« first day (1583 days earlier)   

12:01 AM
@ACuriousMind So that rain turned into 5 inches of nasty, wet snow.
 
@0celo7 Nice...
 
Does parallel transport mean relative to the manifold or the ambient space? But if the former, what does that mean? I always envision parallel as being defined relative to a plane of some sort. So I'm having trouble visualizing more complex structures where you might parallel transport
 
@StanShunpike I'd say it's parallel along the curve you are transporting, i.e. to someone travelling along the curve, the parallel transported vector doesn't turn, shrink or stretch
 
No turn (direction), shrink or stretch (magnitude) so does that mean we define parallel using constant direction and magnitude.
 
Ehhhhh
 
12:13 AM
@StanShunpike It's a little more complicated because the manifold has humps and bumps.
 
I'd say that's the intuition behind the definition, but not the definition
Because, for weird connections, the formal "parallel transport" doesn't look like a parallel transport at all
 
Yeah, the humps and bumps are where I'm confused
 
@ACuriousMind What does Mr. Germany like to drink?
 
@StanShunpike Have you looked at the equivalence of parallel transport and connection? Essentially, you are defining what it means to be "parallel" when you give the notion of covariant derivative - or you are defining what it means to covariantly derive when you give the notion of parallel transport
 
@StanShunpike The definition of parallel transport is the geodesic equation.
@ACuriousMind I explained that in one of his Math.SE questions.
@StanShunpike I don't know the answer to this question.
 
12:19 AM
@0celo7 Well, unsuprisingly, all sorts of beer, but I also like gin and whiskey.
 
@ACuriousMind Martinis are pretty awesome.
I put a little more Vermouth in mine than is considered normal, however.
(1:6 is standard, I like 1:4.)
 
Uggggh. I'll take a gin tonic over a martini any time :P
 
Currently enjoying a nice single malt that my dad found.
No clue if it's any good :D
Seems alright to me.
 
What confuses me is aren't there many kinds of connections? Whereas constant direction and magnitude are intuitive, I don't understand what the properties of a connection (for example as defined in Wald or do Carmo) has to do with commecting tangent spaces
 
@StanShunpike Well, because a connection defines parallel transport, it gives you a way to compare/connect vectors at different points.
 
12:29 AM
@ACuriousMind We call it a connection on the tangent bundle because it connects the tangent spaces, right?
 
Another point of view is that it allows a unique lift of curves on the manifold into the bundle with connection.
@0celo7 What meaning of connect do you have here? (Also, the concept of connection is more general than tangent spaces)
(I don't really know what the original reason for calling it a connection is)
 
@ACuriousMind I'm assuming you can define a connection over a general fiber bundle?
 
@vzn Oh sorry, I was just asking Daniel, and referencing your post
 
@0celo7 Yes
On general bundle, a connection is something like "choosing a horizontal/parallel direction at every point"
 
@ACuriousMind isn't it bed-time for you?
 
12:34 AM
@ACuriousMind Does "affine connection" refer to the covariant derivative connection?
 
And I just learned from Wald that the uniqueness property of the Levi-Civita is crucial because it allows us to pick out a specific directional derivative operator. Or something like that. Why wouldn't there automatically be a unique on on a smooth manifold?
 
@StanShunpike Unique what?
 
@0celo7 Well, you call any derivative induced by a connection "covariant".
Affine refers to the fact that the bundle fibers are affine spaces
 
@ACuriousMind What is the "affine connection" then?
I meant the standard covariant derivative.
 
12:37 AM
@0celo7 directional derivative operator.
I think that's the term he uses. I can go look
 
@ACuriousMind Yeah, looking at some books it appears that that's true. I think it makes no sense though
I think that vectors in Euclidean 3-space with the cross product should be considered a ring
 
@0celo7 There is no "the affine connection", but an affine connection is just a connection on an affine bundle (which the tangent bundle, as a vector bundle, is)
 
@ACuriousMind Too many bundles! What is an affine bundle?
 
@0celo7 Heh. A bundle whose fiber is an affine space
 
@ACuriousMind :/
 
12:39 AM
Yay. I finally understand that.
 
An affine space is essentially just a vector space where you've forgotten that you've got an origin
 
I know what an affine space is.
 
@Danu Why?
 
What's the ramification of not having an origin?
 
@ACuriousMind Because it's one of the best-known examples of 'simple' multiplicaton of well-known objects
Based on my extremely small amount of experience with rings, it appears to me that the essence is 1) Abelian group w.r.t. addition 2) Multiplication plays nice with addition
so no constraints on multiplication per se
 
12:41 AM
@Danu But because it lacks associativity, it is not a good example - you can't easily speak about its ideals, its module theory is ugly, etc.
 
@ACuriousMind I guess I'm influenced by Vinberg :P
 
If the affine connection is defined on an affine bundle, how does this allow us to measure curvature on a manifold?
 
And lacking unity, the allowed morphisms are so unconstrained because you don't have to send 1 to 1!
 
Like why is it on the bundle as opposed to the manifold itself
 
It's really ugly from the viewpoint of most algebraic subdisciplines, I think
 
12:43 AM
@ACuriousMind What are the consequences of not having an origin?
 
@0celo7 Greater symmetry group and stuff
 
@0celo7 Uhhh. Essentially, that you don't have an absolute notion of distance, only a relative one
(Though distance is handwavy here)
 
@dmckee did you see this?
@dmckee shall I vote to close?
 
@StanShunpike Heh. Well, because every bundle has its own curvature, and what you usually call "curvature of the manifold" is really "curvature of the tangent bundle with the Levi-Civita connection".
And because the tangent bundle is somewhat natural to consider for a manifold, one does not really distinguish here
@Danu Well, but, for example, a very interesting construction is localizing a ring, i.e. partially making its elements invertible w.r.t. multiplication. This fails as a nice construction horribly if you don't have associativity, and you can't even say what invertible means lacking a unit
 
@dmckee it has smth. to do with the Green function, s.t. I am not sure.
 
12:48 AM
@ACuriousMind Well, you just have to say "associative ring with unity" then
 
@Danu Yes. But I've never seen any interesting results about non-associative ring without unit
So it is more economic to generally assume associativity and unity when someone says ring
That's not to say that there aren't results about the more general rings, but they don't show up that often
 
I'll get back to you when I read Vinberg's chapter on rings ;)
oh jk, he doesn't have one
 
Huh? :D
 
It's not a very big book (sub 500 pages)
 
B-b-b-but...what do you do in algebra if not rings?
500 pages of group theory?
Or Galois stuff?
 
12:54 AM
Affine & Projective spaces, Tensor algebra, commutative algebra (I guess this is largely about rings?), groups, representation theory, lie groups
Thats chapters 7-12
 
Affine and projective spaces without rings? Oh, well...
But yeah, commutative algebra should be module theory over commutative rings, essentially
How is that just a chapter?
 
0
Q: Does Big Bang Cosmology imply/require infinite space?

Lucy MeadowThe reason I am asking this question is because if all points in space observe recession of galaxies the same as we do from Earth, the universe would have to be infinite (or a closed sphere in 4D or something. I know infinite space isn't a formal position of Big Bang cosmology, but is a non infi...

 
It's not a very advanced book, so that's probably why?
 
The Big Bang does not_require_ that space is infinite, it just so happens that that's the case. Right?
 
@Danu I see. I can then only predict that you'll have to learn everything all over again if you ever need to go into more depth in any of the subjects in there.
 
12:57 AM
@ACuriousMind I think it comes quite highly recommended by math.se users so I don't think it is bad.
 
How can curvature of a bundle inform me about curvature of a manifold? If I am measuring properties about the bundle, is there a one-to-one correspondence between points in the bundle and points on the manifold such that I get info about the manifold from the bundle?
 
Also, the first chapter already introduces the most basic notions so the dfinition of rings fields etc is in there.
 
@StanShunpike The Levi-Civita connection is uniquely determined by the metric, so it carries unique information about the manifold.
@Danu Well, we'll see what you say after you've read it. I can't judge its quality from chapter titles, I guess
 
8
A: Good books for self-studying algebra?

Mathemagician1234The absolute best book for self-study in algebra to me is E.B. Vinberg's A Course In Algebra. A Course In Algebra by E.B.Vinberg This book very rapidly became my favorite reference for algebra. Translated from the Russian by Alexander Retakh, this book by one of the world’s preeminent algebracist...

this is where I got it from
 
@ACuriousMind the connection is defined on the bundle. And the tangent bundle is just the disjoint union of tangent spaces. So what does that have to do with the original manifold? Like if the tangent bundle has the structure of a connection that doesn't mean the manifold does, right?
 
1:09 AM
@StanShunpike The manifold cannot have the structure of a connection, because connections, by definition, live on bundles. "Disjoint union" is a characterization of the tangent bundle I've grown to dislike, because it does not carry the disjoint union topology.
I'd rather say that the tangent bundle arises naturally as the vector bundle on which the Jacobians of the coordinate transformations between the local charts of the manifold act, but for this to make sense, you'd have to internalize that bundles are defined by local patches and transitions between them.
 
Hmm...I see your point. I thought that definition seemed oversimplisitc somehow. Why do they have to live on bundles? Wikipedia does say that is part of the definition but doesn't say why. wikipedia mentions a space of infinitely differentiable vector fields and the connection is a map on this. Can that space not be defined on the manifold itself without the bundle?
I thought smooth manifolds were by definition infinitely differentiable.
 
@StanShunpike If you define "that space" on the manifold, the bundle is what you get
That's what the tangent bundle is - it's where the tangent vectors live
 
Ohhhh
 
And because a connection is a notion of shoving around tangent vectors, it also lives on the bundle
 
1:47 AM
@ACuriousMind One of my earlier questions was, what are we looking for that the manifold doesn't already have?
Like, the manifold doesn't allow us to by itself compare tangent spaces at points
But how do we know what properties we need in order to compare them?
@Danu its a shame topos is so hard. I really was hoping he has some good ideas.
music theory sux. I haven't found any practical ones so far. Most of it I make up.
 
@Danu That's a really good infomercial for that book
 
 
1 hour later…
2:56 AM
On page 31 of Wald, he discusses defining a commutator of two vector fields "in terms of any derivative operator". What does he mean? Does he mean write the vector fields in terms of the derivative operator or the commutator?
 
 
1 hour later…
4:13 AM
Am I wrong in saying that this really isn't much of an answer?
 
 
4 hours later…
7:44 AM
I shudder at the offtopicness of math in this room. Smoking should be done in the smoking room. It's annoying if it isn't.
:D With that said, it's great that everyone here is so well-rounded.
@KyleKanos The answer is in the right direction. There's a huge lack of descriptiveness.
 
@Danu Yes!
My PhD work was an important part of getting that experiment to work :)
 
I have a tiny little question which isn't even worthy of main. Maybe I could ask it out here.
From my understanding of Nulear Fission
^ A heavy nucleus is excited and split into two smaller nuclei.
A while ago, one of my teachers commented that the combination of $^3_2\text{He}$ in the sun to form $^4_2\text{He}$ is fission.
Then, someone in the class said it was fusion. And an unintelligent debate ensued where one party screamed fission, the other fusion.
I know this seems silly but I too am wondering what the following is:
$$^3_2\text{He}\ +\ ^3_2\text{He}\ \rightarrow\ ^4_2\text{He}\ +\ ^1_1\text{H}\ +\ ^1_1\text{H}\ +\ 12.8\text{MeV}$$
All I'm familiar with are $\alpha, \beta, \gamma$ decays and the above is no direct disintegration.
I too feel that this is thermonuclear fusion. But maybe my lack of knowledge on this subject leaves me out of being any judge of it. So, could someone please inform me on what the mechanism behind the above is?
... oh wait! Ive found it
It's in the wiki article for fusion.
Ohk, it's $\beta^+$ decay followed by $\gamma$ decay and then fusion.
So, I guess it's still fusion.
What was my teacher smoking to have said fission!?
The problem here was, I think, a lack of insight into how that final step took place. He must have thought that the heliums combined together and split up again as opposed to a head on collision during which two protons are thrown away.
 
8:48 AM
@DanielSank So... do you have a PhD from 'Google Uni.'? ;)
 
8:59 AM
@Danu No, but I think I deserve one. I grew up on it.
It's still hard to find all the answers though.
 
LOL
 
9:23 AM
@NeuroFuzzy That's what I thought, lol
 
I don't understand what rings are useful for
Whereas fields seem naturally useful, rings seem like a clunkier version without the nice features of being abelian. Are rings used a lot in physics?
 
@StanShunpike You can have commutative rings that are not fields (I assume that's what you mean by abelian, which is normally just used for groups)
 
10:29 AM
@StanShunpike Hm...well, we do define the parallel transport to transport the vectors around, and then we can compare them, right?
@StanShunpike "Useful"...what is that? ;)
@KyleKanos No
 

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