# Frame bundle

lavinia
Gold Member
An Example: The torus is a circle bundle over the circle and has a natural action of SO(2) on each fiber - just rotate. Any vector field that is invariant under fiber rotation and which is never tangent to any fiber circle generates a horizontal space. The connection one form is easy to define from the decomposition of a vector into horizontal and verticle components.

to see that there are infinitely many connections, just take any smooth vector field on a circle that is transverse to the fiber and rotate is around to get a vector field on the whole torus.

lavinia
Gold Member
An example: (Please correct this if it is wrong.)

On a surface (two dimensional Riemannian manifold)

Take a curve that is parameterized by arc length and a unit vector field ,v,along the curve that is perpendiclular to the curve. The covariant derivative of v will be tangent to the curve and so is some multiple of c'(s).

To compute this multiple view v as a map of the curve into the tangent circle bundle (same as the bundle of orthonormal frames). dv(c'(s)) is the derivative of v with respect to c'(s).

Decompose dv(c'(s)) into horizontal and vertical components. The multiple we are looking for is the coefficient of the vertical component of this decomposition.

c'(s )is also a map of the curve into the tangent circle bundle and the argument is the same.
The covariant derivative this time will be in the v direction normal to c'(s).

Extend now by the Leibniz rule and linearity to arbitrary v.

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Bacle2
I will try to clear things up a little for you, if that is possible.

I wouldn't put it that way, because it sounds like you're saying that fibers are made up of basis for the vector space of the vector fields on M, which is false of course. More precise would be that the frame bundle over a manifold M is a principle GL(n,R)-bundle who's fibers are the sets of ordered bases for the tangent space of M at p. (where n=dim(M))

Yes, it is literally an n-tuple of vectors spanning TmM.

You have to be careful here because a fiber bundle (such as a principal bundle) is itself a manifold. So you have to distinguish the dimension of the bundle as a manifold and the dimension of its fibers. Here, 16=4² is the dimension of the fiber. The dimension of the frame bundle as a manifold is 16+4=20. What do you mean "why so many dimensions"? I would think that you know the answer to that since you computed correctly that GL(4,R) has dimension 16!

Ah, perhaps you do not understand what the connection is between the fact on the one hand that the fiber over m are the frames of TmM and the claim that this space is just GL(4,R). This is simply because in a vector space V of dimension n, once you fix a basis (vi), you can then express any vector w in V as a set of n real numbers; namely the so-called coordinates of w wrt to (vi):
$$w=\sum_i a_iv_i$$
In particular, if (wj) is another basis of V, then this corresponds to n² numbers (n numbers for each wj), which you can arrange in a matrix by declaring that the first row is to be made up of the n coordinates of w1 and so on. Then the fact that these (wj) are linearly independant is equivalent to saying that the matrix thus constructed has nonvanishing determinant. That is, it is a matrix in GL(n,R). So you see that there is a (non canonical) bijective correspondance between the frames F(V) of V and GL(n,R). Use that to transfer the smooth structure of GL to F(V). You can verify that this will be independant of the choice of bijection and so puts a well-defined canonical smooth structure on F(V) such that given any choice of basis in V resulting as above in a bijection F(V)<-->GL, this bijection is a diffeomorphism. So that is how each fiber of the frame bundle is (non canonically) diffeomorphic to GL(n,R).

Quasar:
When you refer to the dimension of Gl(4,R) ; algebraic dimension, if I understood well, is equal to 4^2=16. What kind of algebraic structure are you assigning to Gl(4,R)?

morphism
Homework Helper
GL(n,R) is an open subset of ##M_n(\mathbb R) \cong \mathbb R^{n^2}##, so it's a manifold of dimension equal to dim M_n(R) = n^2.

Bacle2
I understand that, but I thought I read a reference to algebraic dimension, tho I may have misunderstood--or misunderestimated what I read.

quasar987
Homework Helper
Gold Member
No reference to algebraic dimension. :)

Bacle2
Sorry, Quasar,
I misunderestimated your question :). Nice post, btw.

Bacle2
Sorry again for my slowness, Quasar. I understand that Gl(n,R)=Det-1(ℝ\{0}), which is open, yada, yada; I realized where my confusion lay.

Anyway, another dumb question; I am also trying to teach myself some bundles: (my apologies, my quoting button is disabled for some reason): say you are considering the case of a vector bundle ( thinking mostly of tangent bundle), in a situation where you have a metric defined. It would then make sense to define the complement bundle to be the ortho-complement (with respect to this metric), right? Is there anything special to this choice?

quasar987
Homework Helper
Gold Member
I'm not sure I follow you. You are thinking of the case where (M,g) is a riemannian manifold. Then you have the projection map pr:TM-->M and its differential pr*:T(TM)-->TM and you want a complementary subbundle to V:=ker(pr*). You cannot just pullback g to T(TM) via pr, and use that to define H:=V$^{\perp}$ because pr*g is very degenerate: pr* vanishes on V and so V$^{\perp}$=T(TM).

Is this what you were thinking?

Bacle2
Yes, got it, thanks. I just need to read more carefully. Sorry.

Bacle2
For the sake of background: I ended up studying bundles by mistake:

My girlfriend wanted to take a class in cosmetology, but she misread the

instructions in the webpage, and ended up registering for cosmology instead

. Since there were no refunds, and she knew no math,I had to help her.

Then I became interested in bundles.

quasar987
Homework Helper
Gold Member
haha, funny story. :)

But how do bundles pop-up in cosmology?!?

Gold Member
Even in graduate GR, I never had to use bundles for cosmology, seems weird. XD

Bacle2