fast ones

Rithaniel

This one is messing with my head a little bit: i.imgur.com/8GpZL.gif

Pentagonal tilings are so weird. The fact that you can almost do it uniformly actually gives rise to some arguably more interesting patterns than uniform tilings would.

Rithaniel

11:45

This one appears to be a tiling of equilateral triangles in disguise, if those flashing green lines are a reliable indication: biorobots.case.edu/personnel/adh/quasicrystal/examples/…

Akiva Weinberger

What the hell did I just watch:

John Whitney "Catalog" 1961

►

(If you do watch this, I would recommend not having the sound on, maybe having some other music on instead)

(Not a fan of the music)

Rithaniel

It's not even music, I'd say. Just discordant noise.

Akiva Weinberger

@Secret It's made in a different way from the Penrose tiling

It's made from superimposing waves, I think

Not 100% sure how they got it to be self-similar but I think it was in a similar way to auditory illusions like the Shepard's tone

(Fun fact: the sound equivalent of "visualize" is "audiate")

Secret

I see

@Rithaniel Well most neoclassical music sounds like that, nothing very surprising

Rithaniel

Is that what this is? (I still wouldn't call it music, though that also begs the questions of how I define music.)

Akiva Weinberger

11:55

I look at the zoom-in GIF and then I look somewhere else and it looks like wherever I'm looking at is shrinking away

(The opposite for some zoom-out GIFs)

It's so weird

Who needs drugs

Rithaniel

Alright, homework question: "If $N$ is a normal subgroup of $G$, show that $G/N$ is abelian if and only if $N$ contains the commutator subgroup of $G$." I'm having some difficulty coming up with a good approach to this. (I think I need to have that if $n\in gN$ then $n^{-1}\in gN$ as well, but I can't figure out how to show that either.)

mrnovice

hey anyone know anything about binary search?

Tobias Kildetoft

@Rithaniel How have you had the commutator subgroup defined?

Rithaniel

It's the subgroup generated by commutators of any two elements in the parent group.

Tobias Kildetoft

Ok, so what does it mean for a subgroup to contain this?

Rithaniel

12:07

It must contain all commutators of any two elements in the parent group. I assume we have to use this to show that $(gh)N=(hg)N$

Tobias Kildetoft

Right. What does that equality mean in terms of certain elements being in $N$?

Rithaniel

Oh, I think I see. On one side you have $(hg)1$ and on the other you have $(gh)[h,g]$

Astyx

hi chat

Tobias Kildetoft

@Rithaniel No, just use the standard result about when two cosets are equal

Rithaniel

They have non-trivial intersection?

Tobias Kildetoft

12:13

No, in terms of something to do with their representatives. I.e. $xN = yN$ iff "something to do with $x$ and $y$ and $N$"

Secret

Fractals are typically not self-similar

►

ok ...

Rithaniel

Well, I know that if $xN\bigcap yN\neq \emptyset$, then $xN=yN$

Tobias Kildetoft

I am looking for the one that does not involve cosets in the end

Rithaniel

I might not have learned that one yet.

Tobias Kildetoft

$xN = yN$ iff $y^{-1}x\in N$

(I refuse to believe you have not learned that btw)

Rithaniel

12:16

Yeah, haven't learned that one yet.

Tobias Kildetoft

what terrible text does not include that one before going into quotients by the commutator?

Rithaniel

Well, I don't know what text we're using, I've been going off the class notes. We've mostly been touching on the notion of cosets forming a partition on the group in lecture.

Tobias Kildetoft

Anyway, prove the statement I wrote (it is pretty easy given the ones you already know), then use that.

Rithaniel

Alright, it seems much more straight forward with that.

Thank you for the help.

Akiva Weinberger

An applet that generates quasiperiodic tilings by taking slices of higher-dimensional, periodic tilings

gregegan.net/APPLETS/12/12.html

@Rithaniel Why is this a normal subgroup?

Rithaniel

12:24

What do you mean? Why does it need to be a normal subgroup or what makes a particular subgroup normal?

Akiva Weinberger

A subgroup $N$ of $G$ is normal if $gN=Ng$ for all $g\in G$

Secret

Unrelated, it actually puzzles me on whether the notion of Lebesgue measure actually have an intuitive counterpart. Because that fact that the ordering of points in an uncountable set can change the value of the measure is something that does not really have a real life counterpart (how often does one saw something that basically change its length or area or mass or whatever purely because of how it is arranged in space)

Akiva Weinberger

In other words, if $h\in N$ then $ghg^{-1}$ is also in $N$

Rithaniel

Yes, I'm familiar with that. (Also, the n=4 tiling on that gregegan applet is cool)

Akiva Weinberger

So why is $a(bcb^{-1}c^{-1})a^{-1}$ a product of commutators

Oh I figured it out

Tobias Kildetoft

12:29

@AkivaWeinberger It is actually a lot easier to write up that the image of a commutator under any homomorphism is again a commutator.

Akiva Weinberger

$a[b,c]a^{-1}=[a,b][b,ac]$

Tobias Kildetoft

@AkivaWeinberger unnecessarily complicated. It is also just $[aba^{-1},aca^{-1}]$.

Astyx

$=[aba^{-1},aca^{-1}]$

Akiva Weinberger

Ohh

Ohh I get you, $f([a,b])=[f(a),f(b)]$

and then let $f(x):=axa^{-1}$

Right yeah that's much simpler

How to get 1D quasiperiodicity from 2D periodicity:

(Note that the diagonal line isn't being cut by the grid lines it intersects, but rather by the projections of nearby grid points onto it)

(so that there are only two lengths of segments that it is cut into)

(If it were cut by the grid lines, the result would be the same as just superimposing two periodic 1D lattices of different sizes.)

Akiva Weinberger

12:50

I'm not sure how this was generated but it doesn't look like periodic lattices rotated and superimposed on each other so I assume it is actually based on Penrose somehow

Or maybe it's from superimposed plane waves again

student

is there notation like big-O to indicate that terms do not depend on a given symbol?

Tobias Kildetoft

@student It is customary to include the symbols that something does depend on, rather than the other way around.

student

13:07

@TobiasKildetoft Yes, but sometimes, you want to say that there are terms in an expression you don't care about. When you complete a square, say, and the term you add is irrelevant for the discussion.

Tobias Kildetoft

@student "don't care about" is not a mathematical thing, so it should be written only in words and not in symbols

student

@TobiasKildetoft I don't write my papers in prose

Tobias Kildetoft

then you are doing it wrong (unless you mean you write them in verse, in which case, awesome!).

Rithaniel

A rhyming proof would be an incredible feat.

Astyx

Map $\Bbb R^n\to M$
Derivative everywhere
It's a manifold

user193319

13:17

Consider the cantor set defined in the following way: if $C_0 = [0,1]$ and $C_n = \frac{C_{n-1}}{3} \cup ( \frac{2}{3} + \frac{C_{n-1}}{3})$, then $C := \bigcap_{n=1}^\infty C_n$ is defined to be the cantor set. Question: from this way of defining the cantor set, how does one show that $\{C_{n}\}_{n \in \Bbb{N}}$ forms a decreasing sequence of sets? I tried induction, but I had some trouble.

Showing that $C_2 \subseteq C_1$ is not hard (this is my base case), but the actual inductive step isn't obvious.

Astyx

If you write $x\in C_n$ in base 3 you'll see why

student

@TobiasKildetoft My point was that there is notation for a reason, when of course you could always write your theorems and proofs without. A symbol akin to $O(n)$ would make it so that one can write $x^2 + 2xy = (x+y)^2 + \ldots$ without having to repeat each and every time that the other expressions do not depend on $x$ (but are not constant, or O(1) or something).

But I take it that there is no widely used thing I wasn't aware of

user193319

How does one do that? Also, is induction not needed?

Tobias Kildetoft

@student The difference is that big O has a precise mathematical meaning. What you are looking for does not, so it has no common notation.

Astyx

You can write any $x\in [0;1]$ as $x=0.a_1a_2\dots = \sum{a_i\over 3^i}$ where the $a_i are 1,2 or 3

The same way you can do it in base 10

Then by induction you find something that characterizes the $C_n$

And that demonstrates inclusion

user193319

13:24

Oh, I see. Thanks for the hints.

Astyx

Where you asked specifically that the $C_n$ form a decreasing sequence of sets ?

Otherwise remember that $C = \bigcap_{n=1}^\infty\left(\bigcap_{k=1}^n C_k\right)$

user193319

Oh...so $C$ doesn't equal $\bigcap_{n=1}^\infty C_n$? It equals that double intersection?

Astyx

Both are the same thing

But it's obvious that the second is an intersection of decreasing sets

user193319

Ooh, very nice! Thanks

Akiva Weinberger

13:42

Weird: a connection between the Penrose tiling and the Koch snowflake, though I don't think the author realizes this

stephencollins.net/Penrose/Default.aspx

Perturbative

14:05

Are the elementary row operations in Gaussian elimination vector space isomorphisms?

Secret

Yeah, as all of them are invertible matrices

Perturbative

Cool thanks!

Akiva Weinberger

14:40

> All of the faces of the five-dimensional integer lattice that lie within a certain distance of the plane are projected onto the plane, and you either wind up with a set of Penrose tiles or a really bad seizure.

> That’s how I created the image above.

> Now that I am able to visualize five dimensional space in my head, I’ve noticed that I get a lot of weird looks from priests and small children.

Secret

15:06

Well, you can kinda made out the envelope of a penteract in those penrose tilings

5-cube

In five-dimensional geometry, a 5-cube is a name for a five-dimensional hypercube with 32 vertices, 80 edges, 80 square faces, 40 cubic cells, and 10 tesseract 4-faces. It is represented by Schläfli symbol {4,3,3,3} or {4,33}, constructed as 3 tesseracts, {4,3,3}, around each cubic ridge. It can be called a penteract, a portmanteau of tesseract (the 4-cube) and pente for five (dimensions) in Greek. It can also be called a regular deca-5-tope or decateron, being a 5-dimensional polytope constructed from 10 regular facets. == Related polytopes == It is a part of an infinite hypercube family. The...

Note the similarities between this and some of the patterns in penrose tilings

nCm

Is that the font of mathjax changed?

$\alpha, \beta$ looks different for me...

Vrouvrou

15:23

hello

i have the family $B_r=\{(x,y)\in \mathbb{R}^2, (x-3)^2+y^2<r^2\}$ is $\mathbb{R}^2=\bigcup_{r>0}B_r$ ?

abenthy

In the world of algebraic groups, a connected algebraic group $G$ is simply connected if every isogeny $G' \to G$, with $G'$ connected, is an isomorphism. This is independent of the field $K$ over which $G$ is defined, right? So its a property of $G$ over the algebraic closure of $K$, right?

Madhuchhanda Mandal

Does any form of \sum_ {n=1}^{\inf} \left (\frac{1}{(kn+1)^{2}} \right ) exists, where k is an integer ? (For example, k=2 we can get a form from basel problem)

Jake Rose

If I’m using Lagrange multipliers, with the constraint being $G(x,y,z)=0$ what stops me from just saying $\lambda=0$ when equating the partial derivatives?

Mike Miller

15:45

@JakeRose Do you get all the solutions?

Leaky Nun

0

Q: Analogy between metric space completion and algebraic closure

I've noticed some similarities between the story of completing a metric space and taking algebraic closure of a field. My question is whether these two stories can be generalized. Metric space Fix a metric space $(X, d_X)$. Consider isometries from it to other metric spaces (i.e. the under cate...

ct.category-theory galois-theory metric-spaces

Jake Rose

@MikeMiller I’m not sure

Mike Miller

do an example?

Jake Rose

what I don’t understand is, I have $G=0$ so why can I say $\lambda =0$

im trying one now

Mike Miller

@JakeRose I do not understand this sentence.

I am a little confused as to how you think Lagrange multipliers go. Will you tell me how you think you're supposed to use it?

Jake Rose

15:56

Say you want to find the extrema of some function $F(x,y,z)$ given the constraint $G(x,y,z)=0$ it can be shown that this is equivalent to finding the extrema if $F-\lambda G$. I.e. when $\nabla F = \lambda \nabla G$. So, $\partial_i F= \partial_iG$. The problem is, as $G=0$ then would $\partial_iG=0$ also?

sorry about the mistakes there

Mike Miller

Huh?? You are misunderstanding the formula $G = 0$. It does not mean the constraint is identically zero for any x,y,z.

A common one is $G(x,y,z) = x^2+y^2+z^2 - 1$.

$G(x,y,z) = 0$ describes the set of points on the unit sphere.

But the gradient of $G$ is $(2x,2y,2z)$. That's not zero anywhere on the unit sphere.

That's almost always what will happen in practice.

Balarka Sen

h

Jake Rose

Ohh yeah

wow

that was dumb

im sorry I even wasted your time with it

Mike Miller

@JakeRose Don't be - the double question marks were just for emphasis

If I can offer a moral: always try an example! (if you are not sure what example to try, grab one from your book's exercise section.)

The sphere is a particularly good fella to cut your teeth on.

@BalarkaSen r

Balarka Sen

I was a bit confused by this question. I can take the vector field on S^2 with one +2 index zero, and flow along that to get a smooth S^1-action, no? The cardinality of the fixed point set is 1.

The right statement is obvious from Poincare-Hopf. Every S^1-action is determined by it's tangent field at identity

I suppose by "isolated" they mean "isolated and transverse fixed points"

If it preserves the AC structure maybe transversality happens. I don't know about that. I thought index wasn't allowed to be negative, that's all

Mike Miller

16:08

I don't think the flow you have in mind has all orbits closed

Balarka Sen

Sure, why not? Take S^2 in R^3, take the tangent plane at a point, and consider planes transverse to that tangent plane cutting the S^2 into circles at various angles

Mike Miller

I believe the following is true. Given a compact group G acting on a smooth manifold M, suppose G preserves the submanifold S. Then there is a diffeomorphism between the neighborhood and the normal bundle taking the action to its linearization.

I think the issue is about global periodicity in your example

Naturally a vf gets a $\Bbb R$ action. To descent to a circle action, there should be a constant $c$ so that the time-$c$ flow is the identity

Balarka Sen

I see, I was worried about it a little. I'm still not convinced you can't parametrize the smaller circles by larger time scales so that it's uniformly periodic

Maybe because the smaller circles keep shrinking

Mike Miller

I think in your example we should have orbits with period in an interval

Balarka Sen

You'll get singularity at the fixed point

Mike Miller

16:13

I feel like I noticed something similar when I once tried to construct circle actions on higher genus surfaces

But not sure

The relevance of saying "the set of periods includes $(a,b)$" is that we can't possibly choose a single real that is an integer multiple of all of these

Balarka Sen

Sounds right. Interesting that I never thought about this.

Mike Miller

My linearization claim above says: if x is a fixed point of the circle action, then a neighborhood looks like S^1 acting on R^2

Probably by the standard action, unless your circle action has stabilizer

Balarka Sen

Right.

I should have thought of that. Just follows from usual G-equivariant tubular neighborhood theorem

You average out to find a G-invariant metric, then use exponential

(G compact is important: otherwise you run into my kind of example)

Mike Miller

Well, I think there is value in ignoring that. In the case you described about planes intersectimg the sphere there was clearly something interesting going on, all orbits being circles but not a global circle action

So it is better to understand what is going on than to handwave the issue away with a bigger theorem

Actually, I now realize that in your example, the orbits are $\Bbb R$! Because the north pole is a fixed point.

Balarka Sen

Ah, heh

Mike Miller

16:19

So then, ok, foliations by circles: do those come from circle actions?

I think the answer is yes in 2D and 3D - it's a big theorem in 3D but I forget how it goes in 2D

Balarka Sen

I think so. Take the tangent field of the foliation, away from the singular points. That integrates by Frobenius theorem

Mike Miller

But it's false in 4D I believe, Bob Edwards came up with the idea and Sullivan a counterexample

Balarka Sen

Now you have to see how singular foliations by circles at singularities look like

Akiva Weinberger

The linear map from $\Bbb R^5$ to itself defined by $(x_0,x_1,x_2,x_3,x_4)\mapsto(x_1,x_2,x_3,x_4,x_0)$ rotates a unique plane by 72 degrees, another plane by 144 degrees, and keeps a line fixed

Mike Miller

Somehow the circles have too many possible periods

Akiva Weinberger

16:21

I think that first plane is the one spanned by $(\cos(0),\cos(72^\circ),\dots,\cos(288^\circ))$ and $(\sin(0),\sin(72^\circ),\dots,\sin(288^\circ))$

but I'm not sure

Are those two vectors perpendicular?

I think they are

Ah, yeah, they are. That's neat

Balarka Sen

Can you do something to that vector field so that the flow actually becomes a circle action? Blowup S^2 at the zero, maybe?

RP^2 has nonzero Euler characteristic so actual blowup wouldn't work, of course

Idk

Akiva Weinberger

Oh, actually, I think I should've given those two vectors in the other order, so that the plane is oriented the right way. Otherwise it's actually a $-72^\circ$ rotation

Mike Miller

@BalarkaSen See: Edwards, Millet, Sullivan, foliations with all leaves compact

Balarka Sen

Thanks!

Akiva Weinberger

In other words, let $i:=(\cos(72n^\circ))_n$ and $1:=(\sin(72n^\circ))_n$

Balarka Sen

16:26

The crucial point here is that the picture I had in mind was not a singular foliation. I didn't realize that.

Stuff limits to singular stuff in a very regular fashion in singular foliations

Mike Miller

Yes, I didn't realize either.

Balarka Sen

Thanks for the paper, I'd need to read it

Ciao for now!

Mike Miller

Tata

Rithaniel

Would it be accurate to say that $\mathbb{Z}[i]$ can represent the vertices of a square tiling of the complex plane and $\mathbb{Z}[\omega ]$ can represent the vertices of an equilateral triangle tiling of the complex plane?

Akiva Weinberger

Yeah

Rithaniel

16:36

Is there a corresponding system for the vertices of a hexagonal tiling? Can you generalize this to any potential tiling, even a non-uniform one? (Or would that be more a question which needs research to answer?)

Akiva Weinberger

@Rithaniel See Gaussian integers and Eisenstein integers

@Rithaniel A hex tiling would not be closed under addition

Rithaniel

Ah darn.

Mary Star

@Secret What do you mean by asymptotic value?

Akiva Weinberger

You could get it as a subset of the Eisenstein integers though I think

Rithaniel

Yeah, that's what I was thinking, initially, but if it's not a proper ring, then it's less interesting.

Lucas Henrique

16:42

@AkivaWeinberger I get dizzy when I try to visualize 4D...

Akiva Weinberger

(I should point out that that's a quote, taken from the link)

Lucas Henrique

16:56

@AkivaWeinberger IKR. But (I think that) some people do actually have a grasp of what's happening in those dimensions and I find this bizarre

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