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5:00 PM
@Perturbative: Balarka was just explaining to you in one word what your mistake was. Lee will talk about group actions on manifolds later. That's different.
 
@LeakyNun Is there a difference? $\ell^p$ is just $L^P(\Bbb N,\mu)$ where $\mu$ is the counting measure after all
 
fair enough
ah those connections
 
@TedShifrin I don't, I have no intuition for the topology on those infinite dimensional spaces :/
 
So unwind the definition for us.
 
Are all normed spaces Hausdorff?
 
5:02 PM
What do you think?
 
I think yes, because of the requirement that $\| x \| = 0 \implies x = 0$
 
Obliquely, I suppose so.
 
@TedShifrin No I didn't mean group actions, I just meant like $v$ acting on $f$, what does that mean? Does it mean something like $\nabla = \langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \rangle$ "acting on $f$ by $\nabla f = \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \rangle$ from typical multivariable calculus
 
@TedShifrin Well a space is locally compact if every point as a compact neighbourhood (there are plenty of definitions all equivalent to this one in an Hausdorff space so this might not be the best choice). I read that no point in $\ell^2$ (let's start with a nice space) has one such neighbourhood
 
is $L^{0.5}$ a thing?
 
5:04 PM
No, @Perturbative. If means $v_p(f) = Df(p)(v)$.
 
what is $D$?
 
@Alessandro: So remember that closed balls won't be compact in infinite dimensions, so what will be?
derivative, Leaky, geez.
 
@TedShifrin what is a more direct way?
 
Well, it's tacit in a complete solution, but it's not a solution.
 
@TedShifrin I don't intuitively see why closed balls aren't compact in infinite dimension so maybe I should understand that first
 
5:07 PM
@TedShifrin so if $x$ and $y$ are distinct then $\|x-y\| > 0$ and you know :P
 
OK. Start with $L^2$ or $\ell^2$.
Right, Leaky.
I should know (remember?) a characterization of compact sets in $L^p$, @Alessandro, but I don't.
But it's pretty clear that you can't enclose a little ball in a compact set.
Heya @Semiclassic!
 
4 mins ago, by Leaky Nun
is $L^{0.5}$ a thing?
 
@TedShifrin Ahh so for some $x \in M$ $v_p(f(x)) = Df(p)(v(x))$?
 
@Perturbative shouldn't it be $v_p(f)(x)$ on the left?
 
5:10 PM
Leaky, $L^p$ is defined for all $0<p\le\infty$, I suppose, but for various reasons one only deals with $p\ge 1$. (The main reason being Hölder.)
 
@LeakyNun To nuke the issue normed spaces are metrizable
 
@Perturbative: What are you doing? No.
 
So I dug around more yesterday, and deduced that a lot of the stuff I was coming up does in fact show up in some papers
 
We fix a point $p$. There is no $x$.
I'm not surprised, Semiclassic.
 
yeah, me either
 
5:11 PM
@AlessandroCodenotti and all metric spaces are Hausdorff, yadda
 
In particular, the connection to quantum correlations (the particular point of contact) is definitely out there already
 
BTW, @Alessandro, Riesz proved that a normed vector space is locally compact iff it's finite dimensional, but we're about to discover that.
 
@LeakyNun Also $T_3$ and $T_4$ if we want to exaggerate
 
@TedShifrin Oh, that's a cool result. Still thinking about $\ell^2$ here
 
5:13 PM
@AlessandroCodenotti why not T5 T6?
also, does anyone have a diagram of the implications between T_n?
 
Hint to Alessandro: In a compact metric space, any sequence has a convergent subsequence.
 
@TedShifrin Ahh sorry, I'm getting lost in notation (my fault I'm not used to the notation yet)
 
So it’s not new, even in the area of application
 
@LeakyNun There's a book by Steenbach that has all the implications
 
Let $\mathscr L$ be the category of all logical statements such that there is an arrow from $\phi$ to $\psi$ if $\phi \implies \psi$
 
5:13 PM
@Perturbative: It's just what I said at the beginning. A tangent vector at a point calculates directional derivatives of functions at that point.
@Semiclassic: Too bad for you, but it's still neat (not that I understood a word of the thing you linked).
 
On the other hand, I rediscovered a log of it in a week without realizing it, so that’s still pretty neat
 
Heya @PVAL.
 
59 secs ago, by Leaky Nun
Let $\mathscr L$ be the category of all logical statements such that there is an arrow from $\phi$ to $\psi$ if $\phi \implies \psi$
omg it's actually a category
 
Hi @PVAL!
 
5:15 PM
@PVAL: Is that a commentary on the politics or on something more personal?
 
My taxes probably just increased by 3-4k
 
@BalarkaSen do you have anything to say about my category?
 
mainly a commentary on that.
 
@LeakyNun $T_k\implies T_n$ iff $k>n$, unless you're one of the people with the weird regular/$T_3$ and normal/$T_4$ definitions
 
Yeah. These corrupt, immoral assholes should be removed from society. Immediately.
 
5:15 PM
I think there are only two classes of objects though
so it's "isomorphic" to the category $X \to Y$
 
There will be lawsuits all over.
 
@TedShfirin Ohhhh ok thank you!
 
Until Trumpolini installs all his judges ...
@Perturbative: I said that right at the beginning when you came in!!
 
Oh that is a nice name
 
@TedShifrin well, the simplest point is just that an elliptope is a set of n-by-n correlation matrices (symmetric with diagonal elements equal to one and with nonnegative eigenvanlues)
 
5:17 PM
@PVAL: I'm about to go on Medicare in two months. Who knows what'll happen. I can afford whatever (he says naively), but all the poor old folks who can't ... And all the folks in nursing homes that Medicaid pays for? Trash.
 
so if you're studying correlations in quantum mechanics, they're an obvious set to study
 
Oh, thanks, Semiclassic.
 
lol, nobody cares about my category
 
Correct, Leaky.
 
@TedShifrin Oh, of course, we're in a metric space where sequential compactness is compactness! So for example in $\ell^2$ the unit closed ball is not compact because of the sequences with a single $1$ in the $i$-th position
 
5:18 PM
:P
 
and when you do 3-by-3 correlation matrices, the (boundary of) the elliptope you get is the Cayley 4-nodal surface.
 
@TedShifrin I'm sorry Ted, I missed what you said while I was typing up a reply to the pointwise multiplication thing, sorry again :(
 
@Leaky: I might listen to categories if you explain how multiplication of integers works using categories.
 
@TedShifrin I can explain LCM though, whose dual is GCD, so multiply them together and you get the usual product :P
 
@Alessandro: Right.
You just said "multiply"?
Nice circular argument.
 
5:19 PM
no, multiply them outside
circular?
 
I don't think that makes sense.
 
what doesn't?
I mean, lcm(a,b) * gcd(a,b) = a*b outside the category
 
Never mind. Not worth my time.
 
outside means externally
 
Then don't bother.
 
5:21 PM
@TedShifrin It's a symmetric monoidal structure or something I guess :P
 
is there any fundamental theorem of Lie algebras?
 
Balarka, I'm just following up on my colleague's comment after I sent her the article Mathei posted about elementary school math and categories.
 
@TedShifrin The same example works for $\ell^p$ of course, in $L^p(\Bbb R)$ it can be fixed by taking the indicator functions of $[n-1,n]$
 
There are several, Leaky.
OK, @Alessandro. Good.
 
@TedShifrin Ah, I see
 
5:22 PM
@TedShifrin like?
also, what is $\mathfrak{gl}_n(\Bbb R)$?
 
Google, Leaky.
It's all $n\times n$ matrices.
 
@TedShifrin Makes sense now, thanks!
 
why isn't it the algebra of $GL_n(\Bbb R)$ ;_;
or is it
 
Glad to be occasionally helpful, Alessandro :)
What are you babbling about, Leaky?
 
any Lie group gives a Lie algebra
 
5:24 PM
That's a very frequent occasionally
 
@LeakyNun Surely what you are reading is telling you what this is?
 
@Narcissusjewel the associative algebra of endomorphisms of $\Bbb R^n$ with anti-commutator Lie-bracket, i.e. $[X,Y] = XY-YX$
 
@LeakyNun What is an endomorphism, as opposed to an automorphism, of $R^n$?
 
never mind, it is the Lie algebra of the Lie group $GL_n(\Bbb R)$ and $SL_n(\Bbb R)$ respectively
@Narcissusjewel the latter is invertible
 
@LeakyNun You are claiming $\mathfrak{sl}_n\cong \mathfrak{gl}_n$?
 
5:27 PM
@Narcissusjewel no
 
@TedShifrin I ended up asking one of my profs for the convergent thing. Makes sense - rearrange a conditionally convergent series until it does what you want.
 
I'm claiming $\mathfrak{sl}_n$ is the Lie algebra of the Lie group $SL_n$
 
What the heck is respectively, when you are referring to only one thing originally?
 
@Narcissusjewel oh... I made a statement about SL and deleted it before I posted it
and thought that I posted it
 
Ahh, right.
 
5:29 PM
@Ted so there is no compact set containing the unit ball
 
Right.
I said that a while ago :P
Well, I said a little ball.
 
I processed it now :P
 
@TedShifrin how many dimensions do you need to embed $SO(n)$?
 
$n\times n$ matrices ... hmm ...
 
as a manifold of course
 
5:31 PM
I now see why wiki says that one might be tempted to think that all closed compact subsets of $\ell^2$ are finite dimensional and the Hilbert cube is a counterexample
 
@KasmirKhaan goddag
 
@Alessandro: Why did you say closed?
hi Kasmir.
 
Hello :D @TedShifrin @LeakyNun
 
@TedShifrin which book would you recommend that goes into things like matrix exponential and Lie algebra?
 
Morton Curtis has a little Springer book on matrix groups which does everything for undergraduates.
 
5:33 PM
oh, and how should one prove that all 1-manifolds are orientable?
 
@TedShifrin Because I like unnecessary adjectives and didn't think this through
 
@TedShifrin thanks
 
(mostly the latter)
 
Put arrows on it, Leaky.
 
@TedShifrin how do you check that it is consistent globally?
 
5:35 PM
You only check that locally, anyhow.
 
how to know that locally implies globally?
 
I don't know what you're talking about with globally.
 
like, you start putting arrows, and then somewhere it crashes
 
@LeakyNun Are you thinking about the actual definition of orientability? Or some intuitive notion?
 
Make a connectedness argument.
I avoided that question, @Narcissusjewel. :)
 
5:36 PM
@TedShifrin Then you travel down the rabbit hole :P.
 
there's not many 1-manifolds, are there?
 
It's not my rabbit hole.
 
@Semiclassical just two
 
@TedShifrin like, if you have a circle sans a point, and you started putting opposite arrows, then you're done in the middle
 
right
 
5:37 PM
upto homeo
 
Connected ... with or without boundary?
 
@Semiclassical but one shouldn't use classifications...
 
fair, connected 1-manifold
 
@BalarkaSen with or without boundary?
 
manifold with boundaries are not manifolds, yes
 
5:37 PM
they are
 
you mean with, Balarka
 
they still are
 
bah
yes
 
No, Leaky, manifolds with boundary are not manifolds.
 
So since no compact set contains a closed ball no compact set contains an open ball either and it's hard for points to have compact neighbourhoods in this situation
 
5:38 PM
locally homeomorphic to $\Bbb R^n$ or a half of it
 
Simplest test case: is [0,1] a manifold?
 
No
not locally homeo to R at 1
 
i figured as much.
 
hmm, where did I see "a half of it" then?
 
(0,1) obviously would be, of course
 
5:39 PM
I did see it in a book
 
@Semiclassical Yup
 
@Leaky: Some (bad) books confuse manifolds and manifolds with boundary.
 
oh ok
 
@LeakyNun In a book defining a manifold with boundary, most likely
 
We are more careful.
 
5:39 PM
@LeakyNun You're thinking about Half-Space? $\mathbb{H}^n$
 
@TedShifrin a paper which talks about this quantum stuff formally is this one: cgm.cs.mcgill.ca/~avis/doc/avis/AI06b.pdf
 
That's used to define manifolds with boundary
 
@Perturbative maybe
I see
but two people shot you in the head
 
@Semiclassic: Do any of them make reference to proofs of the math facts you unearthed?
 
The thing which confused me a lot back in the days are manifold with corners. If you're working topologically, they are manifold with boundaries. But if you're working smoothly, they are very not.
 
5:41 PM
Yup, @Balarka. Not to mention orbifolds. :)
 
I vaguely know about them
Locally modeled on R^n/G etc
 
I remember how interesting it was to realize that the symmetric products of Riemann surfaces are always (smooth) complex manifolds, but that fails for higher dimension. The symmetric square of $\Bbb C^2$ has interesting singularities.
I actually had to talk about that in some talk I gave as a grad student — I don't remember why.
 
Ok, enough functional analysis for the time being, I'll go think about the dinner! Bye everyone
 
Bye, @Alessandro.
 
@TedShifrin well, the convexity is something they take for granted?
but i'm sure that's because they prove it elsewhere
 
5:43 PM
Like all true physicists ... :P
 
@TedShifrin Right, a diagonal's worth of singularities modeled on $\Bbb C/\Bbb Z_2$
 
Oh, I thought you were saying people had found this stuff about the cubics.
 
I think
 
Yeah, Balarka.
And what's the proof that $\text{Sym}^g C$ is a manifold for $C$ a R.S.?
 
5:44 PM
Let me think about it for a bit
 
SUre.
 
$\operatorname{Aut}(\operatorname{Aut})$
 
there's a more general notion of an elliptope of a graph, apparently
not quite sure how my cubic case fits in there, tbh
I want to say it's for the case of $K_{3,3}$ but i'm not convinced yet
 
@TedShifrin if $\operatorname{Aut}(\Bbb R^n)$ is $GL(n)$, then why isn't $\operatorname{Aut}(\Bbb RP^n)$ denoted $PGL(n)$?
(and is PGL a Lie group?)
@Daminark hi
 
@Leaky: It is denoted that, more or less. It's $\Bbb PGL(n+1)$.
 
5:48 PM
@TedShifrin I know, I'm asking about the mismatch in the argument
 
Yes, it's a Lie group (quotient of group by center).
 
maybe it's just $K_3$ here. I'm not sure.
 
@Leaky: What's the group of motions of the $n$-sphere?
 
i.e. rigid isometry?
 
Yeah.
No need for "rigid."
 
5:49 PM
hmm, what is it
I should know
 
Start with $S^1$ and $S^2$.
 
$\operatorname{Aut}(S^n) = O(n+1)$ isn't it
 
SOOO, we were saying...?
 
@BalarkaSen :P
sniped
 
Right, @Leaky. Note the dimension mismatch there too.
 
5:50 PM
cc @Daminark on the pun
 
<---- ignores all puns unless self-made
 
$\operatorname{Aut}(\Bbb R^{n+1} \setminus \{0\}) = GL(n+1)$
 
@TedShifrin That seems like punihilism
 
$\operatorname{Aut}((\Bbb R^{n+1} \setminus \{0\}) / \Bbb R_{>0}) = O(n+1)$
 
Don't let that be a punocide
 
5:52 PM
then what is $\operatorname{Aut}(\Bbb R_{>0})$?
 
@Leaky: At any rate, to avoid this, I called the group $\text{Proj}(n)$ in my algebra book.
 
@TedShifrin so $Proj(n) = \Bbb PGL(n+1)$ :P
 
Right.
 
You want to use the action of $\text{GL}_n$ on $\Bbb A^n$ to induce an action on $\Bbb P^{n-1}$, which gives you the change in index
 
Why $\Bbb A^n$?
 
5:52 PM
what is $\Bbb A^n$ now?
 
Affine n-space
 
so it's fine af
 
Vector space sans distinguished origin.
 
sans origin? Sounds like a font I would use
cc @Daminark
 
You're asking for permanent ignore, you and your cc's.
 
5:53 PM
lmao
 
@BalarkaSen I'd recommend changing to bcc's
@TedShifrin Just so you can use the projectivisation
 
No, @Narcissusjewel: You misunderstand what affine space is.
You want $GL(n,F)$ acting on $\Bbb F^n - \{0\}$.
 
Ok $K^n-\{0\}$
 
You need the origin to define lines through the origin
 
That was very naughty of me :P.
 
5:56 PM
O$K^n$
bcc @Daminark
 
adds Narcissus to the pun-ignore list
 
Hey y'all, need some help with a problem I'm tackling.
What are all the possible Jordan Normal Forms of a 10x10 complex matrix A, satisfying A^4=A^2?
 
Sounds a bit tedious, @JoeShmo, but what's your approach?
 
@TedShifrin Wait, he made the pun!
 
@TedShifrin, yes, quite a bit. I want to find the eigenvalues first.
 
5:57 PM
Ok I am dun with pun today (cc @Daminark). Your move, Demonhead
 
@Narcissusjewel Wasn't this you?
 
@TedShifrin I must be tired, I don't even see the pun I made :O.
 
nought
 
LOL
I wish I could make puns that good.
intentionally
 
@TedShifrin do I use Jordan Normal Form to prove that $\exp$ is total?
 
5:58 PM
What does "total" mean?
 
defined everywhere
 
Ted is like the greatest punist here.
I aspire to be like him
 
@LeakyNun You are moving between questions at the speed of light, won't you forget the previous answers :O.
 
@Leaky: It's always defined on the whole Lie algebra. The question is whether it maps onto the group.
 
@Narcissusjewel I will
@TedShifrin I mean, on $\Bbb R^n$
 
5:59 PM
@JoeShmo: So tell me about eigenvalues.
 
A similar problem I had found suggests that for a matrix satisfying A^3 = A, the minimal polynomial of A must divide x^3-x. Hmm...
 
I have put Leaky on ignore for littering chat with random questions in a span of a few seconds
2
 
What are you talking about, Leaky? What is exp on $\Bbb R^n$?
You talking about Riemannian geometry exp?
 

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