Mathematics

Daminark

9:00 AM

@Balarka is algebraic k-theory an acceptable compromise?

(I don't actually know any so ples don't accept this compromise)

Balarka Sen

no

k theory would be ok though

Just win baby

Moore topology?

Balarka Sen

did Peter May give any more lectures

SteamyRoot

Point-tail and line-fence homotopies? :D

Daminark

His side thing and the main shtick have converged at least for now. I missed the last two lectures, he did lecture today a bit

So apparently in the last two classes they defined CW complexes and then cellular homology

Today, Peter May reviewed Eilenberg-Steenrod and translated it to homology

Balarka Sen

9:02 AM

isee

Daminark

And was doing some other stuff about some differential though I didn't follow

The idea was basically that Dylan had defined homology somehow, and Peter May was trying to show how the axioms basically forced you to define homology like so

Balarka Sen

mhm

Daminark

He had some fantastic quotes today

Balarka Sen

oh?

abenthy

@TobiasKildetoft Right :/ Thank you

Daminark

9:07 AM

"Hang the homotopy on the cone!" -proving that sequence with cofibers is exact

“All this is in Concise. This is unreadable but… oh you can read it!”

-self-explanatory

“It’s a wonderful exercise in understandingd the definitions but there is no way I’m doing it on the board" -re homology and differentials

Balarka Sen

heh

Just win baby

"Quotations" are fun.

Secret

[Brainstorm challenge 001] Details: For each of these challenge, a mathematical object will be given. Your task is to find or come up with creative, useful, and/or interesting generalisations:

Today's brainstorm challenge begins with this:

$$f(x)=x$$

SteamyRoot

That's literally fixed point theory

Balarka Sen

you really like fixed points dont you

SteamyRoot

9:12 AM

Well, yeah, I do research in them.

Daminark

Better than broken points emirite?

SteamyRoot

There's no denying that $f(x) = x$ is all about fixed points, though :P

Daminark

That too

Balarka Sen

i wish i knew more fixed point theory

abenthy

Can someone think of a subgroup of $SL(n,\mathbb{Q})$ which is commensurable with $SL(n,\mathbb{Z})$ but not a subgroup of $SL(n,\mathbb{Z})$?

Tobias Kildetoft

9:14 AM

@Daminark Well, if they aint broken, don't fix them

Just win baby

You should have an "m" there for generality.

Daminark

'tis true

Actually Neves keeps going into this shtick about how you've hit gold if you've found a fixed point for something

tru

T

hi chat

9:22 AM

hi pal

Balarka Sen

Hi Alessandro

BAYMAX

Hi chato-tacko

SteamyRoot

@BalarkaSen I'd recommend Tsai-han Kiang - "The theory of fixed point classes". Requires only some basic knowledge on liftings and covering spaces, homotopy theory, and fundamental groups.

@Daminark You mean my research isn't completely useless? Yay!

Daminark

So it seems

Balarka Sen

@SteamyRoot Thanks! I'll check it out.

Secret

9:28 AM

One interesting question would be: Fixed points under fractal mappings

(Obviously that diagram does not quite reflect that, need to think of a more concrete example)

math.uchicago.edu/~may/REU2012/REUPapers/Natoli.pdf

SteamyRoot

Erm... not sure what you mean with a fractal mapping

But fractals tend to be generated using contractions

So basically all you have to worry about is the Banach fixed point theorem (assuming you work in a Banach space, of course)

BAYMAX

$f(z) = z^2 + c$

fractals tend to be generated using contractions ?

Mandelbrot set ?

is the above a contraction map

?

Secret

I guess one possibility is that the function takes a set of points, and its image is some fractal.

For example, we can consider some kind of map which if the input set is $\Bbb{R^n}$, then the image is the Julia set in dimension n. Now we can then change the input into something else, and the image will be a "Julia set" version of the domain

So in a sense, it is a generalisation of simple functions like $2x$ $e^x$ such that the function itself applies a fractal to the domain x

SteamyRoot

@BAYMAX Well, no, that's a different kind of fractal.

Secret

(Not sure how can that be implemented, but we do know we can generate fractals via iterated function systems)

user84215

9:36 AM

What is wrong with the following command?
f(n) = \begin{cases) n/2, & \text{if $n$ is even} \\ 3n+1, & \text{if $n$ is odd} \end{cases}

Secret

\begin{cases) <------------ } not )

BAYMAX

ok

user84215

Thanks

Secret

Another way to phrase what I have in mind about fractal maps is, if the identity function is plug into it, then it generates the Julia set for example. We can then consider other types of functions, and the output will be like a warped version of that function which should take the form of fractals except in very special cases

SteamyRoot

@aminliverpool It's not wrong, but... first of all, using $'s is bad practise if you're using LaTeX instead of TeX. And second, using mathmode inside text inside mathmode? Ew :P

user84215

9:41 AM

that is, you say that I should not use $ in the above formula?

Secret

$f(n) = \begin{cases} n/2, & \text{if }n \text{ is even} \\ 3n+1, & \text{if }n \text{is odd} \end{cases}$

SteamyRoot

@aminliverpool It's preferrable to use something like \text{if } n \text{ is even}

Because anything that may change the "text" environment would also impact the $n$ inside.

And, in LaTeX, $ ... $ and \[ ... \] are preferrable to dollar signs

user84215

Ok. But I want to use $$ at the first and end of the sentence.

Secret

Hmm, so they have done it, cool
http://maths-people.anu.edu.au/~barnsley/pdfs/nigel4.pdf
http://superfractals.com/wpfiles/fractal-transformations-2/

user84215

I have just tested the alternative. I see no difference between them.

Secret

9:52 AM

Infinite compositions of analytic functions

In mathematics, infinite compositions of analytic functions (ICAF) offer alternative formulations of analytic continued fractions, series, products and other infinite expansions, and the theory evolving from such compositions may shed light on the convergence/divergence of these expansions. Some functions can actually be expanded directly as infinite compositions. In addition, it is possible to use ICAF to evaluate solutions of fixed point equations involving infinite expansions. Complex dynamics offers another venue for iteration of systems of functions rather than a single function. For infinite...

This notation will be useful for my various investigations involving recursive functions

SteamyRoot

You know, a fractal is just the unique fixed point of some Hutchinson operator :^)

Secret

Is it possible for the hutchinson operator to have more than one fixed point, a brief review on its definition seemed to suggest not as you are always contracting the same set?

One thing I have in mind is a set which form a "landscape" such that each unique point is a distinct fractal

However I am not sure if the hutchinson operator is too "one dimensional" to do that

nvm, found the first query here

1

Q: Can't there be more than one fixed points in a contraction? or none?

I was going through the contraction mapping theorem in my book where it says, that if $\phi: G\to G$ is a contraction, then $\phi$ has a unique fixed point $\alpha$ on $G$. Sequence {$x_n$}, $x_{n+1} = \phi(x_n) $ for $n=0,1,2...$ converges to $\alpha$ with $$|x_n - \alpha |\leq\frac{\lambda^n ...

fixed-point-theorems

hindawi.com/journals/joper/2016/5250394

might be relevant

Abcd

10:20 AM

@MaryStar The part before my last to last message

user84215

A strange thing happens:
The following command does not work well.

user84215

$$f(n) = \begin{cases} \frac{n}{2}, & \text{if $n$ is even} \[2ex] 3n+1, & \text{if $n$ is odd} \end{cases}$$

user84215

But when I add "a" at the first and end of this command it works well

user84215

a
$$f(n) = \begin{cases} \frac{n}{2}, & \text{if $n$ is even} \\[2ex] 3n+1, & \text{if $n$ is odd} \end{cases}$$
a

user84215

What is the problem?

Tobias Kildetoft

10:31 AM

@aminliverpool The first one has a \ too few

user84215

No. They are the same

Tobias Kildetoft

No, I just checked the source and they are not

the first one has a single \ instead of a \\ where it needs to break the line

user84215

I just copy and paste the command.

Tobias Kildetoft

which makes the single \ escape the [ and mess everything up

user84215

$$f(n) = \begin{cases} \frac{n}{2}, & \text{if $n$ is even} \[2ex] 3n+1, & \text{if $n$ is odd} \end{cases}$$

user84215

10:34 AM

a
$$f(n) = \begin{cases} \frac{n}{2}, & \text{if $n$ is even} \\[2ex] 3n+1, & \text{if $n$ is odd} \end{cases}$$
a

Tobias Kildetoft

Where are you copying from? It is trivial to see that the first code will mess up

user84215

I first type the first command and then add "a" at the first and end of it

user84215

$$f(n) = \begin{cases} \frac{n}{2}, & \text{if $n$ is even} \[2ex] 3n+1, & \text{if $n$ is odd} \end{cases}$$

user84215

a
$$f(n) = \begin{cases} \frac{n}{2}, & \text{if $n$ is even} \\[2ex] 3n+1, & \text{if $n$ is odd} \end{cases}$$
a

user84215

It is very strange.

user84215

10:42 AM

MathJax automatically changed my first command, but it did not change the second one.

user84215

I put \\ before [2ex] in the first command. But MathJax changes it.

Tobias Kildetoft

$$f(n) = \begin{cases} \frac{n}{2}, & \text{if $n$ is even} \\{[2ex]} 3n+1, & \text{if $n$ is odd} \end{cases}$$

Yeah, same here. That is really strange

user84215

But when I add "a" at the first and end of the command, the problem is solved

Secret

$f(n) = \begin{cases}\frac{n}{2}\\, & \text{if $n$ is even} { [2ex ] } 3n+1, & \text{if $n$ is odd}\end{cases}$

Tobias Kildetoft

@aminliverpool It is probably a problem with the \[...] thing that MathJax does not like, but it is extremely odd that adding the a fixes it

Yeah, that was suppose you have two \\ as well, but the first one was removed for some reason

So in some cases, the chat (or MathJax) will remove a \

Secret

10:48 AM

$f(n) = \begin{cases} \frac{n}{2}, & \text{if $n$ is even} \\ {[2ex]} 3n+1, & \text{if $n$ is odd} \end{cases}$

gxyd

What is the meaning of "solution space of a matrix"?

Secret

\\{ might be misinterpreted as \ \{ (escaped {), I think

Tobias Kildetoft

@gxyd Not necessarily anything without more context

gxyd

@TobiasKildetoft This is a line from a comment I received in a discussion: "The task is to find those linear relations where the coefficients $r_i$ are rational numbers. The computation depends on the existence of some basis over $\mathbb{Q}$ but the result should be a matrix of rational entries ($r_i$'s) such that the rational relations ($\sum_{i=1}^{n} r_i f_i$) are exactly the elements of the rational solution space of the matrix."

@TobiasKildetoft if you think that the comment is not complete in some sense, I can tell you for what the missing thing means.

@TobiasKildetoft that rational relation is $\sum_{i=1}^{n} r_i f_i = 0$ (I missed equals zero in that).

Is its meaning similar to "column space of a matrix $A$"?
That is the linear combination of a columns of $A$ (i.e $A x$).

Secret

11:27 AM

To be googled: parameter dependent intersections of two sets

or more accurately, an intersection that is variable

One possible way to implement this:

Let $S$ be a set of functions $g : A \mapsto A$ and $K_i \subset S$. Then define the map $f = \bigcap_i^n K_i$

gxyd

@TobiasKildetoft what I think rational solution space means here is: just the rational linear combination of $f_i$'s i.e $\sum_{i=1}^{n} r_i f_i$ where $r_i \in \mathbb{Q}$ (i.e for all rational values of $r_i$'s).

user84215

11:46 AM

What is wrong with the following command?
$$\begin{array}{c}\begin{array}{cc}\begin{array}{c|cccc}\text{min} & 0 & 1 & 2 & 3 \\ \hline 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 1 & 1 \\ 2 & 0 & 1 & 2 & 2 \\ 3 & 0 & 1 & 2 & 3 \end{array} & \begin{array}{c|cccc} \text{max} & 0 & 1 & 2 & 3 \\ \hline 0 & 0 & 1 & 2 & 3 \\ 1 & 1 & 1 & 2 & 3 \\ 2 & 2 & 2 & 2 & 3 \\ 3 & 3 & 3 & 3 & 3 \end{array}\end{array}\begin{array}\Delta & 0 & 1 & 2 & 3 \\ \hline 0 & 0 & 1 & 2 & 3 \\1 & 1 & 0 & 1 & 2 \\ 2 & 2 & 1 & 0 & 1 \\ 3 & 3 & 2 & 1 & 0\end{array}\end{array}$$

Simply Beautiful Art

@Secret You are your fixed points lol

user84215

I want to put the third table below.

Secret

That might be true, I am a fixed point of the weirdness function, but there are many more weirdos I want to learn from which are also fixed points. Whether we forma set of ordinals I have no idea, but let's hope not, cause it will be too boring to contain each other in it

user84215

$$\begin{array}{c}\begin{array}{cc}\begin{array}{c|cccc}\text{min} & 0 & 1 & 2 & 3 \\ \hline 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 1 & 1 \\ 2 & 0 & 1 & 2 & 2 \\ 3 & 0 & 1 & 2 & 3 \end{array} & \begin{array}{c|cccc} \text{max} & 0 & 1 & 2 & 3 \\ \hline 0 & 0 & 1 & 2 & 3 \\ 1 & 1 & 1 & 2 & 3 \\ 2 & 2 & 2 & 2 & 3 \\ 3 & 3 & 3 & 3 & 3 \end{array}\end{array}\begin{array}{c|cccc}\Delta &0&1&2&3 \\ \hline 0 & 0 & 1 & 2 & 3 \\1 & 1 & 0 & 1 & 2 \\ 2 & 2 & 1 & 0 & 1 \\ 3 & 3 & 2 & 1 & 0\end{array}\end{array}$$

Secret

$$\begin{array}{c}\begin{array}{cc}\begin{array}{c|cccc}\text{min} & 0 & 1 & 2 & 3 \\ \hline 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 1 & 1 \\ 2 & 0 & 1 & 2 & 2 \\ 3 & 0 & 1 & 2 & 3 \end{array} \begin{array}{c|cccc} \text{max} & 0 & 1 & 2 & 3 \\ \hline 0 & 0 & 1 & 2 & 3 \\ 1 & 1 & 1 & 2 & 3 \\ 2 & 2 & 2 & 2 & 3 \\ 3 & 3 & 3 & 3 & 3 \end{array}\end{array}
\\ \begin{array}{c|cccc}\Delta &0&1&2&3 \\ \hline 0 & 0 & 1 & 2 & 3 \\1 & 1 & 0 & 1 & 2 \\ 2 & 2 & 1 & 0 & 1 \\ 3 & 3 & 2 & 1 & 0\end{array}\end{array}$$

user84215

11:50 AM

I want to put the third table under the others.

SteamyRoot

then end the line after the first two arrays?

user84215

Thanks

Liad

hi, i have $\Bbb Z \times Y$ where $Y$ is a set of two points with the trivial topology.
i want to show that this space is limit point compact.
So , if $A \subset Z \times Y$ is infinite, and $(z,y) \in A $ then if $U$ contains $(z,y)$ , $U$ also contains $z \times Y$ , now i want to conclude that $U \cap \{A - (z,y) \} \ne \emptyset$$

im not sure why this is true. someone can help?

Las Vegas Raiders

12:13 PM

@Secret those particular users are known to test one's patience.

LegionMammal978

Small question: I've been trying (as part of a small project of mine) to solve the equation $fs-\frac12ts+f-\frac12t=0$ for $s$, where $f,t,s\in\Bbb R$. Mathematica tells me the only solution is $s=-1$, but how would I figure that out?

facepalm nvm I'd just isolate it

gxyd

@TobiasKildetoft if you think I need to improve upon my question asking way, I had be happy to hear if that is your opinion.

Simply Beautiful Art

12:30 PM

@AkivaWeinberger @EricSilva math.stackexchange.com/questions/2357336/…