12:02 AM
Check this out.
List the naturals:
$2,3,4,5,6,7, \dots$
Starting at $2$
Then the cross out pattern for deleting all multiples of $2$ is $(1,0)$
repeating. 1 = delete, 0 = keep
But at step prme $p \gt 2$ the cross-out pattern remains $(1,0,0, \dots, 0)$ ($p-1$ zeros).
So at step $p$ you'd result with:
$3,5, 7, 9, 11, 13, \dots$ the odds
But cross out all multiples of $3$
You have to do exactly the same operation on the sequence up to some small shift
So in other words, "deleting $2$ doesn't mess things up in any way for "deleting $3$"
It's kind of strange that it behaves this way the sequence
$3,5,7,9,11,13, \dots \Rightarrow 5, 7,11,13,17,19,\dots$
I guess it breaks down at $p = 5$
Because $5,7,11,13,17,19,23,25, \dots$ delete $5$ is:
Nvm!

@TanMath Do you know what a tangent vector looks like?

12:41 AM
Still... awake... Modular forms lecture at 9 tomorrow morning...
T_T

12:58 AM
@ÍgjøgnumMeg rip

@ÍgjøgnumMeg hast du es getan

das mit der Thetafunktion?

Ich hab's rausgelassen, weil ich mich nicht so gut mit der Poisson'schen Summenformel auskenne :) Man verliert dabei wahrscheinlich nur so 1-2 Punkte, also passt es
Ich hätte es zwar machen können, aber ich könnte die Benutzung der Formel wahrscheinlich nicht rechtfertigen :P
Sonst hab ich ja daraus folgern können, dass $\theta^8$ eine holomorphe Modulform vom Gewicht $4$ ist bezüglich der von $T^2$ und $S$ erzeugten Untergruppe von $\operatorname{SL}_2(\Bbb Z)$!
(wobei $T\langle z \rangle = z + 1$ und $S\langle z\rangle = -\frac{1}{z}$ für $z \in \frak{h}$
(@Leaky btw, man sagt lieber "hast du es geschafft" für "did you manage to do it?")

1:10 AM
@anakhro yah

@ÍgjøgnumMeg danke

Kein Problem :P

1:29 AM
@TanMath it's a good book to read into, get stuck, then come back in a fe wmonths when you've learned more

2:24 AM
@Semiclassical This may be true for math or physics students who might learn these things in their curriculum. But I am reading road to reality on my free time, to have a better grasp of math and physics. Indeed, there are some concepts I need to look up here and there. But there are definitely details I don't understand, and even Penrose says in the Preface that's totally OK.

2:40 AM
"Every non-identity element of Z/3Z has order 3." I'm unsure about the English terms, here. What exactly does that mean? Can someone paraphrase the sentence?

@integralette what words are you unsure about?
I mean basically all of the words are technical terms

What does it mean to "have order 3"?
Also, the "non-identity elements" is the term for 0, 1, 2, right? Because 3 = 0 in Z/3Z

$0$ is the identity in $\Bbb Z/3\Bbb Z$
The non-identity elements are $1$ and $2$

Oh, okay

So take a non-identity element $k \in \Bbb Z/3\Bbb Z$. Then having order $3$ means that the smallest integer $n$ for which $nk \equiv 0 \bmod 3$ is $3$.

2:51 AM
👍 Thanks

German

Dann heißt das einfach Ordnung eines Gruppenelementes auf Deutsch :P

Haha, alles klar. Gut, das hab ich noch nirgends gelesen bisher - also kein Wunder

Ah okey :) bist jetzt im ersten Semester oder wie? :P

2:55 AM
Jup. 🙈
char(Z/3Z) ist 3 und char(Z/5Z) ist 5, oder?

ap
Jap*

Ah, perfekt. Da hab ich einen Satz dazu. Dann benutz ich einfach das um zu zeigen, dass es keinen Körperhomomorphismus Z/3Z \to Z/5Z gibt

Sure, und wieso gilt das?

Wieso gilt das, dass char gleich sein muss meinst du?

ja lol

3:12 AM
Ja, wegen dieser "Ordnung eines Gruppenelements", oder nicht? Zumindest steht das auch so hier im Skript 1+1+1... n-mal = 0
Das ist ja genau das. Nur wird es nicht so genannt

3:29 AM

@TanMath just note that what we mean by "1-form" is what the video means by "1-form field"

3:46 AM
ergh this day is gonna hurt

4:14 AM
What's a good notation for distinguishing between a random variable and the values it can take? Unfortunately I can't use the uppercase-lowercase convention.

@ÍgjøgnumMeg good luck

@Leaky thanks lol, today we'll be defining Eisenstein series I think

cool

We'll define them, show that they are entire modular forms, and probably show they converge absolutely uniformly on some domains that are in the lecture notes lol

@ÍgjøgnumMeg have you done the cool proof of the $f(z) = \displaystyle \sum_{n \in \Bbb Z} \frac1{z+n}$ thing
it's so beautiful

4:24 AM
hm what thing do you mean?

do you know what function that is?

no lo
l

to be clear, that sum does not converge absolutely, and the correct interpretation is 0, 1, -1, 2, -2, ... which makes it $\displaystyle f(z) = \frac1z + \sum_{n=1}^\infty \frac{2z}{z^2 - n^2}$
hint: what are the poles / orders / residues of that function?
I wonder what Poisson has to say about it, since according to this, the Fourier transform of $\dfrac1t$ is $-i\pi\operatorname{sgn}(\omega)$
probably doesn't make sense

How does this universal property become a special case of the universal property of tensor product of two modules?

lol idk, we did show for example that $\sum_{n \in \Bbb Z} \frac{1}{(z-n)^k} = \frac{(-2\pi i)^k}{(k-1)!}\sum_{n=1}^\infty n^{k-1}e^{2\pi i n z}$ for $z \in \frak h$, I'm too tired to do any calculating now

4:31 AM
@WilliamSun you just apply it
@ÍgjøgnumMeg well that's irrelevant because $k=1$
you can read off the poles / orders / residues directly

But shouldn’t it be $S\times N$ instead of $N$ is this case?
Oh I forgot about the bilinearity
Nvm thank you.

oh it's $\pi \cot(\pi z)$ yeah we did show this
christ

@ÍgjøgnumMeg nice
innit beautiful

awake
yeah it's very cool

4:42 AM
analysis also has its beautiful side
who would've guessed

Agreed, I've been discovering it over the past few weeks lol
but it is a course on modular forms
so
I just feel like weird junk happens all the time in analysis
oh actually first we'll be defining the Petersson inner product
actually I don't think we'll get to Eisenstein series today hahaha
there's still a lot to do

3 hours later…
7:33 AM
What is unital ring? Ring of units?
0

Let $F$ be a field, $R$ be a unital ring, and $\varphi: F \rightarrow R$ be a unital ring homomorphism. Show that $\varphi$ must be injective. So what I think is only possible ideal for field are ${0}$ and $F$ ; if ideal is $F$ then it is just $0$ map but you have given unital ring in Image so i...

7:44 AM
@LeakyNun can you please check my approach?

the adjective "unital" describes "ring homomorphism"
it means $\varphi(1) = 1$
so unfortunately your attempt makes no sense

Wikipedia says: "rings with multiplicative identity: unital ring, unitary ring, unit ring, ring with unity, ring with identity, or ring with 1"

8:05 AM
thanks alot

2 hours later…
9:53 AM
Hi, could someone help me with this question of mine?
https://math.stackexchange.com/q/3431465/105100
It's on functional analysis.

3 hours later…
12:36 PM
Ein Sof, or Eyn Sof (, Hebrew: אין סוף), in Kabbalah, is understood as God prior to any self-manifestation in the production of any spiritual realm, probably derived from Solomon ibn Gabirol's (c. 1021 – c. 1070) term, "the Endless One" (she-en lo tiklah). Ein Sof may be translated as "unending", "(there is) no end", or infinity. It was first used by Azriel (c. 1160 – c. 1238), who, sharing the Neoplatonic belief that God can have no desire, thought, word, or action, emphasized by it the negation of any attribute. Of the Ein Sof, nothing ("Ein") can be grasped ("Sof"-limitation). It is the origin...
You know, the more I read kabbalah, the more it reminds of infinite set theory
> In the beginning, there is only the absolute infinite
> Then the law of identity splits apart from this unity, and create equality
> Then, the reflection principle splits apart, giving rise to the hierarchy of infinities
> Further axioms were added, and these infinities take more defined forms as they become more restricted. One of these is modelled by the von neumann universe
> And on the final step, the finite is born, giving rise to the rest of mathematics as we knew it
Perhaps, we have been doing set theory the wrong way around. We should have start with the absolute infinite, and then progressively add axioms to restrict it
typo:
and create equality -> and create tautologies

1:06 PM
Are the normal operators closed in the operator norm topology?"

1:37 PM
I can prove that the normal operators are closed in the norm topology; but I had to assume that the adjoint operation (involution) is norm continuous. Is this true?

What do you think?

I think so, but I was unable to prove it.

* is an isometry

2 hours later…
3:55 PM
I sorted out my orientations related confusion in the meantime, but now I'm confused by differential forms, so I will still annoy you @Ted
And this is how Ted never logged in the chat ever again
Hi @Mike

4:39 PM
If I have an integer polynomial $\sum_{i=0}^{2n+1}a_iT^i$ of degree $2n+1$ such that $p\nmid a_{2n+1}$, $p\mid a_i$ for all $i\le 2n$, $p^2\mid a_i$ for alle $i\le n$ and $p^3\nmid a_0$, where $p$ is a prime, then it is irreducible. How would one go about proving this? I tried looking at hypothetical factorizations mod p and mod p^2; the former seems too imprecise given the hypotheses, the latter has no unique factorization, so I haven't been able to derive anything useful.

@TanMath that's the same as the picture I shared and the one that you were already familiar with. Their ruler is literally just the planes that the vector punctures.

demonic @Alessandro .... I logged in just to spite you!

5:14 PM
@Secret Ein Prosit.

> The void begins with the vanitas, and with the null, naught, nothingness, all shall finally return to Ayin

@Thorgott ah, the Eisenstein criterion

> Existence is but nothingness compared to the grander scale of things

you derive a contradiction from a factorization mod nothing
by considering valuations

@TedShifrin D:

5:19 PM
also this

I'm trying to use "\textasciitilde" or "\usepackage{amsmath}" in my post but its not working. Is there a way to enter latex commands such as these?

@LeakyNun in this specific case, working $\mod p$ is the same as working in the residue field of the $p$-adic valuation. one can prove the Eisenstein criterion like that, but I haven't been able to do much for this (seemingly) more nuanced statement with that method. do I need to consider some valuation other than the $p$-adic one?

5:34 PM
you consider the $p$-adic valuation
working $\mod p$ destroys every information about the valuation so you don't do that

So I'm looking at a reformulation of vector calculus concepts using forms and the musical isomorphisms

Sure, no big deal.
And the Hodge star.

What's the Hodge star?

With a Riemannian metric and an orientation, it is an iso from $k$-forms to $(n-k)$-forms.

5:38 PM
There's an inexplicable increase in my rep .. if I am not mistaken it was 13k+ sth and now it's above 15k+ and I can't see any rep change to account for that .. :/

You should be able to track changes on your profile page, @r9m.

So the $f\mapsto fdx^1\wedge\ldots\wedge dx^n$ we're using is a special case?

Right. That's $\star f$.

@TedShifrin it's not showing any changes there ..

I see

5:41 PM

So I'm confused as to how I compute things explicitely. In general given $X\in \mathfrak X(M)$ we get a map $i_X\colon\Omega^k(M)\to\Omega^{k-1}(M)$ which is "fixing the first argument to be $X$ (pointwise)"

@r9m: If you click activity on your profile page, you can see the profile trend graphically.
Yes, but this has nothing to do with standard grad, div, curl stuff, @Alessandro.

Now I want to know what $i_X(dx^1\wedge\ldots\wedge dx^n)$ looks like "explicitely" for $X\in\mathfrak X(\Bbb R^n)$

ya .. the profile graph seems to have distributed the growth over the past few years .. but I am sure it was sth around 13k+

Write $X = \sum a^i \partial/\partial x^i$.
@r9m: I suspect your memory is faulty.

5:43 PM
@TedShifrin :P maybe .. it's all in the matrix :P

@TedShifrin I'm trying to check that $\operatorname{div}(X)$ is $\star^{-1}(\mathrm{d}i_X(dx^1\wedge\ldots\wedge dx^n))$

Write it down first with $n=2$ or $n=3$, @Alessandro.
Yeah, div makes sense in higher dimensions, but curl doesn't.

Is the proof of Arturo correct? math.stackexchange.com/questions/9532/…

@TedShifrin Hmmm I guess my issue is that I'm not sure what $(dx^1\wedge\ldots\wedge dx^n)(X_1,\ldots,X_n)$ is for a bunch of vector fields $X_i$

btw .. is there a theorem that says if a 2 dim riemannian mfld is has a killing vf then the mfld is conformally flat?

5:46 PM
It seems to me that he just proved that the inverse of the images must only intersect at a single point, u

You have to write them out in coordinates, and you get a giant determinant, @Alessandro. But you don't really need this. First figure out what $\iota_{a\partial/\partial x^1}(dx^1\wedge\dots\wedge dx^n)$ is.
Again, I recommend starting with $n=2$ or $3$ so you don't get overwhelmed.

Hmm so $\iota_{a\partial/\partial x^1}(dx^1\wedge dx^2)(Y)=(dx^1\wedge dx^2)(a\partial/\partial x^1,Y)$ by definition (in the $n=2$ case)

Go on.

Well now I get the determinant you were talking about earlier

Yes, or easier, write $dx^1\wedge dx^2 = dx^1\otimes dx^2 - dx^2\otimes dx^1$ and think.

5:57 PM
@TedShifrin it seems the same issue is there with my MO account as well .. sudden increase in rep .. :|

Hmmm, go to one of the moderator rooms and ask them to check, I guess.

ok .. thanks! :)

Ok so $(dx^1\otimes dx^2)(a\partial/\partial x^1,Y)=dx^1(a\partial/\partial x^1)dx^2(Y)$, right?

Yup, which is $(a\,dx^2)(Y)$.
And the other term disappears.

Right, since $dx^i(\partial/\partial x^j)=\delta^i_j$ being dual basis to each other

6:01 PM
Ayup. And what if you do $\iota_{b \partial/\partial x^2}$?

So now the first term disappears and the second survives giving $-b\,dx^1(Y)$
So I also know what I get for $X=a\partial/\partial x^1+b\partial/\partial x^2$, I see

Yup.
Now you will immediately generalize to your original question :)

Right, I just get an uglier expression because I need to sum over all $\sigma\in S_n$

Well, no, it's totally not ugly. Stop and think about what's going on.

6:22 PM
Ok so if I didn't mess up my calculations and $X=a^j\partial/\partial x^j$ I get that $\iota_X(dx^1\wedge\ldots\wedge dx^n)=a^idx^1\wedge\ldots\wedge\widehat{dx^i}\wedge\ldots\wedge dx^n$

Go back to our $n=2$ case to confirm. Hint: There's something small wrong.

Ah right, I forgot a $(-1)^{i+1}$ factor (and a sum over $i$)

OK, now I'm happy.

So am I :P

6:29 PM
Well now I want to take the exterior derivative of that, then the inverse of the Hodge star and hopefully get the divergence of $X$

You'll get that easily.

But I think I can deal with that

Moral of the story: Start with small $n$ and be concrete.

Sounds like a good hint
I did work out the $n=3$ case explicitely to get the general one

Cool. You know where to find me if you need me (eventually).

6:34 PM
I'll probably have plenty more doubts before the end of this course :P
But for the time being thanks again, I'll go finish this computation!

Okey dokey. Bis später!

Sara Chipps on November 13, 2019

In my very first blog post, I wrote about what a personal experience taught me about the Stack Overflow community. I said we were going to step back and re-evaluate how we deliver feedback, how we can improve content quality, and how we can reduce friction between people. I said that our goal is to have the question asking process be painless and beneficial for new people and Stack Overflow veterans alike.

During this re-evaluation period, we noticed something in our reputation reward system. We give anyone who receives an upvote on an answer ten additional reputation points, but only give five reputation points to people who receive an upvote on a question.  …

The blog only talks about stackoverflow, but there is a notification linking to it on MSE as well so I'm not sure
And my reputation also suddenly jumped on both MSE and MO so I guess that solves the mystery

Oh, @r9m ... that's for you. Since I never have asked a question, I didn't notice anything.
I guess I always thought that an answer deserved more points than a question. Some answers are stooopid, but some answers take a lot of work.

6:54 PM
@AlessandroCodenotti Thanks! Aloizio @ mathmods office said the same thing! :)
@TedShifrin Well I always fish for interesting questions .. so a good question and a good answer are equally 'entertaining' to me :)
@TedShifrin Well it's not always the stooopid answers that I dislike .. there are prompt one that ends up stacking a lot of score compared to well thought or deeper answers that usually come a bit later ..

Let $(X,\mu)$ be a $\sigma$-finite measure space. Is it always true that there exists a function $f \in L^2(X,\mu)$ that is nonzero almost everywhere?

@user193319 cook up one :) if $X = \cup_n A_n$ with $\mu(A_n) < \infty$ and $A_n$'s disjoint then $f = \sum_n \frac{1}{2^n\mu(A_n)}\chi_{A_n}$ should do the job ..

Damn it! That's a nice solution!

7:11 PM
oops .. $L^2$ .. maybe put a square root in there $\sum_n \frac{1}{2^n|\mu(A_n)|^{1/2}}\chi_{A_n}$ ..

Yeah, I made that adjustment too; I also replaced $2^{n}$ by $2^{n/2}$.
Still, very nice.

Even for most spaces which are not $\sigma$-finite it should hold
There might be problems if the $\sigma$-algebra is very small, but the space being very big is not by itself an issue, as long as you can find countably many pairwise disjoint sets of positive measure then you can use r9m's construction

@AlessandroCodenotti but then it isn't $L^2$ right
what's your function if $X = \Bbb R$ and $\mu$ is the counting measure?

There are some criterions of the form $L^p(X,A,\mu)$ is finite dimensional iff stuff but I don't remember the details
@LeakyNun the function which is $1/n$ at $n$ and zero otherwise

what happened to "nonzero almost everywhere"
$\Bbb R \setminus \Bbb N$ gets measure $\infty$ instead of $0$

7:17 PM
Oh lol I forgot about that part
Hmm for some reason I was convinced that as long as $L^2(X,A,\mu)$ is infinite dimensional then there is such a function. Now I'm not sure anymore
Nah that's false

7:31 PM
@LeakyNun I can show that if the polynomial $f$ can be non-trivially factored, then each factor must have constant term with $p$-adic valuation $1$, but that's about as far as I get

7:56 PM
@AlessandroCodenotti I remember one result like ,, closed subspaces of $L^p(\mu)$ are finite dim if it sits in $L^\infty(\mu)$ ($\mu$ is finite measure space) ..
$p > 1$
@LeakyNun counting measure on $\mathbb{R}$ is not sigma finite

8:21 PM
Is anyone else having trouble getting on the main site? I can't get to my profile page.

Ah, now it's back

@robjohn Hi! :) long time no see .. How are you?

@r9m pretty good. I have been busy offline for a while.

@robjohn I see .. :) I am here after a long time too .. so didn't know if you came here inbetween ..
@robjohn did you know Chris's sis finally publisher his book? :)

8:27 PM
@r9m I've been absent from chat for the most part for a while.
@r9m I saw indications, but was never sent anything about it.

@robjohn it's called Almost impossible integrals :) .. and in the mean time he was kind enough to cite my Euler sum article .. first citation of my life :)

@r9m they were definitely interested in complicated looking integrals.

@robjohn you had some crazy awesome solutions too :D

@r9m I should get a copy.
Is it available somewhere online?

ya .. it's available on springer .. lemme get the link
do you have springer access?

8:32 PM
@r9m I may through UCLA.

@robjohn I see .. :) cool .. o/w I can email you an e copy (I have access via my institute's subscription.. )

@r9m Trying...
@r9m The Springer app is busy...

I can email you anyway .. :) btw I don't have you mail id ..

8:56 PM
Here's a notation question
I want to invent a good notation for Gödel's incompleteness theorem
If F is a sentence, then 'F' will be the Gödel number of the sentence
Now consider the sentence "if x encodes a sentence F with a free variable, then y encodes the sentence F(n)"
This is a sentence with three free variables: x, n, and y
Maybe I should write it as x[n]=y or something? I dunno
Hmm
If P(n) says "the sentence encoded by n has no proof"
then the Gödel sentence is

$$P\big(‘P(n[n])’[‘P(n[n])’]\big)$$

@robjohn got it .. I just put it after robjohn?

@r9m yep

Agh why does LaTeX hate quotation marks

9:03 PM
@robjohn ok .. sent it ..

@AkivaWeinberger $\text{Does it "hate" them?}$

P(‘P(n[n])’[‘P(n[n])’])

Oh, they are seen as \prime

If we define [] such that 'F'[n] = 'F(n)' then ‘P(n[n])’[‘P(n[n])’] = ‘P(‘P(n[n])’[‘P(n[n])’])’
which means it solves x=‘P(x)’
sub("P(sub(n,n))","P(sub(n,n))") = "P(sub("P(sub(n,n))","P(sub(n,n))"))"
This is much easier if I think of it in terms of strings

@r9m thanks. Hopefully, all the integral solutions are explained.

9:19 PM
@robjohn I didn't have the time to read it yet .. but I think all the details are there for the solutions .. (unlike his answers here :P .. )

@r9m that is why I asked. The integrals here were quite closely guarded.

@robjohn ya .. but all that hiding was for his book .. it won't make sense to do the same in the book as well ..

@r9m sure, but I would work out one on chat and have to delete it immediately.

Hello all! I am about to start a short notebook where my job is to convert a bunch of polynomials written in trigonometric function in their corresponding vector form. i.e I have expression such as cos(i) \equiv r dot p.... does anyone have experience with this where the vector notation is retained so that my results will show up as explicit vectors and the individual components? Is this possible?

@robjohn ya .. once I collected many of your solutions in a tex file :P (the ones you shared in chat briefly) .. I don't seem to remember where it is now .. prolly lost it when my ext hard disc crashed few years ago along with a lot of other stuff ..
wait .. @robjohn I think he mentioned he tried to contact you sometime .. (in chat) but there was no reply ... I don't remember exactly

9:40 PM
Peano Arithmetic (PA) is a formal proof system. Statements and proofs must follow a very strict syntax.

Step 1: Encode sentences in PA as numbers (regardless of if they have a free variable or not)

Step 2: Encode proofs in PA as numbers

Step 3: If x encodes a sentence F with a free variable and y is a number, let sub(x,y) be the number encoding F(y) (i.e. y substituted into the free variable of F)

Step 4: Let P(x) be the sentence "the sentence encoded by x (has no free variables and) has no proof."

@r9m Thanks, BTW!

@robjohn welcome! :)

Heya robjohn. Long time!!
@r9m: There's a banner on the MSE main page about adding reputation points for questions. We both missed it earlier.
Wow, great that it got published.

@TedShifrin Hey there... I saw you yesterday, but you were deep in conversation, helping someone.

My conversations are rarely deep.

9:47 PM
@TedShifrin "deep in conversation" is different than "in deep conversation" :-p

Point taken. :P

Sidenote: PA can't really do functions, so it's slightly easier to let n encode the following:

∃z, sub(x,y)=z ∧ P(z)

That way, you only need to find a way to write the _sentence_ "sub(x,y)=z" in PA.

10:21 PM
Evening frens

Howdy @ÍgjøgnumMeg.

Hiya @Ted :)

So, how're you enjoying studenting?

I'm really, really loving being at a university with people who are actually interested in mathematics and not just the post-degree job prospects
and my weak background in a lot of areas is slowly repairing itself, which is something I was very worried about hehe

Motivated work does wonders.

10:29 PM
Indeed, I had forgotten a lot of basic stuff in the last year
a surprising amount

Yeah, that's a comment I always made to undergrads who wanted to go to grad school but who were delaying. One forgets far faster than one thinks.

Right, but my financial situation dictated that I needed to work a little before doing a masters :( in fact without the scholarship I probably wouldn't have been able to do it
so i've been very very lucky and I'm super grateful!

I wasn't criticizing you, merely supporting your comment.

I didn't think you were :)