Mathematics

Shine On You Crazy Diamond

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!

anakhro

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

ÍgjøgnumMeg

12:41 AM

Still... awake... Modular forms lecture at 9 tomorrow morning...

T_T

Leaky Nun

12:58 AM

@ÍgjøgnumMeg rip

ÍgjøgnumMeg

:(

Leaky Nun

@ÍgjøgnumMeg hast du es getan

ÍgjøgnumMeg

das mit der Thetafunktion?

Leaky Nun

ja

ÍgjøgnumMeg

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?")

TanMath

1:10 AM

@anakhro yah

Leaky Nun

@ÍgjøgnumMeg danke

ÍgjøgnumMeg

Kein Problem :P

Semiclassical

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

TanMath

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.

integralette

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?

ÍgjøgnumMeg

@integralette what words are you unsure about?

I mean basically all of the words are technical terms

integralette

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

ÍgjøgnumMeg

$0$ is the identity in $\Bbb Z/3\Bbb Z$

The non-identity elements are $1$ and $2$

integralette

Oh, okay

ÍgjøgnumMeg

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 $3$.

integralette

2:51 AM

👍 Thanks

ÍgjøgnumMeg

what's your native language?

integralette

German

ÍgjøgnumMeg

Dann heißt das einfach Ordnung eines Gruppenelementes auf Deutsch :P

integralette

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

ÍgjøgnumMeg

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

integralette

2:55 AM

Jup. 🙈

char(Z/3Z) ist 3 und char(Z/5Z) ist 5, oder?

ÍgjøgnumMeg

ap

Jap*

integralette

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

ÍgjøgnumMeg

Sure, und wieso gilt das?

integralette

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

ÍgjøgnumMeg

ja lol

integralette

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

TanMath

3:29 AM

@anakhro @TedShifrin what do you think about this interpretation of a one-form:
https://www.youtube.com/watch?v=QP-nlfz1yTI&list=PL8erL0pXF3JYCn8Xukv0DqVIXtXJbOqdo&index=19

Leaky Nun

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

ÍgjøgnumMeg

3:46 AM

ergh this day is gonna hurt

user76284

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.

Leaky Nun

@ÍgjøgnumMeg good luck

ÍgjøgnumMeg

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

Leaky Nun

cool

ÍgjøgnumMeg

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

Leaky Nun

@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

ÍgjøgnumMeg

4:24 AM

hm what thing do you mean?

Leaky Nun

do you know what function that is?

ÍgjøgnumMeg

no lo

l

Leaky Nun

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

William Sun

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

ÍgjøgnumMeg

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

Leaky Nun

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

William Sun

But shouldn’t it be $S\times N$ instead of $N$ is this case?

Oh I forgot about the bilinearity

Nvm thank you.

Leaky Nun

glad to have done nothing

ÍgjøgnumMeg

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

christ

Leaky Nun

@ÍgjøgnumMeg nice

innit beautiful

ÍgjøgnumMeg

awake

yeah it's very cool

Leaky Nun

4:42 AM

analysis also has its beautiful side

who would've guessed

ÍgjøgnumMeg

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

maths student

7:33 AM

What is unital ring? Ring of units?

0

Q: unital ring Homomorphism

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...

maths student

7:44 AM

@LeakyNun can you please check my approach?

Leaky Nun

the adjective "unital" describes "ring homomorphism"

it means $\varphi(1) = 1$

so unfortunately your attempt makes no sense

Martin Sleziak

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

maths student

8:05 AM

thanks alot

An old man in the sea.

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.

Secret

12:36 PM

Ein Sof

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

user193319

1:06 PM

Are the normal operators closed in the operator norm topology?"

user193319

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?

Alessandro Codenotti

What do you think?

user193319

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

Alessandro Codenotti

* is an isometry

Alessandro Codenotti

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

Abhas Kumar Sinha

Aja

Thorgott

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.

anakhro

@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.

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$Mathematics$ Mathematics