Mathematics - 2016-11-12 (page 3 of 6)

Alessandro

17:01

For an actually nontrivial example I believe every space $X$ with $|X|>2^{2^{|B|}}$ where $B$ is the set of open sets should work

Adeek

@BalarkaSen nice question.

Balarka Sen

Mhm.

Ali Caglayan

@Alessandro You could probably ask on MO for the technical terminology.

Alessandro

I hope they're called trivially compact or something fitting like that if they have a name

Balarka Sen

@Adeek On a related note, have you become comfortable with transition functions yet?

Adeek

17:05

yeah

Mahmoud

Hello.

I was wondering.

Balarka Sen

Trivialize the moebius strip over two obvious open sets covering $S^1$. What's the transition function?

Adeek

@Balarka I haven't solved this question yet. I will think about it after my exam on tuesday sounds interesting.

Mahmoud

Why is $f(x)=e^x$ The most natural exponential ? Is it because it's slope is equal to the value of $f(x)$ at any given $x$ ? And what do mathematicians mean by the term $\text{natural}$ ?

Balarka Sen

@Adeek Well it's like the simplest example of transition functions so you better work it out. Good luck on the exams.

Ali Caglayan

17:10

@Mahmoud I can give you many reasons, but they probably won't be too useful for you. What level answer are you hoping for?

Mahmoud

@AliCaglayan Just try me $:)$

Ali Caglayan

Actually let me try this way

Adeek

@Balarka yeah I will work it out. By the way I was wondering where did you learn differential manifolds from ?

Lee ?

Ali Caglayan

What other exponential function would you have in mind?

Natural you can think of as easiest/most obvious. Consider every other exponential function $a^x$ and take the derivative what do you get?

@Mahmoud

Balarka Sen

@Adeek Guillemin-Pollack, but it doesn't cover everything about smooth manifolds. I think I learnt more by talking to people than I learnt it from reading a specific book. I like it though.

Adeek

17:15

yeah

Balarka Sen

I prefer G-P over Lee but I haven't read much of Lee thoroughly.

Adeek

yeah it is a nice book.

Mahmoud

@AliCaglayan Thanks, but I still need something else

Ali Caglayan

You can think of taking the derivative as an operator, e^x is an eigenvector

@BalarkaSen interesting question on mo

Mahmoud

What does Linear Algebra tell us about eigenvectors ?

DHMO

17:20

@Mahmoud it is $\displaystyle \sum_{n=0}^\infty \dfrac{x^n}{n!}$ if that isn't special enough

@Mahmoud what on earth do you need?

Balarka Sen

@AliCaglayan Yeah, no idea. It's not true even for finite-dimensional vector spaces that removing a bunch of points is homotopy equivalent to wedge of spheres though.

Eg remove $\{0, 1, 1/2, 1/3, 1/4, \cdots\}$ from $\Bbb R^2$.

Mahmoud

It seems random to me, why exactly the $n!$ and $x^n$ Why not something else @DHMO

Ali Caglayan

Yeah removing a finite number is

DHMO

@Mahmoud consider the fact that e.g. base 2 yields $\displaystyle \sum_{n=0}^\infty \dfrac{(x\ln2)^n}{n!}$

Astyx

Because ${d\over dx}{x^n\over n!} = {x^{n-1}\over (n-1)!}$ as long as $n\ge 1$

DHMO

17:23

it is quite useful in calculus because of this property

Ali Caglayan

Thus as @Astyx says every term in that series becomes the previous

Astyx

Because $\cos x + i \sin x = e^{ix}$

And lots and lots of other very nice properties

DHMO

@Astyx that's another nice property

which makes it useful in complex analysis

Mahmoud

Being it's own derivative ? What is that so special ? Sorry for those questions.

Adeek

@BalarkaSen suppose we consider the tautological line bundle over $\mathbb{P}^n$ complex project space of dimension n. Show that the only holomoprhic section of E over $\mathbb{P}^n$ is the zero section.

DHMO

17:24

@Mahmoud there are only two function that are its derivatives

Astyx

Well de facto this is how $\cos$ and $\sin$ are defined

Mahmoud

TWO ?

Astyx

0

DHMO

@Mahmoud one is the exponential function; one is the zero function

Mahmoud

What's the second ?

DHMO

17:25

to be fair, the first one is $ae^x$ so the second one is just $a=0$

Mahmoud

I'm a bit disappointed.

Adeek

is it true any holomorphic functions from $\mathbb{P}^n$ to $C^{n + 1}$ is constant ?

Mahmoud

That makes it very special.

DHMO

Does this series converge? $$\sum_{n=1}^{\infty} \frac{1}{n^3 \sin^2 n}$$

Ali Caglayan

:33477164 Yeah I mean the finite case is easy, but the OP brings up a good point with removing the basis.

Astyx

17:26

Yes

Balarka Sen

@Adeek Yes.

Adeek

oh wow

lol

Balarka Sen

Any holomorphic function $X \to \Bbb C$, $X$ compact, is constant.

Ali Caglayan

@DHMO Its bounded by $\zeta(3)$ and $-\zeta(3)$

DHMO

@AliCaglayan but does it converge?

Adeek

17:27

lol nice.

1 sec toilet

brb

Astyx

No it isn't

Balarka Sen

This is actually just maximum modulus theorem.

Astyx

Why would it be bounded between $\zeta(3)$ and its opposite ?

DHMO

@Astyx nice point. @AliCaglayan why?

Ali Caglayan

@DHMO because I type before I read

Astyx

17:28

Good answer :)

DHMO

excuses :p

Astyx

I'd say it converges

Because of Abel Integrals

DHMO

@Astyx can you elaborate?

mathworld.wolfram.com/AbelsIntegral.html

this?

Astyx

Ok just give me time :)

Balarka Sen

@AliCaglayan I'm wary of calling the finite case easy. But I guess you can still use Whitehead theorem (a version of it, IIRC, works for infinite dimensional manifolds) - $\pi_1 = 0$ by van-Kampen. $\pi_2 = H_2 = 0$ by Mayer-Vietoris plus Hurwitz. Do this for all $\pi_n$. That should say it's weakly contractible.

meow-mix

17:31

@BalarkaSen false. you ned some knowledge of topology for the first chapter exercises. Luckily i already have it, just saying

Ali Caglayan

@BalarkaSen Yeah, i mean for the fundamental groups its easy. I have no idea about higher homotopies tho

Adeek

@BalarkaSen Given that we have the tautological line bundle defined as $E = \{[z],w) \in \mathbb{P}^n\times\mathbb{C}^{n + 1} : w = uz \ for\ some\ u\ \in \mathbb{C}\}$

Astyx

Wrong link

Balarka Sen

@meow-mix Ah, yes, I recall that. $\text{maxspec} C(X) \cong X$, right.

Adeek

we want to show that any section $\sigma : \mathbb{P}^n \rightarrow E$ is the zero section.

Balarka Sen

17:34

It's just hard to imagine someone would learn commutative algebra without basics of algebra and point-set topology.

Adeek

can we work with $\mathbb{C}^{n + 1}$ instead of E ?

@BalarkaSen

meow-mix

yeah :P

Adeek

where $\sigma$ would be a holomorphic section.

Balarka Sen

@Adeek Yes. Given a holomorphic section $\Bbb P^n \to E$, compose with the projection $\Bbb P^n \times \Bbb C^{n+1} \to \Bbb C^{n+1}$ restricted to $E$.

DHMO

@Astyx sorry for being impatient; any idea?

Balarka Sen

17:41

@AliCaglayan I mean, he's still removing an infinite number of points.

Astyx

I mistook you series for a Hardy series, I'm looking for clues right now !

DHMO

@Astyx I don't have any clues

Adeek

@BalarkaSen so yes given holomorphic function $\sigma : \mathbb{P}^n \rightarrow E$. Then we can consider $\psi := \pi_1 \circ \sigma : \mathbb{P}^n \rightarrow \mathbb{C}^{n + 1}$. By maximum modolus theorem we have that $\psi$ is constant as $\psi$ can be realized as $f_1,...,f_{n + 1} : \mathbb{P}^n \rightarrow \mathbb{C}$, therefore $\psi$ is constant.

Balarka Sen

You mean $\Bbb P^n \to \Bbb C$.

Adeek

yes

Balarka Sen

17:44

You should prove that maximum modulus theorem implies that

Adeek

I have never heard about maximum modulus theorem I will go check it out.

Balarka Sen

Didn't you cover that in your complex analysis course?

Adeek

nop

our course was not very theortical.

I am gonna take complex analysis next year, which will be theortical.

Balarka Sen

Weird. That's one of the essential theorems in complex analysis.

ah, ok

alright, I have to run now

Astyx

See you

ahorn

17:52

How can I prove that a sequence is convergent, when the mapping that defines the sequence is not a strong contraction? E.g. $x_{n+1}=1+\frac 1 {x_n}$

DHMO

@ahorn hint: if it is convergent, it must converge to either $\phi$ or $1-\phi$.

Astyx

@ahorn You can see if $u_{2n}$ and $u_{2n+1}$ have the same limit (both sequence are monotonic and therefore have a limit or diverge to an infinity)

DHMO

@Astyx except that it is false

Astyx

@DHMO I lack ideas to solve your series, please notify me if you find anything ! :)

DHMO

neither sequence is monotonic

Astyx

18:02

Why not ?

DHMO

why?

96

A: Are there any series whose convergence is unknown?

It is unknown whether $$ \sum_{n=1}^\infty\frac{1}{n^3\sin^2n} $$ converges or not. The difficulty here is that convergence depends on the term $n\sin n$ not being too small, which in turn depends on how well $\pi$ can be approximated by rational numbers. It is possible that, if $\pi$ can be ap...

happy april fools

Astyx

Because $f:x \mapsto 1+ {1\over x}$ is decreasing

DHMO

@Astyx no it is not

Astyx

Over $\Bbb R_+$

DHMO

@ahorn what is the domain?

Astyx

18:04

I should have expected this :)

Adeek

maybe someone can help me with this question.

DHMO

I was just thinking that you might have some ideas if you don't know that it is an open problem

Adeek

Show the only holomorphic section of E over $\mathbb{P}^n$ is the zero section.

Astyx

Fair enough

Adeek

where $E = \{ ([z],w) \in \mathbb{P}^n \times \mathbb{C}^{n + 1} : w = uz \ for \ some \ u \ \in \mathbb{C}\}$

DHMO

18:06

If $y=\dfrac{ax+b}{cx+d}$, then $f(y)=\dfrac{(a+b)x+a}{(c+d)x+c}$

This reminds me of the Fibonacci sequence

Adeek

Suppose $\sigma : \mathbb{P}^n \rightarrow E$ is a holomoprhic section.

I proved that $\sigma([m]) = ([u],c)$ where c is constant in $\mathbb{C}^{n + 1}$

DHMO

also, $f^0(x) = \dfrac{1x+0}{0x+1}$

Adeek

now why is $\sigma$ is the zero section.

DHMO

therefore $f^n(x) = \dfrac{f_{n+1}x+f_n}{f_nx+f{n-1}}$ @Astyx @ahorn

Monad

How can one describe a class, highest in hierarchy, that is the super-class of all other classes, all of which derive from said class. How can one describe such class concisely and such that most mathematicians would understand?

Astyx

18:09

You could see that $x_n \in \Bbb R_+$ for n large enough

I think

DHMO

@Astyx ?

Monad

I was thinking of Supreme, because it also originates from "super-" but in most contexts refers to "highest", but I doubt most will understand this as typing it in Google doesn't show any relevant results. Any ideas?

DHMO

@Astyx no, because $1-\phi$ is a fixed point of $f$.

also, $f(-1)=0$ and $f(0)$ is undefined

ahorn

I was busy finding where the Weierstrass M-test is in my old Real Analysis textbook... just to jog my memory what that was about... I'm going to go to have dinner (g2g)

Astyx

But still, sequence are monotonic beyond a certain rank

DHMO

18:13

also, $f^{-n}(x) = \dfrac{f_{-n+1}x+f_{-n}}{f_{-n}x+f{-n-1}}$

so to find the starting point which would yield zero, plug $x=0$...

Astyx

Since either they go to $\Bbb R_+$ after a certain rank and then stay there, either they stay in $\Bbb R_-$

DHMO

the starting point would be $f^{-n}(x) = \dfrac{f{-n}}{f{-n-1}}$

which again converges to $1-\phi$

@Astyx so this statement is true

Hiro

Can anyone have a look at this question:

1

Q: Another Problem Involving Finding the Number of + Integers Under a Given Variable with no even digits

I've already asked a similar question to this, but I didn't quite understand the solution that was proposed. Fix $c\ge2$ an even integer. Find the number of integers less than $c^3$ which are divisible by $c+1$ and do not contain any even digits in their base $c$ representation. (natural general...

number-theory

DHMO

@Hiro It took me a while to realize that + means positive

Hiro

oh, I'll edit it one sec

DHMO

18:17

I misread it and derived another question:

Solve the diophantine equation $c+1 | c^3$

@Hiro e.g. let c = 10

do you know the divisibility test by 11?

Hiro

not quite

DHMO

alright

Hiro

so do we simply plugin a value for c?

DHMO

@Hiro no, it's just for exploration

Hiro

oh okay

So is my initial approach correct:

The positive integers less than $c^3$ that are divisible by $c+1$ can be represented as $k(c+1)$ for $k=1,...,c^2-c-1$

DHMO

18:20

@Hiro no, $(c^2-c)(c+1) = c^3-c < c^3$

Hiro

why did you do $(c^2-c)(c+1$?

DHMO

to show you that $k$ can also be $c^2-c$

Hiro

so it stretches to $c^2-c$?

DHMO

yes

Hiro

and given that information, how am I able to find the number of integers? (Not really sure what to do)

DHMO

18:23

let's go back to c=10

a number is divisible by 11 iff its alternate sum is divisible by 11, e.g. 71809 is divisible by 11 because 7-1+8-0+9 is divisible by 11

now, numbers less than c^3 only have 3 digits...

Hiro

do you mean the maximum number of digits less than c^3 are 3 digits or ..?

DHMO

yes

sorry i have to go now

Hiro

oh ok but thanks for explaining

I think I understand it better now

Astyx

Bye @DHMO

Sophie

I'm arguing with a person that imaginary numbers really do exist. I told them the nomenclature didn't matter much, in fact I like calling them orthogonal numbers. He answered "yeah sure, orthogonal from reality"

Danu

18:35

@Sophie "exist"... The main problem is defining the term.

Astyx

Yeah, do real numbers even exist ?

Sophie

@Danu my volume was too high. I almost had a heart attack

arctic tern

do irrationals exist? what about negatives? you could provide superficial arguments against either. modern math considers sets with operations, one doesn't need an interpretation or real-world application of an algebraic structure to say it "exists." of course, the fun begins when we do actually find interpretations and applications, but they will require looking and reaching beyond the old interpretations and applications.

Astyx

@Sophie see math.stackexchange.com/questions/154/…

Balarka Sen

@Adeek You just proved $([u], c)$ is in $E$ for all $[u] \in \Bbb P^n$, for a constant vector $c$. What does that mean, definitionally?

Astyx

18:39

Sorry I didn't copy/paste what I thought I had

Hiro

is there some sort of formula to calculate the number of integers divisible by a number that are less than a specific number?

Adeek

yeah I understand now I fixed my argument.

it is by definition how the elements of E look like.

Hiro

e.g.--> numbers less than 1000 that are divisible by 11

Balarka Sen

ok

Adeek

@BalarkaSen Thanks.

arctic tern

18:39

@Hiro divide and round down

Astyx

@Hiro $\lfloor {1000\over 11}\rfloor$

Hiro

what's floor?

arctic tern

rounding down

Astyx

Integer part

Hiro

i am not quite sure what that symbol looks like, one sec i'll check

Astyx

18:40

@Hiro See this link

Adeek

hey @BalarkaSen when we define a metric on each vector fiber it then it is same metric for each fiber ?

Balarka Sen

@Adeek No, why would it be the same on each fiber?

Adeek

Balarka Sen

It just varies smoothly fiber-to-fiber.

Adeek

oh ok

that is why we have for each section condition.

Astyx

18:42

What do topologist do nowadays ?

Balarka Sen

That's the formalization of the smoothly varying condition, yes.

Adeek

yeah ok

Danu

@Astyx What does that even mean?

Balarka Sen

You can just say that a metric is a bundle-map $E \otimes E \to \mathbf{R}$ which is fiberwise positive definite.

Danu

@Balarka, wanna help me with something easy?

Balarka Sen

18:45

@Danu I can't help you with most of the things you claim to be easy. But ask; don't ask to ask :)

Danu

That's not true

My standard for "easy" is below your level

for sure

Anyways

So let's consider a holomorphic line bundle on a complex manifold

And open sets over which it is holomorphically trivialized by $\psi_j$

(i.e. transition functions will be holomorphic)

Now Huybrechts has the following bit:

"On $U_i$, the Hermitian structure is given by $h_i:U_i\to \Bbb R_{>0}$, i.e. $h(s(x))=h(s(x),s(x))=h_i(x)\cdot |\psi_i(s(x))|^2$ for any local section $s$."

So the first equality is just the definition of $h(s(x))$, no problem

But the second weirds me out

Adeek

it is so cool I am starting to understand what you guys are talking about @Danu

Danu

because there has to be a compatibility condition for $h_i$ and $h_j$ on $U_{ij}$

Adeek

soon I will be able to understand complicated stuff in this chat haha

Danu

Since $|\psi_i(s(x))|^2$ will transform

Hiro

18:50

I still don't understand floor functions, Wikipedia just talks about how it's related to computer science and shows the different calculations but doesn't explain the steps...

Danu

Because $\psi_{ij}=\psi_i\psi_j^{-1}=\lambda\in \Bbb C$

Hiro

$\lfloor {1000\over 11}\rfloor$ <-- how are floor functions related to the number of integers less than 1000 and divisible by 11?

Danu

So $|\psi_i(s(x))|^2=|\psi_{ij}\psi_j(s(x))|^2=|\lambda|^2 \psi_j(s(x))|^2$ on $U_{ij}$

So he at least needs to say that $\{U_i,h_i\}$ is a family of positive functions satisfying the necessary transformation condition, no?

Because $h(s(x))$ shouldn't depend on trivialization

Balarka Sen

@Danu Is "the" Hermitian structure he's talking about an arbitrary choice of Hermitian metric?

Hiro

@Astyx Why did you say we use this equation: $\lfloor {1000\over 11}\rfloor$ to calculate the number of positive integers less than 1000 but divisible by 11?

Danu

18:54

@BalarkaSen Yeah, I also don't like his usage of "the". Just pretend it says "a" :)

Adeek

Astyx

@Hiro because there are $\lfloor {1000\over 11}\rfloor$ numbers between 1 and $\lfloor {1000\over 11}\rfloor 11$ that are divisble by 11, and none between $\lfloor {1000\over 11}\rfloor 11$ and ${1000}$

Adeek

hey @BalarkaSen ok I understand every part of this proof but why is the summation here makes sense ?

Balarka Sen

@Danu Give me a minute.

Adeek

I mean we are summing over arbitrarily index

Astyx

18:58

gotta go now

Adeek

so why does the summation makes sense ?

Balarka Sen

@Adeek Because we're on a paracompact manifold.

Or rather, on a locally finite cover. Every point is hit by finitely many open sets.

Adeek

can you explain more I never really understood that point I just took it as given.

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$Mathematics$ Mathematics