« first day (2790 days earlier)      last day (2232 days later) » 

3:00 AM
@Secret, induction would only require potential infinity, not actual infinity, right?
 
@AkivaWeinberger the premise is proved by homology :P
 
@LeakyNun Yup :)
 
So it's basically asking for the sum of all the elements of set S, where each element is the product of an integer's nonzero digits, and the cardinality of set S is the number of integers with at most 2018 digits. Right?
 
@AkivaWeinberger basically brouwer's fixed point theorem's proof?
 
@orbit-stabilizer as far we knew, at least in physics, terry Tao has an answer in an MO that mentioned we only need some quotient of the space of all lebesgue integrable function, and not the entire borel algebra as the measurable sets. This rules out the need for $\omega_1$ for example
 
3:02 AM
@DarkRunner no, it means P(1)+P(2)+...+P(10^2018-1)
 
@LeakyNun oh ok, thx
 
interesting question
 
@orbit-stabilizer yes, for the usual induction, only potential infinity is required. Actual infinity is only needed for transfinite induction, which as far user21820 aware we can do analysis without it
 
@LeakyNun I'm actually not 100% sure that the assumption is enough, tbh
 
@DarkRunner the answer should be quite big
 
3:04 AM
@LeakyNun no kidding
hopefully nicely expressible
 
Don't expand this message if you don't want to:
152353627332941873774203697669960802473841745889523601669226418063979178979939033156917166397711482405618235980749582704320062001193307655086235551018411110841255419795054047796959387690975598778599194314410000941423938761130606540092455355384525009710194600578031558950442938533947100173853015652757009018526766007645327076593067267175266872786611716165504825792252465851116515158480789853219298824695447901214663122293613299476858480210629515493830146298011387868665891899773068366571036300723112046789896016964005452975180158153336684045783752677
@DarkRunner that's basically my attempt ^
 
@orbit-stabilizer so yes, it is unclear if all useful analysis can be done by postulating some largest number like in ultrafinitism
 
"Product of all nonzero digits" Aw, I thought I was gonna be able to be clever and eliminate all numbers with zeros in them
 
@0celo7 can you not just use the fact that if the set in question is nonempty then X and Y are isomorphic as Banach spaces to get a Banach space iso between L(X,Y) and L(X,X)
 
@EricSilva I was just thinking that, yeah. The existence of one invertible map gives the isomorphism, yeah?
 
3:07 AM
Yup
 
Nice. One step away from being done.
 
Having said that, while analysis can be done without actual infinities, things related to continuity does becomes less nice so there's a tradeoff
 
@DarkRunner actually, ignore my answer, it's wrong
 
Ah, let's not make continuity less nice.
 
Hmm
Hex not ending in a draw is also equivalent to (2D) Brouwer's fixed point theorem, fun fact
 
3:17 AM
I think in a quartz article quantamagazine.org/… which it went through the relationship between Ramsey theorems and infinitary irreducible sentences. Thus I think the evidence we have so far is potential infinity might be likely to be a convenient shorthand for otherwise a bunch of complicated statements and results that make theorems nicer to work with
 
Unitary matrix as a manifold anyone?
 
Oh wow @akiva, came across your post
 
@JoeShmo what about it?
 
3
Q: Why can't some integral be"found" though they are anti-derivative & exist?

user142971In my book, a list of integrals have been given which the author states ... such anti-derivatives "cannot be found". Some of the members of the list are as under: $\int\dfrac{\sin x}{x} dx$ , $\int\dfrac{1}{\log x} dx$, $\int\sqrt{1 - k^2\sin^2x}dx$, $\int\sqrt{\sin x}dx$, $\int\cos(x...

 
I love that the Hebrew word for infinity is "ein-sof", literally "'there is no'-'end'"
 
3:17 AM
I don't think the proof is 'extremely difficult'. We did it in first year calc...
 
(It is a noun despite the fact that its parts put together make a full sentence)
 
maybe we proved a weaker statement
it was a homework problem
 
@AkivaWeinberger but there's nothing special about that...
I mean, "finite" is derived from the Latin word which means end
 
There's a few philosophical consequences of that choice of phrasal in some discussions in the Kabbalah, @AkivaWeinberger
 
English kinda does the same thingy, but we write it in a different language so it sounds more fancy
 
3:18 AM
and in Chinese it's 無限 : no-boundary
 
At a guess: Japanese borrows it from Chinese (again, same as English: to sound more fancy)
 
ein-sof is a very spiritual entity in Kabbalah that transcends and give birth to most realities including our own
 
@AkivaWeinberger google confirms
 
It's literally god in Kabbalah in a sense
 
@JoeShmo it's even a Lie group
 
3:20 AM
@LeakyNun from what im reading online yes, but i dont know any lie theory
what is it's dimension?
 
from wiki, it's n^2
 
huh? what wiki? link
i wiki'd the crap out of it
 
Just unitary matrices being n^2? Sounds unlikely
 
@AkivaWeinberger over C
 
The set of all matrices makes a manifold of dimension n^2, no? Unitary should be a submanifold
 
3:21 AM
the total dimension is 2n^2
real dimension
 
yuh
 
In mathematics, the unitary group of degree n, denoted U(n), is the group of n × n unitary matrices, with the group operation of matrix multiplication. The unitary group is a subgroup of the general linear group GL(n, C). Hyperorthogonal group is an archaic name for the unitary group, especially over finite fields. For the group of unitary matrices with determinant 1, see Special unitary group. In the simple case n = 1, the group U(1) corresponds to the circle group, consisting of all complex numbers with absolute value 1 under multiplication. All the unitary groups contain copies of this group...
 
so half that right
 
@JoeShmo ^
 
but why
 
3:22 AM
because its lie algebra is the skew-hermitian matrices
and that has real vector dimension n^2
 
Oh. n^2 real dimension, meaning n^2/2 complex dimension @LeakyNun
 
and topological dimension of lie group = vector space dimension of lie algebra
 
2n^2 real dimension ^
 
@AkivaWeinberger right, I also got confused at first
 
n^2 complex dim
 
3:23 AM
U(n) is n^2 real dimension; n^2/2 complex dimension
 
yuh
 
M(n) is 2n^2 real dimension; n^2 complex dimension
 
M_n(C)
 
right
 
U should be < M is the point
 
3:24 AM
....but whats the proof
 
Still. I think it is useful to know how much of the notion of infinity in our maths is actually a shorthand that makes things nice, and how much of it is essential
 
They said the Lie algebra is skew-Hermitian stuffs
Wait hold on what if n is odd
 
2 mins ago, by Leaky Nun
because its lie algebra is the skew-hermitian matrices
 
(Note: "Lie algebra" is different from "Lie space")
 
What do you mean complex dimension
It's a real manifold
 
3:25 AM
@Secret, that's fair
 
hmm
 
But it ain't a complex boi
 
("Lie algebra" is essentially the possible derivatives of unitary-matrix-valued functions)
 
ok so complex dimension doesn't make sense then
 
3:27 AM
im gonna need ted on this
 
@JoeShmo on what?
I already told you the proof
 
what is the proof?
 
@orbit-stabilizer nonono
 
5 mins ago, by Leaky Nun
because its lie algebra is the skew-hermitian matrices
 
thats a nonproof
 
3:28 AM
@orbit-stabilizer No idea but the author can write good
 
proof
 
@0celo7 why
 
learn measure theory from this or don't learn it at all
 
@JoeShmo why is that not a proof?
 
tao can right good
 
3:28 AM
@0celo7 why would u do this to anyone
 
but can he wrong evil?
Are you joking @0celo7
 
first, because i dont understand ( :) ), and second because this is unknown knowledge in this class
intro to multivariable analysis
 
I imagine, find the number of independent variables locally
 
@EricSilva cuz I'm the organizer of the Federer--Schoen--Yau seminar here
 
i dont know lie theory
 
3:29 AM
Pure e v i l
 
$U(n)$ is generated by the $n \times n$ Hermitian matrices. As a real linear space, the Hermitian matrices have a basis consists of $\frac{n(n+1)}{2}$ real symmetric matrices and $\frac{n(n-1)}{2}$ imaginary antisymmetric matrices. This means the real dimension of $U(u)$ is $n^2$. — achille hui Jan 29 '15 at 7:26
 
@JoeShmo computing the lie algebra is literally differentiating a matrix function, it's barely lie theory
 
@JoeShmo does this help?
@0celo7 what is it with those double hyphens
 
sorta kinda not really
 
@LeakyNun latex habit
 
3:31 AM
ok let's see what Ted says then
 
Alternatively, say $f(x)=\Bbb R\to U(n)$ has $f(0)=I$ and $f'(0)=A$. Then $f(x)\approx I+Ax$ for small $x$ @JoeShmo
 
$--$ @0celo7 ?
 
The set of all possible $A$ there should be a vector space
 
to differentiate multiple people vs. double names
like the Choquet-Bruhat--Geroch theorem
 
oh do you actually do that
wtf is that
 
3:32 AM
good style.
 
and the dimension of the vector space should be the dimension of $U(n)$ because it'd be the dimension of the set of all $I+Ax$ and that looks like the neighborhood of $I$ in the space @JoeShmo
 
@LeakyNun what is that theorem?
 
@0celo7 do u know if choquet-bruhat is a good expositor
 
Pretty bad
why?
 
why f'(0) = A?
 
3:33 AM
@JoeShmo What I just said might be a bad idea
 
Are you looking at her new baby book
 
warning
@JoeShmo I defined A to be that
 
the blue one is very hard, I wouldn't recommend it
 
I just found a copy of one of her gr books and was wondering if I should grab it
The one I saw was v big
 
3:35 AM
Yeah that one
 
Well, it costs an arm and a leg so if you can get it for less, do that
But read Wald first
CB has a lot of typos and whatnot
 
Yeah I'm gonna crack open Wald once I take the gre if my quarter is going well
 
experiment: let $H=[0,1]^\Bbb N$ be the cube
 
That's a metric space, fun fact
 
@LeakyNun the CBG theorem says that initial data satisfying the Einstein constraints have a unique maximal globally hyperbolic development
 
3:37 AM
eh... and then...
:43604457 it isn't really, just use the geometric series
 
Right yeah
Rewrite it as $\prod_n[0,\frac1n]$ and use the Euclidean metric
 
and define an equivalence relation on $H$ s.t. $x \sim y \iff$ $x$ and $y$ differ in only finitely many coordinates
 
Are you gonna do $H/{\sim}$ now?
 
right
hypothesis: $H/\sim$ is contractible
 
Isn't it even convex?
 
3:40 AM
@AkivaWeinberger eh, what's the distance between $(0,\cdots)$ and $(\frac1n,\cdots)$?
@AkivaWeinberger is it?
 
unitary matrices aren't hermitian are they
 
At least one of them is lol
 
@LeakyNun No, I meant, map $(x_1,x_2,\dots)\in[0,1]^{\Bbb N}$ to $(x_1,x_2/2,x_3/3,\dots)\in\prod_n[0,\frac1n]$
 
@EricSilva lol
 
@AkivaWeinberger then what is the distance between (0...) and (1...)?
pi/sqrt(6)?
 
3:42 AM
Yeah
 
that's weird af
 
@EricSilva do you have any idea about regularity theory for quasilinear elliptic dudes
 
At least $\sqrt{\sum(x_n-y_n)^2}$ always converges though
In any case, about the contractibility of $[0,1]^{\Bbb N}/{\sim}$, write $f_t(\dots,x_i,\dots)=(\dots,(1-t)x_i,\dots)$
Then $f_0$ is the identity and $f_1$ is the constant map
and it's well-defined even after the quotient
 
whether or not i do depends on what you need @0celo7
 
@EricSilva Well, what I need is probably in Morrey's book but I really can't read it
 
3:45 AM
@LeakyNun I think one consequence of it is that there's a wide class of spaces that can be embedded in $[0,1]^{\Bbb N}$, so it means that everything in that class is metrizable. Forget what exactly the class is though
 
I have a $W^{4,p}$ strong solution of a fourth order quasilinear equation that's probably elliptic
And I just want C^\infty data to imply C^\infty solution
 
@AkivaWeinberger something second countable something urysohn
i'm talking out of my bottom though
 
I might as well just try to do the bootstrapping myself because it's just quasilinear
 
I wonder if there is a word for "necessary and sufficient condition"?
 
> … Urysohn's metrization theorem: … every Hausdorff second-countable regular space is metrizable.
 
3:46 AM
@Secret iff
 
I think it's that one yeah
 
Ah that works
 
lmao
golfing 4 words to 3 characters
 
guys, one more time. how is dim U(n) = n^2?
 
@LeakyNun And yet no way to pronounce it :(
 
3:48 AM
@JoeShmo lol
 
@AkivaWeinberger if and then a long ass f
 
There are many useful spaces that are second countable, but I think spaces involving $\omega_1$ are mainly the first countable ones
 
there ought to be phi : U(n) -> R^(n^2) satisfying rank Dphi(A) = n^2
 
@0celo7 i imagine this is true and probably not that bad but im too lazy to think about it
 
@JoeShmo how about you exponentiate a skew-hermitian matrix
 
3:49 AM
whats a skew-hermitian matrix
 
something A* = -A
 
(Assuming I have not accidentally switch between the two notions)
 
$\bar A^\top=-A$
 
@AkivaWeinberger wtf
 
And then the exponential map of matrices does… something
 
3:49 AM
too many things on top of the A my dude
 
well, exp(A)* exp(A) = exp(A*) exp(A) = exp(A*+A) = exp(0) = I
so exp(A) is unitary
 
$\begin{matrix}\rm no\\\rm it\\\text{isn't}\\\bar A^\top\end{matrix}$
 
iff pronunciation: eiffffffffffff... (and so on) :P
 
@AkivaWeinberger por el amor de dios acabalo
 
$\bar{\bar{\bar{\bar{\bar{\bar{A}}}}}}$
 
3:50 AM
@Secret that's the first time I hear :P from you
 
* is conjugation?
 
O yeah, I use it more commonly in h bar
 
@LeakyNun Dios está muerto.
 
@JoeShmo conjugate transpose
 
@JoeShmo Conjugate transpose
 
3:51 AM
oh shit Akietzsche
 
I use ":P" whenever there is something that is not to be taken seriously
 
And then $A\mapsto e^A$ is… something I guess?
 
what did I just write
 
Though it can be confusing sometimes because I often switch between joking and serious at a flick of a switch
 
@AkivaWeinberger it's the exonential map of Riemannian geometry
 
3:52 AM
If $A$ is skew-Hermitian then $e^A$ should be unitary. But then we have to worry about the inverse map existing locally?
 
${{\overline A}^\top}^* = A$
 
@0celo7 Sure but here I mean the matrix exponential
 
so how does that help me?
 
@AkivaWeinberger refer to Lie theory
 
@AkivaWeinberger I know
 
3:53 AM
I mean it's the same but that's not the point @0celo7
 
Especially when I am actively thinking about something in a problem that can sound a bit eccentric to main streamers
 
Right OK
 
@JoeShmo inverse exists locally around I
 
in U(n), A* = A^-1, not -A
 
and it is a homeomorphism
3 mins ago, by Leaky Nun
well, exp(A)* exp(A) = exp(A*) exp(A) = exp(A*+A) = exp(0) = I
it's exp(A) that is unitary
 
3:53 AM
@JoeShmo But what about $e^A$?
 
but you can compute the dimension of the skew-Hermitian matrices easily
 
And then look at the determinant of that map I guess?? I'm really not sure
 
@AkivaWeinberger what determinant
 
Not determinant, sorry
Jacobian?
Derivative?
 
why?
just construct the inverse by hand lol
not that I will do it
 
3:54 AM
Can you? I though exp isn't surjective on all matrices, how do we know it's surjective on unitary matrices?
 
@AkivaWeinberger this is a very good question. It's locally surjective.
 
As long as the rank of $D\exp$ is what it should be we should be good
 
@JoeShmo $\exp:\{\text{skew-Hermitian matrices}\} \to U(n)$ has a local inverse around $0 \mapsto 1$
and it is a homeomorphism
hence the claim
 
So you know the dimension of this thing locally. But the dimension has to be constant on connected components.
 
cant read that
 
3:55 AM
@JoeShmo that's "exp:{skew-Hermitian matrices}→U(n) has a local inverse around 0↦I"
 
it's not obvious to me how to construct the local inverse
 
ohhh
 
i would just compute the derivative cuz then it's obvious
 
there we go
 
@EricSilva consult any textbook about Lie theory
 
3:56 AM
So, we know $A^*=-A$, and we want to show $(e^A)^*=(e^A)^{-1}$?
 
6 mins ago, by Leaky Nun
well, exp(A)* exp(A) = exp(A*) exp(A) = exp(A*+A) = exp(0) = I
 
that's not a good answer
 
It's kind of a dick answer lol
 
i agree
 
@LeakyNun Ah yeah 'cause they commute so we can do that
 
3:57 AM
it's kind of a "I don't know but I know where to find in case that helps" answer
 
i also know where to find the answer
 
Yeah, telling people to look in "some book" is really not helpful at all
 
whats the dimension of the skew symmetric hermitian matrix?
 
Just use the "local inverse theorem" (I forget what it's called)
 
inverse functiuon theorem
 
3:58 AM
Uh, inverse function theorem? Some shizz
That one yeah
 
it should just be called the rank theorem in more generality
 
are you talking to me?
 
@JoeShmo Well the diagonal elements are any pure imaginary, and then $a_{ij}$ is determined by $a_{ji}$
 
@JoeShmo you need to apologize before I talk to you again >:(
 
so you should be able to show that it's $n\times n$
 
@0celo7 ok ok. i guess analysis is not all that bad
 
(it's a link that will directly download the pdf)
@EricSilva here
section 2.3
 
@JoeShmo tell me this is aesthetic i.gyazo.com/95fcfa7552b9c889d558f92fd810049b.png
 

« first day (2790 days earlier)      last day (2232 days later) »