@MartinSleziak @DavidZhang @BalarkaSen For every prime power $q$ and supernatural number $\bf n$ (formal infinite product of prime powers possibly with infinite exponents) there is a field $\Bbb F_{q^{\bf n}}$, which is the union of $\Bbb F_{q^d}$s over all finite divisors $d\mid\bf n$ within the algebraic closure $\overline{\Bbb F_q}$.

If $K/\Bbb F_q$ is algebraic, for each $\ell$ let $e(\ell)$ be maximal for which $\Bbb F_{q^\ell}\subseteq K$, denote ${\bf n}=\prod_\ell \ell^{e(\ell)}$, then prove $K=\Bbb F_{q^{\bf n}}$. Thus, supernatural numbers enumerate all algebraic extensions of a finite field.

One can verify the countably infinite fields $\Bbb F_{q^{\ell^\infty}}$ ($\ell$ prime) have no infinite proper subfields. But $\overline{\Bbb F_q}$ does have infinite proper subfields, namely $\Bbb F_{q^{\ell^\infty}}$ for primes $\ell$.

This has a nice interpretation in terms of closed subgroups of ${\rm Gal}(\overline{\Bbb F_q}/\Bbb F_q)\cong \widehat{\Bbb Z}\cong\prod_p \Bbb Z_p$ (where $\Bbb Z_p$ means the additive group of $p$-adic integers, not $\Bbb Z/p\Bbb Z$). The closed subgroups are also enumerated by supernaturals.

@Ocelo: For each $p$, take the open geodesic ball of radius $rho_p$ centered at $p$. (Your $\rho_0$ depends on $p$.) This of course gives an open cover.

@0celo7^

Surely five letters are enough for a ping.

Not when the first character is incorrect professor :-)

0 hell

Thank you, sir Skull.

:D

Never mind; stupid question.

@TedShifrin I figured that out...

I figured you would have. But ...

@TedShifrin But why does Jost mention the smoothness property

hi @Antonio

Hey @Ted, just checking in before sleep

@HDE226868 there are no stupid questions, just bad ones

@0celo: I think he wants that to get a different result, namely that in the geodesic ball centered at $p$, every two points sufficiently close are joined by a geodesic staying in that geodesic ball.

@TedShifrin Proving what? That open geodesic balls are open?

Pedro is TorsionSquid's advisor?!

No, what I just wrote you.

@NormalHuman: Different sort of removal, and not necessary and sufficient :P

I don't get it. How does that not follow from the definition of a normal neighborhood.

that referring to what I said, 0celo?

Yes.

The normal neighborhood centered at $p$ only tells you that you join $p$ to points in there by the geodesics and get shortest paths. What about $q,r$ in that normal neighborhood?

Oh, two other points.

right

Ok, I don't see how that follows, then.

(from smoothness)

Because you know it's true for geodesic balls of radius $\rho$ centered at any point in there within a certain distance of $p$.

Sorry, what?

Read what you typed. You get a geodesic ball of radius $\rho$ centered at every $q$, so that tells you that the geodesic from $q$ to $r$ will be the shortest path within the original ball centered at $p$.

> You get a geodesic ball of radius ρ centered at every q

I don't see how that follows from smoothness.

Because of the inverse function theorem, given smoothness of $\exp$. Invertibility is an open condition.

Ok, I'll think about that.

Thanks.

But I really think that stuff was focused on something other than the covering by open geodesic balls.

You might look at other texts (like Spivak, doCarmo, Boothby, Lee, whatever) if you get confused by Jost. I don't know his text personally, but it should be good.

you just did.

Hi @Ted.

hi @MikeM

my students usually caught on faster than that, @user :P asking for another thing puts you back at the same spot you were at.

@0celo7 All I had to do was re-read the Wikipedia article. Kind of a stupid move to not check there again. :P

what's your query, @user?

doCarmo is talking about covariant derivative along a given curve $c$.

This is the same problem you were having before: If you know something makes sense in the large, you can still evaluate it at $p$. To define $\nabla_X Y$ at $p$, you only need to know $X(p)$.

The point is that you only need to know $Y$ along a curve whose tangent vector is $X(p)$, not in a whole neighborhood of $p$.

Yes.

You can look in my undergrad diff geo notes for my favorite example: Differentiating the unit tangent vector along a latitude circle of a circle. You can see it both in formulas and geometrically.

Again, I would urge you to calculate and understand basic examples.

A graduate text like doCarmo or Lee doesn't bother with those.

No, those are the real Christoffel symbols that you have on your manifold. We're just using them to differentiate along a curve. But the Christoffel symbols are there globally to start with.

He's suppressing the notation that you evaluate everything at $c(t)$.

Forget that crap. Think geometrically about vector fields.

My comment stays. :)

Yes, that's what I said too.

Go do some undergraduate level computations before you try to read doCarmo.

NO comment.

I can't magically impart understanding. It comes from working through things — examples — yourself.

Historically I've found "I have an exam soon" to be the statement that least motivates someone to help, where indeed motivation can be negative.

You're welcome to look at my notes for more concrete stuff.

I'm used to removed comments.

I'm not going to keep clicking on these things, but, yes, everything is evaluated at $c(t)$.

@user Hmm, the DG class at my school uses doCarmo. And this is exam time! You wouldn't happen to be going to UTK would you?

This is a grad course, @0celo, not undergrad.

I said that ages ago, @user.

@TedShifrin Huh?

What does that have to do with anything?

You're saying the grad course uses doCarmo?

@TedShifrin I think so, yes.

This is like what you did in calculus all the time, @user. You wrote letters for functions that were really compositions, yes.

Is that a bad thing? My linear algebra TA says it's not a very good book.

@user The Riemannian geometry one.

I'm not a huge fan of doCarmo, but he and I are mathematical brothers.

I don't know that I'm keen on any of the standard Riemannian books, but my tastes diverge.

@TedShifrin So why did you mention undergrad/grad again?

I think I probably would have been happiest just reading Taubes' book...

because doCarmo also wrote an undergrad text that is commonly used, @0celo.

yes, @user

@TedShifrin "curves and surfaces" or whatever?

And you inferred from my profile that I was referring to the undergrad one?

yes, but this affine connection stuff works in general just as it does for surfaces. Hence, getting intuition from surfaces helps the general.

from your discussion about Jost, @0celo, I presume you're ahead of your years, but I wasn't going to conclude that Jost = doCarmo.

I think Jost is more advanced.

yes, @user ... section 2.4

He has chapters on Morse theory and Floer homology, among other cool things.

It's more modern, @0celo, yes.

Oh, interesting: He used my text for the undergraduate course recently, because he contacted me about it.

Huh? That is an undergraduate course, not a graduate course using doCarmo. WTF?

Well, a year ago he hadn't, but he did last spring/summer.

@TedShifrin Ok, I still don't see how the inverse function theorem helps me here...

:/

They look difficult because they're very concrete, not abstract.

It's inverse function theorem plus a compactness argument, @0celo.

@TedShifrin Forget compactness, I don't see why smoothness of the exponential map gives me the condition on other geodesic balls

For each point $q$ you get a ball of radius $r_q$ on which $\exp_q$ is a diffeomorphism. By taking a compact set, you get a minimum for all these $r_q$.

I know that!

Smoothness gives you $r_q>0$ because invertibility at $p$ gives invertibility near $p$.

I have no clue what you need to understand, @user. Seriously.

@TedShifrin I also know this

And don't come crying to us when you're trying to cram for an exam at the last minute. You cannot do that in mathematics.

No, @0celo, I'm making a stronger statement.

I don't understand the sentence "if thus..."

Smoothness tells you that if $\exp_p$ is a local diffeomorphism, then so is $\exp_q$ for $q$ near $p$.

Hmm, are the two $\rho$s there not the same?

@user Jost, Riem. Geom. and Geom. Analysis

@0celo7: Let $f: M \to \Bbb R^+$ be the map assigning $p$ to the largest $\rho_0$ w/ $B(p,\rho_0)$ mapped to injectively by the exponential map. $f$ is at the very least lower semi-continuous. Compactness then implies $f$ takes on an absolute minimum greater than zero.

Well, that's a separate issue, @MikeM.

@user Only 20 pages in.

I thought the question was why there was such a global minimum.

@MikeMiller No, I get that.

No, it's about how we are applying smoothness to get a local injectivity radius at $q$ near $p$.

OK, I'll leave you be and go back to my hole.

Is there room for me in your hole? :D

So what about the two $\rho$s in the proof

Are they the same?

@TedShifrin: Only if you want to read Donaldson.

I've done that before.

@user: I've said that three times.

@user Seems so.

Maybe

Depends on if the two $\rho$s in that proof are the same!!

@0celo: OK, I would agree (and my previous statements alleged) that we don't know the same $\rho$ works everywhere. But we can shrink the original one slightly and then it will work everywhere nearby.

@TedShifrin Right. That's intuitively clear.

@user: I'm getting very tired of this. Go read all my previous comments.

OK, @0celo, so I'm agreeing with you that I don't see why the original large $\rho$ should work everywhere nearby.

But it's not needed.

I'm thinking about this.

This is due to the inverse function theorem?

You're not stupid, @user. But you need to stop talking and read what I say and work on it.

Yes, @0celo, and continuity of determinant.

Being in this room is way more exhausting than teaching a course. I'm beginning to regret retirement :D

Stop being argumentative and go work. Or, seriously, I won't answer your questions in the future.

@TedShifrin Sorry, I don't see that.

Smoothness tells you the derivative varies continuously, @0celo, and so invertible at $p$ (by continuity of determinant) gives invertible nearby.

@user wtf?

@user: You know that's utter nonsense. First of all, I had no idea who you are. Second, I've been openly gay here as long as I've been here.

And you know that ... clearly.

So cut the crap.

A question I answered seems to have showed up in hot network questions. They're going to be disappointed in the answer.

wow, didn't realize that picture was so big

@TedShifrin Continuity of determinant?

That's what I'm hung up on.

Yes, @user, I've deduced that. And you know damn well better.

I thought it was invertible due to the implicit function theorem.

So quit using the gay label as an excuse.

Dude, just shut up.

What is wrong with you?

@0celo, no, it's the inverse function theorem. And that's based on invertibility of the derivative, which is given by nonzero determinant.

@0celo7: Work in charts, so that the derivative is literally a map $U \to \text{Mat}_n$. Then the determinant, the map $\text{Mat}_n \to \Bbb R$, is continuous; and thus it's an open condition for a matrix to be invertible.

@TedShifrin Dude every book says it's a diffeomorphism because of implicit function theorem.

Thanks @MikeM

@0celo, DUDE, false.

I think I've had it with the attitude here tonight. I'm outta here.

wtf?

Is like every book on geometry wrong?

user what is wrong with you man

uh, looks like I misread

4 books on geometry

well, no

Jost says inverse, but Hawking-Ellis and Straumann say implicit

@TedShifrin So $\exp_p$ is a diffeomorphism and $\exp_q$ is a diffeomorphism. But what does Jost want me to do with that?

He's still in the online users list.

Now he really left

Look what you did user

@user just stop being a tard

Hi @Jasper Yes I got it yesterday

For the past few days, I've been browsing math.SE for hours on end and answering questions

Is this an unhealthy addiction

No

So are most users of math.SE undergrads?

Does anyone on here use LyX?

@user That's too bad but just stop being a weirdo

you don't belong here @0celo7

@cheesyfluff no it's great :)

@cheesyfluff do you know anything about matrix norms? math.stackexchange.com/questions/1517103/…

@Huy wtf man

why don't I belong here

I'm a math student

@cheesyfluff ah ok

HI all

I feel that I became dumb

for quick calculation, it take me like 3 seconds when it used to take me like 1 seconds

I don't man. For this online calculation game

calculationrankings.com/game/3

I mean, my score is around 24-ish

no one give a shit

@Jasper Yes I hope so too.

can anyone give me a hand starting on this math.stackexchange.com/questions/1517103/… ?

Hi guys

and girls?

Can anyone recommend a mathematical logic textbook for self learners?

I study math alone at home

@crocket what level?

I was studying "Introduction to Mathematical logic, Sixth edition" by Mendelson, and it is difficult for me.

It assumes I'm an advanced undergraduate student.

and you are not?

I know some highschool math, and I finished reading "how to prove it, 2nd edition" by velleman.

I am not a student anymore.

I'm a jobless man.

ukcatalogue.oup.com/product/9780199587841.do has a philosophical focus

Do I need a philosophical focus?

and philosophy is normally popular for unemployed people :)

I don't care

about?

Is it easy to understand? Is it comprehensive?

it is definitely comprehensive

it is suitable for a bright first year major

Focus on philosophy is not important, but it should be understandable.

what do you want to get from any book?

I want to learn mathematical logic on my own, and I want to learn artificial intelligence and ordinal logic.

I heard ordinal logic comes in handy for AI.

I aspire to be an AI researcher.

modern AI is very statistical

and is called machine learning mostly

You mean machine learning

AGI is not machine learning

I focus on AGI

I recommend getting books from a library

then you can just take a look yourself

what is AGI?

Artificial general intelligence.

It's also called general purpose AI

Anyway, the logic manual seems ok at first glance.

I'm going to have a look.

my pleasure :)

Hi felipa , im doing my CSE school project in which im required to plot the locus of the third vertex of an equilateral triangle whose other two vertices lie on two mutually perpendicular lines . How do i mathematically prove it ?

Please suggest me on how to begin to solve it geometrically . ie: im good at c++ but bad at math

@felipa Can you recommend one more book on logic?

@SujithSizon you could formulate it as a question and pose it on math.se :)

@crocket of course...but as a study at home person you might enjoy maths.cam.ac.uk/sites/localhost/files/pre2014/undergrad/…

@felipa dunno but your link is reported malicious to me

@Agawa001 really??

I think you must have a bug

is fine for me

it's from cambridge university!

im running mozilla

@crocket try mathoverflow.net/a/33102/74651

@Agawa001 you mean firefox ?

yes same

@Agawa001 this must be a plug in of yours?

firefox doesn't by default report links as malicious

and I am using firefox too

oh that says the security certificate is out of date or broken

i didnt fake that

which is almost always the case for universities :)

which means ?

don't worry about it

it just means that whatever data is sent will not be secret

do you care in this case?

it means someone can capture raw traffic from an ssl secured connection ?

really no

don't worry about it!

which version of firefox?

I suspect you need to upgrade

i didnt worry and hat is that article fo ?

as it works fine in my firefox.. which version of firefox are you using?

or just try chrome

@felipa old (41.0)

upgrade to 42

chrome is memory-greedy

im doing so

Is it ok even if I can't solve many exercises in a logic textbook?

I intuitively understood the content with some difficulty, but I couldn't solve the exercises.

@robjohn

@crocket practice!

Should I focus on understanding concepts or solving exercises?

Is it ok to prioritize understanding concepts over solving exercises?

Hi pal @user

Be careful with Prozac, it will make you moody

You should start seeing a therapist.

@skullpetrol dont bother with kids plz, they will start to cry and complain to their moms

@skullpetrol you dont want to hav 1 hour time-out because of children

@crocket for what aim?

@felipa For learning logic?

The purpose of solving exercises is to understand concepts.

Thus, the goal is to understand concepts

It means I don't have to solve every exercise as long as I understand concepts.

how will you measure your understanding?

By trying to solve exercises?

Some exercises are not good for measuring understanding.

A good exercise is not difficult but helps you measure understanding.

Never mind. I will start reading logicmatters.net/tyl

logicmatters.net/tyl

"Teach Yourself Logic 2105: A Study Guide"

@robjohn how are you doing? Did you manage to try that alternating series?

hi @Chris'ssistheartist

@felipa Hi

$$\sum_{n=1}^{\infty}(-1)^{n-1} \frac{H_n^2}{n}$$

sometimes i sit there and begin to have fun solving trivial math problems, i dont know why m i doing this just fun! druggin myself with some mind-ticklking calculations

@Agawa001 It's fun playing with math questions.

For instance, today is sunday, and what can be nicer than doing some research? I wanted to go a bit to church, but I cough, and I don't wanna disturb.

@Chris'ssistheartist for me, today is short vacation, i would go to work instead

when i tell my fiends that i do maths for fun, they start to laugh

@Agawa001 They are nice then. Mine say I went crazy! :-)))))))))

Maybe I go out with them today, not sure. The truth is that I'm very dedicated to my actual work, and no time for silly joke, not now, but later.

hahaha yes they think im bit outa my mind

I have a book to publish, then there are lots of proposed problems, articles, and other research projects.

@Agawa001 it would be great if you could help with math.stackexchange.com/questions/1517103/…

which no one seems to love

@felipa i dont know what are Frobenius and spectral norm