right, that's a good point. I have never bothered about the choice of A in the Yoneda lemma before

that's what they mean with "isomorphic"

(I think)

that's the "structure" that is preserved

i am prepared to believe this.

i am too but it's bothersome to check

i mean

it's easy to write

but it's not really rewarding

you don't feel anything is happening at all

Agreed.

To summarize for @Jake1234: when they mean "isomorphism", they really mean bijection of proper classes, but they use the word "morphism" because the correspondence Nat(h^A, F) ---> F(A) should be (I haven't checked if it is) natural in A.

I suppose it should also be natural in F.

Thanks a lot for the help guys.

I didn't help much, mercio had the interesting explanations.

I feel like all i did was word salad

I feel like I just shouldn't bother thinking about these questions at this point in time xD

@Jake1234 I sure didn't bother about all these, but then I never studied category theory.

I do humble topology, not algebra.

@Adeek So, what are you studying?

I am solving still munkres. I will probably finish it by November completely solving it. I am also solving some analysis book. I am learning also multi-variable calculus from Ted's videos along with a book called pugh. University will start this Thursday for me. I am taking Algebra, differential geometry, and functional analysis that will be fun.

@BalarkaSen I decided for each book that I am solving to keep my solutions so once I finish the book I could bind it as a book. So, that if I am teaching a class like that later in life I can use it.

Nice.

I am not really that organized about solving problems.

I am gonna teach calculus lab in my first semester that will be fun.

@BalarkaSen Yeah I wasn't organized as well. But, I decided it doesn't really take too much time to be organized and you can use it in the future. It also feels really nice when you collect a big chunk of solutions for a book.

haha

I encourage it. I am just lazy.

I just mark down the problems I have done and note down the sketch of solutions on the side of the margin for interesting problems.

I see. It takes quite some time to do everything rigorously in a paper, but it is worth it. Although if I see a complete solution in my head that is trivial I won't write it down.

There are times when I figure out something, think it's trivial, then the next day it's super hard

and I forgot to write it down

There is also sometimes that I think something is trivial, but when I write a solution down for it I would notice that it is not trivial and my reasoning was actually wrong.

Writing helps, yes.

I am gonna learn some exciting math this year. I am really excited.

In the winter semester I am taking introductory course into commutative algebra, so I can prepare for algebraic geometry.

@BalarkaSen Ok, I have made progress on my holonomy.

@Adeek Algebraic geometry is good stuff.

@0celo7 Good to hear.

I'm reasonably convinced of it now.

Yeah I am interested in it. I am interested in algebraic geometry and applications of it to physics.

The trick is to project the curve in the product onto each factor manifold

Now if only I make any progress on what I need to do.

Did you learn algebraic geometry @BalarkaSen ?

Then the parallel transport is the product of parallel transports around the two loops

@Adeek The basics, yes.

Hi

Each loop is of the form $(\gamma(t),q)$ and $(p,\lambda(t))$

cool

I think we discussed that before.

an explicit computation in coordinates shows that each of these parallel transports is basically a parallel transport on the first and second factors

anyway I am gonna go study brb.

I had to write down the connection for the product, which seem unnecessary though

I have to go to sleep. But I didn't prove the thing I wanted to.

:(

what did you want to prove?

Something about bundles. Mike gave a bunch of exercises relating to the question I was asking Danu today.

I was reading en.wikipedia.org/wiki/Division ring and it says both that a division ring is a ring where $$ax=xa=1$$ and that a division ring is an example of a non commutative ring! What's going on here?

That is to say, can you come up with an example of a submanifold which has trivial self intersection but nontrivial normal bundle?

@someonewithpc what don't you understand ?

Half-dimensional submanifold, I should say.

@BalarkaSen Ah. Admittedly I'm not captivated by that problem.

Otherwise there are trivial examples which break dimensionality.

Currently getting back into the Riemannian geometry swing.

@0celo7 You essentially want a bundle which is nontrivial but admits a global section.

E.g. tangent bundle of S^5.

@BalarkaSen For a vector bundle, admitting one global section isn't very interesting, is it?

If you compactify it fiberwise, you get an example.

a ring (R,+,*) is a group with respect to addition and the multiplication operation is compatible with addition.

@Adeek Well, doesn't the fact that $$ax=xa$$ mean it's commutative? I get that it can be non-commutative, but so can all other rings, right?

@0celo7 No, but e.g. there are no examples in 2 dimensions (can you prove this?).

For example $(\mathbb{Z},+,*)$ is a ring that is not a division ring.

No examples of?

Every rank 2 bundle on surfaces admitting a global section is trivial.

@Adeek That's... Not what I asked

Global section with zeros possibly?

no the above doesn't mean commutativity.

Nowhere zero global section.

It only commutes with its inverse.

Compact base?

Yes.

(G,*) is commutative iff xy = yx for all x and y in G.

An easy example is the quaternions.

@someonewithpc en.wikipedia.org/wiki/Quaternion

@Adeek Oh... Is it always the case that for some invertible non-commutative operation $&$ and $a = 1/x$ we have $x & a = a & x$, then?

Yeah

@BalarkaSen No I don't see a proof right now.

the proof I have is a bit involved.

But not for any other pair of numbers... Ok, thanks

yeah exactly

You give the bundle a fiberwise complex structure to make a complex line bundle. There it's a little easier.

Yikes.

I don't know any complex geometry.

Me neither.

I would like to learn complex geometry eventually

Did you complexify the base?

No, no.

Just give it a continuously-varying complex structure fiberwise.

@0celo7 I lied. I meant oriented 2-plane bundle, not 2-plane bundle. Sorry about that.

So...does the section lift to a complex section?

Every oriented rank 2 bundle on surfaces admitting a nowhere zero global section is trivial.

@BalarkaSen "2-plane bundle, not 2-plane bundle" Didn't you just say "I mean foo, not foo"?

Does that suffice to trivialize the line bundle?

@0celo7 Usually, no. Lots of complex line bundles on S^2 which do not admit any global holomorphic sections.

E.g the tautological line bundle.

@someonewithpc Huh?

I said "oriented 2-plane bundle, not 2-plane bundle".

Oh.. Yeah, I misread it :s

@BalarkaSen So what's the gist of the proof once you do that

@0celo7 The fact needed is that any complex line bundle on some n-manifold X is pullback of the tautological line bundle on CP^infty by some map X --> CP^infty.

Lol

where'd you get that from

I don't even know what CP^infinity is

@BalarkaSen

@BalarkaSen

everyone ping Balarka

Idea is that if you have a complex line bundle on X, embed it in X x C^n for some very large n. Then look at the Gauss map (is that what it's called?) which sends X to CP^n by sending each fiber to a 1-dim subspace of C^n.

@0celo7 Union of CP^n's for all n, each CP^{n-1} embedded in CP^n as hypersurface at infinity.

Double yikes!

@Balarka Actually you just need to show it admits a complex vector bundle structure. Then v,iv would give a trivialization of the vector bundle.

"Direct limit" of CP^n's as n \to infty, you can say.

@BalarkaSen Ok, that makes sense. I think.

But still, that fact is not something supporting your claim that you know "no complex geometry"

That isn't complex geometry though.

What is it then

alg. topology.

is it in Hatcher

@PVAL OK, right, true. I guess I was thinking of the more general thing I was asked to prove: every 2-dim submanifold of a 4-manifold which has trivial self-intersection has trivial normal bundle.

So the normal bundle only has a section with trivial Euler number. I haven't yet proved that gives a global nonzero section.

But that's the next thing I have been assigned by Mike to prove :)

You can cancel opposite signed intersection points.

Thats what you need to prove.

Yes, upto homotopy, I can: but after doing that (which requires a large homotopy) I may not be left with a section at the end. Still working on that.

Your homotopies should be fiber wise.

Hmm, OK. I'll think about it.

Any ways you can show any 2-dimensional bundle which admits a metric admits a (almost) complex structure.

You shouldn't need to use anything involving classifying spaces for that.

Misread.

@PVAL Yeah.

@BalarkaSen so where did you find that CP result

Mike gave a sketch of a proof along that lines. My idea was to note that if my bundle is oriented, it's transition functions live in GL^+_n(R). I can homotope that to SO(2). Then the abstract isomorphism SO(2) cong U(1) \cong C* gives a complex structure.

But Mike said one needs to be a little more careful with that. He's written up what I need to do to make that rigorous, I'll have to read it.

In fact if you really want to do it using alg. topology, you can show that the space of complex structures compatible with a given metric over a two dimensional vector space is contractible. But finding a complex structure on a vector bundle is equivalent to finding a section of a bundle of complex structures.

So you just have to find the 2 by 2 matrices J^2=-I which preserve the ordinary inner product g. (i.e g(J_,J_)=g(,).

Actually constructing a section is better though.

It allows you to fix some local constraints on the complex structure.

@PVAL Interesting fact.

I remember needing to actually needing to do the analogous construction when learning about the non-squeezing theorem. The abstract bundle nonsense wasn't enough.

@Balarka You do know that every bundle with contractible fiber admits a section right?

otherwise that was probably hard to read.

Where the heck does one learn about bundles properly

Milnor-Stasheff

is a good start.

Ok, that's on my extended reading list.

Probably Steenrod if you really want to learn about them seriously.

I gave up on that book though.

Also on there.

Why?

I didn't really need to know it all I guess.

It's also kind of an intense book iirc.

and uses outdated language etc.

Yeah, I'm not sure if I need to know that much about vector bundles anyway

@PVAL You mean a nonzero section?

I mean a section

These bundles aren't necessairly vector bundles.

Oh, ok. No, I didn't know this.

Any bundle with a contractible fiber (maybe you need paracompact base) admits a section. For vector bundles you can get this fact by just picking the 0-section.

Right.

Dumb question by wouldn't the base be paracompact

What useful spaces are not paracompact?

It wouldn't ever.

But you need to assume that to prove the theorem.

I know too little about bundles (even vector bundles). Mike has recommended me to read up a few things from Hatcher's VBKT.

Understandable, but when you say fiber bundles I think immediately of a base which is a topological manifold

Theres lots of "useful spaces" that aren't topological manifolds though.

Does one consider fiber bundles over these spaces though?

^ (by ^ I meant PVAL's message)

Yes.

@0celo7 Yes.

Such as....?

Moduli spaces of maps preserving a geometric structure between two manifolds often need to be severely altered before they are manifolds.

@PVAL You wouldn't happen to know about making a Riemannian metric complete by multiplying it with a smooth function, would you?

I personally think Milnor-Stasheff is in every way superior to Hatcher's VBKT

OK, I gotta sleep now. @PVAL I'll look at what you wrote more carefully tomorrow.

@PVAL Perhaps it's not economic in comparison with VBKT, given I am actually learning about smooth manifolds and not about bundles? Dunno.

Well it actually is about smooth manifolds.

It's annoying.

e.g. you get to prove things like every cl. or'd 3-manifold is parallelizable. Various restrictions on when a manifold immerses in another.

Some people use differentiable for smooth.

My analysis prof used smooth for $C^1$.

smooth is universally C^infty

anyone who says otherwise is wrong.

@PVAL Apparently not.

tell that to the cat

are you a furry

You cannot write smooth in an article as a synonym for C^1 and expect to be understood. I doubt this professor ever writes the word smooth in print.

@PVAL I think he was using smooth because $C^1$ functions don't have kinks

they're "smoother" than $|x|$

why are you ignoring someone

You can actually ignore someone without clicking or typing anything.

Yep.

Smash the PC.

Neither is your face

I need a ruler

where can I get a ruler

Hi all

what?

Shao Kahn s house

I posted this question 10 hours ago , but still no answers. Not even comment ... Or even downvote :)

you look for a comment that s cheap, a dvt that is cheaper

I guess

Im an attention ***** :)

Or Maybe the question belongs more on MO ??

Remove that :)

chuck norris never backs down

Soo who is Up for the challenge of giving me an answer on the question linked ?

@Agawa001 neither do I ;)

@mick thats you chuck being on steroids

Unfortunately Steroids dont work for math

smart drugs do

Im gonna read my OWN question again , Hoping guidance from the female Hindu god of math ... The one that helped ramanujan :)

No genius present willing to answer my question( in link ) ? :p

I always try to upvote myself , despite knowing it never works.

:p

Hello

Im thirsty

zbogom