The h Bar - 2017-03-28 (page 5 of 5)

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9:00 PM

@AccidentalFourierTransform I...don't think so

"infinite-dimensional Lie group" is a murky subject, anyway

I think the algebras are far better understood (though I don't know of a Schur's lemma for those, either)

@ACuriousMind Are there any Lie groups whose Lie algebra are Kac-Moody algebras?

@Bass Well, that's the point - the K-M algebras are infinite-dimensional so you need to choose your notion of "infinite-dimensional Lie group" before you can even ask that question

AccidentalFourierTransform

okey dokey

@ACuriousMind Are there different notions, or are there... no notions of infinite-dimensional LGs?

In a sense, for instance, you could say that the Witt algebra (Virasoro algebra with vanishing central charge) is the algebra of the group of conformal transformations of a cylinder

9:05 PM

@ACuriousMind Yes, I thought of that group as a Lie group.

@Bass I'm pretty sure there are some, but I'm also pretty sure people don't really agree on what it should be, just like people don't agree what an infinite-dimensional manifold should be to begin with

After all, one explicitly looks for continuous conformal transformations.

OK I see.

("locally isomorphic to $\mathbb{R}^\infty$" is not a useful statement without further clarification)

"There exists a countable basis" ... better?

@Bass Countable basis of what, though?

A manifold is not a vector space

9:08 PM

I mean "locally isomorphic to something that has a countable basis".

"Admits coordinates with countably many components"

I heard manifold

@Bass Ah, see, but it's already not clear that that's the correct notion. Maybe you want a Banach manifold - an infinite-dimensional Banach space has no countable basis in the sense of linear algebra.

Good, ACM

@ACuriousMind I see, but there are infinite-dimensional vector spaces with countable bases, right? Like $L_2(\mathbb R)$. So that might be a possible definition, which is closer to the finite case.

@Bass Nope

9:12 PM

Hello all

By Baire's theorem, every infinite dimensional normed vector space has only uncountable Hamel bases.

Uhh..

@Bass The Hilbert bases you are thinking of are only countable because you are allowing infinite linear combinations of them, which is forbidden for bases in the sense of linear algebra

$L^2(\Bbb R)$ has a countable Hilbert basis...not a Hamel basis.

@ACuriousMind Oh I see. Then we just relax that requirement.

9:15 PM

23

Q: Let $X$ be an infinite dimensional Banach space. Prove that every Hamel basis of X is uncountable.

Let $X$ be an infinite dimensional Banach space. Prove that every basis of $X$ is uncountable. Can anyone help how can I solve the above problem?

functional-analysis banach-spaces normed-spaces baire-category

@0celouvsky Yep, I was thinking of Hilbert spaces.

@Bass Oh, then you have decided you want a Hilbert manifold! also possible, but again, it's not clear that that's the "correct" notion

@Bass A "Hamel basis" is the one you learned in linear algebra class (finite linear combinations).

Yep I see.

Hilbert manifolds are Riemannian manifolds but scarier.

9:16 PM

It's also not clear it's the wrong notion, either - it just goes to show that you can't just say "infinite-dimensional Lie group" and expect that notion to immediately make sense

@0celouvsky Yep. In QM-related subjects, I often hear "a basis" in the sense of Hilbert bases, because in that context it's much more natural than the requirement of finite linear combinations.

Oh, and I'm also sure the nLab has some funny $\infty$-smooth groupoid that trivially subsumes all these different attempts :P

@Bass So what ACM is saying here is that you want your manifold to be locally homeomorphic to a topological vector space. In finite dimensions, these are always homeomorphic to $\Bbb R^n$, which is quite crazy. As soon as you go to infinite dimensions, this fails.

So there's no "unique" "model space" for your $\infty$-manifold.

Like all those other things that fail when you inadvertently go to infinity.

Yep, so far I understood it :)

Bye guys, nice talking to you.

I wish I could tell you why finite-dimensional TVS are $\approx\Bbb R^n$.

@ACuriousMind Do you know why?

9:20 PM

Since I'm sure you know that all norms on $\mathbb{R}^n$ are equivalent, I'll say no because that's clearly not what you're looking for :P

@Bass cya

Topological vector space, not normed vector space...

Oh, I don't care about vector spaces that aren't normed

@Bass Btw there are some Banach spaces that cannot admit an inner product that generates the norm

I know nothing about general topological vector spaces

So Hilbert $\subset $ Banach is strict

@ACuriousMind But...test functions!

Schwarz functions!

@yuggib Funny you'd stop by

9:22 PM

All just subsets of Banach/Hilbert spaces to me, and you seem to continually forget I don't care about functional analysis all that much to begin with :P

Do you have a script that detects functional analysis?

Maybe

@ACuriousMind Test functions and Schwarz functions do not form Banach spaces...

...they're subsets of $L^2$, are they not?

Bernardo Meurer

There is no system but GNU and Linux is one of it's kernels

9:23 PM

But not in their usual topology

Not in the topology relevant to distribution theory anyways

I can happily say I don't care much about that either :P

"continuous linear functional" on $\mathscr D(\Omega)$ does not refer to the $L^2$ topology

@ACuriousMind fine

@yuggib Is it easy to see that every finite-dimensional TVS is equivalent to $\Bbb R^n$?

Not hard yes

How does it go?

Don't remember exactly, but it's not difficult

Of course it depend on the field you're using for the vector field

9:27 PM

$\Bbb R$ probably

So it's not always $\mathbb{R}^n $

?

If the field of the VS is not the reals, then it is not homeomorphic to a real euclidean space

I just said I want it to be $\Bbb R$

You did not specify before...anyways the proof goes for any field (at least of characteristic zero)

9:31 PM

Is it in Bourbaki?

Yep

It's an exercise in Conway

hmm

Could anyone venture a guess as to why computing the first allowed energy of the infinite square well might plot the energy....upside down???

I guess I'll think about it

Screenshot

It's probably tough to guess why this is happening but I'm hoping someone has tried to compute this themselves in the past

9:36 PM

@yuggib Christ, Bourbaki don't choose the field to be just $\Bbb R,\Bbb C$?

Nope

Why should he?

that's what everyone else does :P

I can't seem to find it in there

"finite dimensional" gets me like 100 hits

It is where it discusses finite dimensional subspaces iirc

@yuggib that's not the name of a section and I can't find the table of contents

The table of contents is at the end

9:47 PM

lol

Is section 2 probably, I will check

Is section 2 probably, I will check

what's a linear variety?

same as a linear manifold?

Yep

French terminology

does this book not have page numbers??

It has, but divided by section

9:50 PM

I think it's Thm. 2 in I.13

Hmm...that's just isomorphic

I don't think they mean in a topological sense

they mean in a topological sense

Wedderburn's little theorem

In mathematics, Wedderburn's little theorem states that every finite domain is a field. In other words, for finite rings, there is no distinction between domains, skew-fields and fields. The Artin–Zorn theorem generalizes the theorem to alternative rings: every finite alternative division ring is a field. == History == The original proof was given by Joseph Wedderburn in 1905, who went on to prove it two other ways. Another proof was given by Leonard Eugene Dickson shortly after Wedderburn's original proof, and Dickson acknowledged Wedderburn's priority. However, as noted in (Parshall 198...

crazyness

AccidentalFourierTransform

Iceland witnesses record-breaking baby boom nine months on from humiliating England at Euro 2016.

hehe

lol

Humans are funny :P

10:11 PM

More like gross

Anyone have any ideas for my problem?

Idk why my graph would plot downwards instead of upwards

I could be sneaky and just multiply the y-values of the graph by $-1$....lol

@loltospoon how did you initialize your problem?

i.e. how did you implement your boundary conditions?

Bernardo Meurer

When you finish the project and the prof releases a reviewed project statement :^)

@BernardoMeurer what?

Bernardo Meurer

@0celouvsky I have a project due, the project statement/guide was on the prof's webpage. I finished it already and he re-released the guide with a bunch of shit changed

10:30 PM

Finance mixing with science: Using relative stock index to track slow slip earthquakes

10:53 PM

@BernardoMeurer aha

Anything big changed?

Bernardo Meurer

He fixed some shitty code he had written that yielded some memory leaks

I had already fixed them though so w/e

He also added some functionality, but it shouldn't be much work to implement

...he said, despair flickering behind his eyes

I got a "Good - Like New" book from amazon

shit has some brown stuff on the cover, huge fold in some of the pages, banged up edges

clearly not read, just abused. wtf?

maybe I can iron the pages

@ACuriousMind can one iron a book?

lol, several ice cream parlors in Tübingen are suspected to have conspired to raise ice cream prices simultaneously this spring (German source)

@0celouvsky Try it, then don't blame me if it catches fire ;)

@ACuriousMind I will try on BBS first to see

11:15 PM

@ZeroTheHero my initial conditions were $psi(0)=0$ and that $\psi(endpoint)$ must be zero.

Also, my initial slope was a guess through a for-loop

Sorry for the delay, they had an event at my school for a late professor

Bernardo Meurer

I gotta implement more stuff

But I'm lazy because I don't like rendering stuff

@loltospoon ugh people who are late are the worst

I'm always early and then people are late

@0celouvsky I...don't think he meant that kind of late

@0celouvsky yea....I meant he died.

maybe

I'm about to die from a string explosion

Bernardo Meurer

11:21 PM

@0celouvsky That's what you get from not using snprintf()

:(

-1

Q: Geodesic Equation Quicky

Text Book GR equation, $$\frac{d^2x^{\mu}}{dq^2}+\Gamma^\mu_{\lambda\nu}\frac{dx^\lambda}{dq}\frac{dx^\nu}{dq}=0$$ Definition of Lambda seems to be tricky to find, what is lambda? Could I get confirmation this expand to 16 equations, ...or maybe 64 equations dpending on what Lambda is?

general-relativity

@ACuriousMind How do people like this happen :(

Entanglement between beryllium ions

@0celouvsky Be Nice.

@ACuriousMind Is Nice a proper noun?

And you know what I was asking

The properest of nouns

11:33 PM

@ACuriousMind you didn't answer the question

Nonuniform distribution of wealth tied to flow physics

hey guys

@ACuriousMind Do you know Cavalieri's principle?

So today i accidentally decided to see what math peeps call qft in hopes of possibly something I easy enough for me to look at and possible enrich my journey a bit. Well, I came across the mit ocw notes, and woit's syllabus. I know about quantum fields and strings, but I promised myself to not go near that text ever again

i see a lot and hear a lot about brst and cohomology . . . someday, I hope someone will tell me the secret

in some very intuitive baby format

without hardcore math,

@0celouvsky @ACuriousMind any suggestions about (soft) directions with baby steps about the brst and cohomology thing? Like real baby steps lol

for cohomology I recommend Bott & Tu, Lee intro smooth manifolds, Hatcher algebraic topology

I don't claim to know BRST

11:42 PM

ok, I am looking up the books

@0celouvsky en.wikipedia.org/wiki/BRST_quantization

I know what BRST is

i just found hatchers's text

who even calls the tangent space $\mathcal T_p(\mathcal M)$

I think @BenNiehoff was right

mathcal is for those things you think are fancy

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