The h Bar

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

0

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...

7:09 AM

let's see ;-)

If it requires Choice I cannot continue reading

why?

I do not accept well-ordering

"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?

yes

I'm going to sleep

cheerio

thanks for the help

7:15 AM

no prob

good night

see ya later

7:55 AM

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

DanielSank

10:01 AM

I'm back home!

@DanielSank from where?

DanielSank

10:18 AM

@Danu Japan and Hawaii.

user116211

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

10:41 AM

Fancy

DanielSank

@MAFIA36790 Nope.

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

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

Physics Meta

11:23 AM

0

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

Nice!

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

12:08 PM

@Danu hmm?

John Rennie

2:08 PM

Still no dark matter :-(

@0celo7 satisfied with your answer? :-P

2:33 PM

@yuggib not really

I'd like a word from @ACuriousMind

Maybe I'm missing something obvious about exact sequences

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

I have big problems with it

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?

@yuggib Banach Tarski, well ordering

so? have you seen the disasters without choice?

2:38 PM

no

@0celo7 A word on what?

Does the first $\simeq$ require AoC?

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

Nooooo

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

2:40 PM

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

You can't do algebra without choice

I know 0 algebra after all

Zorn's lemma is everywhere

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

@ACuriousMind whut

@ACuriousMind Who?

@0celo7 The dude who answered your MSE question.

2:41 PM

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

@ACuriousMind who is he?

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

Like what

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

Forcing me to use AoC is violence

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)

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

16

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...

2:45 PM

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

@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

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

^

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!

@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

2:48 PM

@ACuriousMind so?

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

Not sure what it means

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

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

@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

2:52 PM

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

@0celo7 not looking good without C huh?

@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...

What do I do?

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...

Pop sci loves weird unintuitive stuff

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

2:58 PM

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

physicists don't really care about the AoC

HE uses Zorn, so I stopped reading

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

well some theorems used in physics rely on the AoC

but that's about the extent of it

@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

3:01 PM

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

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

You might also get trouble with classical functionals

Well, you can probably rebuild QM without the AoC

But

That's a pretty major undertaking

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

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

@Slereah I doubt it much

3:02 PM

@yuggib it's a gut feeling, I dunno

@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

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

With quantum logic

It never really went anywhere

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

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

Haha

@yuggib I am not a mathematician nor an engineer

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

3:04 PM

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

@ACuriousMind can you please respond to my Kunneth question?

@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

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

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

@0celo7 THink a bit about that question.

3:08 PM

yeah...it's trivial

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

@yuggib correct, 0

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

cohomology, not homology

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

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

@ACuriousMind I don't remember that

3:09 PM

It was yesterday

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

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

I'm not interested in very abstract math

@yuggib so the answer is yes, right?

@0celo7 yes

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

3:12 PM

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

what

and therefore it is finite only if it is just zero

o.o

finite dimensional very easily

I was considering only linear maps

3:13 PM

by linearity the image is a vector space as well

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

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

@0celo7 yes of course

@yuggib I know what a homomorphism is

@yuggib O.o

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

3:14 PM

Help me

what is happening

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

@ACuriousMind they're pretty different things

I know

what

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

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

3:16 PM

@ACuriousMind I don't know enough linear algebra

You do

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

You should be able to deduce that

From?

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

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.

3:18 PM

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

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

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

@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 $\mathrm{im}\,d^*=\mathrm{span}\, d^*e_i$?

@0celo7 (Dis)prove it yourself!

what?

I don't know enough linear algebra...

it seems reasonable

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

3:28 PM

it is for me, clearly

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

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

Sure you can

I don't know what your problem is

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...

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

3:39 PM

Jesus

thanks

I told you I don't understand linear algebra

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

what is there to get used to

I stared at that for an hour and got nothing

3:59 PM

@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

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

@ACuriousMind exactly my thought, thanks

oh god more linear algebra

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

They mean it's an anti-derivation

The wedge?

Certainly not

it's the exterior derivative that is the antiderivation

So? There can be more than one anti-derivation

4:07 PM

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

@ACuriousMind what are you deriving though

Oh dear

how does $\wedge$ fulfill a Leibniz rule

@ACuriousMind ?

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

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

If you don't see it, prove it

4:12 PM

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

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.

right

Trivial.

now that's pretty swag

that's a lot of things

5:01 PM

@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 $\Bbb R^n$)

You can easily google that!

@ACuriousMind not the first one, apparently

user54412

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

fuck, it's false

@ACuriousMind Do you know anything about "good covers"

Bernard Meurer

5:25 PM

Hello friends

lucas

Hi

Bernard Meurer

@0celo7 T-Pain sings really well

His lyrics are kinda dumb though

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

0

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