Mathematics - 2020-07-08 (page 2 of 3)

Anonymous

4:00 PM

@BalarkaSen Aha! So what exactly is the difference between $\mathbb R/\mathbb Z$ and $[0, 1)$ in terms of topology?

Balarka Sen

They are not homeomorphic. One is compact, one is not, for example.

Amongst many other things.

There is no "THE difference", because there's so many differences.

Anonymous

@BalarkaSen I see. My confusion was basically that I was thinking $\mathbb R/\mathbb Z = [0, 1)$

Thorgott

@Balarka Do you happen to know whether $R[x,xy,xy^2,...]$ is a non-finitely generated $R$-algebra for a commutative ring $R$? For $R=k$ a field, this is the standard counter-example of a non-f.g. sub-algebra of a f.g. algebra, but I don't see the more general version stated anywhere, yet can't figure out where the proof would fail if $k$ is not a field.

Balarka Sen

"=" is correct if you're thinking set-theoretically. Correct if you're thinking measure-theoretically. Wrong if you're thinking topologically.

Anonymous

So could you like explain how to think of $\mathbb R/\mathbb Z$ topologically?

Balarka Sen

4:04 PM

As a circle. I don't know what else to say. It's homeomorphic to $\{z \in \Bbb C : |z| =1\} = S^1$.

Anonymous

@BalarkaSen I mean, how do you define $\mathbb R/\mathbb Z$ in topology? I don't really know much about toplogy.

Balarka Sen

Go read Munkres.

Quotient topology

Anonymous

I was looking for a brief explanation but anyway thanks

Balarka Sen

@Thorgott At a first glance, this feels it should be true for any $R$, no?

Thorgott

I think so too

but it's only stated for $k$ a field everywhere I look

so I'm second-guessing super hard

but I can't find any step in the argument that fails for $R$ not a field either

Balarka Sen

4:08 PM

If it was finitely generated, localizing at (0) would also give you finitely generated as an algebra over R_(0) = Frac R, not?

Assume R is a domain

Thorgott

sure

Balarka Sen

But R[x, xy, xy^2, ...]_(0) = R_(0)[x, xy, xy^2, ...] = (Frac R)[x, xy, xy^2, ...]

Thorgott

localizing is exact

Balarka Sen

So you get the result from that of fields

Thorgott

yeah, that's plausible

but even non-domains shouldn't be an issue, I feel

1 hour ago, by Thorgott

Consider a field $k$ and the $k$-algebra $k[x,xy,xy^2,...]\subset k[x,y]$. If this were finitely generated (as $k$-algebra, now and in the forthcoming) by elements $f_1,...,f_n$, then it would also be finitely generated by the monomial terms appearing in $f_1,...,f_n$, because any polynomial expression in $f_1,...,f_n$ is a polynomial expression in their monomial terms. So assume it is generated by some $xy^{a_1},...,xy^{a_k}$ and pick $n>\max a_i$. Looking at any polynomial expression in $xy^{a_1},...,xy^{a_k}$, we see that any monomial term containing the power $y^n$ necessarily contains …

This only uses the representation of polynomials as linear combination as monomials (afaict) and that holds over any ring

Balarka Sen

4:11 PM

Yeah this honestly looks fine.

Fargle

@S.D. Short version is: you have the canonical map $\varphi : \Bbb R \to \Bbb R/\Bbb Z$. Open sets $U$ in $\Bbb R/\Bbb Z$ are by definition exactly those for which $\varphi^{-1}(U)$ is open in $\Bbb R$. The reason this differs from the subspace topology on $[0,1)$ is that $[0,\epsilon)$ is open in the subspace topology, but not in this quotient topology, as its preimage under the canonical map would be a bunch of half-open $\epsilon$-intervals based at the integers.

Thorgott

thanks man

felt like I was going insane

Fargle

Another way to express this is to say that any open set in $\Bbb R/\Bbb Z$ which contains $0$ must contain some numbers near $1$ as well (if you're thinking of $[0,1)$ as your representative set)---this is the sense in which $\Bbb R/\Bbb Z$ is a circle.

Thorgott

Yeah, the only difference between an interval and a circle is that, on an interval, when you keep going in one direction, you'll just end up somewhere else, whereas, on a circle, when you keep going in one direction, you end up back where you started

and that difference is precisely encoded in the way the numbers "near $1$" behave topologically

Fargle

Yep.

This is how the quotient topology on a quotient set $X/\sim$ is always defined, by the way---it is exactly the finest topology possible on $X/\sim$ such that the canonical map $X \to X/\sim$ is a continuous map.

feynhat

4:17 PM

R/Z is not a circle... its an infinite wedge of circles. /s

Balarka Sen

Question: We talked about soft and hard analysis before. What about a similar classification in algebra?

What is soft algebra? Anything which is functorial, homological algebra-esq?

Fargle

That makes some sense. Maybe you could delineate by saying soft algebra is that which primarily operates at the categorical level, hard algebra primarily "below" that level

Thorgott

I don't really like soft/hard as terminology due to the connotation, but I think, for starters, you could distinguish between categorical and explicit things

Balarka Sen

Yeah that's what I meant.

@Thorgott Agree yeah

Thorgott

like, whether you just think about maps between objects or these objects in their own right (as sets, with elements)

Balarka Sen

4:22 PM

right

Thorgott

say, homological algebra vs. local analysis of groups

Fargle

Sylow's is hard, iso thms are soft

something like that

Balarka Sen

Good examples!

Thorgott

ye, agree

Balarka Sen

Sylow proofs are indeed explicit bashing

Thorgott

4:23 PM

but I don't think this is a holistic perspective yet

need the AG people to tell what algebra actually is about

Balarka Sen

Lmao

Thorgott

the thing is that a lot of universal properties come in both forms

Balarka Sen

Regarding the soft/hard terminology: ihes.fr/~gromov/wp-content/uploads/2018/08/soft_and_hard.pdf

whoops

Thorgott

sometimes you want to think of them purely in terms of the universal property, other times in terms of their explicit construction

Fargle

"Algebra is nothing more than geometry in words; geometry is nothing more than algebra in pictures"

Thorgott

4:25 PM

what does Burnside's theorem say geometrically?

Fargle

Ask Sophie Germain

I dunno

Balarka Sen

the pq-theorem or Burnside's lemma? :P

Thorgott

she didn't know Burnside's theorem

former

Balarka Sen

Burnside's lemma lets you count number of cows from the number of their feet

Thorgott

I'm not sure if it's geometric, but Burnside's lemma is very explicitly combinatorial

I can't tell you a super intuitive interpretation of Burnside's theorem

Fargle

4:26 PM

An intuitive one, probably not

But there's probably some sort of collection of finite geometries you can cook up where Burnside's theorem says something

Balarka Sen

yeah I dunno the pq theorem

Thorgott

I mean, what does it mean for a group to be solvable geometrically

Fargle

That I wouldn't be able to tell you, but a GGT person might

Thorgott

is solvability important in GGT?

Balarka Sen

solvable finite groups don't mean anything in GGT because finite groups don't mean anything

but look up Gromov's theorem on groups of polynomial growth

Fargle

4:28 PM

well shoot

Thorgott

I can tell you why solvable groups are meaningful things to consider, but I'm not sure if there's a super satisfying intuitive view

Fargle

I just wanted to share a funny Germain quote because of your AG comment :<

Alessandro Codenotti

Geometrically solvable means that the group's balls don't grow too fast

Balarka Sen

@Alessandro How fast

I don't know this

virtual nilpotency is the same as balls growing polynomially

Alessandro Codenotti

Ah no wait I was thinking about nilpotent

I always mix them up

Balarka Sen

4:32 PM

yeah

solvable hyperbolic groups should just be virtually cyclic

Thorgott

I mean, I guess I can say that the complexity of isomorphism types of finite groups of a given order depends on the prime factorization of that order and also that solvable groups are, in a sense, made up of abelian groups, so Burnside's theorem says that groups not too complex in some sense are kinda nice in another sense, but this is massive handwaving and not meaningful to anyone who doesn't already understand it.
Also, the complexity depends, roughly, not only on the number of prime factors, but also their exponents, which can still get arbitrarily high for the fixed prime pair in Burns…

Alessandro Codenotti

@BalarkaSen Solvable groups are amenable right?

Thorgott

I guess a morally right way is to think of Burnside's theorem in terms of the reverse Hall theorem, but mathematically that's completely backwards, because you use the former to prove the latter

Alessandro Codenotti

Because they are build through extensions from abelian groups, which are amenable, and extensions of amenable groups also are

Balarka Sen

yeah because it's an extension of abelian things

Nice

Alessandro Codenotti

4:35 PM

Right, so I agree that hyperbolic+solvable should just be virtually Z

Balarka Sen

Magic proof

Thorgott

wow imagine thinking about groups that aren't finite

Fargle

this post was made by Wildberger gang

feynhat

Let H be a load of hogwash

Balarka Sen

So what is a solvable group with exponential growth

Aka what is a solvable group which is not virtually nilpotent

Mike Miller

4:41 PM

@Fargle It didn't seem like it was Germain to this conversation

Fargle

@MikeMiller I was looking for that pun, but couldn't find it. Well done

Vrouvrou

hello, if $X$ is compactly embedded in $Y$, is it compactly embedded in $X\cup Y$ ?

Balarka Sen

GIVE EXAMPLE

Where are the GROUP THEORISTS

Where are they NOW

Solvable with no nilpotent guy of finite index

Pure algebra

Give

Is lamplighter solvable?

Of course it is

LAMPLIGHTERS

@Alessandro Dude it's good to be at the top of group theory

Only pure geometry

Thorgott

what about the trivial group tho

Fargle

Pobody's gerfectly neometric

Thorgott

4:47 PM

oh wait, the trivial group is nilpotent

ok, then no finite group will do

Balarka Sen

G <- finite group

crumples G

throws it in the dustbin

That's what GGT does to finite groups

Every finite group is geometrically a point in GGT

Thorgott

yeah, out of spite

Balarka Sen

No literally

Thorgott

GGT is TOO WEAK to handle finite groups

Balarka Sen

Every finite group is quasi isometric to trivial group

Because finite group theory is trivial

Fargle

4:50 PM

does this make quotienting by finite subgroups basically useless in GGT?

Thorgott

tell me your favorite proof of feit-thompson

Balarka Sen

@Fargle yeah

it doesn't change the quasi-isometry type

$G \times (Monster group)$ is quasi isometric to $G$

Fargle

gotcha

Balarka Sen

and that product can be any extension of groups or whatever

Fargle

I mean the monster group is just made-up nonsense anyway

I don't believe in monsters

$D_3$, on the other hand, that's a real physical thing. I have that set of six troublemakers in a little box in my basement

Balarka Sen

4:55 PM

lmao

4

Thorgott

LMAO

Alessandro Codenotti

5:24 PM

finite groups are all indistinguishable from the ggt point of view

Thorgott

5:52 PM

When is Frac(R) a f.g. R-algebra? Certainly not when R is factorial and has infinitely many primes, but not sure about other cases.

Anonymous

Is there any measure in which all subsets of the reals are measurable?

Balarka Sen

No, this is Borel's theorem

Well, assuming you want [a, b] to have measure b - a

Which you probably want

Thorgott

do you have a reference for that Balarka

I remember you mentioned that a while ago, but I couldn't find it

Balarka Sen

Let me look

Anonymous

@BalarkaSen Interesting, I see

Anonymous

5:56 PM

Like Thorgott says I'd like a reference too

Anonymous

The Wikipedia page doesn't say much

Anonymous

By the way, I wonder why measure theorists don't use a definition (axioms) of measure that allows all subsets of R to be measurable

Thorgott

it does

$\mathcal{P}(\mathbb{R})$ is a perfectly fine $\sigma$-algebra

there just aren't any good enough measures on it

Anonymous

@Thorgott By good enough you mean such that [a, b] has measure b - a?

Thorgott

I mean such that [0,1] has measure 1 and translation-invariant, i.e. Lebesgue measure

cause that's what anyone who is being reasonable wants

asking for something a bit less restrictive will quickly lead to questions that are independent of ZFC, curiously

Anonymous

6:01 PM

@Thorgott Well, then let me rephrase my question as: why don't measure theorists use a definition (axioms) of measure that allows all subsets of R to be measurable with "good enough" measures? :)

Thorgott

I don't understand your question

Anonymous

Or is such a definition not possible?

Thorgott

the definitions allow this

it's just that such a measure does not exist

which is, of course, Vitali's theorem

user2103480

@BalarkaSen most aesthetic thing I've read today

Anonymous

Eh, I was talking more about changing the definition of "measure" so that such a measure exists for all subsets of $\mathbb R$. I mean it would be useful to have a notion of length for all subsets of $\mathbb R$.

anakhro

6:04 PM

Hi

Thorgott

here's something related to the problem mathoverflow.net/questions/103583/…

that doesn't make sense @S.D.

you can't just define things into existence

anakhro

@S.D. the "reasonable" definition for measure precludes this.

Thorgott

that said, there exists a notion weaker than that of a measure, which applies to all subsets of the space, a so-called outer measure

anakhro

You can weaken it, sure. But that doesn't mean what you get in the end is representative of what you want.

Anonymous

I guess I just want a reasonable notion of length for say the Vitali set, such that the length of any interval [a, b] is b-a.

Anonymous

6:07 PM

@anakhro I see, interesting

Thorgott

the Lebesgue outer measure is the closest thing to that

Anonymous

Ah, the Wikipedia page on outer measures seems good

Anonymous

> In the mathematical field of measure theory, an outer measure or exterior measure is a function defined on all subsets of a given set with values in the extended real numbers satisfying some additional technical conditions.

Balarka Sen

@Thorgott math.iisc.ernet.in/~manju/PT/Lectures-part1.pdf

Result 8. See footnote as well

Manju hasn't given a reference but I can ask him sometime

Or maybe you can figure it out somehow. How hard can it be?

Gotta get dinner, seeya

Thorgott

very hard, I assume

I think this is closely tied to some difficult set theory

cya

I found this in Bogachev: "1.12.40. Theorem. If a finite countably additive measure μ is defined
on all subsets of the set X of cardinality ℵ1 and vanishes on all singletons,
then it is identically zero."

due to Ulam

so this takes care of it if we assume CH

"A cardinal κ is called real measurable
if there exist a space of cardinality κ and a probability measure ν defined
on the family of all its subsets and vanishing on all singletons."

so the set-theoretic phrasing of this question is whether the continuum is a measurable cardinal

"1.12.44. Theorem. The supposition that measurable cardinals do not
exist is consistent with the ZFC. In addition, if either of the following assertions
is consistent with the ZFC, then so are all of them:
(i) two-valued measurable cardinals exist;
(ii) real measurable cardinals exist;
(iii) the cardinal c is real measurable;
(iv) Lebesgue measure can be extended to a measure on the σ-algebra of
all subsets in [0, 1]."

the fact that this is a theorem stated this way seems to imply that the measurability of c is open, no?

Alessandro Codenotti

6:50 PM

What's the question @Thorgott

Thorgott

if we know whether c is measurable

Alessandro Codenotti

It isn't

But it can consistently be real-valued measurable iirc

Thorgott

I mean real-valued measurable by measurable, I think

existence of probability measure on the entire powerset that vanishes on singletons

orientablesurface

is the image of a regular submanifold under a diffeomorphism a regular submanifold?

Alessandro Codenotti

Ok so, $\mathfrak c$ isn't measurable because measurable cardinals are inaccessible. It can be real-valued measurable, but that has the consistency strength of a measurable cardinal (as the theorem you're citing says), which is strictly higher than the consistency strength of ZFC

Thorgott

6:56 PM

So is result 8 stated in these notes math.iisc.ernet.in/~manju/PT/Lectures-part1.pdf wrong or am I missing something?

Alessandro Codenotti

We don't ask $\mu((a,b))=b-a$ if $\mu$ is to witness the real valued measurability of $\mathfrak c$

Ah but wait, we can construct such a measure if $\mathfrak c$ is real valued measurable

Yeah I think result 8 needs a translation invariant or something similar thrown in

But I might be missing something, those things are too tricky

Thorgott

oh, duh, you're right

that's an additional requirement

so the non-existence of such a measure is actually not a claim about the measurability of c

this is all perfectly consistent

Alessandro Codenotti

@Thorgott Right but (iv) here says that result 8 is wrong as stated, doesn't it?

Thorgott

also true

theorem 1.12.44. is pretty surprising

so I guess there is something missing after all in result 8 after all

Alessandro Codenotti

I'm not sure, I've seen it with translation invariance before, but then the proof is not hard like the footnote suggests, because you can do the usual Vitali set argument

Thorgott

7:06 PM

yeah, that's clear

if result 8 is true as stated, it would definitely be hard

Alessandro Codenotti

Right

Thorgott

but since I'm willing to believe Bogachev, this would imply that measurable cardinals don't exist, which should be open?

or is it actually undecidable

Alessandro Codenotti

Hm I'm not sure where the mistake is. But 1.12.44 and result 8 can't both be true unless I'm being stupid, and I think Bogachev is a reliable source

@Thorgott It's independent of ZFC

Actually the situation is a little subtler than that

Thorgott

independence would mean that both it and its negation are consistent with ZFC, no?

Alessandro Codenotti

Yes but that's not quite true

Thorgott

7:09 PM

but then we would have result 8 by theorem 1.12.44

Alessandro Codenotti

Let me explain, CH is independent of ZFC meaning that ZFC proves Con(ZFC) implies Con(ZFC+CH) and it also proves Con(ZFC) implies Con(ZFC+~CH)

So ZFC+CH or ZFC+~CH have the same consistency strength of ZFC, if one of those theories is consistent so are the others

But ZFC+there is a measurable cardinal has higher consistency strength, if ZFC is consistent it cannot prove Con(ZFC) implies Con(ZFC+a measurable cardinal)

So we believe that a measurable cardinal is not inconsistent with ZFC, but that cannot actually be proved in ZFC itself (unless ZFC is inconsistent of course)

Thorgott

so it's undecidable in ZFC?

Alessandro Codenotti

We believe that, but it cannot be proved in ZFC

Thorgott

can we prove that it cannot be proven within ZFC or is that a claim true in a stronger system?

Alessandro Codenotti

ZFC proves that ZFC does not prove Con(ZFC) implies Con(ZFC+a measurable)

Man I hope I'm not messing this up, it always confuses me

The thing is that if you have an inaccessible cardinal you can use it to build a model of ZFC, and then Gödel bites you

Thorgott

7:19 PM

ok, I think that makes sense

so that and theorem 1.12.44. means that ZFC proves that ZFC does not prove result 8, no?

Alessandro Codenotti

yes, if theorem 1.12.44 is true, then result 8 isn't (unless having a measurable cardinal is inconsistent with ZFC, which would be HUGE lol)

Thorgott

it would definitely be huge, but a priori not impossible if I understand this correctly, right?

Bogachev cites 1.12.44. from Jech's Set Theory btw, so that's most certainly true

Alessandro Codenotti

yes, it is possible that a measurable cardinal is inconsistent with ZFC, but measurable is very weak in the large cardinal hierarchy

Leaky Nun

@AlessandroCodenotti what do you think about this large cardinal?

Alessandro Codenotti

...

Thorgott

7:42 PM

thank you Leaky

user2103480

en.m.wikipedia.org/wiki/Thomas_Wolsey

Still my favorite large cardinal

Stolen from a quora answer

Leaky Nun

@Thorgott you're welcome

Ted Shifrin

Howdy, @Leaky, @Thor, demonic @Aless

Leaky Nun

hi

Alessandro Codenotti

Hi Ted

Thorgott

7:58 PM

hey Ted

Balarka Sen

rehi

Ted Shifrin

howdy, a @Balarka ... what is this with Burnside dividing by 4 for you?

Balarka Sen

yeah its something dumb

Thorgott

@Balarka dunno if you backread, but result 8 from the notes you cite is most likely stated incorrectly

Balarka Sen

I'd be surprised if it is incorrect

Manjunath is usually a trustworthy source for these kind of things

Thorgott

8:12 PM

it would imply that the existence of real-measurable cardinals is inconsistent with ZFC

Balarka Sen

ok lol

shoot him an email and ask

Thorgott

(which isn't a contradiction, if I understood Alessandro's explanation correctly, but it's believed to be untrue)

Balarka Sen

I very much doubt Manjunath is wrong on these sort of things; he's a big probabilist. You can ask though

Drathora

$\mathbb{R}^d$ valued functions $f$ in $L_n(\mathbb{R}^n)$ whose distributional partial derivatives of order $1$ belong to $L_n(\mathbb{R}^n)$. Furthermore, the Lorentz norm $\lvert \lvert \triangledown f \rvert \rvert_{L_{n,1}}$ is finite.

Is anyone aware of any classification of functions that satisfy this?

Or at least some of it

Mike Miller

What is $L_n$

$L^n$?

Drathora

8:17 PM

Ye

I hope anyway, otherwise I've misunderstood the paper in a big way

Thorgott

maybe I should email him

Ted Shifrin

Sounds like a Sobolev space to me, Drat.

Thorgott

I'll have to check Jech's set theory for a relevant result before to make absolutely sure tho

maybe there's a slight loophole

Balarka Sen

Seems like this has something to do with Ulam

Mike Miller

Yeah this is called $L^{n,2}$

Balarka Sen

8:21 PM

Did you check the original Ulam reference

Mike Miller

in your notation

I'd call it $L^n_2$

Thorgott

which Ulam reference?

Drathora

That's referring to the first two properties right? I assume that the Lorentz norm property isn't part of that

Thorgott

the $\aleph_1$ result seems irrelevant as long as we don't assume CH

Mike Miller

@Drathora What would you call the Lorentz norm?

Balarka Sen

8:22 PM

what is the N_1 result man

does CH imply result 8?

Mike Miller

"A function $f \in L^n$ whose distributional partial derivatives belong to $L^n$" is called something in $L^n_1$

The $L^n_1$ norm is $\|f\|_{L^n_1} = \|f\|_{L^n} + \|\nabla f\|_{L^n}$

Thorgott

CH implies result 8

Mike Miller

Oh, Lorentz norms are complicated

I don't know this stuff. Sorry for saying nonsense

Balarka Sen

@Thorgott Oh, so done

What's the big deal

Thorgott

there is no measure defined on the entire powerset of a set of cardinality $\aleph_1$ that vanishes on all singletons

Drathora

8:25 PM

Yeah I'm struggling to decipher this Lorentz norm business.

Thorgott

except the trivial one

Balarka Sen

I'm sure every sane person assumes CH

Thorgott

do they??

Balarka Sen

In any case, CH is consistent with ZFC (as in non-CH) so I don't understand "most likely result 8 is incorrect"

seems like you're doing a set theory prank to me

of course man he's a probabilist. he doesn't give a shit about CH

lol

Thorgott

I said incorrectly stated

Balarka Sen

8:29 PM

Yeah but that makes you sound like an annoying pedant

But more seriously I think he'll appreciate it if you tell him this result depends on CH.

You should email

Thorgott

cause my understanding was that it implies a set-theoretic claim that is not known, but believed to be false

I think my understanding may be flawed though

here's a question

if we had such a measure satisfying \mu[a,b]=b-a for all the intervals, then it would necessarily agree with the Lebesgue measure on all Borel-measurable sets, but does it have to agree with the Lebesgue measure on all Lebesgue-measurable sets?

Balarka Sen

Honestly don't know

Thorgott

I think not without some regularity assumptions

Balarka Sen

This stuff is too technical for me

Thorgott

cause I'm not sure if theorem 1.12.44. from Bogachev (which is the crux here) is about the Lebesgue measure on the Borel- or the Lebesgue-algebra, but I'd assume it's the latter

in which case the implications I mentioned actually don't exist

Mike Miller

8:32 PM

@BalarkaSen [stars this meme] reject modernity. embrace tradition

Thorgott

ahh I'm getting a headache from trying to decipher set theory statements

Balarka Sen

Never been much of a masochist

wouldn't know

Thorgott

I don't wanna figure out what a $\kappa$-complete nonprincipal ultrafilter is

Mike Miller

good

Drathora

So wait how would such a measure measure the set of irrationals in [a,b]?

Balarka Sen

8:39 PM

@CalvinKhor LMFAO

Drathora

What are you saying is $\mu (X)$ where $X$ is the intersection of the irrationals and [a,b]

Thorgott

the set of irrationals has full measure

Drathora

Alright good

Balarka Sen

Full Measure Analyst: Borelhood

sorry

Alessandro Codenotti

@Thorgott just a nonprincipal ultrafilter closed under intersections of cardinality $\kappa$ rather than finite ones

Balarka Sen

8:44 PM

@Alessnadro just a LKJJHJGFHLFJKHLKJFHLKJFH

thats how i read your sentence

Alessandro Codenotti

the connection with measures is that if you have an ultrafilter $U$ on $X$ you get a finitely additive two valued measure by saying that stuff if $U$ has measure $1$ and stuff outside has measure $0$

Balarka Sen

Ah

No that makes sense

Alessandro Codenotti

And if you have such a measure the set of sets of measure $1$ is an ultrafilter

So what do you get if you have an actual two valued measure rather than a finitely additive one? An ultrafilter closed under countable intersections

For a measurable cardinal we usually ask that $\kappa$ has a $\kappa$-additive measure, but that makes no difference, ZFC proves that if there is a cardinal with an $\omega$-complete nonprincipal ultrafilter, then there is also a cardinal $\kappa$ with a $\kappa$-complete nonprincipal ultrafilter

Thorgott

@BalarkaSen lol that caught me off-guard

so it seems that Jech talks about two-valued-measurability but Bogachev talks about real-valued-measurability

I'm big confuse

Alessandro Codenotti

Jech talks about real-valued measurable cardinal in a later chapter

Chapter 22

Thorgott

9:08 PM

urgh, I can't figure any of this out

I'll just ask for a reference for result 8 and if there is one, the result is true and if there is none, it isn't

Balarka Sen

CH implies result 8 man you just said it

Assume CH

Goddamn

Thorgott

why would I

Balarka Sen

why would you not

Thorgott

cause I don't like assuming significant axioms out of nowhere

Balarka Sen

I don't like you

rekt

Thorgott

9:12 PM

but do you like the fact that residue fields of finitely generated Z-algebras are necessarily finite?

Alessandro Codenotti

There's actually some strong forcing axioms that imply not CH that set theorists like a lot

Thorgott

see, that's what I'm talking about

I'm all about those forcing axioms

Balarka Sen

@Thorgott sure man. You have a finite type morphism $X \to \text{Spec}\, \Bbb Z$ of schemes, and you're saying a geometric point lifts to finitely many goemetric points along this.

Pure geometry

Thorgott

nice

orientablesurface

9:42 PM

Let $U$ be an open subset of $M$. Consider the inclusion map $i$.Then, $di_p$ is the identity on $T_pM$ for $p\in U$.

right?

Alessandro Codenotti

Yes

Thorgott

no, it's a map $T_pU\rightarrow T_pM$, so it can't be the identity

orientablesurface

$T_pU=T_pM$

tangent space is defined on germs

hence if $U$ is open in $M$ then $T_pU=T_pM$ for $p\in U$.

Ted Shifrin

Yes, orientablesurface, you're right.

Restricting the range, it's the identity map $U\to U$.

Thorgott

well, not literally equal, but if you make that identification, then yes

Ted Shifrin

9:47 PM

Yes, literally equal.

orientablesurface

What do you mean? yes, literally equal

$C_p^{\infty}U=C_p^{\infty}M$

Balarka Sen

Thorgott shut up

Ted Shifrin

If we're in $\Bbb R^n$, are we going to debate whether the tangent space at any point is "literally" $\Bbb R^n$? I am not.

votes for a Balarka

Thorgott

I'm not saying it's a problematic identification, I'm just saying they're not the same set

Ted Shifrin

I say you're wrong. We're done.

Thorgott

9:51 PM

I am not

orientablesurface

If $f: V\subseteq M \rightarrow \mathbb{R}$ is smooth, then $[(f,V)]=[(f,V\cap U)]$, no?

Balarka Sen

Yes you are

Thorgott

yes, but you're looking at equivalence classes for equivalence relations defined on different sets

so the sets of equivalence classes aren't the same, even though they're in canonical bijection

Ted Shifrin

I have no use for this sort of pedantry. Go away and do your algebra.

orientablesurface

I'm not completely sure what you mean by the first sentence, @Thorgott

elements in $C_p^{\infty}N$ are defined by $(f,U)\sim (g,V) \iff f\equiv g $ on a neighborhood $p\in W\subseteq V\cap U$

Thorgott

9:58 PM

$C_p^{\infty}(U)$ is the set of equivalence classes of smooth functions defined on a neighborhood of $p$ in $U$ by the equivalence relation you give, whereas $C_p^{\infty}(M)$ is the set of equivalence classes of smooth functions defined on a neighborhood of $p$ in $M$ by the equivalence relation you give. There are more (in the sense of proper set containment) neighborhoods of $p$ in $M$ than there are neighborhoods of $p$ in $U$ and therefore also more smooth functions defined on neighborhoods of the former type than smooth functions defined on neighborhoods of the latter type. So you are…

(all assuming $U$ is a proper subset of $M$, of course, otherwise they are literally equal and trivially so)

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