12:02 AM
@EdwardEvans good luck

12:17 AM
How do I flag three posts of the identical question — allegedly by different people, but I don't believe it? What's worse, the offender first emailed me and then posted a question and never had the politeness to acknowledge my comment/correction (even in the subsequent post). This behavior should be censured.
If a mod sees my question, here are the links. 1 2 3
I posted to the Mods chat.

Hi. Could you please look at the question I just posted? Thank you
0

The problem I am looking at asks to evaluate the moment generating function of the multivariate normal distribution $N\sim(\mu,\Sigma)$ using moment generating function of the standard normal distribution. I have the followings in my notes: For $k$-dimensional $\mathbf{X}$ on $(Ω, F , P)$ with d...

12:33 AM
I'm not a probability expert, but can't you just apply $e^{u+v}=e^u e^v$ and separate the integrals?

@Agi Ted is correct

Score 1 for Ted.

@TedShifrin I know you're rusty on measure theory, but just thought I'd give this a shot: define the Borel sets in $\mathbb{R}$, $\mathcal{B}(\mathbb{R})$, as the $\sigma$-algebra generated by all open sets in $\mathbb{R}$. How would you show that it has cardinality equal to the continuum? I don't even know where to begin.
(not homework, just studying)

It's not just that I'm rusty. I don't like the stuff. But there are several people in here who can do this in a second (e.g., @Thorgott @Alessandro).

I'm just looking forward to finally doing some probability stuff... it has been a headache to learn about regularity of measure over the last few weeks

12:38 AM
@Clarinetist Do you know how to build it "layer by layer"?

@AlessandroCodenotti No idea what you're talking about.

Countable union of countable unions, etc.

uhh, transfinite induction?

So when you say "the $\sigma$-algebra generated by the open sets" you have the whole thing at once

@TedShifrin Even if I do that, I'm not sure about the $dF(x)$ part, since I think the variables are not considered independent from one another. I also don't get where should I use the mgt of std normal

12:39 AM
I looked up what transfinite induction is, but I wouldn't know where to apply it
@Agi They must be independent for that factorization to work with the MGFs

Why are the variables $(x_1,\dots,x_k)$ for $\Bbb R^k$ not independent?

But you can start with open sets as the first layer, take union of complements as the second layer, unions of complements of things in the second layer as the third one and so on

OK, Clarinet can help Agi and Alessandro can help Clarinet, and Ted — thankfully — can be lazy.

@Clarinetist so I can assume they are independent?

Then the proof boils down to showing that layer one has cardinality $\mathfrak c$, moving up from one layer to the next doesn't increase cardinality and the slightly harder part is that there are at most $\omega_1$ layers

12:40 AM
The only explicit description of the Borel-algebra I know is by transfinite recursion, so I'd try proving the cardinality doesn't exceed that of the continuum by transfinite induction - I haven't actually done this before, tho

(the proper name is the Borel hierarchy for what I'm talking about)

@Agi Yes, I believe it can be shown that product factorization on the MGF of the sum holds if and only if those variables are independent

(considerably harder is showing that, in this case, there are exactly $\omega_1$ distinct levels, but luckily that is not needed)

@AlessandroCodenotti Never heard of it until now, thanks. My professor had given it to us as an exercise, and I suspect he may throw it on our exam. As far as I can tell, no one in my chat group for my class knows where to begin.

@Clarinetist Oh yes for product factorization they should be independent but I am not sure if they are here since there is no statement saying they are independent.

12:43 AM
@Agi Actually, it does - it says in the problem that $X$ has independent components

@Clarinetist That was just what I found in my notes, the question is just to find the mgf of multivariate normal dist using standard normal mgf

that sounds like a rather hard exercise, unless you're doing something more set-theory oriented

@Thorgott Definitely not. That was brought up as an exercise in our first measure theory lecture.
@Agi The very definition of the multivariate normal distribution requires that each component of the vector can be written as a linear transformation of independent $N(0, 1)$ random variables
@Agi Let me give this some thought quick

I was wondering if someone can provide intuition as to the difference between $f(x+1)$ and $f(x)+1$? If you have a function $f(x)$ and it's inverse $f^{-1}(x)$ is it possible the inverse of $f(x+1)$ is $f^{-1}(x)+1$?

have you tried some examples

12:50 AM
@Thorgott I'm not sure where one would even begin by way of "examples"

that was to northerner
as for your exercise, I can not think of a way of tackling it that isn't built on using ordinals/transfinite induction or something along those lines
which is why that exercise strikes me as rather odd in a general measure theory course

@Clarinetist This gives some insights but I'm still confused on how to begin. courses.washington.edu/b533/lect4.pdf

@Agi Sorry, I realize now that question isn't as trivial as I thought it'd be. So first things first, you're probably going to be dealing with using the fact that multivariate normal distributions can be written as a linear transformation of standard-normal random vectors

@Clarinetist yes
@Clarinetist In the document I sent, would proof def3 to def2, in the second page, suffice?

@Agi Oh. I think this would be the method of proof: let $\mathbf{Z} \sim \mathcal{N}(\mathbf{0}, \mathbf{I})$. Find its MGF. Then apply linear transformations on MGFs. Then you'll reach the result in that document.
@Agi Yeah, it is more or less that.
I haven't had to think about multivariate normal distributions from first principles in a while, thanks for asking that.

12:56 AM
@Clarinetist Sure thing, thank you too!

@Thorgott I think I'm just going to bet on the professor not asking that on the exam at this point. I agree it's silly to put as an exercise.
The solution I found deals with uncountable ordinals... I don't even know what that means

"Book". PDF

@Agi One of the things I've learned when it comes to multivariate normal distributions is that 95% of the time, direct integration isn't the way to do work with them.

Well, you actually can get a physical copy here if you want bookstore.ams.org/mbk-135
but it's mostly pictures

@Clarinetist indeed

1:00 AM
Also

1:55 AM
This is a question about transformation matrix. After watching 3blue1brown videos, I've assumed that a 2x2 transformation matrix has a 2 vectors—i hat and j hat. i hat lies on the x axis and j hat lies on the y axis. what about 3x3 matrix?

Order 2 branched cover of the trefoil knot:

do i hat lies on the x axis, j lies on the y axis and k hat lies on the z axis? Also what is the convention for this axises? Example: the x axis is pointing to my right, y as my up and z as my front. Is that correct or z and y are swap?

2:17 AM
hey chat
when it comes to matrix Lie groups, can you generally speak about the connectedness of $\mathrm{GL}_n(\mathbb{C})$?

> There seem to be two kinds of geometers: those who imagine that their
mathematical objects are something that could fi t on your desk, and those
who imagine their mathematical objects are room-sized, something you
could travel through.

I read a few questions about topological aspects of this group, but I'm not sure if they hold independently of the topology

- Sebastian Bozlee (though the idea might be older)
It's connected in the usual topology induced by $\mathbb C^{n\times n}$, and I believe it's connected in the Zariski topology, but you can certainly give it the discrete topology or something and make it disconnected
Right?

I don't know, because a Lie group requires that the group is a differentiable manifold

Oh
(I like to use movement - have z be my thumb, but curl the rest of my fingers together so they all start at x and end at y)
@LucasHenrique I think the topology is induced by the Lie algebra
Hm, maybe not

2:24 AM
I feel like this might be a subtle question
No wait, it's easy
No wait, it's not

lol
I mean, different topologies give different tangent spaces and thus different Lie algebras, right?
If you have the Lie algebra you have the Lie group
But maybe you can have two Lie groups that are isomorphic as groups but not (topologically) homeomorphic

you mean different smooth structures?
it's a fact that if a topological group has a Lie group structure, that Lie group structure is unique up to diffeomorphism
but you might be able to equip a group with two different topological group structures, which in turn admit different Lie group structures

The question is about different topologies
Yeah

discrete topology would give an easy counter-example if we don't require manifolds to be second-countable, but alas, we do
this is awkward

@AkivaWeinberger So z does seem pointing up instead of y. What about these 3 vectors i hat k hat and k hat? I mean not lie groups. In this rotation matricea wikimedia.org/api/rest_v1/media/math/render/svg/… do i hat represent the x axis, j hat as y and k hat as z?

2:29 AM
i is x, j is y, k is z
You can rotate your hand so that y is up
"x towards me, y up, z right" follows the right hand rule
So does "x right, y up, z away from me"
"x east, y up, z north"
Wait did I do that right
No I didn't
Swap x and z all of those

@AkivaWeinberger that confuses me as y is pointing left.
in the picture

If you rotate your hand it still follows the right hand rule
There are only two possible orientations for a coordinate system: right handed and left handed
You can't rotate your right hand to look like your left hand, so these really are different
All three of these are right handed (since they're rotated by 120 degree rotations about the line x=y=z)
There are 6 logical possibilities for permuting three objects
3 are right handed, 3 are left handed

the question reduces to: given a group, are there two topologies both turning the group into a topological group and into a topological manifold, yet such that the resulting structures are not homeomorphic

0

Question in title. Said another way, can you have two Lie groups that are isomorphic as groups but not homeomorphic? If so, the group isomorphism map will not be continuous (and thus not a Lie group isomorphism), and there will be no natural map between their tangent spaces (Lie algebras). I susp...

It seems to me that there are no convention, as far as I know, where x, y and z is pointing in the real world. But I'll stick to the right hand rule. @AkivaWeinberger thanks for you accompany.

2:43 AM
nice

I want a 2D video game that takes place in the universal cover of a disk minus the origin (perhaps stylized as a top-down view of a spiral staircase). The branch cut can be located opposite the player
Spiral staircases solve VR's locomotion problem, actually
the problem of walking in the virtual world while not crashing into walls in the real world

1 hour later…

1 hour later…
4:54 AM
Good video:
Klein bottle hidden in space of 3x3 high-contrast patches found in natural images

2 hours later…
7:09 AM
Can Jordan Form always diagonalize a matrix, assuming matrix is not diagonalizable?

if matrix not diagonalizable, nothing can diagonalize matrix

7:23 AM
@LeakyNun well then what is the benefit of Jordan matrix?

the powers of the Jordan blocks are easy to compute

So Jordan form handles the case where the geometric multiplicities is not equal to algebraic one? right?

right

I see. Thanks.
I thought for a moment that Jordan Form will do it for any square a matrix.
@LeakyNun do you know an example where the matrix is not diagonalizable at all?

$\begin{pmatrix}1&1\\0&1\end{pmatrix}$
you sound like you're treating putting a matrix into Jordan form as "diagonalization"

7:34 AM
@LeakyNun no but if I can compute the jordan form, I can come up with the actual matrix that carries out the diagonalization
right?
S^-1 A S = J , where J is the Jordan block matrix. Then it is easy to compute S if it exist. This my understanding.

yeah but we don't call that diagonalization
@CroCo out of curiosity, what's your native language?

Arabic
So I should treat Jordan without associated it with the notion of diagonalization, right?

right

My English is not perfect. Maybe you caught me with an accent.
lol

well I'm just interested in languages

7:40 AM
@AkivaWeinberger cool vid!
hi chat

@LeakyNun I see. Good luck. I can help if you want

How would I write a measure as a "limit" of a sum of Dirac masses?

@LeakyNun are you aware of other applications for Jordan form other than computing the powers ?

It should be clear that discrete measures are dense in the space of all measures in eg $\Bbb R$, right?
You just look at the the CDF, break $\Bbb R$ into $1/N$-pieces, localize measure at the center of each interval, and done
This is in the weak topology
I guess this goes through for any Polish space for sure

7:57 AM
@BalarkaSen are you assuming any reglaurity?

I think it's true that the space of finite-support measures is dense in the space of all sigma-finite measures or something
for any Polish space
Hm but i am not sure what i really want
Nvm I see what I want

@AlessandroCodenotti ty

8:48 AM
Let D be a division ring and M_n(D) a matrix ring over D. Then M_n(D) is finite n^2 dim. It can be shown that it is D-Artinian, is this simply from the fact that M_n(D) is simple?

9:15 AM
@Hawk what is D-Artinian?

9:31 AM
@BalarkaSen you can even restrict to linear combinations of rational centered deltas
That's the usual argument to show separability in the proof that the space of Probability measures on a Polish space is Polish itself

10:18 AM
@LeakyNun i meant to write Artinian as a D-space

does that follow from the fact that Mn(D) is finite dimensional

wel that's what i want to know

10:30 AM
Since Mn(D) is simple, it has only two ideals, so no infinite descending chain of ideals, thus it is Artinian
Or am I missing something ?

what's a good reference for group cohomology?

@EdwardEvans cassels frohlich

orly

idk

hahaha

11:08 AM
ken browns book is good for a first pass
Adem's book is good for a second pass

11:37 AM
[0, 2] equipped with the discrete metric is compact.
i think this is true but my brother says its false
can anyone tell me why

Every point is open
And there is an infinite number of points

but isn't the points
restricted to [0,2]

What do you mean ?

i mean by the defn of compact
i can find an open cover of [0,2]
and a subcover of [0,2]
so what does being open
have to do with anything here

Well the covering $\{x\}, x\in [0,2]$ is an open cover of [0,2]
Which has no finite subcover

11:46 AM
i see
so do u think this defn is not a good one to work with
since there might be counterexamples
i cant think of

What definition ?

the defn of compact being that E is compact iff every open cover of E has a finite sub cover

That's exactly how I proved [0,2] with the discrete topology is not compact
I don't understand what you mean

yes but before that
u said every point is open
and infinite number of points
did u use the same defn there

Yes, that's what lets me say {x}, x in [0,2] is an open covering

11:50 AM
Definition: a property is called "good" if there aren't counter-examples that Joseph Rock can't think of

lol
thats a good defn
diu lei
thx astyx

np

hey astyx

Thanks @MikeMiller

@JosephRock ...

11:54 AM
when u think of these type of problems
do u visualize it
or how do u think of them starting

yes

@LeakyNun lol
@Astyx but how do u visualize discrete things
it seems a little awkward

How do you think of $\{1,2,\dots,n\}$ ?

lol
just like that it comes to you?

Anything with the discrete topology is exactly that, except you can get an infinite number of points

11:56 AM
@Leaky is that super offensive?

Think of them as points in a line, but that all separated by a minimal distance
gotta go bye

@EdwardEvans depends on context

@LeakyNun erm it was not directed at nun tho
more like an expression to myself lol

are you aware what it means

11:58 AM
erm sort of

@EdwardEvans i don't mind if someone I know say that to me

ah so it's like saying "f*ck u" to your buddy or smth

my friend told me it means like f*** this problem lol

"lei" means "you"

11:59 AM
hahaha maybe your friend is screwing with you
tryna get you in trouble

lol that i didnt know
@EdwardEvans ahh nth new..

alright, anyway,

12:18 PM
@LucasHenrique @Thorgott
9

$\mathbb{R}^n$ and $\mathbb{R}^m$ are abstractly isomorphic (assuming the axiom of choice) for $n \neq m$ but not homeomorphic and so not isomorphic as topological groups. I think this might be the only thing that can go wrong, though; e.g. it seems plausible that for, say, compact semisimple Lie...

I'm struggling with a question regarding topology, specifically Tietze's extension theorem.

Suppose you have a finite collection of pairwise disjoint closed subsets ${A_i: i < n}$ of a compact Hausdorff space $X$. By Tietze's extension theorem and Urysohn's lemma, one could create a function $f:X \rightarrow \mathbb{R}$ where $f(A_i) = 0, f(A_j) = 1$ for any two $A_i, A_j$ in the finite set defined earlier.

My question is, how could one define a continuous function such that $f(A_i) = i$? I was thinking perhaps something like $f(X) = f_1(X)(f_2(X)(f_3(X) + 1) + 1)$ etc for $n$ many subset

12:51 PM
@Remana Given two disjoint closed sets $C_1$ and $C_2$ in a normal space $X$ there is a function $f:X\to\Bbb R$ which is contant $1$ on the first and constant $0$ on the second, do you know this fact?

1:17 PM
Can we conclude that as Hawaiian earring has no universal cover so that it's not semilocally simply connected?

Sure
But it's easier to check the definition of sl sc

1:35 PM
From the definition? Well, let U be a nbd of x where x is the origin. Then, there is a circle that completely contained in U. Then a loop that wrap around that circle once will give a nontrivial element so that the inclusion map will make a nontrivial element
Am I right? @MikeMiller

Almost
You need to justify why that element is homotopically nontrivial

@MikeMiller How can I? I don't know the fundamental group of Hawaiian earring..

So the Hawaiian earring is a really nasty space
But you know fundamental groups have induced maps
If you start with one of those circles (call it $\gamma$) you think of a way to cook up a map $r: H \to Y$, where $Y$ is a nice simple space, so that $r_(\gamma)$ is *clearly nontrivial?

What do you mean? I can't understand what you're talking

2:24 PM
I wanted to ask, How is modular arithmetic extended to complete real domain?
eg. $a=b\pmod c$ but $a$ being rational...

You look at the subgroup generated by c, and quotient R by that
a=b (mod c) iff a=b+kc for some integer k

Everything is "divisible" by everything $\neq 0$ in $\Bbb R$ so
congruence mod a real number is not very interesting

3:06 PM
@Astyx I did not understand...does that mean 6.7876 is congruent to 0.7876 mod 2?

yes
But as Edward said, it's not really interresting

Yes perhaps it isn't. It's more like {x} fractional part of x which reduces mod 1

yes
Note you also have 3 = 7.6 mod 2.3 for instance. c does not have to be an integer

3:22 PM
@Astyx 1.6? why 7.6?

because $3 + 2.3 \times 2 = 7.6$

3:35 PM
Everything is congruent to everything: $a \equiv b \bmod x$ for any real $x \neq 0$ because $a - b = \frac{(a-b)}{x}x$