Anyone still awake?

@LeakyNun In mine I mostly talked about the areas of math that interest me and my background/which courses and stuff I took in them

My application also asked for an excerpt of my masters thesis, so I talked quite in detail about the stuff I was doing for my thesis as well

are tensors taught in linear algebra?

depends

Do you ever just stumble upon a matrix of order 6? i.imgur.com/NPw8lhO.png

@Sophie Quite special too. I believe every order 6 matrix of determinant 1 with integer entries is conjugate to that one.

sell these to make Easy Money during the pandemic

What is a simple example of a finite degree covering $p\colon E\to B$ with connected $E$ for which the deck group is not cyclic?

@Alessandro Simple as in?

I would like to propose $X = (\Bbb R^2)^3 \setminus \Delta$ where $\Delta$ consists of all the "diagonals", so that $X$ is the space of three distinct points, ordered, in $\Bbb R^2$. Then $S_3$ acts naturally on $X$ by permuting all three coordinates, this is a free and properly discontinuous action.

I don't know, some stuff with nice manifolds I guess

The quotient $X \to X/S_3$ defines a covering space with $S_3$ deck group.

The base $X/S_3$ is configuration space of three points on $\Bbb R^2$.

Uhm why do you need $(\Bbb R^2)^3$ instead of using $\Bbb R^3$ directly?

Points in $\Bbb R^3$ correspond to three (not necessarily distinct) points on $\Bbb R$

I wanted three distinct points on $\Bbb R^2$, so one copy of $\Bbb R^2$ for each point.

Yeah I mean $\Bbb R^3$ without diagonals

And then minus diagonals to make them distinct

@Alessandro Ah but that is a silly disconnected space.

Three points on $\Bbb R$ doesn't have any freedom.

They always preserve order if you move them around

@BalarkaSen Wait, what is disconnected?

$\Bbb R^3$ minus diagonals. Space of three points, ordered, on $\Bbb R$.

It has $6$ components.

Maybe this is not the simple example you wanted. Nevermind!

Ahh of course. Diagonals are bigger than I thought

Yeah it's much better to think of three points moving around in $\Bbb R$, or $\Bbb R^2$ (clearly the latter is connected but the former isn't!)

This is like your 3-transitivity problem

So whenever $G$ acts propertly discontinuous $X\to X/G$ is a covering map, right? And then if $G$ acts freely the deck group is $G$ again?

Covering iff the action is free and properly discontinuous :) Deck group is always $G$ whenever this happens.

Ah I see

Ok I like the "moving around $3$ points in $\Bbb R^2$ example"

Thanks

Cool!

Also I don't quite understand how the correspondence between deck of universal cover and $\pi_1$ of the base is used in practice. I'm being told that it makes computing $\pi_1$s easier, but determining the universal cover and the deck group looks quite hard in general to me?

So for example, if $W$ is simply connected, and $G$ acts properly disct. and freely on $W$, then you get $\pi_1(W/G) \cong G$

Without any effort.

Hm wait I'm confused again by the $S_3$ example

:) I am so sorry

Shouldn't I get fibres with $6$ elements? Because deck has the cardinality of a fiber when it acts transitively on the fibers and the base space is somewhat reasonable I think?

Yes, and you do. The map $X \to X/S_3$, is forgetful; it takes an ordered triple of points in $\Bbb R^2$ and sends it to the unordered triple.

The fiber over an unordered triple of points consists of all possible orderings on it. There are $3! = 6$ orderings on an unordered triple of points aka a set.

Aha, now it makes sense!

Here is one possible way to make the distinction, conceptually, because it confuses me as well. You can think of $X$ as the space of 3 distinct colored points on $\Bbb R^2 = \Bbb C$, red green and blue. Moving in this space creates a braid with colored strands, red green and blue.

So paths in this space are colored (open) braids. Loops are colored closed braids.

Downstairs in $X/S_3$, you forget colors.

Yeah that makes sense, thanks

There is a slick proof using this example, and what you're being told about how computing $\pi_1$s become easier, of the fact that braid groups are torsion free.

Ah interesting

I can tell you after I am done with my stupid 2 problem complex analysis assignment. Maybe then we can start thinking about how hyperbolic braid groups are... they contain $\Bbb Z^2$'s, but maybe they are relatively hyperbolic mod these torii.

I'll leave for dinner soon unfortunately

Ah ok maybe some other day. Enjoy!

Sure, it seems interesting!

Braid groups also happen to be MCGs of punctured disks, and we know MCGs act on curve complexes, which are Gromov hyperbolic. But maybe for braids it's actually a better situation.

I don't really know anything about Braid groups

I don't know, cool stuff. Me neither!

I know there was something weird going on with Braid groups inspired by large cardinals

Oh huh

en.wikipedia.org/wiki/Dehornoy_order this thing

OH WHAT

Oh yes I know this order!

This is extremely geometric

Yeah I think what happened is that this order was discovered through a connection to some problem relying on large cardinals assumptions, but once it was known to exist people figured out simpler ways to obtain it

Note that a left-invariant total order autoforces $B_n$ to be torsion free

Which is another way to see it is torsion free

Akiva taught me this proof

@BalarkaSen Ah makes sense

Linearly ordered groups are excellent

Is it still open whether $B_5$ is linear?

Or was it the more general question that was open?

Linear as in?

has a finite dimensional faithful representation (let's say real)

Wow.

$B_3$ naturally sits inside the universal central extension (read: cover) of $\text{PSL}_2(\Bbb R)$, which is not linear, right?

ahh, never mind, that has been settled for a long time

@AlessandroCodenotti ok thanks

How much do we know about representation theory of braid groups in general?

Is there some text I could look into

personally I know nothing about them

Not sure about a good source for braid groups

Hm thanks

This reminds me, I need to read: pdmi.ras.ru/~arnsem/Arnold/arnold_MZ69e.pdf

I have mainly studied Coxeter groups as the closest thing to them, and those tend to be treated very differently

Ah right

Coxeter groups are great but I don't know anything about them

I just know lots of examples of triangle groups

I feel like I won't learn half the math I want to learn about in my lifetime so what's the point

Upload my consciousness in a quantum computer already

It is quite obvious that (𝑋∘)∘⊂𝑋∘

does anyone know how to show this?

nvm, explain in the next line lmao

@robjohn In the spirit of Halloween (almost two weeks from now), will you transform your identicon (with or without face-mask, to a mean orange circle/pumpkin?? ;D I challenge you to that! ;D

@amWhy I'm orange already. There is no "mean circle", so I don't know if that will happen.

@robjohn I wanna see you a pumpkin!! pwetty pwease? Think about it. Consider it a trick, instead of a treat! ;D

Does Fubini theorem applies to continuous functions over unbounded domains? I'm trying to come up with a counterexample of the form f(x,y) continuous and defined say on R but such that the iterated integrals don't match

I was able to find only Non-continuous counterexamples to Fubini

(of course the function has to change sign otherwise Fubini-Tonelli applies)

f defined on R^2

@LuigiM I do not have much time as its pretty late here, but if you want to look at continuous counterexamples maybe you can try to construct measures that are not sigma finite

@SayanChattopadhyay Are you saying that if we take the standard measure on R2 (which is sigma finite as far as I can tell) the iterated integrals of a continuous function always coincide?

Oh I see, this is exactly Fubini Tonelli! I've been misreading the statement of the theorem

@LuigiM as long as the integrals converge absolutely.

@LuigiM I see that you already knew that.

@robjohn Circling the square?