Mathematics - 2020-10-15

Faust

05:30

Anyone still awake?

Alessandro Codenotti

13:28

@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

Stupid question inc

13:58

are tensors taught in linear algebra?

Thorgott

14:19

depends

Sophie

14:39

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

Mike Miller

15:34

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

skullpatrol

16:02

sell these to make Easy Money during the pandemic

Seth Taddiken

16:16

Question about probabilities:
If we have a continuous random variable X and P(a<X<b) then is this equivalent to P(a<X)*P(X<b) since we want a<X AND X<b? I'd guess no because P(a<X)*P(X<b)=F(b)*(1-F(a)) but the original inequality should be equal to F(b)-F(a) where F is the cdf of X

Alessandro Codenotti

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?

Balarka Sen

16:35

@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.

Alessandro Codenotti

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

Balarka Sen

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 $\Bbb R^2$.

Alessandro Codenotti

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

Balarka Sen

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.

Alessandro Codenotti

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

Balarka Sen

16:40

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

Alessandro Codenotti

@BalarkaSen Wait, what is disconnected?

Balarka Sen

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

It has $6$ components.

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

Alessandro Codenotti

Ahh of course. Diagonals are bigger than I thought

Balarka Sen

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

Alessandro Codenotti

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?

Balarka Sen

16:44

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

Alessandro Codenotti

Ah I see

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

Thanks

Balarka Sen

Cool!

Alessandro Codenotti

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?

Balarka Sen

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.

Alessandro Codenotti

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

Balarka Sen

16:52

:) I am so sorry

Alessandro Codenotti

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?

Balarka Sen

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.

Alessandro Codenotti

Aha, now it makes sense!

Balarka Sen

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.

Alessandro Codenotti

Yeah that makes sense, thanks

Balarka Sen

16:58

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.

Alessandro Codenotti

Ah interesting

Balarka Sen

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.

Alessandro Codenotti

I'll leave for dinner soon unfortunately

Balarka Sen

Ah ok maybe some other day. Enjoy!

Alessandro Codenotti

Sure, it seems interesting!

Balarka Sen

17:01

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.

Alessandro Codenotti

I don't really know anything about Braid groups

Balarka Sen

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

Alessandro Codenotti

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

Balarka Sen

Oh huh

Alessandro Codenotti

en.wikipedia.org/wiki/Dehornoy_order this thing

Balarka Sen

17:02

OH WHAT

Oh yes I know this order!

This is extremely geometric

Alessandro Codenotti

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

Balarka Sen

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

Alessandro Codenotti

@BalarkaSen Ah makes sense

Balarka Sen

Linearly ordered groups are excellent

Tobias Kildetoft

Is it still open whether $B_5$ is linear?

Or was it the more general question that was open?

Balarka Sen

17:05

Linear as in?

Tobias Kildetoft

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

Balarka Sen

Wow.

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

Tobias Kildetoft

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

Leaky Nun

@AlessandroCodenotti ok thanks

Balarka Sen

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

Is there some text I could look into

Leaky Nun

17:10

personally I know nothing about them

Tobias Kildetoft

Not sure about a good source for braid groups

Balarka Sen

Hm thanks

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

Tobias Kildetoft

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

Balarka Sen

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

Joseph Rock

19:18

It is quite obvious that (𝑋∘)∘⊂𝑋∘

does anyone know how to show this?

Charlie

19:33

nvm, explain in the next line lmao

amWhy

19:49

@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

robjohn

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

amWhy

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

Luigi M

20:20

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

Sayan Chattopadhyay

20:35

@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

Luigi M

@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

robjohn

21:51

@LuigiM as long as the integrals converge absolutely.

@LuigiM I see that you already knew that.

skillpatrol

22:13

@robjohn Circling the square?

skillpatrol

23:22

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$Mathematics$ Mathematics