Mathematics - 2020-07-14 (page 2 of 3)

4:05 PM

what you say makes 0 sense to me, not sure about edward

Read the question carefully first mr balarka sen

i did

it makes no sense

I divided the sphere into slices and consider them as cylinder than volume comes out to be correct but surface area doesn't ?
The height of cylinder at an angle x from centre is rcosxdx . If I use this height volume comes out to be correct but for surface area I have to consider RDX as height ?
Did you understand the problem ?

Still not

Can you produce your calculations? What are you integrating in each case?

I can but I don't know how to add image in the chat ? Please help

4:09 PM

You can't just do a first order approximation of the spherical rings by straight cylinders and expect everything to work out

That's not how math works

There is a classic picture of approximating the circle by vertically and horizontally straightline segments which gives length of a circle as 4

seems like you're doing some nonsense calculation like this

But by dividing the sphere into many cylinders and sum their values give exact formula In limit when the no. Of cylinders tend to infinite.

See picture above

The area converges, the perimeter does not.

Can you tell me how to add image so that I can show what's my problem easily ?

there's an upload button right next to the part where you write the message...

Wait let me see

4:13 PM

I think what's your problem is the same as the pi = 4 thing above

Sorry but I am using mobile so I am not able to see any such option

@thorgott you were asking about pseudocompact spaces a while ago, right?

No problem is not really like this .

Why not? You're just doing a 3D version of the same thing

yeah, I did

4:14 PM

Looks like the same mistake to me

unexpected callback

Ah here it is

Wait for some time I try to upload image. Of problem

I'm going off, someone else can help maybe

Here is your random topology fact of the day then. A space is pseudocompact iff for all decreasing sequences of open sets $U_n\supseteq U_{n+1}$, we have $\bigcap\overline{U_n}\neq\varnothing$.

4:17 PM

I think I need to earn some more reputation for upload button .

interesting

so compact => pseudocompact then is Cantor's intersection theorem

no wait, that also needs them to be closed

Sure but compact => pseudocompact is also the fact that continuous image of compact is compact which is much easier

are we assuming Hausdorffness?

@Thorgott There's a closure in the intersection

oh, duh

4:20 PM

Got smth here that says $\operatorname{Gal}(L_1/K) \cong \operatorname{Gal}(L_2/K)$, i.e. $L_1 = L_2$

that seems like a weird conclusion to me lol

of course the closures are closed...

@Thorgott Hm good question, there might some assumptions

@BalarkaSen Is there a topology on piecewise $C^1$ curves so that the sequence of regular n-gons converges to the circle?

In the book where I found the result all spaces are assumed to be completely regular (it's a book on compactifications pretty much) but I'd need to unravel the proof to see which assumptions are being used

and sequentially compact => pseudocompact in this sense is actually the standard proof of Cantor's intersection theorem

4:22 PM

@MikeMiller If you're fine confusing curves with their images in $\Bbb R^2$ the topology induced by the Hausdorff metric on compact sets should work

Gromov Hausdorff convergence yes

doesn't the Frechet norm do this?

Is arclength the same as 1-dimensional Hausdorff measure

Fréchet distance

In mathematics, the Fréchet distance is a measure of similarity between curves that takes into account the location and ordering of the points along the curves. It is named after Maurice Fréchet. == Intuitive definition == Imagine a person traversing a finite curved path while walking their dog on a leash, with the dog traversing a separate finite curved path. Each can vary their speed to keep slack in the leash, but neither can move backwards. The Fréchet distance between the two curves is the length of the shortest leash sufficient for both to traverse their separate paths from start to finish...

I want arclength to be continuous

4:25 PM

@MikeMiller I think you need some assumptions (injective curves maybe? But the answer should be yes)

@AlessandroCodenotti :54942729 If not, then $X \setminus \overline U_n$ gives an open cover of $X$, and hence if $X$ is compact has a finite subcover. Since this sequence of opens is increasing, we see $\overline U_n$ is empty for large $n$

So $U_n$ is eventually empty, contradiction

Doesn't use Hausdorffness. Is there an error?

$X$ is pseudocompact

The claim is that compact implies pseudocompact without further assumptions

Theclajm is like a good Hungarian name

Oh ok sure, but that's clear because continuous image of compact sets is compact without assumptions

4:27 PM

Or maybe Nordic

@MikeMiller See section 3.2 of Maggi's "sets of finite perimeter" for a proof

I was just objecting to giving it a name

@Balarka Þekleim

when can I deduce an equality of fields just from an isomorphism of their Galois groups? Is there smth obvious that I've forgotten? lol

Why would you expect to be able to do that ever

I wouldn't

lol

But the paper I'm reading concludes that two fields are equal after proving that two Galois groups are isomorphic

Guess there's something more in there

4:33 PM

Presumably!

Gal(\bar{F}/K) ~ Gal(\bar{F}/L) => K ~ L is like a rigidity statement

$\operatorname{Gal}(K^{\text{ab}}/E_{\sigma}) \cong \operatorname{Gal}(K^{\text{ur}}E_\sigma/E_\sigma)$, i.e. $K^{\text{ab}} = K^{\text{ur}}E_\sigma$

lol

I think function fields of hyperbolic curves have this

Isn't that Mochi's theorem

the arithmetic Mostow rigidity

fuk knoes

Anabelian geometry

Anabelian geometry is a theory in number theory, which describes the way in which the algebraic fundamental group G of a certain arithmetic variety V, or some related geometric object, can help to restore V. The first traditional conjectures, originating from Alexander Grothendieck and introduced in Esquisse d'un Programme were about how topological homomorphisms between two groups of two hyperbolic curves over number fields correspond to maps between the curves. These Grothendieck conjectures were partially solved by Hiroaki Nakamura and Akio Tamagawa, while complete proofs were given by Shinichi...

Mochizuki and rigidity have the same number of syllables

just noticed

4:35 PM

publish it

oh god, why is this chat anabelian geometry now

lol I'm just tryna live my life (and prove kronecker-weber)

I think the bracketed part somewhat increases the difficulty

living life is the greatest difficulty of all

- Edward Evans, 2020

colourised

truly the greatest philosophist of our times

4:48 PM

philosophicler

Perturbative

5:02 PM

Is the Hopf invariant only defined for even $n$ in maps $f : S^{2n-1} \to S^n$?

It's defined for any $n$. Consider the complex $D^{2n} \cup_f S^n$ and look at the cup-square of the generator of $H^n$

that's some multiple of the generator of $H^{2n}$

That number of hopf

Perturbative

Ah okay I see, I was skimming through some stuff in a couple books and it seemed on the surface like it was defined only for even $n$

5:17 PM

Well, what Balarka just said is always $0$ when $n$ is odd.

Right, it's only interesting when $n$ is even

Perturbative

@TedShifrin Oh yes you're right, I was making some silly error which made me think it would be undefined for odd $n$

My usual way is (with forms) to pull back the volume form $\omega$ of $S^n$, note that $\pi^*\omega = d\eta$, and then compute $\int_{S^{2n-1}}\pi^*\omega\wedge\eta$.

Obviously all fine with cup.

This makes it transparent that it's the linking number of the fibers of $f$

Well, transparent to the cognescenti.

5:24 PM

Everything is unclear until it's clear

An irrefutable statement cannot be refuted

Wanna bet?

how much

Two refutations.

I could use some refutations

Well, then.

5:31 PM

I bet you a dollar than I will give you a dollar.

I bet you a dollar you that you will lose this bet.

I know better than to bet with a probability expert.

I don't know any probability party tricks to stun you, I'm afraid

You can probably do the $\pi_1(SO(3))$ belt trick, though.

Yup I love that trick

I have actually done that at a party

I saw that in a magic saloon in San Francisco with a Kirby student friend of mine when we were in grad school. :)

5:34 PM

Haha nice

Oh, you probably know his name. Turned out a famous topologist. John Harer.

Harer stability guy?

MCG has stable homology after some point, or something like this

Yes.

Niiice

What do you think of this? I obviously don't have the patience to write everything out. What does someone who posts such a question actually know/understand?

5:37 PM

yikes, I'm not sure. You could write out the definition of Gysin homomorphism out for them. Seems like they are indeed talking about de Rham

Oh wait I actually do know a fantastic card trick

Take a full deck of cards (half a deck will do but the more cards the better the chances). Pick three cards from the top, ask your friend to choose a card.

I have no idea what the OP knows or doesn't; that's my complaint.

The game is as follows. The cards are numbered usually, Ace is 1, 2-10, Jack is 11, Queen is 12, King is 13. Suppose your friend picked x, then ask them to count starting to x as you lay down cards one by one

That was on MO and for some reason people said it's more appropriate for MSE. Rather a reference is most appropriate

Ask them to stop when you reach their number x.

Suppose they got y this time. Repeat same procedure.

@MikeM: On MO? The poster is deluded.

5:41 PM

Keep doing this until the deck is over; suppose the last number they made you stop at was some z. You can guess this z

Most of us are, really

I actually don't know a good reference. I actually put an appendix about this in my thesis because I didn't know a good reference and I needed it everywhere.

I guess nobody cares about the card trick

I'll bugger off

I'm trying to understand, @Balarka.

I'm just too tired for it

5:42 PM

@BalarkaSen I don't quite follow

Are we using all 52 cards, or just 13?

Use all 52

What does "Suppose they got y this time." mean?

@Tobias So say you laid down Ace, 2, 6. Friend picked 6. You start laying cards down one by one and ask friend to stop at 6th card.

Hi @Tobias

5:43 PM

@TedShifrin Hi

@BalarkaSen So you reveal three cards and ask them to choose one?

That's correct.

Oh, I see. You keep going five, six, seven times until you exhaust the deck.

And then you lay down cards. Face up or face down?

Face up. Lay it down one by one, tell your friend to not tell you when he mentally stops at the 6th card.

When he stops at the 6th card, it has some number. Say Jack

No, I still don't get it.

5:45 PM

Then he'll stop again at the 11th card, mentally

ahh, so you actually just slowly lay out all of the cards in the deck

Yeah exactly

You don't know what he mentally picked at the beginning

You'll have to livestream it for us

and you friend keeps track of one number at a time

Exactly, you don't know his tracked numbers

5:45 PM

and at the end, you can tell which number he got last?

YES

which then leads to the other numbers I suppose

But if you count how many cards you're turning over, don't you know his "secret" numbers?

You never stop laying down cards

@TedShifrin All cards are laid down face-up, and you're laying it out slowly one by one, and your friend is mentally keeping note of the numbers he's getting one at a time. The initial number he chose out of the three is unknown to you, so the whole sequence is.

5:46 PM

wait, no, you don't get the other numbers from the last one

@TobiasKildetoft Right, you don't :)

I see, so he updates but doesn't tell you.

Right, precisely

@BalarkaSen I assume it is important that you gave him three to choose from at the beginning

Or do you not yourself get to see the three original cards?

It was important that I gave him a small number of cards to choose from in the beginning, because that makes the probability of me guessing the last card in his sequence very high :)

I do see the three original cards. Everything is face-up

5:49 PM

Are you guaranteed to guess the card?

Nah, it's just that the probability of me guessing it wrong is vanishingly small for a 52 deck with 3 initial cards

Here's the point. The process is a Markov chain: You're doing $X_0$ (initial choice), $X_0 + X_1$, $X_0 + X_1 + X_2$, so on.

Markov chain converges to something deterministic regardless of the initial choice. So it doesn't matter if he picked 6, or Ace, in the beginning -- with high probability the tail of the sequence will eventually be the same

So just do the same process he's doing with a card of your own choice in the beginning

The number you both terminate with will be the same with high probability

Bizarre.

How high?

I have to recalculate but I remember it being massive(ly close to $1$)

With some practice, you can probably even know if the trick has failed by doing it for all three

5:54 PM

The point is if your numbers match at one point in the sequence, they'll be identical afterwards, right?

Because suppose you and him end up at Ace (1), then the whole sequence after that becomes the same

right

with a 52 deck card it's very unlikely it won't match even once

So that's the intuitive reason

But potentially the number o places to have a match could be quite small

We are not talking about a Markov chain of 52 steps after all, but potentially as few as 4

Yeah but that's also a very small probabilistic event

Because I mean come on, 4 Kings?

So the expected length is about 8, right?

5:57 PM

damnit you're going to make me recalculate. let me think

1 sec

Closed nbhd of a point x is subset of a) closed space and b) nbhd of x?

Is this correct?

To me "neighborhood" is always an open set.

Yes, same

I don't even understand your first question.

Irritatingly some people say a neighborhood is a set containing an open set containing x

5:58 PM

What is closed nbhd of x?

I would never say that.

@flowian I mean...

@MikeMiller kelley does

A closed set containing an open set containing x

@MikeMiller I saw a rationale for that definition in the fact that it makes the definition of continuity at a point easier to state

5:58 PM

Which need not be the closure of an open set.

It does?

@TedShifrin it's from a problem in GT book

Ask the book what its definition is.

It's an irritating definition imo

No definition in the book.

Sadly

I think it's the definition you want in order to axiomatize a topology by neighborhoods

Q: Advantage of the more general notion of "neighborhoods" in topology

6:01 PM

19

The notion of "open sets" is a fundamental concept in topology. I have been puzzled by (the use of) another slightly more general but closely related one: neighborhoods. Given a topological space $(X,\tau)$, and a point $p\in X$, a neighbourhood of $p$ is a subset $V$ of $X$ that includes an open...

I think Mike's right, that's the only other definition of nbhd that I've ever seen

I think it's useful language. Sometimes you want to talk about things such as compact neighborhoods and these will often not be open themselves

I made it through almost 50 years of being a mathematician and never having to deal with this contortion.

Agree, though less time

LOL ... for your sake, I hope so. :)

6:05 PM

Thank you guys

I used this fact recently: Let $X,Y$ be topological spaces, $Y$ Hausdorff. If $f\colon X\rightarrow Y$ is a continuous bijection such that every $x\in X$ possesses a compact neighborhood $K$ such that $f(K)$ is a compact neighborhood of $f(x)$, then $f$ is a homeomorphism.

just a variation of the usual inversion theorem, of course

Reading Moishe's response, I guess the point is defining continuity at a point. I guess I think of this in the analytic setting but not in the topological setting.

@TobiasKildetoft So let's see. The expected length of the thing is 1*1/13 + ... + 13*1/13 = 7, right?

sounds reasonable given my guess of 8

I don't even follow that.

Couldn't it have length $52$?

6:16 PM

that would require 52 aces

Ah, right. I'm a dummy. So if I use 4 aces, 4 two's, etc., in order, how long?

I'm not allowed to repeat?

18 (A A A A 2 2 2 2 3 3 3 3 4 4 4 4 5 5)

All the cards are being laid face-up in sequence for you, and you're just mentally tallying which one is "yours" as you go along, so no repeats are allowed

I'm inclined to write it out in symbols but there should be some trick argument, let me think

@BalarkaSen You have 49 cards to go, and the average card has value 7

so you should expect length 7

Yeah essentially.

6:20 PM

since the values are evenly distributed

Now I'm trying to find the probability of my trick succeeding

I guess I needed to have taught probability a few more times in my life to get good at this.

Probably

There's a 0.3 chance of my initial card being the same as yours, right?

Throw some more threes in there and I agree ;)

6:23 PM

And your cards are like 1/7 dense in the 52-deck sequence.

@BalarkaSen When you are done with this calculation, I want to know how many decks I should use to reliably do this without revealing any cards to start with

So more or less my sequence not matching yours will have probability (1 - 0.3) * (1 - 1/7)^7?

Sounds right

LOL @Tobias

~0.23 chance of failing

So success rate is nearly 77%

I do respect Moishe a lot

6:26 PM

Absolutely. He started out with a different name, right? studiosus ... but I still see someone with that name in modern days.

If your calculation is correct, 196 cards gets you a better than 99% chance

Damn nice

It's almost like the birthday problem, but Markovian

$\left(\frac{2}{3}\right)\left(\frac{6}{7}\right)^{28} < 0.01$

Yeah, when you first started with this, I thought of the birthday problem.

@TedShifrin Yes, the other guy is just unrelated I think. Maybe a fan?

6:28 PM

@TedShifrin Yeah it's just less of a dad magic trick

And again, assuming a correct calculation, 191 cards is the minimum for which success is >99%

Who knows. I'm amused by people who choose names like Chern and C.F.G.

@Fargle: How many people will have the patience to endure this for 191 cards? :D

Or the counting skills?

@MikeMiller Moishe was studiosus back in the day, right? There's another studiosus we found

That's what I just said, @Balarka :P

Come on Ted I have done this trick many times

People love it

6:29 PM

With 191 cards?

52 of course I am not nuts

That was my point, silly goose.

Just do it like you're a casino dealer. 8 decks

There are definitely a nonzero number of people with the counting skills to handle that---they generally get kicked off tables for it.

@TedShifrin I once answered a question by this imposter studiosus

Oh dear.

6:31 PM

Seemed like a nice guy, and had a good question

Moishe is scary, he knows everything

Sorta like Robert Bryant.

@TedShifrin First day of my probability-1 course in freshman started with the instructor collecting our birthdays

I mentally went "Oh dear"

Nobody is amused by it

Just do this trick instead

Way cooler

I did that in a Calc II class many years ago.

:(

"Never tell me the odds!" -Han Solo, algebraist

6:39 PM

"Of course it is happening on a homology sphere, Harry, but why on earth should it mean it's not nullcobordant" - Dumbledore

3

Do you think I should pull a Yudkowsky and write Harry Potter fanfics if math doesn't work out

Harry Potter and The Homotopy Sphere

Well, since JK Rowling has turned out to be a disappointment to the LGBTQ community, go for it.

Yeah haha that was awful

I can't stand JKR and her writings

I actually have looked at none.

Besides being a virtue signaler and politically stupid, whatever she's written is copy pasted from much older and better fantasy texts

like Lord of The Rings

that one is a masterpiece

Well, she's sure made many fortunes.

I never read LOTR, either, although one of my best friends in high school used to recount the story to me as we walked to/from school.

6:47 PM

yeah. people do read junk. remember the time fifty shades of grey became a cultural phenomenon?

your friend has good tastes

Haven't seen said friend since we graduated high school.

Lord of The Rings (henceforth assumed to be commutative with unity)

One Ring to rule them all

Free algebra

In mathematics, especially in the area of abstract algebra known as ring theory, a free algebra is the noncommutative analogue of a polynomial ring since its elements may be described as "polynomials" with non-commuting variables. Likewise, the polynomial ring may be regarded as a free commutative algebra. == Definition == For R a commutative ring, the free (associative, unital) algebra on n indeterminates {X1,...,Xn} is the free R-module with a basis consisting of all words over the alphabet {X1,...,Xn} (including the empty word, which is the unit of the free algebra). This R-module becomes an...

Oh and remember the time Twilight became a cultural phenomenon

people read horseshit man

its sickening

I remember when my algebra prof told me the trick to construct determinants over general rings by working with a universal matrix and called it "one matrix to rule them all"

lmao universal matrix

that is honestly a useful construction to understand for example Cayley Hamilton over general rings

you take determinant of a matrix of operators

6:52 PM

Ugh, reminds me of the argument that appeared where someone very literal complained repeatedly about the "formal" definition of cross product using a determinant.

it's very useful

no its not very useful

its useful

you get naturality and all the desired properties for free