Mathematics - 2020-12-09 (page 1 of 2)

Balarka Sen

00:00

yeah I should sleep. Maybe

Dubias

Is anyone familiar with DFAs? I have a small question about one of their characteristics

Balarka Sen

Suppose $f : X \to M$ is a smooth map, together with $F : TX \oplus \varepsilon^k \to TM \oplus \varepsilon^k$, an isomorphism, such that $F$ restricts to $f$ on the zero section. Consider $f \times \text{id} : X \times \Bbb R^k \to M \times \Bbb R^k$; then $F$ gives an isomorphism on the tangent bundle of these manifolds which restricts to $f \times \text{id}^k$ on the zero section. By Smale-Hirsch theorem, this means $f \times \text{id}$ can be homotoped a little bit to a diffeo

So we have a general diffeomorphism $G : X \times \Bbb R^k \to M \times \Bbb R^k$. Restricting to $X \times \{0\}$ and projecting to $M \times \{0\}$ I expect will give me a map $g : X \to M$ which is almost a covering map... except at various places where $G(X \times \{0\}) \subset M \times \Bbb R^k$ is vertical. Maybe at worst a fold singularity.

One can probable surgery this "bad locus" out by hand and produce a bordant map $X' \to M'$ which is an actual covering map

I don't think people have written down such a proof by hand, but maybe it's known to Eliashberg.

Anyway gotta sleep

Edward Evans

00:52

I love this

"Where in the first step we have used the second isomorphism theorem"

Lukas Heger

this looks like global CFT and maybe Iwasawa theory? not sure about the latter

Edward Evans

It's in Sharifi's notes on Iwasawa theory

Lukas Heger

I see

Sharifi generally has good notes

Edward Evans

Yeah I'm enjoying them atm

Lukas Heger

we used his CFT partially in our ANT lectures and his group cohomology notes for the group cohomology seminar

Edward Evans

00:55

I just skimmed over this and the sentence made me laugh though

nice

I'm gonna write to Venjakob tomorrow and see if he can suggest me some stuff to read about Iwasawa cohomology

Ted Shifrin

01:09

howdy @Lukas @Edward

Lukas Heger

Hi @Ted

Edward Evans

Grüß dich, @Ted

Ted Shifrin

Vielen Dank, @Edward

Edward Evans

:D

skullpatrol

01:24

:-)

Dubias

01:38

Is anyone familiar with DFAs? I have a small question about one of their characteristics

skullpatrol

Askaway

Dubias

I was just wondering if when we have a DFA that accepts any language L1L2 and another DFA that accepts only L2 then could we make a new DFA using those two to accept L1 only?

also just to clarify, by a DFA that accepts L1L2 I mean a DFA that accepts the concatenation of the languages L1 and L2

skullpatrol

is this what you mean by DFA?

Dubias

01:54

Yes exactly

Are you familiar with that?

Edward Evans

02:31

Under what conditions are zero divisors always nilpotent?

Lukas Heger

@Edward it's true for a local Artinian ring

if you assume that the ring is commutative, then this is true iff there is a unique minimal prime ideal

Edward Evans

yay

Rithaniel

My first thought was that the nilradical is then the set of all zero divisors

Edward Evans

e.g. $\mathcal{O}$ of a local field

Rithaniel

I suppose that the first step to getting to "unique minimal prime ideal"

Lukas Heger

02:41

that doen't have any zero divisors, does it?

Edward Evans

Oops, I mean an $\mathcal{O}$-algebra

Lukas Heger

not every $\mathcal O$-algebra will have this property. For example $\mathcal O\times \mathcal O$ doesn't

@Rithaniel if $ab$ is a zero-divisor, then $a$ and $b$ are zero divsiors. This shows that the nilradical is actually a prime ideal under this assumption. But as the nilradical is the intersection of all prime ideals, it must be the unique minimal prime ideal

Rithaniel

Yep, pretty straight forward

If everything else is a unit, then you have a pretty interesting situation

Edward Evans

durr I'm being dumb

Lukas Heger

I was wrong, I think, I can't show that having a unique minimal prime ideal implies that all zero-divisors are nilpotent

at least one implication holds

Unknown x

02:55

Is it possible to apply divergence theorem here?

$\nabla \times \vec{G}$ is continuoius. since $\vec{G}$ is smooth.

Is it possible to guarenteed the existance of continuous first partial derivatives of $\nabla \times \vec{G}$?

Lukas Heger

$k[x,y]/(xy,y^2)$ is a counterexample, the unique minimal prime ideal is $(y)$, but $x$ is a zero divisor that is not nilpotent

Unknown x

If it could be possible, then I could have apply divergence theorem.

Lukas Heger

@Unknownx clearly if $\vec{G}$ is smooth, the components of $\nabla \times \vec{G}$ will be smooth functions and have in particular continuous partial derivatives

Edward Evans

@Lukas thanks for your efforts. I was just being dumb, but I guess I know something new now lol

Unknown x

@LukasHeger smooth means only first derivative continuous. right?

Lukas Heger

03:03

smooth means $C^\infty$, infinitely differentiable

Unknown x

is that the definition
?

Lukas Heger

yes

Unknown x

okay.

thanks

Ted Shifrin

03:56

@Unknownx Yes, you can use either Divergence or Stokes.

If only $C^1$, need Stokes. If $C^2$, then Divergence. Smooth means either.

123

Hi All..

Leonhard Euler

05:18

@robjohn thank you very much for citing your answer

@robjohn woah you gave three answers to that question!

Yes I am sometimes impatient but nowadays I am rarely getting answers here so I thought to ask somewhere else.

Seems like only a few people like analytic NT

Algebra questions get more answers here

@EdwardEvans elementary number theory is not necessary, and I don't see why it is useful. For example, apostol's book requires no elementary NT, and even says that it doesn't require that

Edward Evans

05:35

Because it's the basis of literally all of number theory

you wouldn't start studying ANT if you didn't know what modular arithmetic is

Leonhard Euler

Ah yes

But still

Edward Evans

What I mean is: you don't need to study elementary number theory in depth to start studying more advanced number theory, but it's helpful and motivational

Leonhard Euler

Yes

I think algebraic NT gets a but hard than analytic NT

(For me)

So it has more prerequisites

I have some books on advanced analytic and advanced algebraic NT. They're probably the hardest books ever written

Edward Evans

that's

a bold sweeping statement to make

Leonhard Euler

Opening their first page is the most demotivational thing. EVER

The advanced analytic NT book says that its "elementary" prerequisites are abstract harmonic analysis and other eredristkxtitdd subjects

Don't know about the "advanced" prerequisites

There is no use of humans if they can't solve RH after reading it

This is not hypocrisy. I am serious

Edward Evans

05:45

you can't claim to be serious if you don't know what you're talking about

Leonhard Euler

Please believe me. I know what I am talking about

Edward Evans

You literally said you don't know about the advanced prerequisites. It wasn't meant disrespectfully.

Leonhard Euler

I mean I didn't read them because I was frustrated after reading the elementary prerequisites. There were three pages of prerequisites

Haha finally convinced them that I am serious :p

Hey I recently found out that

$$\pi(x)\sim\frac{x}{\psi(x)}$$

Ted Shifrin

06:21

You're making ridiculous statements, Euler. You have no clue.

Study all aspects of math and a lot of advanced books before you offer statements like that.

Leonhard Euler

Ok

But what is ridiculous in that?

Wait for a year. I would have learned many aspects of math. You are right, @TedShifrin

Edward Evans

lol

Leonhard Euler

Lol*2

Most of the people here are a lot smarter than me

They can surely understand that book

Edward Evans

Or just older

Leonhard Euler

Yes

In the case of age, most can be replaced by all

Edward Evans

06:33

which is precisely why I "lolled" a second ago. One year is not a long time at all.

unless you're like Scholze or smth

Leonhard Euler

One year goes very fast

Like 2020

Edward Evans

You can't learn "many aspects of math" in one year

Leonhard Euler

Man learning many aspects can take five years for me

Or ten

But I have much time

Edward Evans

e x a c t l y

yo @user2103480

user2103480

yo

Do you ever sleep

Leonhard Euler

06:35

That's an advantage. I won't end up learning topology at the forties

Edward Evans

no

user2103480

Nice

Edward Evans

wtf you can't repeat words

user2103480

Why not

Edward Evans

try typing Why not again

user2103480

06:37

Why not again

Edward Evans

weyyyyyyyyyyyyyyyyyy

user2103480

Sheeeeessssh

Leonhard Euler

Ohhhh @EdwardEvans is teaching spamming

Just joking

Edward Evans

good one

user2103480

I think no one needs to teach me how to spam

Leonhard Euler

06:38

So you are already a spammer?

user2103480

I started early, just as you

Edward Evans

If you heat up spam it almost tastes like Leberkäse

user2103480

Urgh

Edward Evans

oh man I could go for a Leberkäse rn

Leonhard Euler

I never spam. I make people think that I myself am a spam

user2103480

06:39

People eat spam

Leonhard Euler

People eat food

user2103480

Unpleasant fact of the day

Leonhard Euler

Yes

user2103480

Leberkäse can be nicer than I'd like to admit though

Edward Evans

my austrian ex-gf who now lives in the UK made it known to me that warm spam almost tastes like shitty Leberkäse from Rewe

user2103480

06:40

@ed

Edward Evans

Mate

user2103480

Still swake or already awake again?

Edward Evans

Käsleberkäs from Spar is amazing. Even better is Käsleberkäs from Sutterlüty. That's top qual Leberkäse

user2103480

@EdwardEvans ........

Edward Evans

I'm still awake

Leonhard Euler

06:41

Edward Evans never sleeps. Just like math 55 students

user2103480

I only know a german phrase to reflect my sentiment when I reas that she eats Rewe leberkäse

Die ist kein mensch

Leonhard Euler

Math 55

Math 55 is a two-semester long first-year undergraduate mathematics course at Harvard University, founded by Lynn Loomis and Shlomo Sternberg. The official titles of the course are Honors Abstract Algebra (Math 55a) and Honors Real and Complex Analysis (Math 55b). Previously, the official title was Honors Advanced Calculus and Linear Algebra. == Description == The Harvard University Department of Mathematics describes Math 55 as "probably the most difficult undergraduate math class in the country." Formerly, students would begin the year in Math 25 (which was created in 1983 as a lower-level Math...

Edward Evans

Ein unmenschliches Wesen scheußlichster Art

user2103480

@EdwardEvans but do you mean like "frischetheke" leberkäse or shelf leberkäse? And why am I starting to make supermarket leberkäse distinctions?

Edward Evans

Frischetheke

user2103480

06:43

Oh thats fine

Edward Evans

There's a supermarket in Vorarlberg called Sutterlüty that genuinely serves the best Leberkäse in the universe

bold

user2103480

Thought it was genuine shitty shelf leberkäse

Edward Evans

Nah we bought some of that and cooked it and it tasted like fehlgeburt

not that I know what that tastes like

user2103480

lmao understandable

@EdwardEvans a man's gotta try

Especially when something's done in a south park episode

Tinkeringbell

→ 5 messages moved to Trashcan

Just a sec, it's still early. Continue :)

user2103480

06:55

Haha no worries and thanks

Edward Evans

tyty

user2103480

Finally, messages that really belonged to the trashcan

Tinkeringbell

@NoName dismissed the mod flag, you can use standard r/a one for that...

You really should: They stack and if there's a bunch of em no mod is needed to handle them :)

NoName

Oh I see. Thanks.

user2103480

@EdwardEvans "banter > morals" is the principle that makes it impossible for me to ever have a career in politics, at least for the spectrum of people that care about past remarls of politicians

My whole career could easily be derailed already on a communal level lmao

Edward Evans

07:12

@user2103480 same tho

also, got leberkäse

user2103480

@EdwardEvans proof that britons really lack taste

Edward Evans

what

user2103480

Or proof that you're drunk at 8am, choose

Edward Evans

both

user2103480

us

moment

I'd try to contribute something to the chat here but we really have 0 math to talk about

Edward Evans

07:18

yeah I belong to an interesting area of mathematics

user2103480

I belong to a funded area of mathematics

Edward Evans

fair one

user2103480

Yeah, would be more useful if I was good at a funded area

Still better than being no good in an unfunded area, eh?? lmao gottem

NoName

$\text{funded} \cap \text{interesting} =\emptyset ? $

user2103480

@NoName I'mma go tell lurie his work is shite

NoName

07:26

I take it back.

Leonhard Euler

07:53

Hey, please forgive me

I was just going to delete that

I have a good tongue twister

Song the ievan polka correctly

*sing

user777

08:34

Hi, I read an article that said in the first paragraph the following:

Randomness is powerful. Think about a presidential poll: A random sample of just 400 people in the United States can accurately estimate Clinton’s and Trump’s support to within 5 percent (with 95 percent certainty), despite the U.S. population exceeding 300 million. That’s just one of many uses.

My question is: how did he know that from sample of 400 persons, then we can know with 95% certainty that 5% of the U.S. populations are supporting the ones who have the majority in the sample?

Leonhard Euler

I have noticed that math has produced more prodigies than any subject. Why is it so?

user2103480

@user777 oh to hell I accidentally flagged that I think

Sorry

I'm on my phone and it doesn't do what I want

I think it's inaccurate to take polling as an example

As we have seen two elections in a row

user777

@user2103480 It is Okay!

Leonhard Euler

You should be careful @user2103480

Sayan Chattopadhyay

@user2103480 Lol I did that to Ted once, now I am dead to him

user2103480

08:41

I think this assumes that our samples are actually distributed in the right fashion

If you say that the probability that someone votes trump is p, and the probability that someone votes clinton is 1-p

And you assume that you ask people in a way such that the probabilities of their respective answers have the same distribution

Leonhard Euler

Becoming a mathematician takes ~7 years

Lol that means I can learn most branches at the age of about 20 if I follow that pace

user777

@user2103480 So they made a mistake in their article

Leonhard Euler

And at the age of 30...

user2103480

Then, and this is a bit tricky if you haven"t seen it yet, you know that the normalized, shifted sum of the amount of people should be approximately normally distributed with variance equal to the variance of a bernoulli R.V. with parameter p (by the central limit theorem), and so you can estimate the parameter p from the answers of people pretty accurately, if our assumptions are right

@user777 they made the mistake of choosing a bad example. The mathematical principle is correct. If the assumption are right.

You can also go a less sophisticated way and say that the sum of bernoulli random variables is binomially distributed and then do some calculations with stirling's approximation

Leonhard Euler

09:13

"Studying is fun and learning will be fast, if there were no exams and tests, and also rude teachers"

This is why can learn math very fast

there is no pressure on me so math is fun for me

a website says calculus 1 takes ~180 hours in college man I completed in 90 hours

I could have done faster

*not boasting, anyone can do that

weird sierpinski

NoName

09:55

@LeonhardEuler Are you 13?

Leonhard Euler

lol yeah

gonna turn 14

nothing special in that

skullpatrol

:-D

nvm

Leonhard Euler

ok

@skullpatrol your edit is the best edit in the history of edits

moderators should be as kind as you

5

skullpatrol

10:14

thnx, pal

Leonhard Euler

np, pal

skullpatrol

cya

skullpatrol

11:54

@Dubias nope, sorry

Lucas Henrique

12:42

hey chat

Leonhard Euler

Hello @LucasHenrique

Lucas Henrique

say $C$ is any closed path in the complex plane and $f$ a continuous function. is it true that $\oint\limits_C f(z)\, \mathrm{d}z = 0 \implies \oint\limits_C f(z)^2\, \mathrm{d}z = 0$?

if $f$ is analytic it's clear. but I'm having trouble if that's not the case

Astyx

What about something like $re^{i\theta} \mapsto r\cos \theta$ ?

And C the unit circle

Lucas Henrique

does not satisfy the hypotheses. :(

Astyx

Why not?

Lucas Henrique

12:48

The closed integral is $i \pi$

the first one, I mean

Astyx

huh?

Lucas Henrique

@Astyx this function, restricted to the unit circle, is simply $\cos \theta$, isn't it?

Astyx

Yes

So you're integrating cos over $[-\pi, \pi]$

Lucas Henrique

and $\oint\limits_C f(z)\, \mathrm{d}z = \int\limits_0^{2\pi} i \cos \theta e^{i \theta}\, d \theta$

Astyx

13:07

Ah yes you're right, I was thinking of |dz|

Sorry

Lucas Henrique

yeah, my professor made up the question in a wrong way

he wanted a fixed function $f$ s.t. the closed integral is 0 for every path

thus analytic

in this case, the implication is obviously true

I need a counterexample, otherwise I won't get a full mark

(Even though... yeah, I didn't know any counterexample when I wrote that this implication is false lol)

Mike Miller

Are you certain it's false? I'm very hazy but doesn't Morera imply it is true that f is analytic?

Oh I misunderstood your assumption.

Lucas Henrique

if Morera hold, yes, $f$ is analytic. But in the original way he wrote the question, you simply fix a $C$, closed path. So the closed integral being 0 needs not to hold for all closed paths

Mike Miller

Right exactly you fix C

I misread

Lucas Henrique

that's my problem and sincerely, this teacher is an asshole. we're having this complex analysis course with the physics people and the best people from pure math have already gave up from it

it's not supposed to be a course that hard, the professor is just a beetlehead and doesn't care about the ridiculous results we're having

love_sodam

13:24

If a regular surface is diffeomorphic to S^2 then is it possible for the region on which $K\leq 0$ to have greater area than the region on which $K\geq 0$? $K$ is a Gaussian curvature

anakhro

14:03

@love_sodam do you know Gauss-Bonnet?

Balarka Sen

Why does that help, @anakhro?

I don't think it does.

Take a standard pseudosphere. This has area $4\pi$. The base is a circle of radius $1$, let it go upwards for a long time until the waist is $\varepsilon > 0$, a very small number.

Thorgott

my intuition tells me it should be possible, no?

Balarka Sen

Cap it off at the base by a disk and at the top by a disk. You have a surface (with singularities) which has negative curvature area very close to $4\pi$ and zero curvature area very close to $\pi + \pi \varepsilon^2$

Then round off near the corners

You get some positive curvature but mostly negative curvature dominating

Yeah, @Thorgott, I think my example works

Thorgott

you just need small negative curvature in a lot of places and large positive curvature in a few places

Balarka Sen

Right

Leonhard Euler

14:08

Hello I am back again

Sorry for so many errors

myassignmenthelp.net/blog/…

anakhro

@Thorgott <-- it helps get intuition for this, Balarka

Leonhard Euler

This website is mainly saying that every branch of math is hard

love_sodam

@anakhro Yes. From that, I know that $\int\int_S K dA = 4\pi$

Balarka Sen

I came up with my example without thinking about Gauss-Bonnet.

anakhro

One of the things I have learned about teaching is that students learn in more ways than I know how to teach.

Thorgott

14:11

you just internalized GB already

anakhro

What works for one of us might not work for every student.

Thorgott

I also said what I said without GB in mind explicitly, but that's just cause GB has very intuitive content to an extent

Balarka Sen

OK, I found it more interesting to just try to come up with an answer to the question than making @love_sodam suffer :)

It's a cool question, wasn't clear how to build an explicit example.

Thorgott

the intuitive content is that if curvature goes down somewhere, it has to go up somewhere else

the unintuitive part is why on earth the total is the euler characteristic

anakhro

I don't think helping someone figure out the question for themselves rather than giving them the answer is "making them suffer".

Balarka Sen

14:13

That was facetious. I didn't know the answer so I didn't feel qualified to guide them through the answer.

Nihar Karve

@BalarkaSen hey man, aren't you active on the h bar anymore?

Balarka Sen

@Thorgott Well the first thing I thought was GB but immediately discarded it because it didn't say anything explicit (though I like your high +ve curvature low area summary). Then I was thinking about capping off ends of a hyperboloid by spherical caps

Cuz if you are hyperbolic you expect area to grow very fast

And then you stop the thing after some time and cap it off.

But who knows what area of a hyperboloid of one sheet (truncated) is? Pseudosphere was obviously the right choice.

Thorgott

I was thinking of like denting in the pole of the sphere and then denting more and more in, but I don't think I can actually get a majority negative curvature that way

Balarka Sen

Yeah

Maybe you can if you wiggle around with those inward-tendrils

Get it to almost space fill

love_sodam

I thought it's impossible but suddenly people try to find counterexample

anakhro

14:18

@love_sodam are you sort of following the ideas being mentioned?

love_sodam

Not really.

Balarka Sen

I gave an explicit example, maybe understanding that will help.

anakhro

@love_sodam Thorgott mentioned "you just need small negative curvature in a lot of places and large positive curvature in a few places". Do you see how this idea sort of makes sense when you consider either Gauss-Bonnet or "diffeomorphic to the sphere"?

love_sodam

Yes

Balarka Sen

@NiharKarve Nope, I go there occasionally though

(Sorry, lost your message in the discussion above)

anakhro

14:22

@love_sodam So how one can start is by trying to force this to be the case. That is, you want a large region with only slight negative curvature, then you want to add areas of harsh positive curvature.

love_sodam

I see

anakhro

That's kind of the intuition that's being worked out in the example Balarka gave. He starts with a portion of the pseudosphere, and then completes it to a surface that is diffeomorphic to the sphere with small enough positively curved caps.

It probably helps to look at a picture of a pseudosphere if you do not know what that is.

love_sodam

I know that

Balarka Sen

Do you know the area of the pseudosphere?

anakhro

because the smaller the positive caps are, the more curved they are.

Balarka Sen

14:25

Also, what is the set it intersects the $xy$-plane in?

love_sodam

Well I know the Gaussian curvature of it

Balarka Sen

Yeah, that's a good summary @anakhro. Surprising I didn't think of it this way though

anakhro

Most things are easier to reason in hindsight.

Balarka Sen

@love_sodam That's a good start. Find the area.

anakhro

Now I should get back to my homework I suppose.

love_sodam

14:32

@BalarkaSen It's $2\pi$

$\gamma(t)=(sin(t),0,cos(t)+log(tan(t/2)))$ for $t\in(\pi/2,\pi)$

the generating curve of pseudosphere I have

Balarka Sen

Oh, my bad then. But still OK.

I was thinking of the doubled pseudosphere, which is $4\pi$. Thanks.

What's the base of the pseudosphere? Circle of radius $1$, am I correct?

love_sodam

What do you mean base

Balarka Sen

The pseudosphere abruptly stops in the $xy$-plane, right? It intersects the $xy$-plane on some subset -- what is that subset?

love_sodam

Does it intersect?

$sin(t)\neq 0$ on $t\in(\pi/2,\pi)$

Balarka Sen

Correct, but you can take a half-open interval.

It certainly "limits" to the $xy$-plane. What is in that limit?

love_sodam

14:41

unit circle you mean

Balarka Sen

Right.

When people draw the pseudosphere they draw a symmetrized version. The equator is what I am speaking of, along which the symmetrized surface has singularities. It's the unit circle.

In your version you just took the open upper half, the "upper hemisphere of the pseudosphere"

love_sodam

Yea my diff. geo. text is strange

Balarka Sen

It's ok, this convention varies text to text anyway.

So here is my plan. The surface above has curvature $-1$ everywhere, except the equator where it is singular. Area of the whole thing is $4\pi$

Sandwich this symmetric pseudosphere between the hyperplanes $z = -N$ and $z = N$ for some large $N \gg 0$

Chop off everything outside these hyperplanes. Add appropriate caps to the top and bottom so that you have a smooth surface there anyway

Finally, smoothen the equator. You can of course remove a very tiny cusp-like neighborhood of the equator and replace it by a spherical annulus.

This new surface is diffeomorphic to $S^2$. Agree?

love_sodam

Yes

Balarka Sen

The positive curvature pieces are the two spherical caps top and bottom and the spherical annulus at the equator. All of them have extremely small radius so they contribute nothing to the area.

But note that these pieces all have HIGH curvature. Because radius is small, they have curvature which is like $1/\varepsilon$. Massive!

Low area high positive curvature regions, high area low negative curvature region, as @Thorgott and @anakhro promised.

love_sodam

14:54

Yes I got it. Thanks :)

Transcript for

$Mathematics$ Mathematics