Mathematics - 2021-04-13 (page 2 of 3)

dc3rd

05:00

@EdwardEvans I had a question about programs in Germany. I was just looking at TUM's website with regards to Masters in Statistics. Are the two intake periods for new students the Winter and Summer semesters in Germany? They don't do fall in Germany?

Edward Evans

Usually just winter and summer semester

In my university the winter semester starts in October and summer semester in April :)

dc3rd

Ok. Something to be aware of then........wait what?.....lol

Edward Evans

whut

dc3rd

That's kind of how I envisioned it.

Edward Evans

TUM might be different because it's in Bavaria but afaik it's the same everywhere, best to just look it up on their website

dc3rd

05:02

I'm looking it up again. I had it open

Edward Evans

Okie dokie

Leonhard Euler

I bethink not yond the univ'rse is a fractal

dc3rd

SO this means that You would start in October of 2022?

@EdwardEvans

StudySmarterNotHarder

@LeonhardEuler it appears at small scales like it's not, but it's so grandiose and huge that we can't make out its structure from 4D spacetime

Edward Evans

@dc3rd if you apply now for the winter semester you'll start in October 2021. The winter semester will usually be called smth like WiSe 21/22.

Or whenever their semester dates are

Leonhard Euler

05:07

@StudySmarterNotHarder yea

Edward Evans

On their site it says winter semester 21/22 starts on 01.10.2021

dc3rd

Ok....that's what I wanted clarified. I'm not thinking about it until next year, but I just wanted to get my bearings straight

StudySmarterNotHarder

@LeonhardEuler what if you took $\lim_{n \to \infty}\sum_{i = 1}^{\pi(n)} \dfrac{p_{i+1} - p_i}{n^2}$

Edward Evans

Okay nice. Then winter semester starts around the end of September/start of October

dc3rd

Got it and then if I were to apply for summer I would start summer of 2022. Now I got an idea of how to plan out my application for the following year.

Thanks for the clarification.

Edward Evans

05:10

No worries, make sure you get all of your application documents PERFECT. The Germans are ridiculous bureaucrats.

dc3rd

I'll definitely have more questions closer to time. Just getting the ducks lined up from now.

Leonhard Euler

@StudySmarterNotHarder I'm not on a computer right now

Edward Evans

No worries, lemme know

Leonhard Euler

Lemme switch it on

StudySmarterNotHarder

@LeonhardEuler I think the sum just goes to $(p_{\pi(n)} - p_1)/n^2$, so it's probably a divergent series

But if there are are only $c_k$ consectutive prime pairs differing by $k$, then there is finite summation for (all of) those pairs, then oops a giant leap into $c_k = \infty$.

After you group terms

What about taking all differences $p_i - p_j, i,j \leq \pi(n)$

Then what denominator $n^k$ would you need to make the series converge instead

Define $k \gt 1$ to be the least real number such that the series converges

Actually you'd have to square the gaps $p_i - p_j$ other wise they all would sum to zero

Alessandro Codenotti

07:15

@StudySmarterNotHarder Departure Songs - We Lost The Sea

StudySmarterNotHarder

@AlessandroCodenotti thx

Andrew Micallef

Neat, although $75 seams like a lot, I hope I didn't somehow manipulate you into purchasing another book that won't get read :P
(I'm just reading through the responses from @leslietownes and @TedShifrin re my question earlier, I appreciate the pointers!)
For the moment I feel like my maths is the weakest link in my understanding, hence asking here not physics :P

StudySmarterNotHarder

Nice

Andrew Micallef

also, my copy of griffiths came without the front and back covers so I have no idea what the integrals are in the tables!

StudySmarterNotHarder

Yeah, it's fun to ask never-before asked questions or at least not posted yet

and since you purchased it, it would be ethical

Um.. yeah, I removed that

Andrew Micallef

07:25

bad boy

StudySmarterNotHarder

don't want to get banned

Andrew Micallef

better

:P

StudySmarterNotHarder

There's nothing like a hardback book copy, it seems that flipping through pages though primitive is more efficient than adobe reader

I wanted to get the purple 2nd edition but I clicked the link on amazon and the book turns blue

I thought purple was an unusual math book color

The third edition has some chapters deleted and some chapters added

Andrew Micallef

@leslietownes I realise now on reflection that my strategy may not be particularly profitable in simplyfying the question, but I think my goal is to get to an understanding by following as many blind paths as possible, not efficiently getting the right answer

StudySmarterNotHarder

That's how you learn

And possibly one day stumble upon a fruitful path that no one has taken

Andrew Micallef

07:28

I've got a digital edition of I think the latest one, borrowing my partners library card (so dunno how ethical that is all things considered) but the html version has no integral tables :(

Yeah, maybe

Makes me think of the knowledge mines in Greg Egans novel "Diaspora"

I'm a huge fan of hardback textbooks too, but they are a bit too a) heavy and b) conspicous for everyday consumption

StudySmarterNotHarder

If $0 \leq f(n) \lt 1$ does $\sum_{n = 1}^{\infty} f(n)^{n}$ converge?

@AndrewMicallef they're nice for sitting on the couch with

Andrew Micallef

Gotta go agian now, have ddinner to cook

thanks again though :p

StudySmarterNotHarder

great. Enjoy your dinner!

Studying math is all about taking good breaks

robjohn

@StudySmarterNotHarder Not necessarily. If $f(n)=\left(1-\frac1n\right)$, then the terms tend to $1/e$, not to $0$.

StudySmarterNotHarder

@robjohn that is okay

with me

:)

I let the numbers do what they want and just observe

robjohn

07:42

Just answering the question. The answer was "no".

StudySmarterNotHarder

@robjohn does $$\sum_{n = 1}^{\infty} \dfrac{ p_{n+1} - p_n}{n^s}$$ converge for all $\text{Re}(s) \gt 1$ or something like that?

robjohn

$p_n$ is the $n^\text{th}$ prime?

StudySmarterNotHarder

Yes

I think we could even include $(p_{n+1} - p_n)^s$ into the power

robjohn

@StudySmarterNotHarder I believe so.

Though prime gaps are tricky.

StudySmarterNotHarder

Dirichlet series

In mathematics, a Dirichlet series is any series of the form ∑ n = 1 ∞ a n n s , {\displaystyle \sum _{n=1}^{\infty }{\frac {a_{n}}{n^{s}}},} where s is complex, and a n...

According to that subsection of the article we can safely take $\text{Re}(s) \gt 2$ since $p_{n+1} - p_n = O(n)$

and it will converge absolutely there

robjohn

07:50

@StudySmarterNotHarder sure, for $\mathrm{Re}(s)\gt2$

StudySmarterNotHarder

It's real part

so $s$ can be in the complex numbers

robjohn

In Dirichlet series, imaginary parts usually aid convergence.

$\sum\limits_{n=1}^\infty\frac1{n^{1+i}}$ converges conditionally.

StudySmarterNotHarder

So collect terms and write: $$\text{the sum above} = \sum_{ k = 1}^{\infty} k\sum_{n \in \Bbb{N} \\ p_{n+1} - p_n = k} \dfrac{1}{n^s}$$

I honestly don't know where I'm going with this :)

Just exploring, new to analysis

robjohn

08:06

primes are not an easy place to start, but if you can get a feeling for them, it might be good

porridgemathematics

good day, I am seeking verification of the following :

Let $X$ be a CW complex and $Y \subset X$ a subcomplex such that all cells of $X \setminus Y$ have dimension at least $n$, where $n \geq 2$. Then $X^{n-1} \subset Y$ and $X^{n-1} = Y^{n-1}$.

frankly i also don't see why $n \geq 2 $ is necessary

username

08:36

Suppose I were to start a mathematics related project, some sort of rating of professional mathematicians in departments around the world. Suppose rhat I was starting to program it on github, built a proof of concept, and then felt it was time to ask for some help and gauge the level of interest in such a thing.
It would be a very controversial thing to do, as it could be read to mean this mathematician is better than that mathematician, which is a very big no no,
By nature, it would not ask said mathematicians' opinion, or their department's submission, like funding agencies. It would ju…

Anyhow, suppose I did it anyway. Would this chat room be the right place to recruit volunteers to help me make it better?

Prithu biswas

09:03

I seached for 'what is a tensor' in the internet and found this interesting answer from William Oliver in Quora (it is the first answer). Can someone take a look at it?
https://www.quora.com/What-is-a-tensor

I found it interesting because this answer goes "kind of" contrary to almost all 'introduction to tensors' type videos in the internet.

Astyx

09:43

@ShaVuklia @TedShifrin Sorry I vanished yesterday, I saw Ted got involved and had some business to attend to. Is everything settled?

porridgemathematics

ooo I have a question if you have some time

Astyx

idk CW complexes sorry

porridgemathematics

ok np

StudySmarterNotHarder

10:09

Say you have $f(s) =\sum_{n = 1}^{\infty} \dfrac{\liminf_{k \geq n} (p_{k+1} - p_k)}{p_k^s}$

Can we somehow bring this liminf outside the summation?

Btw, by Zhang's result, $f(s) = C + Z(P(s) - \sum_{k=1}^{Z - 1} \dfrac{1}{p_k^s})$ for some $Z \lt $ 70 million where $P(s)$ is prime zeta

I mean it should be $f(s) = \sum\limits_{n=1}^{\infty} \dfrac{\inf_{k \geq n}(p_{k+1} - p_k)}{p_n^s}$ i.e.

The twin prime conjecture is that $f(s) = 2P(s)$ for all $s$ I guess

$\text{Re}(s) \gt 1$

robjohn

@StudySmarterNotHarder assuming the twin prime conjecture, your liminf would be $2$.

StudySmarterNotHarder

Yes that is correct

Instead it's $Z$ by Zhang

so far

robjohn

@StudySmarterNotHarder doh!

porridgemathematics

10:36

the suspension of a space minus one of the tips of the cones is contractible, right?

70 million prime conjecture

Andrew Micallef

10:57

Okay, after following this path through I can conclude that this is exactly one of those cases. By integrating with respect to $x$ I wind up injecting a second derivative with respect to $t$, which in this particular case is much less good to work with.

But this dead end has given me an idea for where the profitable path may lie. which will have to wait till my lunchbreak to explore.

satan 29

hi'

Andrew Micallef

hi

@TedShifrin Just as I am about to go to bed I realise I am denser than a lead brick. The freaking problem was of the form $\int{f(x,y) \cdot g'(x,y)}dx$...I can use integration by parts to solve it!

satan 29

11:12

some test examples were:

(a,b): (f/inf)

10,10 inf

1,10 f

6,9 inf

7,3 f

i tried to come up with a solution by purely looking at these examples, and it worked lmao

the ans is inf if gcd(a,b) >1, and f if gcd(a,b) =1

now, I tried to come up with a rationale for this

after messing around for a bit, it is clear that, for a no. to be coloured white, it must be of the form

$$ax+by$$

now, if gcd(a,b) is not 1, we can come up with infinite numbers such that they are nonzero mod( (gcd(a,b)))

which means there will be an inf no. of (non white), i.e black squares.

However, I am not sure if we can simply use this argument to conclude that if gcd(a,b) is 1, then we are guaranteed to have finite solutions?

Rover

Hey Hi Guys Can anyone please tell what really is Maths? What significance it has of it's own and is it just about numbers and no visualisation of what that equation is really signifying ?

Ashish Ahuja

11:40

@satan29 codeforces.com/blog/entry/71080

satan 29

@AshishAhuja oh I see

Ashish Ahuja

@Rover maths does have a lot of visualization, I personally like 3b1b videos; they show plenty of visualization, and I really liked the linear algebra series since it showed the visual side which usually isn't introduced in high school.

satan 29

@AshishAhuja wait

oh no nvm

Prithu biswas

12:00

(commenting again because it got behind a lot of comments) I seached for 'what is a tensor' in the internet and found this interesting answer from William Oliver in Quora (it is the first answer). Can someone take a look at it?
https://www.quora.com/What-is-a-tensor
I found it interesting because this answer goes "kind of" contrary to 'almost' all 'introduction to tensors' type videos in the internet.

geocalc33

12:18

Hello

Rover

@AshishAhuja Thanks the series are cool ! Focuses on visualisation!

Srijan M.T

13:00

@AshishAhuja Amazing bro.

qwerty uiop

13:26

hi

hello

Sha Vuklia

@Astyx No worries! I still have to figure sth out, but for now I'll leave it as it is

Alessandro Codenotti

@EdwardEvans can you ping me when you're around if you want to help me sort out an ANT problem?

geocalc33

13:41

$X_2=\big\{(x,y)\mapsto\big(ax,\frac{y}{a^2}\big)\big\}.$ Does this transformation have a name?

copper.hat

14:01

jimmy?

Srijan M.T

When finding trigonometric ratios for 90+theta. Why don’t we make the diagram like this ? I see we are not getting an angle of 90 degree. So , can’t we say there is no values of trigonometric ratio possible for 90+theta.[![enter image description here][1]][1]

[1]: https://i.sstatic.net/Lm9O9.jpg

https://www.math-only-math.com/trigonometrical-ratios-of-90-degree-plus-theta.html. So , on this site I checked the proof for this. Below is the passage needed to understand this proof. What I didn’t get is how is sin theta = FE/OE

Rover

14:17

The two methods of same integration and I think both are correct then also I am getting different answers !!

How's that possible?

@robjohn @copper.hat

Lucas Henrique

14:31

either they agree everywhere or you made a mistake.

or ZFC is inconsistent but I do doubt that

:p

robjohn

@Rover They are the same answer.

$\frac{\frac{2(x-1)}3}{2\left(\left(\frac{x-1}3\right)^2+1\right)}\cdot\frac{9/2}{9/2}=\frac{3(x-1)}{(x-1)^2+9}$

leslie townes

one of the most accursed lessons of integral calculus is how many different ways there can be of writing the same thing.

rational functions, expressions with trig functions in them, on and on. it's maddening.

Lucas Henrique

and that can be quite interesting. like proving that $\frac{\pi^2}{6} = \sum 1/n^2$

robjohn

The most commonly worrisome "different" answers are $\arcsin(x)+C$ and $-\arccos(x)+C$

Lucas Henrique

ok, I probably need help. my book says that the derivative $f^(k)$ of a function $f\colon U \subset \mathbb{R}^m \to \mathbb{R}^n$ can be inductively defined as a $k$-linear form

robjohn

14:40

Is $f^{(k)}$ the $k^\text{th}$ derivative?

or the coordinate $k$?

Lucas Henrique

for $k = 2$, I did the math and it's ok. how about the general case? how do you apply the canonical isomorphism $\mathrm{Hom}(\mathbb{R}^m, \mathrm{Hom}^k(\mathbb{R}^m, \mathbb{R}^n)) \cong \mathrm{Hom}^{k+1}(\mathbb{R}^m, \mathbb{R}^n)$?

robjohn

As soon as you wrote Hom, my eyes glazed over.

Lucas Henrique

lol

@robjohn: it's the $k$-th derivative

ok, so $f^{(k)}(x) = \sum\limits_{i=1}^m \underbrace{\frac{\partial f^{(k-1)}}{\partial x_i}(x)}_{k-1 \text{ linear}} \overbrace{dx^i}^\text{linear}$

robjohn

That looks like the chain rule for one derivative.

Oh, I see. You are doing that one derivative at a time

Lucas Henrique

my book talks about derivatives in any finite-dimensional normed spaces

the norm doesn't matter since they're all equivalent

Thorgott

14:55

should just be $\sum_{i_1<\dots<i_k}\frac{\partial^kf}{\partial_1^{i_1}\cdot\dots\cdot\partial_k^{i_k}}dx^{i_1}\wedge\dots\wedge dx^{i_k}$, methinks

Lucas Henrique

so the "actual" definition of $f^{(k)}$ is $(f^{(k-1)})'$ which belongs to the functions from U to Hom(R^m, Hom(...))

however, the author says that you can inductively define $f^{(k)}$ as a function from $U$ to the $k$-linear forms

my question is: how?

(this definition of derivative makes sense since $\mathrm{Hom}(E, E')$ is also a normed space when $E, E'$ are normed spaces)

Lucas Henrique

15:08

ok, so if you define $df$ as the (unique) linear transformation s.t. (blablabla...), then you can define recursively $f^{(k)}(x_1,\ldots,x_k) := d(f^{(k-1)})(x_k)(x_1,\ldots,x_{k-1})$

copper.hat

even at the simplest level (Boolean functions) its a big deal in (digital) circuit design to check if two functions (finite automata really) are the same.

Edward Evans

15:30

@Alessandro hey I just woke up lol, ask the Q and I shall think about it over a guilt-coffee

Leonhard Euler

15:49

whom'st'd've'dist'd'n't'st'd've'll's'd've're'n't'y'all'll'ven't is the father of topology?

Alessandro Codenotti

16:12

@Edward Still here? I was at a seminar

Edward Evans

Yeah I'm here, I can't think about it in real time though, I'm still trying to work through probability lol

Alessandro Codenotti

No problem, I need some setup anyway, let me write it down

I have $K$ a number field with its ring of integers $O_K$ and for $P\in\mathrm{Spec}(O_K)$ I defined the valuation $v_P\colon K\to\Bbb Z\cup\{\infty\}$ in the usual way. Now I fix $Q\subseteq\mathrm{Spec}(O_K)$ cofinite and $\Sigma=\mathrm{Spec}(O_K)\setminus Q$. I consider the ring $A_Q=\{x\in K\mid v_P(x)\geq 0\,\forall P\in A\}$ and I want to show that its group of units $U_\Sigma$ is finitely generated and that $U_\Sigma/O^\times_K\cong \Bbb Z^{|\Sigma|}$

I showed that $x\in U_\Sigma$ iff $xO_K$ factors as a product of stuff in $\Sigma$

So now the natural map $\varphi\colon U_\Sigma\to\Bbb Z^{|\Sigma|}$ is to take $x\in U_\Sigma$ and send it to $(v_{P_1}(x),\ldots,v_{P_{|\Sigma|}}(x))$, where the $P_i$'s enumerate $\Sigma$

This is a group hom with kernel $O^\times_K$, so I get an iso $U_\Sigma/O^\times_K\to\varphi(U_\Sigma)$

Unfortunately $\varphi$ doesn't look surjective at all to me in general

But I still want to argue that it's image is a maximal rank subgroup of $\Bbb Z^{|\Sigma|}$ which I don't see how to do

@AlessandroCodenotti woops in the definition of $A_Q$ I meant $\forall P\in Q$ rather than $\forall P\in A$.

Ted Shifrin

16:36

@LucasHenrique This is just multilinear alg. $Hom(E,Hom(F,G))\cong Hom(E\otimes F,G)$.

Edward Evans

@Alessandro Err this is a generalisation of Dirichlet's unit theorem for what you would call the $\Sigma$-units of your number field. I've seen this result before in the form of a not-short-but-not-long exact sequence lol

anakhro

Is definiteness and symmetry of elements of a Lie algebra well-defined if we go about defining the terms through Ado's theorem?

Edward Evans

@Alessandro shouldn't.. the fact that the $v_{P_i}$ are surjective do it?

$K^\times \to \Bbb Z$

Rover

@robjohn The two answers are 1/54[Tan^-1(x-1)/3 + 3(x-1)/((x-1)^2+9)] and -1/(x-1)^2+9.

robjohn

@Rover Oh, I looked at the bottom of both columns and assumed those were the answers. I see that the right column ended short.

Rover

16:51

@robjohn ok, ya the right was short ..

Alessandro Codenotti

@EdwardEvans hmm I'm not seeing it

robjohn

@Rover The mistake in the right hand answer is where you take $\int\frac9{(t^2+9)^2}\,\mathrm{d}t$ to $9\frac{-1}{t^2+9}$

Rover

@robjohn 9 is not there in numerator in integration?I thought so, there is possibility of mistake in that part and checked in many times , I will check again .

Edward Evans

@Alessandro alright I'll think about this lol, I only know this as a corollary of a bigger thing

robjohn

@Rover It comes from $\color{#C00}{-}\int\frac{\left(t^2+9\color{#C00}{-9}\right)}{\left(t^2+9\right)^2}\mathrm{d}t$

Alessandro Codenotti

17:04

But wait, am I misunderstanding you or what you're saying would imply that my $\varphi$ is surjective? Because I don't think that can be true for arbitrary $\Sigma$

geocalc33

Just following up about: $X_2=\big\{(x,y)\mapsto\big(ax,\frac{y}{a^2}\big)\big\}.$

robjohn

@Rover: The reason images of your work are so hard to correct is not only trying to read someone else's handwriting, but also we cannot copy and paste from what you give us. We need to generate latex from the images to put here.

Please try to learn latex so that you can use MathJax here to make things easier on others.

Rover

@robjohn okay I will.

leslie townes

i also had to turn my monitor on its side.

geocalc33

0

Q: About a collection of linear maps $L=\bigcup_{i=1}^\infty X_i$

I recently asked: About this transformation, and I wanted to follow up with another related question: Consider a collection of linear maps $L=\bigcup_{i=1}^\infty X_i.$ Where $X_1=\big\{(x,y)\mapsto\big(ax,\frac{y}{a^1}\big)\big\}.$ $X_1$ is called a "squeeze mapping" for parameter $a.$ I don't k...

any thoughts, concerns, hints?

robjohn

17:09

@leslietownes I copied the image to my computer and rotated the image with software, but yes, that was a problem

images are not the best way to present math here

geocalc33

1

Q: About the transformation $(x,y)\mapsto (ax,\frac{y}{a^2})$

Suppose $f(x)=\frac{1}{x^2}$ and $g_k(x)=kx$ for parameter $k\in\Bbb R_{>0}.$ $g_k(x)$ maps onto itself as $k$ increases. The transformation maps a point on $f(x)$ to another point on $f(x).$ How would you write down this transformation more mathematically? Is it related to another, more well-kn...

linear-algebra linear-transformations transformation

leslie townes

i used to have a hotkey on one of my computers that flipped orientation. e.g. control-alt-left would put make the left edge of the screen the bottom and the right edge the top. it was a lot of fun.

Rover

@robjohn oh yes I got where I went wrong!

Thanks @robjohn @leslietownes I will definitely try to learn latex .

leslie townes

i think math.SE has a show latex option where you can see what people are doing to get what they get on posts. it could be helpful. i learned mostly from examples, which thankfully were good examples. there's a lot of bad latex out there.

robjohn

17:25

@leslietownes something other than right clicking on the rendered latex?

leslie townes

oh no, just that. i knew it was in there but had forgotten what it was because my tex is so masterful now that i don't need to look at that of others.

geocalc33

so you might say that you're a tex master?

leslie townes

yes. speaking of that i think your "where x_1 = " should be a "where x_i =" and again in the formula.

a set theorist could tell you better than i whether you want L to be the set $\{X_i : i \in I\}$ as distinguished from the union of the $X_i$.

Ted Shifrin

@leslietownes Far easier just to suspend yourself sideways!

leslie townes

i will rig up some kind of hoist system. i've been meaning to do that.

robjohn

17:30

@TedShifrin I don't know if you saw my image rant.

We have several people who post images of their work here. I can see once or twice, but for heaven sake, learn LaTeX!

If it's worth your time to post several times here...

geocalc33

@leslietownes maybe that's an issue. I was going by what StudySmarterNotHarder was saying

Ted Shifrin

Well, it's not just LaTeX, but laziness to type up all the scribble in a legible fashion.

I support your rant. (Plus we had our friend RK who was posting image after image of test booklet.)

robjohn

@TedShifrin Same page, several views :-p

leslie townes

@geocalc you refer suggest that maybe L is a one-parameter group, and appear to mean group in the algebraic sense, but you don't explain what the operation is. it isn't clear to me what $X_i \star X_j$ ought to be, or that so defined it would satisfy an addition law. i'm not saying it isn't, i just can't see it. there's also some X_1 vs. X_i ambiguity later.

there's also the issue of the issue of the potential group structure if you fix i and let a vary. people who downvote may be confused by that. or they could just be jerks.

geocalc33

@leslietownes yeah, I see what's making it not a great question

Ted Shifrin

17:41

Confusingly written questions which make it clear that the OP has no idea what he/she is talking about ... deserve downvotes.

Speaking in the abstract here.

geocalc33

@leslietownes no I was saying that the collection of $X_1$ (squeeze maps) is a one parameter group isomorphic to the multiplicative positive reals

not $L$

Edward Evans

@Alessandro How about considering elements of $U_\Sigma$ that are a power of exactly one prime in $\Sigma$

i.e. $(x) = \mathfrak{p}^e$

then the image of this guy is $(0, \dots, e, \dots, 0)$

and guys of this form generate a rank $|\Sigma|$ subgroup in $\Bbb Z^{|\Sigma|}$

leslie townes

17:59

geocalc33 I sort of get it, but "the collection of X_1" is already a bit confusing. as i read it, a is fixed, so X_1 is just one map. you get a collection if a is allowed to vary, but then maybe the parameter should be built into the notation.

Edward Evans

18:22

@Alessandro alright that was slightly too simplistic but I have an answer for you now lol

Alessandro Codenotti

That's what I wanted but how do you guarantee that $p^e$ is principal for some $e$ for all $p\in\Sigma$?

Edward Evans

Right you want to take $e = h_K$ (the class number of $K$)

Alessandro Codenotti

Ahhh right

Nice, thanks

Edward Evans

that took longer than I wanted it to rofl

@Alessandro Here's a thing: there's an exact sequence

$$0 \to \mathcal{O}_K^\times \to U_\Sigma \to \bigoplus_{\mathfrak{p} \in \Sigma} K^\times/(\mathcal{O}_K)_\mathfrak{p}^\times \to \operatorname{Cl} K \to \operatorname{Cl}_\Sigma K \to 0.$$

Since $\mu_K \subseteq \mathcal{O}_K^\times \subseteq U_\Sigma \subseteq K^\times$, the torsion of all these guys is $\mu_K$. Also, the class groups are finite, so you get $\operatorname{Rank} U_\Sigma = \operatorname{Rank} \mathcal{O}_K^\times + \operatorname{Rank} \bigoplus_{\mathfrak{p} \in \Sigma} K^\times/(\mathcal{O}_K)_\mathfrak{p}^\times$. …

which is the nuclear option lol

Thorgott

18:39

famous SES, of course

every child knows this

Edward Evans

did it in primary school bruv

leslie townes

i didn't even go to primary school. i learned it on the streets.

Edward Evans

lmao with chalk on the pavement outside

Ted Shifrin

19:28

pictures leslie writing math graffiti on the sidewalks

leslie townes

no geometrical diagrams though. just letters and matrices and stuff

Ted Shifrin

Of course. You would have been spanked if there were a geometric picture.

leslie townes

i wasn't allowed to play four-square or hopscotch. too close to interacting with geometric figures. tetherball, forget about it.

Ted Shifrin

Wouldn't want you to turn out square.

Andrew Micallef

20:34

Morning

Andrew Micallef

20:45

So integration by parts seamed like a great idea last night,
and I found myself in a kind of infinite recurssion.
I had $\int{UdV} = UV - \int{VdU}$. Then realised that I didn't know how to solve, $\int{VdU}$, so I used integration by parts again, to get $UV - \left(VU - \int{UdV}\right)$ at which point I decided to go to sleep because performing another integration by parts was clearly not going to help.

And I suspect now is probably a good time to study Taylor Series.

satan 29

20:58

what are U and V in your case?

Ted Shifrin

@AndrewMicallef no, @Andrew, this is the trouble with differential notation. Write $uv’$. Then it turns into $vu’$, which in turn can turn into $Vu”$, with $V’=v$, etc.

Andrew Micallef

Huh, okay, I'll try again at lunchtime (in 4hrs time)

Balarka Sen

Hi, @Ted.

Andrew Micallef

Thanks

Ted Shifrin

Heya a Balarka!

Balarka Sen

21:09

I got an interview call from TIFR (aka I passed the graduate program entrance test).

Thanks, all, without years of conversations with whom I would not have learnt enough math to qualify.

Nike Dattani

My first attempt at answering a representation-theory question :) Let me know if you have any comments or think there's any way to improve my answer!

2

Q: What is meant here, when it is said that "diagonal generalized Gell-Mann matrices are not unique"?

According to the answer given here, the diagonal generalized Gell-Mann matrices are not unique. But what exactly is meant by this? Are they just saying that we can multiply the diagonal matrices by a constant and still have a valid generator of the group since the matrices will still be orthogo...

group-theory representation-theory lie-groups physics

Ted Shifrin

21:25

Wow, a Balarka. I know not about programs in your country. It seems like yesterday that you were merely Balarka 1.0!

Heartfelt congratulations. Maybe Mike gets $2\epsilon$ credit and I get $\epsilon^{3/2}$, but all credit goes to you!

Balarka Sen

@TedShifrin: Thank you! I would argue about the credits, but maybe some other day :)

Ted Shifrin

@Thor taught me something second-hand from you yesterday!

Balarka Sen

Oh?

Ted Shifrin

Suspension is only a top manifold for spheres. I guess I know about links of singularities, but I still never realized this.

Balarka Sen

Aha!

Avra

21:32

Hello all. Cam you please think of any affine equivariant location estimator for $X_n$ matrix

Balarka Sen

It has always intrigued me, however, that if you suspend a non-manifold, it can become a manifold.

Avra

any affine $m(X_n)$ location estimator.

Balarka Sen

In fact, suspending the Poincare homology sphere twice gives a sphere.

Ted Shifrin

Crazy. Good thing I never pretended to be a topologist.

Balarka Sen

Haha. Hopefully I'll learn some topology in graduate school.

Ted Shifrin

21:34

But I should thought about links and pseudomanifolds when I screwed up and Thor meducated me.

And some complex geometry, Balarka.

Balarka Sen

Yeah that'd actually be nice. There are some serious complex algebraic geometers in TIFR, so I think I will have the opportunity.

Narasimhan, Seshadri, Ramanan, et al are all TIFR production

@TedShifrin Speaking of complex analysis and crazy topology, do you know polynomial mating?

Ted Shifrin

Nope!

Balarka Sen

I don't know too much about it, but what the people there do is something like so: Take two, let's say quadratic, polynomials $z^2 + a$ and $z^2 + b$. These give rise to certain dynamics on $\Bbb C$, given by iterating these maps.

Ted Shifrin

Ah, instead of iterating one only.

Balarka Sen

Oh, no, sorry. I meant you get a pair of dynamical systems by iterating each, which is familiar to you of course.

On a vast part of $\Bbb C$ the system is "totally discontinuous", the Fatou sets, which is usually the basin of attraction of a critical point. The complement of the Fatou set is the Julia set, where the system is chaotic.

People glue pairs of filled-in Julia sets (Julia set with "everything inside") for each of these dynamical systems (one for $z^2+a$, one for $z^2 + b$, say), along the Julia set. By topology magic, the resulting object is $S^2$.

Equip $S^2 = \Bbb{CP}^1$ with a complex structure. Now you can ask if the map $S^2 \to S^2$ you have given by mating these dynamical systems is uniformizable to a rational map. Often they are.

That's what these people do.

Ted Shifrin

21:51

Not my tea of cup, but interesting.

Balarka Sen

Yeah, very different flavor from complex geometry. I just like the topology part; who knew gluing pairs of solid Mandelbrot sets along the Mandelbrot set is $S^2$?

Thorgott

are they doing that for fun or is it for something else

robjohn

@BalarkaSen Does that include monomial husbandry?

Balarka Sen

Lol

Ryan Unger

@BalarkaSen why aren't you coming to the us?

robjohn

21:54

@TedShifrin is that like teaching combined with substance abuse?

Balarka Sen

@Thorgott Maybe for fun. The point is mating dynamical systems which are not semigroup actions and actually group actions is more natural, and has been studies throughout centuries, starting with Fricke-Klein. Bers simultaneous uniformization for example says that a vast number of discrete subgroups of $\text{SL}_2(\Bbb C)$ come from mating pairs of Riemann surfaces, in an appropriate sense.

Ted Shifrin

I'm not much for substance, robjohn.

Balarka Sen

The precise theorem is that "quasiFuchsian subgroups of genus $g$ of $\text{SL}_2(\Bbb C)$ are in 1-1 correspondence with $\text{Teich}(\Sigma_g) \times \text{Teich}(\Sigma_g)$"

This has some serious consequences in 3-manifold topology (Thurston et al)

@RyanUnger I didn't think I should to be honest. I had only three years of undergrad, with no publication. It'd be kind of hard to get into a good school.

I might after Masters (this is a MSc+PhD program), but I haven't decided.

Leaky Nun

@TedShifrin as-tu deja alle a un seminaire bourbaki?

Ted Shifrin

I think you'd do fine, @Balarka. With good GRE. You will have strong letters.

Oui, une fois, je crois, Leaky. 1980.

Balarka Sen

21:59

I was not so sure. GRE wasn't there this year because of COVID. Maybe I'd have strong letters, but less in number - I haven't actually done serious reading projects with too many people.

It seems dishonest to ask random people for recommendations!

Leaky Nun

@TedShifrin as-tu quelques histoires de cette experience pour partager?

Ted Shifrin

By the way, Leaky, it's être + allé, but you'd say assister à, not aller.

Leaky Nun

thanks for the corrections

Ted Shifrin

Nah. It was just a lecture.

Ryan Unger

@BalarkaSen you didn't solve a gromov problem as an undergrad?

Balarka Sen

22:00

Nah man

Ryan Unger

I owe people money

you let me down

Balarka Sen

lol

I have things in the drafts. I'll see in my MSc. Don't keep high hopes

Ted Shifrin

I could write for you, Balarka, but I'm not at the Harvard-Princeton level.

Balarka Sen

@TedShifrin Haha, thank you. It's quite alright though, I'm happy with where I am right now.

Ted Shifrin

But don't stress. Right, no SATs or GREs.

They don't. Remember my allegiance is to MIT.

Balarka Sen

22:03

TBF, between you and me, I don't think I'd make a good research mathematician. But I'll get back to you on that after I drop out soon.

Ryan Unger

@BalarkaSen very few people here consider your situation when trying to abolish the gre

not sure where I stand on the issue

the GRE is certainly bad

Balarka Sen

Yeah, GRE is annoying

Ryan Unger

I am unsure if that means we should just get rid of it though

@TedShifrin I have some collaborators and a roommate who would disagree

Balarka Sen

@RyanUnger Mohan and co have been doing some good stuff.

Ted Shifrin

Harvard turned me down for college and grad school, but I turned them down for postdoc . So there!

Balarka Sen

22:05

I talked to him yesterday

Ryan Unger

I live next to mohan

Balarka Sen

Lol @Ted!

@RyanUnger Oh!

Ryan Unger

I didn't apply to harvard for college or grad school

Ted Shifrin

And I turned Princeton down for grad. Very glad I did.

Ryan Unger

eye roll

Ted Shifrin

22:07

Can't possibly have done better than the adviser I had.

Roll away.

Ryan Unger

idk who was working at princeton in the 50s to be fair

papa?

Ted Shifrin

Very funny.

Balarka Sen

Anyway, academic things give me a headache. Just gonna try to make enough money so that I can move north, to a colder place.

Ted Shifrin

Asshole.

Ryan Unger

?

Ted Shifrin

22:08

How far north, Balarka?

Balarka Sen

Iceland would be good, but extremely costly!

Maybe Sibera

Ryan Unger

svalbard

Balarka Sen

lol

Ted Shifrin

Verrrry cold!

Balarka Sen

@TedShifrin: You would want that too if you lived where I do. Daytime temperatures are nearing 40 degrees C out of the gates!

Soon we'll be scorched by the sun

Global warming is probably a conspiracy to @Ryan though

Ryan Unger

22:10

why?

Ted Shifrin

Yeah, I can't take heat and humidity. But my body doesn't do well in super cold after cancer.

Another hoax, yes.

Balarka Sen

@RyanUnger Just kidding. But it seems the right temperament for you. Wasn't the Earth flat or something?

LOL

Ryan Unger

hm, that doesnt say much

Balarka Sen

@TedShifrin Ahh yeah I forgot.

Ryan Unger

I'm not saying the earth is flaty

I don't like being framed this way

Ted Shifrin

22:12

Just shows that math talent is not correlated with intelligence.

Ryan Unger

I just have no reason to think it isnt

Balarka Sen

@RyanUnger lolol

Nah, Ryan is fine. He just read too much Hawkings-Ellis and played too many Fallout games as a kid so that messed him up

Ryan Unger

what

Idt I played fallout until college

but yeah too much hawking ellis for sure

Balarka Sen

Gotcha

Transcript for

$Mathematics$ Mathematics