The h Bar

yuggib

06:26

@0celo7 what's the prob?

finite dim vec spaces are boring, but infinite dimensional ones are fun

0celo7

One sec

@yuggib ok. so in the finite dimensional case, suppose we have $f:V\to W$, a homomorphism

Then $\mathrm{dim}\,V=\mathrm{dim}\,\mathrm{im}\,V+\mathrm{dim}\,\mathrm{ker}\,V$

consequently, $V\cong \mathrm{im}\,f\oplus\mathrm{ker}\,f$, right?

In the infinite-dimensional case I know the dimension thing is too much to hope for/wrong

yuggib

yep

0celo7

But is the isomorphism still good?

yuggib

I think so, yes

0celo7

I know that $\mathrm{im}\, f\cong V/\mathrm{ker}\, f$

Is it true that $V/\mathrm{ker}f\oplus\mathrm{ker}\,f\cong V$?

I has to be true in finite dimensions

But that proof uses a basis

yuggib

06:44

well, $V/\text{ker} f=\{[x], x\in V\}$ and $[x]=\{v\in V, x-v\in\text{ker}f\}$

0celo7

Yes.

yuggib

so you can map $([x],k)\in V/\text{ker}f \oplus \text{ker}f$ in $V\ni v=x+k$

0celo7

That depends highly on the representative

yuggib

and that mapping is one-to-one and surjective

0celo7

I don't think that map is one-to-one.

yuggib

06:45

why not?

0celo7

Because what if you pick $y\in[x]$

Same equivalence class, different representative

You then get $v=y+k\ne x+k$.

It's not even a function

yuggib

but you can't pick a representative

you pick the equivalence class, and fix a representative for each class

0celo7

What is going from $[x]$ to $x$ besides "picking a representative"

Oh, that's some bullshit

Some bullshit I had on my paper two hours ago

...do you need AoC for this?

yuggib

probably yes

0celo7

I reject the AoC, we've been over this.

@yuggib you probably need a choice function on the set of equivalence classes, right

yuggib

06:49

it's hard to do infinite dimensional shit (with arbitrary dimension) without C

@0celo7 right

0celo7

FUCK

Are you telling me cohomology requires C?

yuggib

almost surely does

0celo7

yuggib

also, have you ever used Zorn?

0celo7

Not without heavy protest.

I think the first isomorphism there requires choice

yuggib

06:51

maybe, I don't know what $H^q$ stands for...

0celo7

$q$-th de Rham group

yuggib

ah, that I don't know at all

0celo7

can you use forcing or whatever to show that it's equivalent?

You've read Jech, right

yuggib

@0celo7 what is equivalent to what?

C to Zorn?

0celo7

@yuggib the isomorphism thing above and C.

yuggib

06:52

Well, not exactly

0celo7

$V=ker\,f\oplus im\,f$

yuggib

as long as you need an arbitrary choice function, then you need choice (I would not be able to prove the equivalence however)

you should ask the logicians on math overflow

it may be that for such result full C is not necessary

0celo7

I doubt this is MO material

yuggib

it may suffice to use DC

0celo7

and I don't know how to ask on there

yuggib

06:55

@0celo7 maybe it is not, you can try on math.SE

0celo7

ok, question title?

I'm always bad with those

yuggib

@0celo7 "Equivalence of vector space decomposition and axiom of choice" ?

0celo7

@yuggib what's the inverse of that map above?

why is it surjective?

@yuggib this one

yuggib

it is surjective because every $v\in V$ belongs to an equivalence class ($[v]$), and therefore it can be written as the fixed representative $x$ plus an object of the kernel (if $x=v$, then $k=0$)

in addition, the decomposition is unique because of the properties of the linear space

once the representative is fixed

the choice function fixes a representative for each equivalence class

(there you need full C in my proof, but maybe you can have a proof using a weaker form of C)

((I doubt you can))

0celo7

0

Q: Is a vector space isomorphic to the kernel $\oplus$ image of a map out of it?

Let $f:V\to W$ be a linear map of finite-dimensional vector spaces. By simply counting dimensions and using rank-nullity, it is clear that $V\cong \mathrm{im}\,f\oplus\mathrm{ker}\,f$. I want to know if this holds on general vector spaces. In fact, the first isomorphism theorem tells us that $\m...

yuggib

07:09

let's see ;-)

0celo7

If it requires Choice I cannot continue reading

yuggib

why?

0celo7

I do not accept well-ordering

yuggib

"the axiom of choice is obviously true, the well-ordering theorem is obviously false, and who can tell about Zorn's lemma"

is that your position?

0celo7

yes

I'm going to sleep

cheerio

thanks for the help

yuggib

07:15

no prob

good night

see ya later

yuggib

07:55

@0celo7 Ha! I exactly thought of Asaf Karagila for someone that could answer your question about equivalence of axioms... ;-)

DanielSank

10:01

I'm back home!

Danu

@DanielSank from where?

DanielSank

10:18

@Danu Japan and Hawaii.

user116211

@DanielSank Ah! @yuggib also went there for a scientific visit... did you two meet each other ;P

Danu

10:41

Fancy

DanielSank

@MAFIA36790 Nope.

Danu

@0celo7 math.stackexchange.com/questions/1869041/…

Physics Meta

11:23

0

Q: August 23 Physics AMA with Daniel Sank: question pool

I am Daniel Sank, guest for the Physics Stack Exchange AMA taking place in the hbar chat room on August 23 at 16:00 UTC [a]. I am looking forward to discussing my experience in the USA physics education system from lower education schools up through the PhD program, my research in experimental qu...

discussion

3

Danu

Nice!

I corrected the spelling of Super Smash Bros. Melee ;)

0celo7

12:08

@Danu hmm?

John Rennie

14:08

Still no dark matter :-(

yuggib

@0celo7 satisfied with your answer? :-P

0celo7

14:33

@yuggib not really

I'd like a word from @ACuriousMind

Maybe I'm missing something obvious about exact sequences

yuggib

@0celo7 I am pretty sure essentially all of mathematics nowadays uses C without caring so much

0celo7

I have big problems with it

yuggib

anyways, if you restrict to countably-dimensional spaces (or functions with countable dimensional images), then you don't need full choice but just CC (countable choice)

@0celo7 why?

0celo7

@yuggib Banach Tarski, well ordering

yuggib

so? have you seen the disasters without choice?

0celo7

14:38

no

ACuriousMind

@0celo7 A word on what?

0celo7

Does the first $\simeq$ require AoC?

ACuriousMind

If the dimension of $H^q$ is not bounded, sure

0celo7

Nooooo

ACuriousMind

BTW, AsafKaragila is far more qualified than I to tell you that :P

0celo7

14:40

I figured there might be some exact sequence thing I'm missing

ACuriousMind

You can't do algebra without choice

0celo7

I know 0 algebra after all

ACuriousMind

Zorn's lemma is everywhere

0celo7

Although I feel somewhat comforted I was able to figure out myself that you need choice for that

@ACuriousMind whut

@ACuriousMind Who?

ACuriousMind

@0celo7 The dude who answered your MSE question.

vzn

14:41

@DanielSank awesome whered you go in hawaii/ japan? wanna hear details sometime :) ... reading this more lately luv it hawaiimagazine.com

0celo7

@ACuriousMind who is he?

ACuriousMind

He also has a nice answer about the horrors that ensue when you reject choice.

0celo7

Like what

I don't know if I can continue with this book

Forcing me to use AoC is violence

yuggib

then you should stop reading any mathematics book that goes beyond calculus :-P (and even for calculus it is not obvious you can do without choice ;-P)

ACuriousMind

16

A: Soft Question: Why does the Axiom of Choice lead to the weirdest constructions?

This is a very natural question, even if a little subjective (due to the inherent subjectivity of what does it mean "weird"). But the answer is that the axiom of choice is not the cause of weirdness. The axiom of choice is just a tool with which we can prove there is some uniform "mess" througho...

yuggib

14:45

....and nobody forces you, you're very welcome to find alternative proofs without choice ;-)

0celo7

@yuggib What are some common things that go wrong when Choice is gone

forget infinite dim vec space stuff

I'm not sure I believe in those anyway

ACuriousMind

@0celo7 The reals then can have more disjoint subsets than they have elements.

yuggib

^

ACuriousMind

And all sorts of things can go wrong in measure theory, read the post I linked. You might even destroy the Lebesgue measure and hence integration!

yuggib

@0celo7 you should not believe, but be aware of the things that can be proved/disproved by means of the chosen axioms

then choose the ones that fit better your purposes

usually, choice is a desirable axiom to prove results

0celo7

14:48

@ACuriousMind so?

ACuriousMind

@0celo7 How can you find Banach-Tarski bizarre and not that?

0celo7

Not sure what it means

vzn

this is all very interesting & not an expert on this but afaik some mathematicians would argue the idea that something goes "wrong" is incorrect, ie its just a different universe so to speak. but maybe its more outlandish from a real world/ physics pov, dont know

0celo7

Up until now, I was enjoying this book...

ACuriousMind

@0celo7 It means that you can find non-empty sets $U_i \subset \mathbb{R}$ such that $U_i\cap U_j = \emptyset\forall i\neq j$ such that the set of all $U_i$ has greater cardinality than $\mathbb{R}$.

I.e. you can divide the reals into more parts than they have elements.

Which, to me, is far more obviously absurd than Banach-Tarski

0celo7

14:52

I cannot tell you why, but using AoC gives me a pit in my stomach

@ACuriousMind ...what

Wait, the set of them has greater cardinality?

That's not too far-fetched...oh, disjoint

That's fucked

yuggib

@0celo7 not looking good without C huh?

vzn

@0celo7 lol, dramatic

to me banach-tarski so called "paradox" is a paradox in topology but topology itself is an abstraction of the real world...

0celo7

What do I do?

vzn

there is a lot of popsci writing on AoC... seems to inspire awe/ wonder sort of like math version of "ripleys believe it or not"... not sure what to make of all of it...

Slereah

Pop sci loves weird unintuitive stuff

Nobody is making a pop sci article on the axiom of union

vzn

14:58

am guessing physicist pov on AoC may be different than mathematician pov. wonder if any top physicists have expressed opinions on it... (surely?)

Slereah

physicists don't really care about the AoC

0celo7

HE uses Zorn, so I stopped reading

vzn

@Slereah the physicists in here are expressing strong opinions lol... does it affect/ impact physics-math/ math-physics? dont know. does physics require it?

Slereah

well some theorems used in physics rely on the AoC

but that's about the extent of it

yuggib

@0celo7 what's the big fuss about C, I think that the only thing that matters is to be aware of which axioms you need and you're using

vzn

15:01

@Slereah so are they major thms? what is physics without AoC? does it collapse somehow? wondering

ACuriousMind

The whole functional analysis needed for QM collapses if you take choice away.

You might also get trouble with classical functionals

Slereah

Well, you can probably rebuild QM without the AoC

But

That's a pretty major undertaking

yuggib

anyways, this is always a matter of our way of describing nature

vzn

@ACuriousMind ok, really? have any "major authorities" pointed this out in papers/ books? re one of your own exprs [citation needed]

yuggib

@Slereah I doubt it much

0celo7

15:02

@yuggib it's a gut feeling, I dunno

ACuriousMind

@vzn I don't think so. Physicists rarely care about the fully rigorous math to begin with, let alone about the intricacies of set theory underlying it

Slereah

I know that redoing QM with different axioms has been tried before

With quantum logic

It never really went anywhere

yuggib

@0celo7 it's not a suitable feeling neither for a mathematician nor for an engineer...you should get rid of it ;-P

0celo7

> Obama's half-brother says he's voting for Donald Trump

Haha

@yuggib I am not a mathematician nor an engineer

yuggib

@Slereah the algebra of observables of CCR are non-separable always

vzn

15:04

@yuggib think its quite plausible for an engr to question/ be suspicious of the whole thing

0celo7

@ACuriousMind can you please respond to my Kunneth question?

yuggib

@vzn an engineer wants things to work, it does not matter why ;-P

@0celo7 well, those are the two things you are majorly interested in

ACuriousMind

@0celo7 You honestly waited this long for me to respond instead of simply googling whether that's a general theorem in homology?!

0celo7

@yuggib can the image of a map from an infinite dimensional vector space be finite?

ACuriousMind

@0celo7 THink a bit about that question.

yuggib

15:08

yeah...it's trivial

0celo7

@ACuriousMind no, I'm asking specifically about de Rham

@yuggib correct, 0

ACuriousMind

@0celo7 That doesn't mean anything, since deRham is the same as every other cohomology theory

0celo7

cohomology, not homology

@ACuriousMind well that's news to me, sorry!

ACuriousMind

...I told you about deRham's theorem, that's not news to you!

0celo7

@ACuriousMind I don't remember that

ACuriousMind

15:09

It was yesterday

0celo7

and I don't know what "simplicial cohomology" is, anyway

vzn

@yuggib few engrs are interested in GR/ very abstract math etc...

0celo7

I'm not interested in very abstract math

@yuggib so the answer is yes, right?

yuggib

@0celo7 yes

0celo7

Sigh...@ACuriousMind after the AoC thing on page 43, why do those end spaces being finite dimensional imply the middle one is too?

yuggib

15:12

but if you restrict to homomorphisms, then the image is a vector space as well

0celo7

what

yuggib

and therefore it is finite only if it is just zero

0celo7

o.o

yuggib

finite dimensional very easily

0celo7

I was considering only linear maps

yuggib

15:13

by linearity the image is a vector space as well

vzn

@0celo7 makes no sense (eg wrt this chat transcript etc)

0celo7

so you can't linearly map an infinite-dimensional space to $\Bbb R$?

yuggib

@0celo7 yes of course

0celo7

@yuggib I know what a homomorphism is

@yuggib O.o

yuggib

but not to $\{1,2,3\}$

0celo7

15:14

Help me

what is happening

ACuriousMind

lol, @yuggib, I don't think he got your distinction between "finite" and "finite-dimensional"

yuggib

@ACuriousMind they're pretty different things

ACuriousMind

I know

0celo7

what

can you map $\Bbb R$ linearly to an infinite dimensional vector space?

ACuriousMind

@0celo7 It's obvious. Stare at it until you get it.

0celo7

15:16

@ACuriousMind I don't know enough linear algebra

ACuriousMind

You do

0celo7

I don't know if $\mathrm{im}\,d^*$ and $\mathrm{im}\,r$ have to be finite-dimensional or not

ACuriousMind

You should be able to deduce that

0celo7

From?

Suppose $H^{q-1}=\Bbb R$

ACuriousMind

I refuse to believe that you are able to understand advanced Riemannian geometry but can't figure out how to show that the image of a finite-dimensional vector space is finite-dimensional.

0celo7

15:18

I don't see why you can't map that into some huge Hilbert space or something

ACuriousMind

...you need to recall the definition of image.

0celo7

$\mathrm{im}\,d^*=\{x\in H^q\mid d^*v=x,v\in H^{q-1}\}$

@ACuriousMind I do not understand this...

I can't even find it on google

@ACuriousMind Can you give a hint?

And even if that is true, I don't see what to do with $\mathrm{im}\,r$

Maybe I have to take a basis?

@ACuriousMind If $e_i$ is a basis for $H^{q-1}$, do we have $\mathrm{im}\,d^*=\mathrm{span}\, d^*e_i$?

ACuriousMind

@0celo7 (Dis)prove it yourself!

0celo7

what?

I don't know enough linear algebra...

it seems reasonable

ACuriousMind

Linear algebra is not that hard, there is not much to know

0celo7

15:28

it is for me, clearly

ACuriousMind

If you know what a linear map and a basis is, you can (dis)prove that statement.

0celo7

I do not understand it

I know what a linear map and a basis are

Let $v=v^ie_i$. Then $d^*v=v^id^*e_i\in \mathrm{span}$

uh

what am I doing

@ACuriousMind Inclusion both ways works, they are equal

so the dimension of the image is at most the dimension of the source space

i.e. finite

@ACuriousMind ok but I really don't understand $\mathrm{im}\, r$

can you not map an infinite dimensional space to a finite dimensional one?

that totally has to be possible

ACuriousMind

Sure you can

I don't know what your problem is

0celo7

uhh...projection onto a finite-dim subspace?

That's a linear map from an infinite dimensional space and has finite dimensional image

@ACuriousMind Let $f:V\to W$ be a linear map with $V$ infinite-dimensional

if we let $fe_i=0$ on all but a finite number of basis vectors, then the image will totally be finite dimensional

so finite-dimensional image does not imply finite-dimensionality of the source

so how does the Mayer-Vietoris thing work?

it has to be wrong...

ACuriousMind

$H^q = \mathrm{im}(d)\oplus\mathrm{im}(r)$. Why do you need that the source is finite-dimensional?

0celo7

15:39

Jesus

thanks

I told you I don't understand linear algebra

ACuriousMind

@0celo7 I think you just need to get a bit more used to it

0celo7

what is there to get used to

I stared at that for an hour and got nothing

0celo7

15:59

@ACuriousMind In a good cover, is each element of the cover supposed to be diff to $\Bbb R^n$?

BT talks about intersections, but what about just one of them

ACuriousMind

@0celo7 That's an intersection of an element of the cover with itself, so it's a special case of the intersections being contracible

0celo7

@ACuriousMind exactly my thought, thanks

oh god more linear algebra

@ACuriousMind What do they mean by "the wedge product is an antiderivation"

ACuriousMind

They mean it's an anti-derivation

0celo7

The wedge?

Certainly not

it's the exterior derivative that is the antiderivation

ACuriousMind

So? There can be more than one anti-derivation

0celo7

16:07

I think what we need is that $[\omega]\wedge[\tau]=[\omega\wedge\tau]$

@ACuriousMind what are you deriving though

ACuriousMind

Oh dear

0celo7

how does $\wedge$ fulfill a Leibniz rule

@ACuriousMind ?

ACuriousMind

It's probably some form of typo, they mean that d is an anti-derivation w.r.t. the wedge product.

0celo7

@ACuriousMind ok, but why does that mean the wedge decends to cohomology?

ACuriousMind

If you don't see it, prove it

0celo7

16:12

prove what? the $[\omega]\wedge[\tau]$ thing, right?

ACuriousMind

Yes, you have to show that $\omega\wedge \tau$ is closed for closed $\omega,\tau$ and that its class does not depends on which representants $\omega$ and $\tau$ you chose.

0celo7

right

Trivial.

now that's pretty swag

Slereah

that's a lot of things

0celo7

17:01

@ACuriousMind Is the intersection of a finite number of open balls homeomorphic to an open ball?

assuming it's nonempty

@ACuriousMind Is a contractible open set homeomorphic to an open ball?

(all of this is in $\Bbb R^n$)

ACuriousMind

You can easily google that!

0celo7

@ACuriousMind not the first one, apparently

user54412

@ACuriousMind Your fault for modeling physics with uncountable sets in the first place :p

0celo7

fuck, it's false

@ACuriousMind Do you know anything about "good covers"

Bernard Meurer

17:25

Hello friends

lucas

Hi

Bernard Meurer

@0celo7 T-Pain sings really well

His lyrics are kinda dumb though

0celo7

0

Q: Why are geodesically convex sets diffeomorphic to $\Bbb R^n$?

In the construction of a good cover for a manifold $M^n$, Bott & Tu use the fact that each point in $M$ is contained in a geodesically convex set (after picking a Riemannian metric). They then claim that the intersection of geodesically convex sets is a geodesically convex set, and that makes sen...

differential-geometry algebraic-topology differential-topology riemannian-geometry

@BernardMeurer everyone has seen that

Bernard Meurer

@0celo7 Still cool

0celo7

@ACuriousMind Ok, are convex open sets diffeomorphic to balls?

10

Q: A star shaped open set in $\mathbb{R}^n$ is diffeomorphic to $\mathbb{R}^n$

After googling this fact (which is used in Bott and Tu to show the existence of good covers of manifolds) I've gotten the impression this is somewhat difficult to prove. But I also came across this homework problem (problem 0) with a hint: http://www.math.columbia.edu/~thaddeus/geometry/hw10.pd...

geometry differential-geometry differential-topology

Jesus christ

Bernard Meurer

17:41

@0celo7 Why are closed sets not the opposite of open sets?

0celo7

because Jews invented topology @BernardMeurer

whew, that proof is killer

Slereah

Wasn't topology invented by Euler

He was a german fellow

A lot of jews worked on QM, though

0celo7

Topology was invented by Poincare.

Slereah

That's why it was considered a JEWISH SCIENCE

Along with relativity

0celo7

Was Heisenberg a jew

@ACuriousMind Does star shaped imply bounded?

Slereah

17:55

He wasn't

But

The nazis called him a "white jew"

Due to his very ungerman physics

This didn't last too long, though, since they soon realized that they needed QM for their nuclear program

0celo7

Why did they use Jewish science though

Slereah

A lot of people in academia were jewish

Bernard Meurer

@0celo7 What's a clopen set?

Slereah

clopen set is both open and closed

Bernard Meurer

@Slereah Whats the definition of open and closed?

Slereah

17:59

Depends on the topology

Bernard Meurer

Oh god

Transcript for