Mathematics - 2024-01-04

Jakobian

00:50

@ShaVuklia answer should depend on how you define $|\mu|$

Sha Vuklia

Jakobian

By Hahn decomposition theorem there exist disjoint positive and negative set $P, N$ for $\mu$ with $P\cup N = X$. Let $A = P\cap \{f\geq 0\}\cup N\cap \{f < 0\}$ and $B = X\setminus A$, then $|h\mu(A)|+|h\mu(B)| = \int |h|\mathrm{d}|\mu|$

@ShaVuklia

the justification for last equality might depend on how you defined integrals over signed measures

Sha Vuklia

actually $\mu$ is complex

but I will first see if I understand your argument for signed $\mu$

Did you mean $h$ instead of $f$?

Jakobian

yes

Sha Vuklia

01:08

aha okay, that is smart

We can always refine our partition by intersecting with $A$ and $B$

the case where $\mu$ is positive is the most pressing case for me, so for now I'll consider myself content

thanks for your help!

now I feel less guilty to go sleep

Ted Shifrin

Sleep well!

Sha Vuklia

wait, actually... $h$ has complex values as well. I'll continue thinking about it tomorrow. thanks Ted

Jakobian

01:31

@ShaVuklia First write $\mu = \alpha\cdot |\mu|$ for some function $\alpha$ with $|\alpha| = 1$, $|\mu|$-a.e., then use Radon-Nikodym theorem

$\mu << |\mu|$ so that $\alpha = \frac{d\mu}{d|\mu|}$ exists

one potato two potato

06:14

21 hours ago, by one potato two potato

@Jakobian @AlessandroCodenotti If $\alpha$ is an ordinal and $n\in\Bbb Z_+$, then is
$$(\omega^\alpha\cdot n+1)' = \begin{cases} \omega^\beta\cdot n+1 & \text{if }\beta+1 = \alpha\\
\omega^\alpha\cdot n+1 & \text{if }\alpha\text{ is a limit ordinal}\end{cases}$$
true?

@Jakobian What if I change $=$ to $\cong$ (homeomorphic)? Still no?

Jakobian

07:02

If $\alpha = 1$ and $n = 1$, then $(\omega^\alpha\cdot n+1)'$ has one element, but $\omega^\beta\cdot n + 1$ has two

one potato two potato

07:21

@Jakobian $(\omega+1)' =\{\omega\}$ so only one element but $1 = \{0,\{0\}\}$ so two right?

it should be $\emptyset,\{\emptyset\}$

Jakobian

@onepotatotwopotato You write two in a really weird way

First time I see someone write two with $1$

one potato two potato

I mean $\beta = 0$ if $\alpha = 1$ so $\omega^\beta\cdot n +1 =1$

Jakobian

no, $\omega^\beta\cdot n + 1 = 1+1 = 2$

one potato two potato

oh

Jakobian

$2 = \{0, 1\}$

one potato two potato

07:31

$2$ has two consists of $0$ and $1$

alright thanks

one potato two potato

11:20

@Jakobian If $\alpha\neq 1$ then no?

I guess $\alpha =1$ can be treated as a "singular" case

If $\alpha =2$ and $n=1$ then $(\omega^2+1)' = \{\omega^2,\omega,\omega\cdot 2,\ldots\}$ and $\beta=1$ so $\omega+1$

Sha Vuklia

11:37

@Jakobian Ah, nice, thanks. I should really remember that result when working with complex measures and absolute values

Sha Vuklia

13:11

btw @Jakobian I've been struggling to finish the proof of $\Vert h\mu\Vert=\int\vert h\vert d\mu$ for complex $h$ (and positive $\mu$), because the square root in the modulus makes it not possible for me to use your initial argument for real $h$.

oh waitttt

nvm actually

oh, maybe I can write $h(x)=e^{i\theta(x)}r(x)$, and then absorb $e^{i\theta(x)}$ in the measure

I could partition the plane such that $\theta$ is well-defined (inclusion a block where $h=0$)

so that $\theta$ poses no problems (inclusing a block where $h=0$)*

I think this works!

Sha Vuklia

13:46

oh huh, I realise now I should have been able to show the identity at once, since I know that $\vert f m\vert=\vert f\vert m$

So we have $\Vert fm\Vert=\vert f m\vert(X)=\int \vert f\vert dm$, since $\vert f\vert$ is the Radon-Nikodym derivative of $\vert f m\vert$

I hadn't properly realized what $\vert fm\vert=\vert f\vert m$ actually means. I guess I'm glad I've spent some time on this (and that you were there yesterday, Jacobian, that really helped), because my understanding of Radon-Nikodym is clearly not on a workable level yet

one potato two potato

14:09

Let $M$ be a closed orientable 3-manifold and suppose there is a fibration $K\to M\xrightarrow{f}S^1$ and consider a closed 1-form $u = [f^*(d\theta)]\in H^1(M;\Bbb Z)\simeq Hom(H_1(M),\Bbb Z)$ such that $u(\pi_1(M)) = \pi_1(S^1)$. If $S_0$ is a component of $S\subset M$ in which represents $u$, then $\pi_1(S_0)\in\ker u$ @Thorgott ?

Thorgott

14:25

1. This seems to have nothing to do with fibrations. It is just a question about continuous maps $M\rightarrow S^1$.
2. Such a closed 1-form need not exist, for the record.
3. I find the notation here messy. The morphism $H_1(M)\rightarrow\mathbb{Z}$ induced by the 1-cohomology class $u$ that classifies $f$ is simply equal to $H_1(f)\colon H_1(M)\rightarrow H_1(S^1)=\mathbb{Z}$.
4. The Poincaré dual of $u$ is represented by a generic fiber of $f$ (after homotoping it to be a smooth map wlog), so $f$ restricted to this submanifold (or any component thereof) is constant, hence induces $0$ on …

To clarify 2., you always have the cohomology class $u\in H^1(M;\mathbb{Z})$ classifying $f$ and what I said in 3. and 4. always holds, but the condition "$u(\pi_1(M))=\pi_1(S^1)$" as you put it is equivalent to saying that $H_1(f)$ is surjective, which simply may or may not be true.

one potato two potato

Is 3. something straightforward?

Thorgott

Yeah, this is naturality of the Hurewicz homomorphism $H^1(-;\mathbb{Z})\rightarrow\mathrm{Hom}(H_1(-;\mathbb{Z}),\mathbb{Z})$ together with the fact that the canonical class in $H^1(S^1;\mathbb{Z})$ (the one you call $d\theta$) corresponds to the "identity" $H_1(S^1;\mathbb{Z})\rightarrow\mathbb{Z}$ (this is either tautological or easy to see, depending on how exactly you define that class)

one potato two potato

Ah, yeah I was thinking that naturality. Is it true that every representative of $u$ can be realized as a fiber of $f$? I think you implicitly used this in 4.

Thorgott

14:49

ah no, I just assumed you were thinking about a generic fiber of $f$

certainly, not every smooth submanifold representing a Poincaré dual of $u$ is realized as a fiber of $f$ (you can always wiggle a bit), but I think $f$ restricted to such a representative should be nullhomotopic anyway

one potato two potato

why that's true?

monoidaltransform

If $X\rightarrow Y$ is a fiber bundle, $X$ paracompact and $Y$ paracompact, is it a locally trivial fibration?

nickbros123

15:15

is $\vert P(\mathbb{N}) \vert=\vert \mathbb{R} \vert$

Sahaj

15:32

I upvoted a comment on accident yesterday and only saw now. Anyway I can remove it?

Thorgott

15:46

@onepotatotwopotato Hmm, I'm not so sure anymore. The claim is equivalent to $i^{\ast}(u)=0$, where $i\colon S\rightarrow M$ is the inclusion, and we can calculate $i_{\ast}(i^{\ast}(u)\cap [S])=u\cap i_{\ast}[S]=u\cap(u\cap [M])=u^2\cap [M]=0$ (geometrically: $S$ intersects a generic fiber $f^{-1}(x)$ transversely in $(fi)^{-1}(x)$, so this intersection is Poincaré dual to $i^{\ast}u$ in $S$ and Poincaré dual to $D[S]\cup D[f^{-1}(x)]=u^2$ in $M$), but $i_{\ast}$ need not be injective.

Alessandro Codenotti

@nickbros123 yes

Thorgott

I recommend trying some examples and see what happens.

@monoidaltransform you only need $Y$ to be paracompact (Hausdorff included) and yes

monoidaltransform

Thanks!

psie

16:18

Consider the Fourier series of a $2\pi$ periodic function $f$, that is $f(x)\sim\sum_{k=-\infty}^\infty c_k e^{ikx}$, and suppose it converges for some $x=t$. Does then $\sum_{k=-\infty}^\infty c_{-k} e^{-ikt}$ converge to the same value or is this considered to be a rearrangement?

nickbros123

16:32

@AlessandroCodenotti do you have a link to a proof? im not able to find material on thi

this

Alessandro Codenotti

37

Q: The set of real numbers and power set of the natural numbers

I have learnt that the cardinality of the power set of the natural numbers is equal to the cardinality of the real numbers. What is the function that gives the one-to-one correspondence between these two sets? I have also learnt that there exists no set whose cardinality is strictly between the ...

nickbros123

thankss

one potato two potato

$Hom(H^1(M;\Bbb Z);\Bbb Z)\otimes\Bbb R$ is somehow isomorphic to a $Hom(H^1(M;\Bbb R);\Bbb R)$? ($M$ is just a manifold)

psie

17:01

@psie I guess it is a rearrangement, but one that is guaranteed to exist, because the partial sums are equal...let me know if you think otherwise :)

Xander Henderson

I read a question this morning on snowflaked metric spaces. Snowflaking is a process which is near and dear to my heart, and I recognized most of the references in the question---indeed they are all on my bookshelf (or on my hard drive, in the case of one of the sets of cited lecture notes). I was kind of excited, and started compiling notes for an answer.

Only to realize, while Googling, that 99% of the question was an unattributed copy-paste of an answer I wrote a couple of years ago. The only new text was something to the effect of "Can this be made more rigorous?"

leslie townes

xander: well, can it?

ephe

17:17

What is the reduced row echelon form of $ \begin{bmatrix}
1 & 1 & 0 \\
1 & 0 & 1 \\
0 & 1 & 1
\end{bmatrix} $ where the entries are from a field of characteristic 2? I've tried row reduction but after a while it is just repeating. How can this be solved?

leslie townes

ephe: replace R2 with R2+R1 to get 110/011/011, replace R3 with R2+R3 to get 110/011/000, done?

ephe

bu isnt that just row echelon?

Xander Henderson

@leslietownes Sure. A lot of the answer is about how the details are in the cited papers. Though, honestly, the argument I quoted from Tyson and Wu seems really straightforward...

leslie townes

oh, that's not reduced. then replace R1 with R1+R2, 101/011/000.

Xander Henderson

(And is rigorous, if a bit lacunary.)

leslie townes

17:20

xander: when beginning to read your comment, i was expecting "then i realized the question was my own question" or "then i saw that i had answered this question," but fun twist in having someone else do that for you.

Xander Henderson

Heh.

ephe

@leslietownes oooh wow i totally misunderstood reduced row echelon form. thank you so much

Xander Henderson

I'LL REDUCE YOUR ROW ECHELON FORM!

ephe

heheh

Thorgott

18:08

@onepotatotwopotato yes, cause $H^1(M;\mathbb{Z})\otimes\mathbb{R}=H^1(M;\mathbb{R})$

Jakobian

19:12

@psie I think otherwise

We can call it a rearrangement, sure

But guaranteed to exist should really be, guaranteed to converge to same number

Ted Shifrin

I don't think of writing the same thing backwards as a rearrangement, but technically it is. Indeed, the symmetric partial sums are identical.

smth

19:33

If H and K are two Sylow p-subgroups of G prove that then HK is not subgroup of G

H \ne K

p is prime and G has finite order

I know that from second Sylow theorem we have that H and K are conjugated (gHg^-1 = K , g\in G)

Also, HK is subgroup of G iff HK = KH

but I am stuck with that

can someone help me please

Thorgott

20:26

think about the relation between normality, conjugacy and HK=KH

smth

I am not sure what with that

Ted Shifrin

Doesn't one of the subgroups need to be normal in order to get $HK=KH$?

Xander Henderson

22:00

My new prediction for the election in November: Trump will not be on the ballot in Colorado or Maine (this is a current fact; I would put even money on SCOTUS upholding those rulings---Roberts and Kennedy, at least, are likely to be swayed by "states rights" arguments), and may not be able to run at all. Some other Republican will run on the platform of "TRUMP WAS ROBBED! PACK THE COURTS!", and win. American democracy will fall, and I will try to find a job in Scotland.

Are there community colleges in Scotland?

Will any of them hire me?

Sine of the Time

why politician can vote for themselves and we can't upvote our own posts?

Xander Henderson

@SineoftheTime The two things are in no way related?

businessinsider.com/…

Sine of the Time

🙄

@XanderHenderson LOL

Sine of the Time

22:50

$I_k=[4k-1,4k+1]$, $Q(x)=\sum_{k\ge 1}\chi_{I_k}(x)$, how to prove that $Q*Q$ is well defined ? I tried with Young inequality but $Q\notin L^1$

Xander Henderson

23:00

Gross.

Jakobian

@SineoftheTime $\int_{-\infty}^\infty Q(y)Q(x-y)dy = \int_0^x Q(y)Q(x-y)dy$

this is an integral on a finite interval

Sine of the Time

let me think about it

if $y\ge x$ then $x-y\le0$ and the integral is $0$

Sine of the Time

23:19

but then how to deal with the sum?

Jakobian

23:44

@SineoftheTime $Q$ is merely $\chi_{\bigcup_{k\geq 1} I_k}(x)$

since $I_k$ are disjoint

Sine of the Time

yes

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