Mathematics - 2016-10-17 (page 2 of 4)

Jessy Cat

04:00

Having some difficulty here, algebra peeps.

arctic tern

@JessyCat do you have any guesses?

usukidoll

04:17

I need a hint to prove this x.x prntscr.com/cv8les

it's like if the degree values aren't the same then the sum of degree f and g is equal to its max...whatever that is xx.xx

0celo7

what's confusing?

you can expand both $f$ and $g$ in terms of powers of $x$

then compute the sum

check which terms survive

usukidoll

$f(x) = a_{0}+a_{1}x+a_{2}x^2+...+a_{n}x^{n}$
$g(x) = b_{0}+b_{1}x+b_{2}x^2+...+b_{k}x^{k}$\\
add these two?

0celo7

right

pick $f$ to have higher degree, wlog

usukidoll

$f(x) = a_{0}+a_{1}x+a_{2}x^2+...+a_{n}x^{n}$
$g(x) = b_{0}+b_{1}x+b_{2}x^2+...+b_{k}x^{k}$
$f(x) + g(x) = a_{0}+ b_{0}+x(a_{1}+b_{1})+x^2(a_{2}+b_{2})...x^{n+k}a_{n}b_{k}$

0celo7

uh

how did you get $x^{n+k}$ o.o

it was all good until that last term...

usukidoll

04:22

$x^{n}x^{k}$ oh O_O! I should leave the last term like this

0celo7

what

the last term is $a_nx^n$

usukidoll

ohhhh $a_{n}x^{n}b_{k}x^{k}$

0celo7

no

We are assuming $n>k$.

So we get

usukidoll

so somehow the $b_{k}x^{k}$ doesn't survive when we add f(x)+g(x) because n >k

arctic tern

@usuki Add the following two polynomials:
x^4-2x^3+x+2
x^2+1
What is the degree of the result?

0celo7

04:24

$f(x)+g(x)=a_0+b_0+(a_1+b_1)x+\cdots+ (a_k+b_k)x^k+a_{k+1}x^{k+1}+\cdots +a_nx^n$

Jessy Cat

@arctictern whoa, you're here

@arctictern I have guesses, check out what I wrote.

usukidoll

$x^4-2x^3+x^2+x+2+1 =x^4-2x^3+x^2+x+3$ the highest degree is 4

Jessy Cat

I was actually planning on going to bed when I notice you pinged me

0celo7

I guess what I said has been ignored

arctic tern

@JessyCat sure, Q is a homomorphic image of G if and only if G has a normal subgroup N such that G/N\cong Q. That's just the first isomorphism theorem, which is true for all groups and has nothing specifically to do with cyclic groups.

Generate a specific hypothesis for the question at hand about cyclic groups.

usukidoll

04:26

well we're assuming n >k with wlog

Jessy Cat

It's either going to be $\mathbb{Z}$ or $\mathbb{Z_{n}}$ then

arctic tern

What is either going to be Z or Zn?

0celo7

@usukidoll What is the degree of the polynomial that I wrote above?

Jessy Cat

$G/N$?

is going to be isomorphic to either of those

arctic tern

@Jessy There is a unique cyclic subgroup of every possible order, including infinity right? So suppose I pick two values in {1,2,3,...,infinity}. How do you propose to tell if one is a homomorphic image of the other?

Don't just say something algebraically equivalent to "one is a homomorphic image of the other." Say something about the actual values chosen from {1,2,3,...,infinity} and how they're related.

Jessy Cat

04:29

interesting

Something about divisibility perhaps?

arctic tern

indeed

Jessy Cat

Huh.

One is a homomorphic image of the other iff the orders of the groups divide each other?

usukidoll

n?

arctic tern

@JessyCat well, one of the orders divides the other. they can't both divide each other unless they're equal. anyway, any ideas how to argue it?

0celo7

@usukidoll yes!

done

usukidoll

04:32

yay :3

Jessy Cat

But not an iff

0celo7

now make sure you understand it

Jessy Cat

@arctictern not right now, my brain needs sleep.

It's 12:32 am here. I was planning on getting up at 4 and continuing then.

0celo7

jesus

Jessy Cat

@0celo7 hello there, smiley.

0celo7

04:34

hi

arctic tern

good night

0celo7

what does "smiley" mean

Jessy Cat

It means you're a happy, friendly guy.

usukidoll

smiley as in :3

0celo7

I think that's a fair and valid characterization of me.

Jessy Cat

04:35

@arctictern goodnight. I hope after sleeping for a little bit, it makes more sense. Otherwise I'll be back.

usukidoll

offers fries

Jessy Cat

Chicken fries?

usukidoll

yes

Jessy Cat

Mmm

Okay, g'nite.

arctic tern

@usukidoll you some kind of chef?

usukidoll

04:36

sometimes I cook

arctic tern

seem to offer a lot of food around here

usukidoll

in real life

0celo7

probably laced with poison

ok I need to sleep too

bye people

usukidoll

-.-

my head hurts ... too much math ;p
past two days

Ted Shifrin

04:52

@arctictern: I'm some kind of chef :)

usukidoll

Ted is back yay

Ted Shifrin

where? where?

usukidoll

in this chat

Ted Shifrin

But I don't throw food :P

usukidoll

hmmm stares

sniffs

Ted Shifrin

04:57

rehi @chell

@AndrewT: You're awake early in the morning.

Andrew Thompson

Yup. Weekday after all.

Ted Shifrin

Still ... in this country, grad students don't rise early.

Well, most don't.

Andrew Thompson

They usually don't here either, unless for courses. I started doing it just to make my days longer, surprising how much an hour or two a day helps.

Ted Shifrin

Not if you waste another hour or two here :D

Andrew Thompson

Well, I wake up early so that I can both work and procrastinate.

Ted Shifrin

05:04

You and @MikeM have been running a joint seminar on procrastination.

usukidoll

>.< ugh

Andrew Thompson

We have. I'm mostly joking about it, my thoughts go more in the direction of "It'd be nice to be able to work 80 hours a week without getting tired."

I should start walking to uni, bye!

Ted Shifrin

bubye :)

user227867

Hello @ted. I switched to my autogenerated avatar because it looks nice, like me, lol.

Ted Shifrin

Of course, Jasper.

user227867

05:11

I finally bought a copy of Hartshorne's Algebraic Geometry from Book Depository at a good price.

Ted Shifrin

Well, have fun with that. Not my favorite.

user227867

Sometimes, it has better prices than Amazon and Springer.

usukidoll

book depository ... I ordered from them once

Ted Shifrin

I actually hadn't heard of it, but I'm not buying any more books.

user227867

@usukidoll They sell nothing but books, as the name says.

usukidoll

05:12

mhm

user227867

Amazon sells everything, including my undies.

TheGreatDuck

05:40

@MartinSleziak I misread your post and thought it said to post a question on meta every time one has a formatting trouble (which would be horribly inefficient). XD

user228700

07:50

Hi everyone :-)

user228700

Is anybody familiar w/ the parametric eqn. of a straight line?

NotAGenie

08:10

I might be, what about it?

user227867

@0celo7 Have you seen Zheng's Complex Differential Geometry? I can't see if it is suitable for me because there is currently no copy on Russian servers...

user227867

@Kaumudi You should immediately ask instead of ask to ask. If anyone wants to answer and can, they will. =)

user228700

Never mind, I think I got it :-P

NotAGenie

I've having some trouble with partial differential equations, how would I find $\frac{df}{dy}$ and $\frac{df}{dy'}$ where $f = y'^2 + yy' + y^2$

user227867

@Kaumudi Many people get answers by simply typing in this room. No need for any response. =)

user228700

08:18

@WillHunting Yeah! It's eerie, how that happens a lot o.O

Secret

@NotAGenie are y and y' functions of each other or independent?. If they are independent, you can treat all non variable terms as constants

NotAGenie

@secret y' is meant to be the derivative with y with respect to some x

specifically I'm trying to solve the Euler-Lagrange equation

Balarka Sen

@Kaumudi Not strange, only natural.

user227867

I am still trying to solve 1+1, hmm...

Tobias Kildetoft

@WillHunting Well, that is certainly tricky as it is not something that can be "solved"

Secret

08:27

if that's the Euler lagrange, then y' and y are independent of each other thus if you compute $\frac{d}{dy}$, you can treat y' as constants wrt y

user227867

@TobiasKildetoft I was waiting for you to say that. =P

NotAGenie

@secret alright, so that would mean that $\frac{df}{dy} = y'+2y$ and $\frac{df}{dy'} = 2y'+y$ because I can treat them as being independent of one another

and putting that in the Euler-Lagrange equation I'd get $y'+2y - \frac{d}{dx}( 2y'+y) = 0$

Secret

you can further simplify that d/dx term and you should get a y" term, which is the generalised acceleration in the equation of motion

NotAGenie

$y'+2y - 2y'' - y' = 0$

Secret

yup, and that is a linear ODE with constant coefficient, in fact, I predict the solutions will be sinusodial

or exponential, depending on the IVP

NotAGenie

08:34

specifically I'm trying to find a path from (0,0) to (1,1) which makes the path integral between those points of f stationary

$\int (y'^2+yy' + y^2)dx$

but I should be able to solve for y using
$y'+2y - 2y'' - y' = 0$, and then find constants such that y(0)=0 and y(1)=1 correct?

Secret

then you are in effect minmising the variation of the path from (0,0) to (1,1). On minimsing that, the function $\int f dx$ will be stationary, which (given that your function is a smooth and nice one), means f solves Euler Lagrange, thus you seemed to be on the right track

NotAGenie

alright well thank you very much, its been a bit since I've done differential equations so the idea of $\frac{d}{dy'}$ sort of hit me like a brick wall.

enthdegree

08:49

Hey can anyone spot the constant error here? dropbox.com/s/hsouvysj0gybi6h/…

I have picked up or lost a factor of -i somewhere in the (short) analysis but I can't figure out where

Lozansky

09:29

Aren't polynomials entire?

Tobias Kildetoft

@Lozansky Yes

Lozansky

So shouldn't they be analytic at $\infty$?

Tobias Kildetoft

write up the definition of what that means

Lozansky

That it's differentiable in some neighborhood to $\infty$

Tobias Kildetoft

are you sure that is the definition? That would make any differentiable function analytic at infinity

Lozansky

09:39

That it can be represented by a convergent power series at infinity?

Tobias Kildetoft

what does "at infinity" mean there?

Lozansky

I wouldn't know how to define it formally

Tobias Kildetoft

then you probably want to look that up

Lozansky

It seems that differentiability at a point does not imply analyticity at that point

Tobias Kildetoft

right

(though being differentiable as a complex function at the point does)

user228700

09:53

Is anybody familiar w/ the properties associated w/ the centroid of a triangle? I wrote down/drew this but now I've got no idea what it means:

user228700

user228700

(G being the centroid. I've no clue as to what E and D are :|)

Lozansky

@TobiasKildetoft What do you mean "differentiable as a complex function"?

Tobias Kildetoft

@Lozansky I mean holomorphic

Lozansky

My book doesn't actually mention holomorphic

I conjenctured it was the same as analytic

enthdegree

10:01

the sign error was at the very end, from line 24 to line 25

(in ref to my last post)

Tobias Kildetoft

@Lozansky Well, complex analytic yes

Lozansky

There is a distinction?

Rahul Joon

ju

Tobias Kildetoft

Not sure both real analytic and complex analytic can make sense for the same function

Lozansky

Well I'm pretty sure my book defines a function $f$ to be analytic in an open set $G$ if it has a derivative at every point in $G$

Tobias Kildetoft

10:06

Well, for complex functions this is the correct definition

Lozansky

@TobiasKildetoft How would you prove $f(z) = z$ is not analytic at $z=\infty$ using that definition?

Tobias Kildetoft

this is a nice feature of complex functions. That all these various degrees of differentiability are the same

DHMO

@Kaumudi E is the mid-point of AC. Centroid is the intersection of the three medians. A median is a line connecting a vertex (corner) to the mid-point (middle point) of its opposite edge (side).

Lozansky

Oh we can't have an opet set containing infinity

Tobias Kildetoft

we can't have any set containing infinity in this context

Lozansky

10:08

Right

Is that how you would argue?

Using the definition

Tobias Kildetoft

which definition are we using now?

Lozansky

The def. with the open set $G$

user228700

@DHMO Nah, D is definitely not the midpoint of AB. And dude, of course Ik what a median is! ._.

Tobias Kildetoft

@Lozansky that definition makes no sense for infinity, but since there is a notion of being analytic at infinity, it means that the definition must be something else.

DHMO

@Kaumudi Sorry, my fault. They should just be random points somehow restricted by the fact that ADB and AEC and DGE are straight lines.

Lozansky

10:13

@TobiasKildetoft Yeah okay, $f(z)$ is analytic at $z=\infty$ if $f(1/z)$ is analytic at $z=0$

user228700

@DHMO While that is true, there is more to it...

user228700

But I can't remember what! :/

Tobias Kildetoft

@Lozansky Ok, that looks more like something we can work with

DHMO

@Kaumudi that the right hand side identity is true?

Lozansky

@TobiasKildetoft I guess it then follows that any polynomial with degree $\geq 1$ is not analytic at infinity

Tobias Kildetoft

10:20

right

user228700

@DHMO Lol, yes, that too, but something else too!

Lozansky

11:41

We say $$\frac{1}{z+z^2/2!+z^3/3!+...}$$ has simple poles at $ z=2 \pi n i, n = 0, \pm 1...$

If we instead had $$\frac{1}{z-z^2/2!+z^3/3!-...}$$ would the point $z=0$ be considered an essential singularity?

Tobias Kildetoft

11:57

@Lozansky Well, as always first thing to do is look up why we say that the first one has simple poles and what all the definitions mean

DHMO

12:52

@Lozansky $\displaystyle \frac{1}{z+z^2/2!+z^3/3!+\ldots} = \frac{1}{e^z-1}$, so the simple poles are when $e^z=1$, namely $z=2\pi n i$

Similarly, $\displaystyle \frac{1}{z-z^2/2!+z^3/3!+\ldots} = \frac{1}{e^{-z}-1}$ would have the same poles

Lozansky

13:04

@DHMO You should reverse the signs in the denominator in the expression but I get your point

DHMO

@Lozansky no, it's written as plus

even if the next term is negative

because it precedes the ellipsis

proceed, precede, because English

Lozansky

$e^{-z}-1 = -z + z^2/2! - z^3/3!+...$

DHMO

oops, sorry, can't delete now

let's see...

Lozansky

It's okay

DHMO

$\displaystyle \frac{1}{z-z^2/2!+z^3/3!+\ldots} = \frac{1}{1-e^{-z}}$ right

still the same poles

Lozansky

13:08

Yes

DHMO

alright

Ramanujan

@DHMO hi

Lozansky

Say we have $$f(z) = \frac{1}{a_1z+a_2z^2+...}$$ with infinitely many poles and a finite limit $\lim _{z \rightarrow \infty} f(z)$ , would we classify $z=0$ as an essential singular point or a pole with infinite order?

DHMO

@Ramanujan hi

@Lozansky why is it a pole with infinite order?

A pole with infinite order would require $a_1,\ a_2,\ \cdots$ to be all $0$.

Lozansky

Okay say there exists a point $z = c$ for which $a_1z + a2_z^2+.... = c$ has infinitely many solutions, then $$g(z) = \frac{1}{c-(a_1z+a_2z^2+...)}$$ would have a pole of infinite order at $z=c$

Hm actually I don't want to use that as an example

DHMO

13:17

@Lozansky and $c-(a_1z+a_2z^2+\cdots)$ would be $0$

Lozansky

Right

What I want is to construct a function with a pole of order infinity at some point where the function is also bounded

DHMO

@Lozansky you can't

Lozansky

Yeah that's what I figured

@DHMO If $f$ and $g$ have a pole at $z_0$, do $f+g$ have a pole at $z_0$?

So my answer would be, not always

DHMO

@Lozansky not necessarily I reckon

but let me think

Danu

@BalarkaSen eyyo

Lozansky

13:29

I don't think it's true if the order is different

DHMO

@Lozansky $f=\dfrac1z$, $g=-\dfrac1z$

Lozansky

Yeah

Huh

Okay but can we prove that in the general case?

Danu

If I have some holomorphic line bundle that is globally generated by sections $\{s_j\}_{j\in J}$, then why is its dual a subbundle of the trivial bundle of rank $|J|$?

DHMO

@Lozansky What is the general case?

I already disproved it

Lozansky

Say $f(z)$ has a pole of order $m$ at $z_0$ and $g(z)$ has a pole of order $n$ at $z_0$

DHMO

13:33

@Lozansky looks like it's true

Lozansky

Then $f(z) = \frac{f_1(z)}{(z-z_0)^m}$ and $g(z) = \frac{g_1(z)}{(z-z_0)^n}$

True that it is false?

DHMO

@Lozansky true that it is true

Lozansky

No

DHMO

?

Lozansky

In your example we have $f+g = 0 $ which doesn't have any poles

DHMO

13:34

I thought you let $m\ne n$

Lozansky

Oh that's the case I'm looking at now

Sorry

DHMO

WLOG $m>n$

Your turn, brb

$f(z)+g(z) = \dfrac{f_1(z)+g_1(z)(z-z_0)^{m-n}}{(z-z_0)^m}$

Lozansky

$f+g = \frac{f_{1}(z)(z-z_0)^n + g_1(z)(z-z_0)^m}{(z-z_0)^{m+n}}$

DHMO

Since $f_1(z)$ is coprime with $(z-z_0)$, so is the whole numerator

Therefore it is a pole with order $m$

Lozansky

There is a lemma that says if $f$ has a pole of order $m$ then $|(z-z_0)^{k}f(z)| \rightarrow \infty$ as $z \rightarrow z_0$ for all integers $k<m$

DHMO

13:39

@Lozansky so?

Lozansky

So the sum $f+g$ has a pole of order $m+n$ but if you look at $f_1(z)(z-z_0)^n$ it's obvious this goes to infinity if $n<m$

DHMO

@Lozansky not $m+n$?

Lozansky

No, I'm looking at $(z-z_0)^{m} \frac{f_1(z-z_0)^n}{(z-z_0)^{m}}$

DHMO

@Lozansky but they cancel out?

Lozansky

Wait I might have to write it down on paper

@DHMO It seems to be difficult to use the lemma

DHMO

13:54

@Lozansky what difficulty are you going through?

Lozansky

I want to look at $\frac{f_1(z)(z-z_0)^n}{(z-z_0)^{m+n}}$ and conclude that this part has a conjenctured singularity of order $m+n$ and then show that $|(z-z_0)^k \frac{f_1(z)(z-z_0)^n}{(z-z_0)^{m+n}}|$ blows up for sufficiently small $k$ but this is not a worthwhile approach

But I have another idea

DHMO

@Lozansky why don't you cancel?

@iwriteonbananas So you write on yourself?

iwriteonbananas

@DHMO Since I am a banana, yes, I write on myself.

DHMO

@iwriteonbananas I see

Lozansky

Uh there's a plus

Which doesn't matter

Danu

14:02

Hi @MikeMiller

Mike Miller

h

iwriteonbananas

Sup Mike

Lozansky

So $h(z) = f_1(z)(z-z_0)^n + g_1(z)(z-z_0)^m = 0$ for $z=z_0$

DHMO

@Lozansky yes

Lozansky

But that would mean $f+g$ never have at $z=z_0$ if $f$ and $g$ have a pole at $z_0$

That can't be right

Danu

14:05

I'm trying to do an exercise in Huybrechts @Mike ;)

(awaits cookie from master)

DHMO

@Lozansky why would this imply that?

0celo7

@BalarkaSen Turns out strong differentiability is somewhere in between differentiable and $C^1$

Mike Miller

good man

Lozansky

If $f(z) = \frac{g(z)}{(z-z_0)^m}$ then $f$ has a pole of order $m$ at $z_0$ iff in some punctured neighborhood of $z_0$, $g$ is analytic and $g(z_0) \neq 0$

0celo7

@BalarkaSen The main point is that if your function is strongly diff at a point, it's a homeomorphism of some ball around that point onto its image.

Danu

14:07

@MikeMiller But of course I'm getting hopelessly stuck haha

Mike Miller

what problem

DHMO

@Lozansky yes?

Mike Miller

@iwriteonbananas WHAT DO YOU WANT

0celo7

@BalarkaSen The problem is that $C^1$ is infinitely better while str. diff. is only marginally better than diff. So you hear about diff or $C^1$, but not str. diff.

Danu

@MikeMiller Something that's used in the main text, also. He has the following: Consider a Hermitian line bundle $L$, globally generated by global sections $s_1,\dots,s_k$. Its dual inherits a Hermitian structure (I'm taking that to just be the thing that dualizes and then uses the already given structure, I hope that's correct). Then I'm supposed to show that $L^*$ also receives a Hermitian structure from inclusion in $\mathcal O^{\oplus k}$ and that it coincides with the other.

One first point, which I think should be clear but isn't, is that $L^*$ is a subbundle of $\mathcal O^{\oplus k}$

Lozansky

14:11

@DHMO In this case $h(z)$ is the numerator which is $0$ for $z=z_0$

DHMO

@Lozansky so?

iwriteonbananas

@MikeMiller My parents always told me I could be anything I want to, so I decided I want to be a microwave oven.

Lozansky

So then we can't say $h(z)/(z-z_0)^{m+n}$ has a pole

DHMO

@Lozansky So $\dfrac{h(z)}{(z-z_0)^{m+n}}$ does not have a pole of order m+n

Lozansky

Right

DHMO

14:12

(it does have a pole of order $m$)

Mike Miller

@iwriteonbananas You misunderstood. They said you can be anything I want.

Danu

@iwriteonbananas Neptr, is it you?!

iwriteonbananas

Fuck. I'm scared now.

Lozansky

@DHMO Well not necessarily a pole of order $m$

DHMO

@Lozansky yes, but you can't say that it doesn't have a pole

Lozansky

14:14

Right

iwriteonbananas

@Danu Only if Mike wants....

Lozansky

@DHMO Well then I'm out of ideas

DHMO

@Lozansky you just have to follow what I wrote above

39 mins ago, by DHMO

$f(z)+g(z) = \dfrac{f_1(z)+g_1(z)(z-z_0)^{m-n}}{(z-z_0)^m}$

38 mins ago, by DHMO

Since $f_1(z)$ is coprime with $(z-z_0)$, so is the whole numerator

37 mins ago, by DHMO

Therefore it is a pole with order $m$

Lozansky

I'm not sure that's necessarily true

Unless I'm misinterpreting you

Mike Miller

@iwriteonbananas what are you doing

DHMO

14:21

@Lozansky Well, which step is doubtful?

Lozansky

That since $f_1$ is coprime with $(z-z_0)$ the whole numerator is

Are you saying that's true because the other term in the numerator is not coprime with $(z-z_0)$?

DHMO

@Lozansky yes

Lozansky

Okay

DHMO

@Lozansky just ping me as I'm not always focusing on this window

Lozansky

@DHMO Right, will do

iwriteonbananas

14:25

@MikeMiller Trying to understand the spectral sequence for homotopy colimits

Do u know anything about that?

Lozansky

@DHMO Suppose $f$ has an essential singularity at $z_0$ and $g$ has a pole at $z_0$, can we then say $f+g$ has an essential singularity?

iwriteonbananas

(my battery will die within 5 minutes)

DHMO

@Lozansky Let $f=e^{1/z}$ and $g=1/{|z|}$

I believe that $\displaystyle\lim_{z\to0}\frac1{f+g}$ exists

Lozansky

@DHMO This is the chapter before the whole point at infinity stuff

DHMO

@Lozansky alright

Lozansky

14:31

@DHMO So for $f+g$ to have an essential singularity, we know that $|f+g|$ neither is bounded nor goes to infinity at $z \rightarrow z_0$

So we show either and the other one follows

DHMO

@Lozansky But in this case it does go to infinity

Lozansky

@DHMO That's what I'm thinking as well

@DHMO The answer is that $f+g$ does have an essential singularity at $z_0$ by the way

DHMO

@Lozansky According to Wikipedia, essential singularity $\iff$ neither $\displaystyle\lim_{z\to z_0}f$ nor $\displaystyle\lim_{z\to z_0}\frac1f$ exists

Lozansky

Okay

ondrejsl

Hello. If I count 1,1,2,2,3,3,... |natural numbers|, do I say two times omega numbers or omega times two numbers?

Mike Miller

14:36

@iwriteonbananas no

Lozansky

@DHMO But we haven't been given that "theorem" yet

DHMO

Ordinal arithmetic

In the mathematical field of set theory, ordinal arithmetic describes the three usual operations on ordinal numbers: addition, multiplication, and exponentiation. Each can be defined in essentially two different ways: either by constructing an explicit well-ordered set which represents the operation or by using transfinite recursion. Cantor normal form provides a standardized way of writing ordinals. The so-called "natural" arithmetical operations retain commutativity at the expense of continuity. Interpreted as nimbers, ordinals are also subject to nimber arithmetic operations. == Addition... ==

@ondrejsl ω·2 = ω+ω ≠ ω = 2·ω

in your case it is 2·ω

@Lozansky then just use your definition

ondrejsl

@DHMO Thank you. I tried to read that page but it was too much for me.

Lozansky

@DHMO Then I have to show $|f+g|$ neither is bounded nor goes to infinity as $z\to z_0$

Which I suspect is tricky

DHMO

@Lozansky I just told you that $f+g$ goes to infinity

so it is not an essential singularity

Lozansky

14:38

@DHMO That contradicts the answer in the solutions manual

DHMO

@Lozansky what does your manual say?

Lozansky

@DHMO It says it's true that $f+g$ has an essential singularity at $z_0$

DHMO

@Lozansky for which $f$ and which $g$?

Lozansky

$f$ with essential singularity and $g$ with pole

At $z_0$

DHMO

I have no idea then

0celo7

14:52

@TedShifrin Turns out the Taylor estimate approach was completely wrong lol

The proof is actually trivial

Thomas Rot

15:27

Does every bounded invertible selfadjoint operator have an eigenvector?

Balarka Sen

hi @iwriteonbananas

Danu

Heyo Balarka

Balarka Sen

hi

iwriteonbananas

Hi B.

Balarka Sen

what does it mean to say a line bundle is generated by global sections, @Danu?

Danu

15:39

There are no points where they all vanish

Balarka Sen

oic

Danu

I'm currently mostly interested in the example of $\mathcal O(1)$ and the coordinate functions $z_j$

But the more general thing I asked you about is an exercise

Mike Miller

@ThomasRot Of course by the spectral theorem it suffices to see if multiplication by a function bounded away from $\infty$ and $0$ on some measure space has an eigenvalue. Other than the trivial observation I have no idea what to do, though. I can ask an operator theorist in about an hour.

I assume you're not actually working with an elliptic operator or anything so lucky?

Balarka Sen

@Danu I see, this sounds like we're trying to find the analogue of the proof of the real version. If $E$ is generated by $s_1, \cdots, s_n$, $E$ trivializes over the complement of the zero sets of each of them. Let that be your open cover $U_1, \cdots, U_n$.

Danu

@BalarkaSen Sure.

Balarka Sen

15:46

You have maps (trivializations) $p_i : p_i^{-1}(U_i) \to U_i \times \Bbb C$. I think you should define $E \to \Bbb C^n$ componentwise as $v \mapsto (p_1(v), \cdots, p_n(v))$, where $p_k(v) := 0$ if $v$ is not in $U_k$.

That's how the proof for real bundles go at least. This should be a proper injective immersion.

Danu

Your notation is... weird

$p_i:p_i^{-1}(U_i)\to U_i\times \Bbb C$?

Balarka Sen

I know, please bear with me.

I meant $p_i$ is projection $U_i \times \Bbb C \to \Bbb C$ composed with that

Semiclassical

hi chat

Danu

@BalarkaSen Take your time to correct typos :)

Balarka Sen

@Danu Or, no, just define $p_i$ as it is, and make the map $E \to B \times \Bbb C^n$ (B the zero section), sending $v$ to $(f(v), (p_1(v), \cdots, p_n(v)))$ where $f$ is the bundle map $f : E \to B$.

Because you want a bundle embedding after all.

Danu

15:51

Sorry, I can't parse your messages. Can you type it out?

Balarka Sen

Let $p_i: f^{-1}(U_i) \to U_i \times \Bbb C$ as it is. No projection. Do you understand the map I wrote down?

i.e., the map $E \to B \times \Bbb C^n$?

Danu

Your map $p_i$ makes no sense.

Balarka Sen

Edited.

Danu

$U_i$ is not something in the codomain of $p_i$, if $p_i$ is supposed to map into $X\times \Bbb C$

So $f$ is essentially projection?

Balarka Sen

$f$ is the bundle map $E \to B$.

bundle projection, yes.

Danu

15:55

You want to call the "base space" the zero section, right?

Balarka Sen

Yep.

Danu

fine

What are $p_i$'s?

Can we just call the trivializations $\phi_i$? :)

Balarka Sen

I'm calling them $p_i$, but sure. They are the trivializations of $f^{-1}(U_i)$.

Danu

Yeah, okay. Got it.

So you want to say that this is an embedding or what?

Balarka Sen

Yes, that's precisely what I want to say.

Danu

15:58

Note that I wanted to know about the dual bundle

Balarka Sen

Once up embed it in the trivial bundle of rank $n$, can't you give it a Hermitian metric or something? That's how I do it for the real case.

I am not familiar with the complex story.

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