In particular, the $y$ coordinate varies smoothly.

Howdy @Xander.

Hello, @TedShifrin. I'm not really here; I was just looking for the ChatJax link.

Bye.

LOL ...

If $L_{s_0}$ and $L_{s_1}$ are leaves, we define the distance function $d(L_{s_0},L_{s_1}) := |s_0 - s_1|$. Hence we wish to show that for any $s_0\in [0,1]$ and for any $\eta>0$, there exists $\zeta > 0$ such that for any $s\in[0,1]$, $|s_0-s|<\zeta$ implies that $d(\ell(s_0),\ell(s))<\eta$.

Does this seem to define the condition I want?

I'm suggesting just making it less abstract by looking at the canonical coordinates where the foliation is (locally on the left) $y=const$ and (locally on the right) $y=const$. Then look at the map of the interval to the interval, rather than actual geometric distance.

Hey @Ted, sorry I was out for a bit, I'll check this out

Demonark, it was Poline's question, but I don't see it ... so you can help her if you'd like. :)

And I'm in a fight on main with a guy who posted a complete solution to one of my text homework problems. When the student asked for guidance. Drives me nuts ... makes me want to quit again.

@TedShifrin I am not quite sure how I'd go about that. Am I taking two charts separately, one on the left, and then one on the right? Or am I taking a chart of the entire region?

Also "again"? have you quit before?

Or just wanted to?

Hi Mike!

Oh your topologist expert showed up.

Gently diverting my lame question to Mike. ;)

I've wanted it numerous times, and I have disappeared for a few pauses, yeah.

Pauses being sabbaticals, or you literally going AWOL?

You don't necessarily have a global chart that flattens out the leaves. Do you have any specific hypotheses on this foliation?

Not particularly. It's initially the horizontal chart on the x-y plane

And then I deform it with an isotopy.

I showed that my isotopy does indeed "move" the leaves as explained.

Oh, so you do have a global flattening chart. So, yeah, just use the same $y$-interval on both sides.

And the map is the identity outside of the interval affected by the isotopy.

Basically my isotopy is my chart, right?

No, no.

You are using the identity as your chart on the plane.

Only possibly at time 0, no?

I don't need a chart inside $D$. I'm looking on the left and on the right.

Are you thinking of the isotopy as coming from the flow of a vector field?

Then it's the smoothness I mentioned earlier.

Otherwise, you might have to work a little harder.

I explicitly give my isotopy. It's not a result of vector field flow though.

And it's a little bit difficult because I am obscuring one small detail.

Which is how the foliation is defined.

But you are saying, take a chart of the plane setminus some closed set containing D?

Yeah, and I guess you can just apply the isotopy to a point on the right-hand vertical interval to determine your map.

Hmm, take a function which dies at infinity and then chop it off at larger and larger intervals

Or no nvm

Yeah, it is right.

Demonark: I don't see how to make unbounded functions fit into a Cauchy sequence!

I think that just amounts to my previous suggestion with the eta-zeta

Honestly I'm not sure that it's not complete. Like okay technically it's not even a metric space

Yeah, Demonark, that was my complaint (in French). I said you had to look at the bounded functions, and then she said she had a proof that it's complete.

So, I could see a case for making this an almost metric space, and saying that whenever you do have a Cauchy sequence, it converges

Yeah, it's definitely not a metric space. But any Cauchy sequence of bounded functions must have bounded norms. So we have to look at ones where the norm is $\infty$ but the differences are tiny?

Like if you just have $x + \frac{1}{n}$, those guys all have infinite norm but then $|f_n - f_m|$ is small

She made it sound like she was looking for a counterexample to your statement, but I dunno. I've never seen this sorta thing.

Right.

Can you make $\|f_n-f_m\|$ small without getting convergence?

@anakhro give me half an hour

I mean you have uniform convergence, clearly.

@MikeMiller that's fine, I will be here for longer than that.

Yeah exactly, there's going to be a limit function, and it's continuous, that fact doesn't rely on compactness at all

I guess uniformly Cauchy makes for uniform convergence. Done.

I have no idea what is going on in her exercise.

So yeah the only barrier to it being "a complete metric space" is that it's not a metric space

I guess it's just the usual proof. Get the pointwise limit and then use the famous $\epsilon/3$ result.

Bah.

The usual thing is to do seminorms (expanding compact sets) ....

That's all I've ever seen in my long life.

Really, Ted? You don't look a day over 30!

30 semi-decades?

Ted looks like at least in his 80's

:DDDDDDDDDDDDDDDDDD

Yup.

haha no jk ofc

hmm i think you pass for 50

I feel older tonight. I got two vaccinations this morning ....

or 47

oh for what?

No, Kasmir, no longer true.

But thanks.

influenza and pneumonia

I did not see you irl to judge but based on those lectures =p

oh I hope you get well soon !

I've never seen Ted before.

Ted is famous

Better just to see him after.

just google his name :D

Ted are you a smoker?

pneumonia is kinda bad if you are

Let $A=C[x] $ prove there is no norm on A in which a C* algebra

can anyone help with this?

Hi Faust

i thinks its cause the spectrum of a is infinity

I hope you are feeling better today

if $a\in A $ hey @KasmirKhaan

Start by checking all the axioms for a norm.

Hell, no, @Kasmir. Never ...

@TedShifrin Good :D

But because of my heart disease and cancer, my immune system is weaker, so don't need to take chances.

wait what?

you have cancer?

:(((((((

I had it ... fine for about 7 years now.

omg you never told me this

I hope you get cured !

Iv known ya for like 2 years now and you never said anythign

omggggggg

@anakhro yeah i kinda tried that it was more of what does a* look like

It isn't usually relevant to math in here, Kasmir.

Does anyone understand what the deduction theorem is really saying?

i know but still :(( anyway sorry

Nothing to be sorry about, Kasmir. But thanks.

praying for you Ted! <3

nah

I never pray.

@user525966 that you are actually allowed to use deduction.

not good Ted!

I don't understand the real difference between $A \vdash B$ and $\vdash A \to B$

everyone belives that there is greater power than a humain

nope, Kasmir, not so.

You may believe that, but don't insist that I believe it.

thats diffrence between agnostic and atheist or something

am not Ted

I respect your opinion

agnostic isn't sure, atheist is sure :P

yeah

im one or the other

Gnosticism is a claim about knowledge, theism is a claim about belief (four permutations here!)

but it is logic to say that we ( all mankind ) did not create our selves

do you not belive in evolution?

ie there is .... higher than us

Nope, you've been brainwashed.

Good thing I'm leaving to go to dinner. :)

the only part i get alittle funny on is how the universe came to be

why would I ? would you belive that you iphone

after billion years of evolution

was created by sand and other materia ?

like at random ?

evolution isn't random

I'm outta here. Bye, folks.

theres very logical course for small organisms to be created out of nothing

@user525966 Do you understand the difference between the syntax $\rightarrow$ and $\vdash$?

when if you look at a single cell, is more complex than whole earth

we have done it in a lab with chemicals

@anakhro $\vdash$ is a metalogical symbol meaning "proves" andd $\to$ is a logical connective, which if we're talking about syntax alone doesn't "mean" anything but it's used in axioms and modus ponens to that extent

how faust

out of nothing

if it can happen in a lab

if i link you the article will u actually read it?

maybe nothing for you is soemthign microscopic that you cant see?

sure

@TedShifrin Bonne apetit!

@Faust You referring to Miller-Urey?

I did not mean to start this convo but even atheits are well aware that there is something stronger than them

@KasmirKhaan My iphone was created by random evolution for sure

i meant somethign liek

is it possible that you went to africa

in an island and that you were first to be there

and found an iphone or anything of that matter

A billion years ago there were block phones, and only the ones with better specs would end up breeding, bigger speeds, fewer headphone jacks.

After all of that selective breeding the iphone 13QX.33 was born.

say you have all the pieces of a watch you put them in a box and shake it, is it possible that you will get a complete watch when u take it out? @KasmirKhaan

that's not how it works though @Faust

yeah sure 0.0000000000000000000000000000000000000000000000....

but comparing a cell with a watch with limited pieces

@user525966 then it should be fairly clear what the difference is.

am not talking about a full organism yet

@anakhro Well yeah, but my question is more about the utility of it

What the advantage is of turning $A \vdash B$ into $\vdash A \to B$

That's not what you are doing.

just 1 single cell , it is more complex than i dont know what to say !

well the materials for RNA are everywhere on our planet and its pretty big so it can do many millions of these experments every few seconds so its actually not improbable that u get an RNA chain that useful eventually

not a cell

let alone a humain being

not an organisim

let me take the 1 / infinity case of yours and ignore the rest

those rna

just a random string of chemicals

from where did they come ?

i know what they are

RNA still exists today

ACGT

Ofc

there chemicals

otherwise none of us will be here

my question is from where did the rna come

if your asking where did all the stuff come from?

like matter?

yes lets talk about the first rna code

ok the matter from where?

yeah nfi

the thing is, say some higher power created it, then what created him?

we can go into circles but at the end there is someone who made this happen

exactly

that is the only logical question

@anakhro What's the Q?

its just as valid\silly to belive some created it as it is to belibe it just appeared

Now we can proceed, our brains cannot accept / belive this very easy , who created God

which is why i dont know if im agnostic or atheist

but lets say somone did creat God, than what ? that God number 2 is then the Real GOd

etc... we keep going in circles again

exactly if you belive god just appeared then why not just belibe matter just appeared its alot simpler

but what if we just accept , the god who created the god etc,ect to be the creator

because there's no reason to believe that

Can $\lim_{n \to \infty} (n \sin(2\pi/n)$ be done without using l'Hopital's?

it is not , at some point one should cut that circular reasoning and accept existing of a creator

of course

@jess

who was a first

I respect your opinions am not trying to convert anyone, just logical talk

we could just as easily say cut the circular reasoning by accepting that it came out of nowhere, uncaused

@user525966 that way we did not cut it, we just entered another loop

no...

no

its the exact same argument ur making @KasmirKhaan

litterly replace word god with word stuff

it is not, the thing is

ok we will have to agree to disagree then

if you enter a room, and you put ur phone there on a table and then leave and dont find it on the ground say

we could just as easily argue a creator is "not cutting / entering a loop" because now we have to explain what created that creator

do you ask your phone why are you on the ground ?

the point is that we just don't know yet

or you suspect someone who moved it ?

and until we know more we can't say for sure one way or the other

which why im open to both sides

eithier we came from nothing or a creator

okay so far so good , if you are open to both options then it is good

i dont have proof eithier way so why would i belive otherwise

both are logical

okay , for me I dont see them both logical but that is just me =P

anyway good talk !

Kasmir gotta go do his math ._.'

@MikeMiller I have a horizontal foliation of the plane. I have a smooth isotopy $\psi_t$ of the plane which is only non-identical in a region $D$.

I fix a point to the left of $D$ (say at height =0) and call the leaf going through that point $L$. I then look at the map $\ell(t) = \psi_t(L)$.

I want to show that it is "continuous" in the following manner: after time $1$, during the deformation, my fixed leaf $L$ is deformed so it is flowing out of $D$ into some new leaf on the right at height $\varepsilon$, call it $L_\varepsilon$. I want to show something like the intermediate value theorem where each leaf between $L_0$ and $L_\varepsilon$ on the right corresponds to $\ell(s)$ for some $s \in [0,1]$.

Sort of picture of what I mean

So L' = L_0, and psi(L) = \psi_1(L)

Remind me what horizontal means again?

Like the picture, the red lines

parallel to x asis

oh... the horizontal foliation

I guess I'm confused by the formulation

$\psi_t(L)$ doesn't make sense to me; that's now some curvy line. What is the codomain of $\ell$?

That was Ted's problem with my statement too.

I just mean " the leaf after time t"

you mean that curvy line?

or do you mean the horizontal line through $\psi_t(x)$ for some specific $x$, or something?

How about I remake the function like this. It spits out $\ell(t) = s$ where $L$ is now connected (after time $t$) with $L_s$.

it looks to me like you want something like: $\psi_t(L)$ is the second coordinate of $(x,y)$, where this is the point in $L \cap \partial D$ with greatest $x$-coordinate

Or maybe rather you orient your lines and choose it to be the height of the "second point"

It will correspond to the height of the leaf it connects to on the right outside of D

Be careful about statements like "right". If $L$ connects up near the top of the domain, and the isotopy is large, you could end up such that $L$ enters up top and then winds back around to leave below where it started

then the $x$-coordinate where it leaves is actually lower

That's true.

that's why I switched to the language of orientations

I also think $D$ should be such that $\partial D$ intersects each horizontal line in at most 2 points

that way lines don't get cut into multiple pieces

Yeah it is exactly a disk of radius r

@MatheinBoulomenos final update: the proof is all bogus

@LeakyNun which proof

And indeed, centred at 0

@Faust here

@anakhro Consider the map $g_t: \partial D \to \partial D$ defined as follows: through any given point there is exactly one line $L_x$. If $L_x$ is not tangent to $\partial D$, there is a unique direction it points inside $D$, and the same is true for $\psi_t(L_x)$ since $\psi_t$ is the identity on $\partial D$. Now let $g_t(x)$ be the other boundary point of $\psi_t(L_x)$.

When $x$ is the north or south pole, $g_t(x) = x$.

Now you want to know that this is continuous.

In fact it's a well-defined homotopy. I was wrong earlier: the fact that $\psi_t = \text{id}$ on the boundary means that a curve that enters on the left side leaves on the right side.

I guess it's probably best thought of as an isotopy $I \to I$.

But yeah I should figure out why it's continuous

I think I have an argument but I feel too lazy to write it down

Want to know that $t\mapsto g_t$ is continuous, right>

not that $g_t(x)$ is continuous with regards to x

or am I confused

Well, you want both. But neither seems too hard to me.

a little cheap of me to say without proving anything

Oh so that's what you meant as an isotopy.

$(t,x)$

yeah

1) You can write down a smooth path of vector fields $X_t$ so that $\psi_t(L)$ are its integral curves 2) the function of $I$ which says "how long it takes to enter and leave $D$ on $L$" is continuous (it's twice the $x$-value of the point in $\partial D$); this remains true for $\psi_t(L)$. call that function $f$

eg set $X_t = (\psi_t)_* \partial/\partial x$

(then the function described there is the same for all $t$)

If you consider the vector field on $\Bbb R^2 \times \Bbb R$ given by $(v,t) \mapsto X_t(v)$, then the time-$f(x)$ evaluation map on $\partial D \times \{t\}$ is our map $g_t$

So it seems to me this is smooth on the interior of $I$ and continuous on all of $I$

@MikeMiller John Lee's Introduction to Riemannian Manifolds, Second Editon, will be released next week.

@user525966 The origin of life is mentioned in the Agganna Sutta in the Digha Nikaya, and the origin of the universe is mentioned in the Brahmajala Sutta in the Digha Nikaya, one of the four main Nikayas in Theravada Buddhism.

@MikeMiller would I have to find the vector fields explicitly?

Maybe I don't understand $g_t(x)$ so well. When you say that $g_t(x)$ is the other boundary point of $\psi_t(L_x)$, you are defining $L_x$ to be the unique leaf that enters $D$ at $x\in\partial D$?

@anakhro Yeah

@anakhro No, I mean to say they are $(\psi_t)_* \partial/\partial x$

@MikeMiller what do you mean by "time-$f(x)$ evaluation map"?

Oh, nevermind I didn't see you define f before

@LeakyNun you know anything about C* algebras i stilll cant figure this out

@MikeMiller so you are saying $g_t(p) = X_{f(p)}(p)$?

Yeah, I guess so.

Why is it not smooth on the entirety of $I$?

Isn't it smooth everywhere because $X_t$ was a smooth path of smooth vector fields?

Or does $f$ mess it up possibly, @MikeMiller

@anakhro I was worried about $f$. Remember that if we write $x \in [-1,1]$, it's $f(x) = \sqrt{1 - x^2}$.

I thought $f$ was $\partial D\to I$

Maybe I overloaded notation somewhere, but $f$ was supposed to be the amount of time it takes to leave $D$.

Oh, so you are taking only points on the left half of the disk, indexing them by $y$?

Maybe I am just confused. I thought both $g_t$ and $f$ were defined on the entirety of $\partial D$. Should it have only been on the left side of $\partial D$ for both of them?

Probably.

$g_t$ I tried to define everywhere by saying "the other endpoint", but the problem is really we're going from left-to-right.

You can define it on one half then extend it to be the inverse on the other side

Yeah, does it still work if you extend everything to the entirety?

Including f?

well, $f$ is definitely not smooth at the north and south pole, right? it would be $\sqrt{1-x^2}$

You said it was twice the x component, so I was imagining it to be f(x,y) = |2x|, which isn't smooth still at the north and south, right?

sorry

we're parameterizing the left interval by $y \in [-1,1]$, which has corresponding left $x$-coordinate $-\sqrt{1 - y^2}$

And in the case we extend $f$ to take into account all of $\partial D$ rather than just the left?

So, including "time going backwards" from exiting to entering

"unstable" exit time? :P

yeah

that seems right

The f(x,y)=|2x| or did I misunderstand what you said again?

I'm really sorry about the bad organization here

It's fine, I think we are just thinking in two different pictures.

Let's start by writing the left half of the circle, including north and south poles, with its $y$-coordinate; $[-1,1]$

We can write $x(y)$ for the corresponding $x$-coordinate

Then the time-to-exit by flowing along $X_0$ is $-2x(y)$

Here $x(y) = - \sqrt{1-y^2}$

Alright, that makes good sense.

Uhm

I just realized I should object rather strongly to the premise

If $\psi$ is supported in $D$ then $\psi$ fixes $\partial D$

So $\psi_t(L)$ will always have the same endpoints

Question: Why does $\displaystyle\int_{-1}^1\frac1{\sqrt{1-x^2}}\operatorname d\!x=\pi$?

Answer: reddit.com/r/3Blue1Brown/comments/9h9u54/suggestion/e6aix1l

The k-th rectangle equals the k-th triangle. The sum of the rectangles approach the area under the curve; the sum of the triangles approach the area of the semicircle, $\pi$.

@MikeMiller you mean to say that (as a function $\partial D\to\partial D$) $g_t(p) = p$ for all $t$ and $p$?

This is what happens when you take the usual proof by trig substitution and "unravel all the lemmas" until you get something purely geometric.

On unrelated news: I have not slept, and it is now 6:20am

@anakhro yeah

I don't see why.

if $\psi_t(x) = x$ for $x \in \partial D$, then certainly the endpoints of your line $L$ are fixed

Wait, something's gone wrong

the picture you showed me to define $\psi$ doesn't allow the isotopy to be the identity on the boundary

Oh crap

Wait, I can fix this

I don't see why, @MikeMiller

Duh, the semicircle has area pi/2

Because I am fairly certain of my proof that it does change the endpoints.

Actually, here let me give an explanation of my smooth isotopy.

It might be the issue

Do you remember characteristic foliations?

from contact geometry

Take a contact manifold (say R^3) and embed a hypersurface in it (for a 3-manifold, that's a surface, e.g. the x-z plane). Then look at the singular foliation resulting from the intersection of the tangent planes of the surface with the contact field of hyperplanes.

Traces out a singular foliation on the surface

So what I am dealing with here is not just the x-y plane, but the x-z plane embedded in R^3.

It has a characteristic foliation for the usual contact structure of horizontal lines.

Just like our x-y plane example we have been working with

And my smooth isotopy is just a bump function in D, bumping the x-z plane along the y-coordinate.

And this changes the foliation.

And I proved quite explicitly that the leaves do not exit at the same place in t=0 as in t=1.

(if you wanted, I can email you my proof)

But I am fairly certain it's right.

Because otherwise I think contact geometry is bunk because it seems to be implicit that I can do this.

Maybe the fact I am isotoping it in R^3 changes your argument though

But also maybe I am crazy and I am wrong.

@anakhro I mean, you're not flowing along some isotopy in the plane. I'm not paying close attention, but what I was saying is formal. If $\psi$ is the identity on the boundary, then $\psi(L \cap \partial D) = L \cap \partial D$. So the point you enter is the same height as the point you exit.

So you must be doing something slightly different.

Apparently, but I am not sure what.

Well thanks for your help nonetheless. I need to wake up early tomorrow!!!

Thanks again, @MikeMiller!

Goodluck!

O_O

@MikeMiller @TedShifrin

Atiyah is usually a guy who knows what he's talking about, yeah?

was

(Unrelatedly, proof that $\displaystyle\int_{-1}^1\frac1{\sqrt{1-x^2}}~{\rm d}x=\pi$, proof-without-words edition)

(Also, correct edition)

I heard that some time back he had a proof that there was no complex structure on $S^6$ and... well... he didn't actually prove anything

indeed, Atiyah was a great mathematician

Can someone please explain point (ii)

It says: if A is singular matrix and (adj A) D = O then the system of equations given by AX = D is consistent with infinitely many solutions. Why is that?

it's somewhat sad to see atiyah becoming like this

:(

@Abcd your linear algebra questions are so hard

I feel ashamed

$\operatorname{im} A \subseteq \ker \operatorname{adj} A$

$\Bbb R^n \xrightarrow A \Bbb R^n \xrightarrow{\operatorname{adj}A} \Bbb R^n$

well adj A A = A adj A = 0

rank A <= null adj A = n - rank adj A

rank adj A <= null A = n - rank A

no that’s nonsense

@Abcd counter-example: A=0, D=(1,1,1)

Can somehting have length infinity in a norm?

or it need to be finite?

it needs to be finite

though so thanks