Could you change "at least two points." to "infinitely many points"

Seems as soon as you want a simply connected region in $\Bbb CP^1$ which admits atleast two points, you are already omitting infinitely many points

You can without loss of generality assume one of the points you omit is at infinity

And then if there are only finitely many points omitted from $\Bbb C$ you'll have a big fundamental group

Sure, since $\pi_1(\Bbb C \{x_1, \cdots, x_n\})$ is free on $n$ generators

The point they're making is much more explicit though: if you delete one point, that's biholo to the plane, and is not biholo to the disc

And deleting no points leaves something compact

Any other simply connected region is biholo to the disc though

Good point

Thanks

hey

A q..

and I now need to find $f^{-1}\circ g^{2}$

where $g^{^{2}}=g\circ g$

How'd I go about that?

Sorry

messed that up

need to find $g^{^{-1}}\circ f \circ g^{^{-1}}$

without the where clause

Or generally, how to do this kind of thing?

@RandoHinn Not sure what you're stuck on? If you know how to apply these, the inverse should be clear?

$(15432)$ has inverse $(23451)$

In which case $g^{-1}\circ f\circ g^{-1} = (23451)(132)(45)(23451)=(123)(45)$

ooh, that seems.. logical

Can be reduced to $(15432)$

Which you can read as "one goes to 5 (which) goes to 4 (which) goes to 3 (which) goes to 2 which goes to 1"

I just read as "1 goes to 5 goes to 4 goes to 3 goes to 2 goes to 1"

And $f=(132)(45)$

Where you break them up if they don't interact

1 goes to 3 goes to 2 goes to 1, and 4 goes to 5 goes to 4

Then you can read them right to left $(23451)(132)(45)(23451)$ following $1$ from right to left

right most says $1$ goes to $2$ and then $(45)$ doesn't interact, then $2$ goes to $1$, and then the left most term $1$ goes to $2$, so all in all $1$ goes to $2$

Then you repeat for $2$

Take this again: $f^{-1}\circ g^{2}$ Why do I think it is 45213 and the test thinks it's not?

You follow his whole path through $(23451)(132)(45)(23451)$ from right to left $2\mapsto 3\mapsto 3\mapsto 2\mapsto 3$, i.e. $2\mapsto 3$

deer god

okay, I need to learn this...

$f^{-1}\circ g^2 = (312)(45)(15432)(15432)=(15)(243)$

$f^{-1}\circ g^2=\left(\begin{matrix} &1 &2 &3 &4 &5 \\ &5 &4 &2 &3 &1 & \end{matrix} \right )$

In your giant notation

@tigre and this thing in my notation...

lower row 23154?

no?

probably no

The (123)(45) in my notation, I mean

@tigre?

or

54231 ?

Hm?

You want to write $(123)(45)$ in your notation?

That would have lower row $2\,3\,1\,5\,4$

Just read downwards in your notation

and the second column says $2$ goes to $1$

The third $3$ goes to $2$

The fourth $4$ goes to $5$

The fifth $5$ goes to $4$

So in my notation this is just $(132)(45)$

does anybody knwosq the second power series of $4x^2y''(x)-4x^2y'(x) + (1-2x)y(x)=0$ the first one is $\sum_{k=0}^{\infty}1/k! * x^{k+1/2}$? (this is the first time I'm in a chat can I just interupt?

You can just interrupt, that's fine Mari3, although I have to go now, so someone else will have to help you

thanks

$\asymp$

sum of eigenvalues of an infinte matrix are a convergent infinite series?

Not necessarily (I'm guessing that infinite matrix means compact operator on a separable Hilbert space)

yes, compact operator on a seperable hilbert space. But actually I meant to say if you require the sum of the eigenvalues to be a convergent infinite series

Look at the operator $e_n\mapsto \frac1ne_n$ on $\ell^2$

Silly question: if I wanna show $\sum_{n=-\infty}^\infty e^{\pi i n^2 z}$ is holomorphic on the upper half plane I need to show convergence of $\sum_{n=-\infty}^\infty \pi i n^2 e^{\pi i n^2 z}$ ye?

$(z \in \Bbb H$ is the variable lol)

Hi @ÍgjøgnumMeg

hey @Alessandro

How's the semester going?

@ÍgjøgnumMeg yeah sure

I see we're doing the same thing!

@Alessandro it's getting a lot better, the first few weeks were really really tough

@Leaky yeah? :)

I'll recognize that function from a mile away :P

lol yeah

I'm supposed to be showing that the 8th power of the theta function is a weight 4 meromorphic modular form wrt the subgroup generated by T^2 and S (where those are the usual integer matrices in this context) lol

@ÍgjøgnumMeg I think that's normal, German's bachelors are a bit crazy compared to rest of Europe so everybody seems so much more well prepared

@Alessandro indeed, the first year analysis/complex analysis modules here are comparable to master level modules in the UK lol

the theta function is beautiful

Also I dropped a lecture course because I wasn't really understanding anything :(

but that's fine because I still have 3 courses this semester lol

Same here, the first algebraic topology course of the bachelor program was equivalent to the only master course on the topic at my old uni. Except that here they have another bachelor and two master courses as follow up lol

@Leaky yeah it's a very cool exercise

@Alessandro saaad, I'll be taking alg. top. 1 in my 3rd semester alongside alg geo 1

Nice

does anybody knwos the second power series of $4x^2y''(x)-4x^2y'(x) + (1-2x)y(x)=0$ the first one is $\sum_{k=0}^{\infty}1/k! * x^{k+1/2}$?

algebraic number theory is going super well, getting marks between 80 and 90% in the exercise sheets

@Alessandro how's your semester going? :)

I'm doing some really nice set theory and a couple of courses I don't really care much about but I need the credits so...

Faaair

I also officially have a supervisor for my thesis now so I'm starting to work on that as well

Nice! Congrats, what's your topic?

Some stuff about asymptotic dimension of proper metric spaces. It's somewhere between geometric group theory and geometric topology

Sounds nice, is one expected to produce original results here?

It's expected but not strictly necessary

ble

h

I think I might end up needing 5 semesters for my masters tbh

My supervisor proved a nice result using a big theorem by another guy, so he wants me to understand this big theorem and write down a self-contained exposition of the theorem and all you need to get to it (which will already be enough for a thesis), ideally I should then try to simplify my supervisor's proof by using the other's guy idea directly, without invoking this big theorem, but we'll see how that goes

Sounds nice :)

$\between$

@Ultradark matter

Did you solve that transcendental number prob?

Hey, remember this?

Is that the thing where the smallest solutions have dozens of digits?

Yeah

And this

which is also a parody but less subtle

I had an idea, maybe someone can make a cleaner version

Express your answer in terms of apples, oranges, and blueberries please

@shi What if you have a square matrix with zeros everywhere except on the diagonal s.t. the trace equals the riemann zeta function?

@shi no it's a very difficult problem

@Ultradark idk, lol, it's a complex problem

pun

For $$ \int f(x).g(x) dx = \int f(x) dx \times \int g(x) dx $$ find all the possible $f(x) \, and \, g(x)$....

^this problem has given me lifeless sleeplessness for nights...

Also given, $$f(x), g(x) \in \mathbb{R}^{+} \, \forall x \in \mathbb{R}$$ and they are continuous...

@AbhasKumarSinha don't you have a group of such functions?

Since $f(x) \in \Bbb{R}^+$

@Leaky do you know how to check meromorphy for a modular form? The lecture notes say it's enough for the modular form to be "meromorphic at the cusp" but I'm not sure what this means in the case that your congruence subgroup isn't all of $\operatorname{SL}_2(\Bbb Z)$ (because there's more than one cusp class in this case)

well, actually what I mean is it's enough for your function $f$ (that satisfies the relevant invariance under the action of $\operatorname{SL}_2(\Bbb Z)$) to be meromorphic at $\infty$ for it to be a modular form*

wrt. SL_2(Z)

@ShineOnYouCrazyDiamond you there?

@Ultradark yep

Started ANT book just now

I have an equation: $$ \exp(1/\log(A))= A$$ where $A$ is a matrix that has a logarithm

@Shine alg or analytic?

algebraic

yaaa

s

which book?

@Ultradark don't you get $\log^2(A) = 1$?

yeah

Neukirch

Nice

I know what an algebraic integer is, lol

or that's what I'm going over now, the start of the book

@shi what's the solution?

@Shine cool :)

@Ultradark what ring do your matrix entries lie in?

Neukirch isn't very soft if that's your first time learning ANT

do you think it's possible to solve it?

@Alessandro it's complete af tho

@Ultradark looks possible, but I don't know the technique

@Ultradark introduce $\sqrt{A}$

@ÍgjøgnumMeg Without doubt, but because of that he does much more stuff (and in a more general setting) than a usual first course

define that using some series maybe?

@Alessandro right, my first run-through I used the book "Introductory Algebraic Number Theory" by Saban Alaca and Kenneth Williams

but it didn't use any Galois theory at all rofl

LOL!

josuke?

Possible candidate for deletion: math.stackexchange.com/questions/3428482/…

@stressedout I disagree, it's missing context but it's technically a "good question" because it's the kind of thing beginner topology students might find useful

@ÍgjøgnumMeg I remember there was a time that I lost 300 points answering questions that were allegedly missing context but were far better than this question.

This is a very easy question. The question is ambiguous (it is not clear whether the OP is asking about one particular set or every subset of the topological space) and the question is very typical. So it has probably been answered a million times before.

And it smells of hypocrisy because one of the people who has answered the question was the main person who voted for similar questions, which were far-better worded, to be deleted.

Anyone here familiar with solving continuous MDPs?

@Rietty If the probability of A happening is x, what is the probability of A not happening?

1-x

I believe he wanted to say "the chance that at least one of them beats you..."

So if I wanted the chance that none of them beats me, it would still be $0.85^N$?

Yes. The chance that you don't get beaten (:P) after N rolls is 0.85^N :P

Ok. Just was a bit puzzled on wording. Thank you.

You're welcome. I too think that the wording is confusing.

it's a $d$-fold direct sum of the $m\times m$ matrix you got for the same operation in the special case $L = K(x)$

@ÍgjøgnumMeg yes that's what I can't see

@Shine Say you have $K$-vector spaces $V_1, V_2, V_3, V_4$ and $T_A : V_1 \to V_2$ and $T_B : V_3 \to V_4$, then $T_A \oplus T_B : V_1 \oplus V_3 \to V_2 \oplus V_4$ is a linear transformation sending $(v_1, v_3) \mapsto (T_A(v_1), T_B(v_3))$. If $A, B$ are the respective matrix representations of $T_A$ and $T_B$ then $A \oplus B$ is a block diagonal matrix representing $T_A \oplus T_B$.

In the example in the book you have $L$ is a $K(x)$-vector space with basis $a_1, \dots a_d$

so interpret $L = \bigoplus_{i=1}^d K(x)a_i$

$(=K(x)^d)$

and for each of the $K(x)$ you have a $K$-linear transformation given by the matrix you wrote in the question

so altogether you have a $K$-linear transformation of $L$ given by a $d$-fold direct sum of that matrix

Yes, but how is that equal to $T_x$

?

@ÍgjøgnumMeg

you know why $T_x$ has the matrix you wrote down for $K(x)$?

I didn't write down any such matrix !

:D

you did lol

And I don't know what you mean "for $K(x)$"

as in, the multiplication by $x$ map $: K(x) \to K(x)$

$T_x : L \to L$

I'm trying to prove that $T_x$'s matrix is a block-diagonal of the form shown

I'm pretty sure it's straightforward

I'm just not seeing it

I know lol, the point is that the matrix shown is the matrix representation of the multiplication by $x$ map from $K(x) \to K(x)$, and then you extend it to $L$

Oh, that makes more sense what you just said

and then extension to $L$ is what gives you the block diagonal matrix

so then $T_x' : K(x) \to K(x)$

is $y \mapsto xy$

right

Then to extend to all elements of the form $z = \sum_{i=1}^{d} z_i a_i$

where $a_i \in L$

And $z_i \in K(x)$

I'm not sure how to do that

Yeah if the $a_i$ are your basis elements for $L$ as a $K(x)$-vector space

So how do you compute $T_x(a_i)$

I told you how to do it above

Not sure how that works

Can you explain not using block-diagonal

Do you mean just define: $T_x(a_i) = a_i$?

$T_x (z) = \sum_{i=1}^d T_x(z_i) a_i$

Then $T_x(z + z') = T_x(z) + T_x(z')$

and $T_x(cz) = \sum_{i=1}^d cT_x(z_i) a_i$

for $c \in K$

Is that the extension you're talking about of $T_x$ to all of $L$?

@ÍgjøgnumMeg

So then $T_x(a_i x^j) = a_i x^{j+1}$

How does that weird $P$ matrix come into play?

Sorry went downstairs

Hi @Ted

@Shine the matrix $P$ represents the linear transformation $T_x : K(x) \to K(x)$

it's called a "companion matrix" btw

for some terminology

@ÍgjøgnumMeg thanks for teaching

I will come back to this later

coding now

On BananaCats

which you may very well use one day or your students

If you don't know it, look up "the matrix of a linear transformation" in whichever algebra/linear algebra book you want

@Shine perhaps :P

Yes, I know it's $T(\text{column}_j)$

I mean colum_j = T(basis_j)$

so compute that for $T_x$ and you'll see it's the matrix $P$

and then the fact that $L = K(x)^d$ gives you the block diagonal guy

Is that compositum of fields?

Or vector space cartesian product?

Sorry I meant as vector spaces

Thx

np

= p

:D

P = NP all day!

haaa

Are you a teacher?

(Note: the exact passage in Neukirch that you're talking about was on my first problem set for algebraic number theory this semester lol)

@Shine no I'm a master student

You are a master yoda

Do you know any Python?

This has been useful for me too, I was having problems understanding it myself a couple of weeks ago

Not really, I know rudimentary programming stuff for testing hypotheses

Cool! Don't hack us all

:>

I know some sympy

but I mostly do PyQt5 gui apps

Fair, I mostly use maple

but not even that much

It's difficult to support real math in an app

each time has to be a full fledged CAS essentially

lots of parsing

my bachelor's thesis supervisor is an expert on computer algebra lol

he writes homological and differential algebra packages for maple

Nice. Lucky - he gets to play with chain maps and stuff

*they

yeah it's a he

How do you know?! bad joke

lol

well he certainly identifies as a he

@ÍgjøgnumMeg do you know about diagram chasing?

as in, proving things by running around diagrams?

Yep

meh I don't have any experience with it

I haven't done any category theory

I'm trying to support it in my app called BananaCats, but you end up having to code a full-fledged math / logic system from scratch

it's difficult

fair

I'm willing to do that under the conditions that there is not successor function for natural numbers, i.e. integers will just be coded at a higher level not at the lambda calc level

The users ideally will be able to code the library for the thing visually

The type expression $a : A$

will be a node "a" within a larger node "A".

And there will be a way for communicating these transformation rules to the system

and then back to their actions on the diagrams

But to support all that cohesively

I needed to parse out variables from general LaTeX strings

bleh

That led me down the rabbit hole of parsing the expressions down to operators *, +, -, etc

Then that led me to the full fledged thing

So the grammar for user input is now 2 pages long

and not finished

Anyhow, the end result is hopefully something mathematicians will want to use

sounds like it'd be useful :)

Drawing large diagrams and doing chases on chalk board or paper will seem like a lesser experience than using this tool

It does searches for

you, of various theorems. As these are just encoded as possible diagram transformations that are valid.

In logic you have elimination and introduction rules that cancel one another. So for instance $a : A \Rightarrow $ a diagram with a node in a node

The inverse of that is to take a node within a node and convert it back to textual

So I tie the logic system to the diagrams directly by drawing the diagram "in the rule specification".

Lots to code

sounds

hard

It is hard. But in theory it will work

Features are easy to talk about but take months to impl

I just blew my mind thinking about it for a second. Best to focus on one thing at a time lol

Good news is the graphics system works and is 98% done

That's not supposed to make any logical sense. That was a screen cap for a post I had a question on