12:41 AM
What's the name of the field that studies how global properties of manifolds are determined by properties of solutions of differential equations on those manifolds, or something along those lines?

Do you have an example in mind because I don't know what you're thinking of

12:53 AM
Hodge theory?

1:25 AM
Sounds like parts of global analysis.
Hodge theory is too specific, methinks.

1:36 AM
@TedShifrin Hello Sir!
If a periodic function have a period $T \in \mathbb Z$, then it's Fourier series is
$$f(t) = a_0 \sum_{n=1}^{\infty} \left( a_n \cos ( \omega n t) + b_n \sin(\omega nt) \right)$$
Where $\omega = \frac{2\pi}{T}$. Considering our function continuous, we have
$$a_0 = \frac{1}{T} \int_{0}^{T} f(t) dt$$

Grumble I can't sleep. Also I have to read probability

probably the name they're looking for is "geometric analysis"
that's not really what i would call the thrust of that field, but i don't think there's any field by that description

For $a_n$ we have
$$a_n = \frac{2}{T} \int_{0}^{T} f(t) \cos(\omega nt) dt \\ a_n = \frac{2}{T} \cos(\omega n t) \int_{0}^{T} f(t) dt + \frac{2}{T}\int_{0}^{T} \sin(\omega n t) \int_{0}^{T} f(t) dt ~~dt \\ a_n = 2a_0 \cos(\omega n~t)\bigg|_{0}^{T} + 2a_0 \cos(\omega n t)\bigg|_{0}^{T}$$
What's the mistake?

2:02 AM
I got my mistake

Hey guys ^^

2 hours later…
3:42 AM
@Drathora hi

2 hours later…
5:29 AM
Hello!! Let $1\leq m,n\in \mathbb{N}$ and let $\mathbb{K}$ be a field.
For $a\in M_m(\mathbb{K})$ we consider the map $\mu_a$ that is defined by $$\mu_a: \mathbb{K}^{m\times n}\rightarrow \mathbb{K}^{m\times n}, \ c\mapsto ac$$

Let $\lambda$ be the eigenvalues of $\mu_a$ then we have that $\mu_a(c)=\lambda c$ so we get that $ac=\lambda c$, or not? But what does this mean? That $\lambda$ is also an eigenvalue of $a$ with eigenmatrix $c$ ?

@MaryStar: Where are you getting the equation $\mu_a(c)=\lambda c$? This only holds if $c$ is a matrix whose columns are all multiples of the eigenvector of $a$ with eigenvalue $\lambda$.

If Fourier series for $f(t)$ is $$f(t)= a_0 + \sum_{n=1}^{\infty} \left[ a_n \cos (\omega n t) + b_n \sin(\omega n t) \right]$$ then what will be the Fourier series for $f(t+a)$ ?

@TedShifrin Do we not use this definition : en.wikipedia.org/wiki/… ?

What is the period of $f$ and what is $\omega$, @Knight?
I know what an eigenvector is, @MaryStar. Where are you getting your equation?

5:44 AM
@TedShifrin $f$ is a continuous function with period $T$ and $\omega = \frac{2\pi}{T}$

We have in this case $T=\mu_a$ and $v=c$, or not? @TedShifrin

Oh, I misread, @MaryStar. I apologize. I thought $\lambda$ was an eigenvalue of $a$.

@TedShifrin clearly $f : \Bbb R/(\frac{2\pi}\omega)\Bbb Z \to \Bbb C$ /s

@TedShifrin Ah ok! So is it correct that from $\mu_a(c)=\lambda c$ we get that $ac=\lambda c$ ? Does this mean that $\lambda$ is also an eigenvalue of $a$ with eigenmatrix $c$ ?

So, yes, @MaryStar, if $\lambda$ is an eigenvalue of $\mu_a$, then there is some nonzero $c\in\Bbb K^{m\times n}$ with $\mu_a(c)=\lambda c$. You didn't state quantifiers.
Then, as I stated, the columns of $c$ are all eigenvectors of $a$ with eigenvalue $\lambda$. I don't know what an eigenmatrix is.

5:50 AM
Ah ok! What I want to show is that $\text{Sp}(\mu_a)=n\text{Sp}(a)$. With the above we have shown that an eigenvalue of $\mu_a$ is also an eigenvalue of $a$, right? What is left to show to get the desired result?

What does that equation mean?

I also a little confused with that, I haven't seen that notification before in the notes. Is it maybe meant the number of $\text{Sp}(\mu_a)$ is equal to $n$ multiplied by $\text{Sp}(a)$ ? @TedShifrin

Sp is the spectrum, i.e., the set of eigenvalues?

@TedShifrin Yes

Ted when you get free please give your views on how we can go on for simplifying the Fourier Series for $f(t+a)$

5:58 AM
Then the equation makes no sense to me.

@TedShifrin Ok, so it cannot be meant the cardinality, can it?

You never answered my question ages ago, @Knight. You need to translate your variables and use addition formulas for sin and cos. But I'm leaving now.
I have no idea, @MaryStar.
Good night, all.

Ok.. Thank you!! Good night!! @TedShifrin

15 mins ago, by Knight
@TedShifrin $f$ is a continuous function with period $T$ and $\omega = \frac{2\pi}{T}$

Just one thing.... Now I looked again... "sp" is not the spectrum..it is the trace @TedShifrin

6:16 AM
Die Spur (Spurfunktion, Spurabbildung) ist ein Konzept in den mathematischen Teilgebieten der Linearen Algebra sowie der Funktionalanalysis und wird auch in der Theorie der Körper und Körpererweiterungen verwendet. == Die Spur in der linearen Algebra == === Definition === In der linearen Algebra bezeichnet man als die Spur einer quadratischen n × n {\displaystyle n\times n} -Matrix A {\displaystyle A} über einem Körper K {\displaystyle K} die Su...
aha that's why

Yes.. But how is the trace of a linear transformation defined? Again the sum of eigenvalues? @LeakyNun

sure
it's also the sum of the diagonal

For $a\in M_m(\mathbb{K})$ we consider the map $\mu_a$ that is defined by $$\mu_a: \mathbb{K}^{m\times n}\rightarrow \mathbb{K}^{m\times n}, \ c\mapsto ac$$ We have that an eigenvalue of $\mu_a$ is also an eigenvalue of $a$. So what do we need to get the result? Do we have to check the multiplicity? @LeakyNun

6:48 AM
Is the multiplicity of an eigenvalue equal to the number of eigenvectors that correspond to that? @LeakyNun

Can we simplify the Fourier series of $f(t+a)$ to get something like the Fourier Series of $f(t)$. $f$ is a continuous function with period $T$ and $a \in \mathbb R$.

7:03 AM
Because if the multiplicity of an eigenvalue is equal to the number of eigenvectors that correspond to that, then since the columns of $c$ are all eigenvectors of $a$ with eigenvalue $\lambda$, we would have that the multiplicity is equal to the number of columns, i.e. n, and so we would get the desired result, or not? @LeakyNun

1 hour later…
8:18 AM
@Knight yes its a simple change of variables, you end up needing to multiply by a complex exponential so its not only the fourier series of $f(t)$ but that's life for you

@CalvinKhor That’s really what I found !
We should write Fourier Series in form of exponential not in trig form
But what’s the exponential form of Fourier Seires lol

8:34 AM
instead of $c+\sum_{n=1}^\infty a_n \cos(n x) + \sum_{n=1}^\infty b_n \sin(n x)$ (I'm ignoring the period) you write $\sum_{n\in\mathbb Z} c_n e^{inx}$. You can convert exponentials to trig using $$e^{\pm inx} = \cos(nx) + i (-1)^n \sin(nx)$$ and trig to exponentials using $$\cos(nx) = \frac{e^{inx}+e^{-inx}}2, \ \sin(nx) = \frac{e^{inx}-e^{-inx}}{2i}$$
@Knight
lmao oops instead of $(-1)^n$ i should have wrote $\pm$, that was some weird autopilot typing

@CalvinKhor So, we have $$f(t)= \sum_{n=-\infty}^{\infty} c_n e^{i ~n ~t} \\ f(t+a)= \sum_{n=-\infty}^{\infty} c_n e^{int + in a} = \sum_{n=-\infty}^{\infty} c_n e^{int} e^{ina}$$
Am I right?

yes

But I still cannot see how to write $f(t+a)$ in form of $f(t)$, that sigma operates on the whole product.

yes, you can't, but you can write one in terms of the other like above, that's probably as far as you can go in general
sorry misunderstood you

No, you understood very correctly. Thanks for your help Cal

8:43 AM
if you prefer the computation directly for the coefficients
$$\mathcal F[f(t+a)](n) = \int_0^{2\pi} f(t+a)e^{-int} dt = \int_a^{2\pi+a} f(\tau) e^{-in(\tau-a)} d\tau = e^{ina} \mathcal F f(n)$$
i forgot to divide by $2\pi$

What would $a_0$ correspond to in this exponential expression.
Is it $c_0$ ?

that's just $n=0$
yeah

Okay

np!

9:15 AM
@CalvinKhor Are coefficients $c_n$ same for both $f(t+a)$ and $f(t)$?
36 mins ago, by Knight
@CalvinKhor So, we have $$f(t)= \sum_{n=-\infty}^{\infty} c_n e^{i ~n ~t} \\ f(t+a)= \sum_{n=-\infty}^{\infty} c_n e^{int + in a} = \sum_{n=-\infty}^{\infty} c_n e^{int} e^{ina}$$

the symbol $c_n$ there is the same, but the fourier coefficients of the tranlated function are not those of the original function
rather its $\mathcal F[f(t+a)](n) = e^{ina} \mathcal F[f(t)](n)$
I write it this way instead of $c_n$ so that its clear which function's fourier coefficient you're talking about
but i gtg hope thats clear ttyl

Bye
Thanks you so much

9:52 AM
@BalarkaSen is that formal enough? and how would i procceed for n>2

@MikeMiller I agree. The join model of EZ/2 gives S^\infty. But I want to construct EZ/2 with the simplicial model.

Hello. Could anyone please help me with understanding the Proof of change of variables theorem from Spivak's calculus on manifolds?

@BalarkaSen How though? EZ/2 has 2 0-simplices, 1 1-simplex, and all higher simplices are degenerate. For higher simplices, say, [0, 1, 1], we identify glue the face [0, 1] with the 1-simplex [0, 1], right?
If so, then EZ/2 has no higher simplex. How can the quotient BZ/2 have a non-degenerate simplex in each dimension?
i m confooz
@ManolisLyviakis What are you doing?

10:07 AM
im trying to prove munkres 53.1 theorem
of his topology book
leme posted
but its esay to waave hands in homotopy theory i want to be formal i think i proved it for n=2
for n greater than 2 my $g$ function will be the same its restriction to the different n pieces of $I$ $g=p_i( of something )$ where $p_i$ is the linear positive map of $I$ to $[a_{i-1} , a_i ]$
im running the positive lineara maps faster thats all im doing
what you think @feynhat if you have time check my proof if it makes sense and it is formal enough.

Why is this not clear? $f$ is the same thing as the concatenation $f_1...f_n$ up to a piecewise linear reparametrization that fixes the endpoints, so homotopic

its clear
im asking if what i wrote above is an ok way to formulate what you just said

11:04 AM
hello
If $g\circ r^1$ is smooth, where $r^1$ is a standard coordinate on $\mathbb{R}^2$, does it follow that $g:\mathbb{R}\rightarrow \mathbb{R}$
is smooth?

$r^1$ restricts to a diffeo $\mathbb{R}\times\{0\}\cong\mathbb{R}$

aha , thanks
so yes
$\mathbb{R} \times \{0\}$ is a smooth manifold?

11:21 AM
it's a product of smooth manifolds

oh right, singletons of $\mathbb{R}^n$ are smooth

If $f$ is a positive continuous function, and if we have $$\int_{0}^{1} f(x+a) -f(x) dx \lt 0$$ $a \in \mathbb R$ then can I conclude that $$f(x+a) \lt f(x)$$ for all $x$ in domain?

does a function having negative integral over some interval imply that it's everywhere negative?
draw a picture in case the answer isn't clear

11:35 AM
@Thorgott Okay! I got it, all we can conclude is that area below the $x-$ axis is more than the area above the $x-$ axis under the graph of $f(x+a) -f(x)$ in the interval $[0,1]$. Right? But how to write this information analytically?

$\int_0^1f(x+a)-f(x)dx<0$
that's precisely what the integral means

@Thorgott Can’t we get anything more than that?
I mean from that can we conclude anything more? $f$ is always positive is given

What do you want to conclude?

$f$ is a continuous periodic positive function with period 1. I want to prove that the average value of $f(x+a)$ is more than that of $f(x)$
More than or equal to, I meant

they will in fact be equal, so try proving that

11:50 AM
So, I attempted to arrive at a contradiction, $$\int_{0}^{1} f(a+x) dx \lt \int_{0}^{1} f(x) dx$$
@Thorgott They can be equal

I'm saying they always are
the periodicity assumption is necessary for this, of course

Really? Can you show me the path for proof? I’m pretty much excited to see and it’s about to dawn and we would have enough sunlight

The claim is that $\int_0^1f(x+a)dx=\int_0^1f(x)dx$. Try starting with a substitution and then use periodicity.

Let $x+a=u$, we have $$\int_{0}^{1} f(u) du = \int_{0}^{1} f(x) dx$$ What’s next?

That would already be the conclusion

11:59 AM
@feynhat You still haven't explained why you thought the result from some other method is RP^n.

Yes I understand $u$ and $x$ Are dummy variable but if antiderivative will exist then I think $u$ will appear as $x+a$. I believe in you Thor but can you please help me a little more in convincing myself about the conclusion?

I still don't understand the claim "EZ/2 has no higher simplex".

Well, you need to perform an actual calculation to convince yourself of this conclusion

:)

Actually, I'm going to bow out, because it's not very interesting to me. Sorry.
Shouldn't have pulled the bait and switch.

12:23 PM

@MikeMiller Sorry. I thought all RP^n (n > 1) were K(Z/2, 1). My bad.

Would you explain this one?

Any n-simplex of $EG$ can be written in the form $[g_0, g_0g_1, \dots, g_0g_1\dots g_n]$, where $g_i \in G$. But our $G$ has only two elements. So any 2-simplex will degenerate into a 1-simplex right? Like [0, 1, 0] or [0, 1, 1]
@Knight Its the graph of y=tan(x).
By 'no higher simplex', I meant all higher simplices are degenerate.

12:39 PM
@feynhat yes but why only the graph of tan not any other function?

uhh... tan line?

Oh okay!

Let $\mu$ be a Haar measure on a locally compact group $G$, let $U$ be any open subset with finite measure, and let $H$ be an $\sigma$-compact open subgroup. Why does $U$ intersect only countably many cosets $xH$?

@feynhat Well, $S^n$ is not contractible.
I don't really follow why [1,1,1] is a degenerate simplex.
I tend to feel like if you want something concrete than build a concrete model. The simplicial model is abstraction meant to work for formal reasons. So you should work with it formally.

Wouldn't the geometric realization of [1, 1, 1] be a 0-simplex?

1:05 PM
@MikeMiller I just want to see it for one simple case (that's why I chose Z/2).

1:21 PM
??? The 2-simplices are by definition triplets $[g_0, g_1, g_2]$. So their geometric realization (before quotienting etc) are 2-simplices. But
If you want to argue that the image of that in the geometric realization (after quotienting) is a single point you'd have to make an argument.

2:16 PM
What the name given to functions that are $\in C^\infty(\Bbb R)$, it's an "X function" where X is the name of someone, I can't quite remember

smooth

There's some other name I swear
Something like "Jacob functions", with someone's name

never heard of that

damn
this is going to drive me crazy
I must be wrong about them being smooth
fwiw I was thinking of Schwartz functions

2:32 PM
ah

Can we use the differential equation adx=vdv for calculating freefall velocity in case of large height?
Can we use the differential equation adx=vdv for calculating freefall velocity in case of large height?
Like putting a=-Gm/(R+h)^2 and then solving by taking limits of integral as h to h-delta h to calculate velocity after particle has travelled delta h height

@feynhat I don't understand why this means every higher simplex is degenerate
I mean, I know for a fact they aren't. But what is the argument?
I mean, [0, 1, 0] should be a 2-simplex which is non-degenerate.
The geometric realization of the 1-skeleton of EZ/2 is an interval (0 ----> 1), and [0, 1, 0] is a 2-simplex which you glue to this by pasting the edges [0, 1], [1, 0] and [0, 0] to the edge 0 ---> 1, to the edge 0 ---> 1 but oriented oppositely, and to the degenerate edge 0 ---> 0 (i.e. collapse to 0) respectively
The triangle is very much not degenerate. You might be forgetting what a simplicial complex is
You have simplices of the form [0, 1, 0, ba-dee-daa, 0, 1, lalalala, 0, 0, 1], sure, there are lots of vertex repetitions. That means when you actually attaching the simplex in the process of making EZ/2 it's faces are glued in a very complicated fashion to the lower skeleta
The interior of the faces aren't getting squished
Or rather, note that this EG is a delta-complex not a simplicial complex so it's faces doesn't determine what the simplex is
I have to run and read probability to prepare for a meeting tomorrow; ping me if you have q I'll respond later

2 hours later…
4:54 PM
I have a math/programming question (mainly maths), can I ask here?

yes

I am given a number n, I want to find the next number after n where the sum of the digits in even positions in its number is equal to the sum of the digits in odd positions.
But looping through all the digits one-by-one is too long, apparently there is a quicker way to do it
n can go up to n to the power of 100
My current code loops through every number after n, splits into 2 lists (1 even index numbers and one odd), checks if they’re equal. If yes it returns the number if not it keeps going

5:22 PM
find the number of 7 digit numbers the sum of whose digits is even.

Does a continous peridoic function have a convergent Fourier Series?

> Does 'a' continous peridoic function have a convergent Fourier Series?
Yes. Many do.
But many also don't.

This is a famous question of Fourier. The answer is no in general.

depends on your notion of convergence

5:29 PM
By a famous theorem it converges almost everywhere anyway

@BalarkaSen Thanks a lot for this.
This cleared up a lot of things
I was actually confusing the nerve construction with the delta complex construction

@feynhat Yeah actually I think I simply misspoke there. I think 0 --> 1, 0 --> 0, 1 --> 0, 1 --> 1 are all non-degenerate edges in the 1-skeleton.
Everything is nondegenerate
The homotopy gets killed by how higher simplices are attached at each stage: It's a massive contractible complex.

@BalarkaSen I'm degenerate

Every Lean coder is

if your function is integrable and has left- and right-hand limits at a point, then it's Fourier series converges to the average of these limits in Cesaro mean

5:32 PM
Nobody cares, Thorgott

this is actually important

No
In fact, you are unimportant.

@BalarkaSen , btw, that's mean

I should study simplicial complexes. I actually don't know much beyond the definitions.
...and stop thinking about concrete example for simplicial construction of EG.

I think it was a fine thing to think about

5:37 PM
Thanks for your time though @BalarkaSen @MikeMiller

@Arjun Everyone is unimportant though

@Arjun Cesaro mean, yes.
3

Lmao
Thats pretty good
I have to read probability Jesus Christ

The reason why this is important is cause it tells you that if you want the Fourier series to converge pointwise, its limit has to be the average of its one-sided limits (granted, I have no clue what happens with functions where these don't exist, but that probably is actually unimportant). This explains part of the behavior of Fourier series at jump discontinuities. This result can also be adapted to give a rather quick proof of uniqueness of Fourier series (for sufficiently nice functions).
If you impose additional conditions on the one-sided derivatives of the function, you can improve c

Do you seriously think anyone is going to read that paragraph Thor

5:39 PM
@BalarkaSen ,depends on perception, if you feel so, good.don't demotivate others

no, but that also didn't stop me from proving the projection formula in here
so why would it stop me now

you tricked me to

@Arjun If that sentence demotivates you, then you've ascribed too much importance to Balarka
2

LMAO

Taking his statement seriously means that he, too, is unimportant, his words leaves in the wind

5:41 PM

Too far, Thorgott, too far
Calling me a dirty algebraist

that's what you call me all the time

Because thats what you are
I am an honest man

@LeakyNun Our function is periodic

2/3.5 things I'm doing this semester are analysis/topology

5:43 PM
@Thorgott Point wise convergence
@BalarkaSen Really? But you said above that in general it is not true.

"almost everywhere pointwise convergence" is different from "pointwise convergence"

yeah, won't happen in general

what is up with these probabilists by the way

also Balarka forgot to mention that the famous theorem has an additional hypothesis

a random metric space??? the hell
yeah I did mean L^2

5:46 PM
L^p for p>1 suffices

NOBODY CARES

Doesn’t periodicity ensures convergence?

lol, Hunt sure did care

you think a man named Hunt is important?
totally ridiculous, just look at his name

I'm not saying he's important, I'm saying he cared

5:48 PM
Who is senior Thor or Bala?

yeah because he was a nobody
nobody cares
checkmate atheists

curiously, for functions on the real number line, you need L^p for some 1<p<=2
I guess p>2 isn't trivial anymore cause you don't have the L^p space inclusion anymore, but I wonder why 2 is the exact barrier

@LeakyNun Drawnish Giri drew 3 perfectly winning games

6:33 PM
Why does it matter who is senior
If they're talking about something purely mathematical then you can read the justifications and figure out which is correct or incorrect
If it's non- or pseudo-mathematical then you can read their informal arguments and decide which informal argument you find more convincing and take that one more seriously

Anyone got any ideas?

given earlier conversation I think it's safe to say we are both 8th graders calling each other names

2 hours ago, by Daniil
I am given a number n, I want to find the next number after n where the sum of the digits in even positions in its number is equal to the sum of the digits in odd positions.

6:51 PM
Hm. Suppose $(\Omega, \mathcal{A}, \mu)$ is a probability measure space, $T : \Omega \to \Omega$ is an ergodic transformation and $f \in L^1(\Omega, \mathcal{A}, \mu)$.
Think of the iterates $X_n = f \circ T^n$ as random variables. What's the probabilistic interpretation of $T$ being ergodic in terms of the stochastic process $\{X_n\}_{n \in \Bbb Z_+}$?
It must be something like, $X_{n+k}$ becomes asymptotically independent to $X_k$ as $n \to \infty$
That's like mixing, which is stronger, I guess
$\Bbb P(X_n \in A, X_0 \in B) = \mu(T^{-n}(f^{-1} (A)) \cap f^{-1}(B)) \to \mu(f^{-1}(A)) \mu(f^{-1}(B)) = \mu(T^{-n}(f^{-1}(A)) \mu(f^{-1}(B)) = \Bbb P(X_n \in A) \Bbb P(X_0 \in B)$
Ergodicity implies strong mixing so yeah
Any function on some chunk of the stationary process and some other function on a shift of the chunk taken earlier become asymptotically independent as we shift more
Weak mixing would translate to asymptotic uncorrelation

7:06 PM
I have an infinite group generated by $\{t_1,t_2,t_3,t_4,t_5\}$ and also $t_1^{10}=t_2^{10}=t_3^{10}=t_4^{10}=t_5^2=e$ and also things like $t_2t_1t_2t_2t_1t_2=e$. Does anyone know if this group might be isomorphic so something interesting?

What a horrendous group

it's the orientations obtained by placing regular pentagons side by side in the plane
which means I can make pretty things with it like this i.imgur.com/jJoku7L.png

Ah yeah i was wondering if it's something like that

the pretty picture more than makes up for the ugly algebra imo

these x^p = y^q = z^r = 1 etc stuff appears when you take oriented isometry groups of geometric figures on the plane
i was thinking of what you get if you do it with a triangle, eg the von Dyck group
these are not crystallographic groups on the plane though

7:13 PM
Conjecture: Let $G$ be a topological group. If $C \subseteq G$ is path connected subset, then $\langle C \rangle$ is a path connected subgroup. I believe I have a counterexample: Let $G = \Bbb{Z}$ with the discrete topology and let $C = \{1\}$. The set $C$ is trivially path connected, because it is a singleton, but $\langle C \rangle = \Bbb{Z}$ is not path connected.
Does this sound like a genuine counterexample?

ye

yeet

@Thorgott you know youre in big trouble when the name talagrand pops up in what youre reading

yeet bois

I just spent an entire minute thinking about "minimal maximal ideals" until I realized what I was doing
@Balarka I only stumbled upon his concentration inequality once, I think

7:43 PM
please someone have a reference for this :math.stackexchange.com/questions/3745349/…

What's the definition of a compact neighborhood? Is it an open set which is compact?

a neighborhood which is compact
neighborhood being any set containing an open set about the point
being compact and open is a somewhat unlikely property

@Thorgott Ah, okay. I see. How do I prove that in any locally compact group, there is a compact unit-neighborhood?
Nevermind...

@Vrouvrou Clearly, $|E|\le|G|$. On the other hand, $|G|\le\sum_{J\subseteq E,J\text{ finite}}|\mathbb{Z}^J|=|\{J\subseteq E\colon J\text{ finite}\}|\cdot\aleph_0=|E|$. By Cantor-Schröder-Bernstein, $|E|=|G|$.

8:12 PM
Hey guys
Thorgott, did you get any time to look into weakening the constraints on the coarea formula?
I've been looking around a lot but struggling to get away from the need for the function to be Lipschitz

not yet, I'm afraid
been kinda busy
there was a comment in Federer about only requiring bounded limsup of the difference quotient a.e. or something

8:32 PM
Yeah, a lot going on here also >_<
I found a lot of things about rectifiable sets but that condition still required a Lipschitz map

@Thorgott I don't know algebra and I want to help someone that's why I'm sourcing a reference
why $|E|\le|G|$?

you can easily come up with an injection from $E$ into $G$

8:47 PM
I don't know. how ?

the most natural choice of map will work

9:04 PM
$G=\{f:E\to Z, |supp(f)|<\infty\}$ can i define the identity $id:E\to G$ ?
but how it can be defined ?
@Thorgott

no, an identity map always goes from a set to itself, but $E$ and $G$ are different sets

ohh yes I don't see it
I don't know what I can define

9:22 PM
I give you an element of $E$. Can you give me a map $E\rightarrow\mathbb{Z}$ that has something to do with this element?

the absolute value ?

what's the absolute value?

for $a\in E$ then |a|\in Z

that doesn't tell me what the absolute value is

9:39 PM
$|.|: E\to Z, a\mapsto |a|$

those are symbols, not definitions

a norm if the element of f. are vectors in general

I do not understand this sentence

9:54 PM
the map that define from E to Z is the norm map

what is the norm map?

sorry I can't choose the norm map because ||a||\in R^+
I don't how to define the map from E to Z

I give you an element $x\in E$. Can you give me a map $E\rightarrow\mathbb{Z}$ that does one thing with $x$ and a different thing with all the elements of $E$ that are not $x$? (The reason for this requirement is that we want to have different maps for different $x\in E$ in order to get an injection from $E$ to $G$)

no I don't know

10:14 PM

I need a book
or I can define $\chi_a(x)= \begin{cases}1 \, if\, x=a\\0\, if\, x\neq a\end{cases}$
here $a$ is fixed on E

10:30 PM
yup, that works!

10:56 PM
Hm, I forgot how to couple random variables. Let $X, Y$ be random variables with $\Bbb P(X \leq t) \leq \Bbb P(Y \leq t)$ for all $t \in \Bbb R$, I want to modify them without changing the distribution such that $X < Y$ (pointwise on the probability space). What do I have to do again?
Oh nevermind I just simulate
$U$ be uniform on $[0, 1]$, $X = F^{\leftarrow}_X(U)$, $Y = F^{\leftarrow}_Y(U)$ where $F_X, F_Y$ are the distribution functions
That does the trick
This trick never gets old

11:35 PM
The inequality is flipped above but ok