
#1
Nov211, 11:28 PM

P: 2,045

Suppose that I have a differentiable manifold M on which I don't define a metric. I wish to define an affine connection on that manifold that will allow me to parallel transport vectors from one tangent space of that manifold to another tangent space.
How arbitrary can I make my choice? I do know that connections must transform correctly under coordinate transformations in order to keep my vector still a vector if I transport it, but is that it? Are there other restrictions on how I can choose my connection? For example, we see pictures of parallel transport on the two sphere and because we can see the 2 sphere embedded in 3D space, we can "intuit" what parallel transport would be like on that sphere. However, am I confined to that choice? Could I define an affine connection on a two sphere (given no metric, so I don't have to worry about compatibility issues) that would parallel transport vectors completely unintuitively to how I "would" do it from my 3D perspective? Could I make the vectors twist and turn in weird fashion? It seems to me that the affine connection is quite arbitrary; however, I have also seen equations that link it to how basis vectors "twist and turn" as we move throughout the manifold, so I am confused on really how arbitrary it is. Perhaps I am too reliant on bases? @_@ 



#2
Nov211, 11:59 PM

Sci Advisor
P: 1,563

Generically speaking, a connection is a matrixvalued 1form [itex]\omega^a{}_b[/itex] (actually, Lie algebravalued). Then the covariant derivative of a vector field [itex]X = X^a \, e_b[/itex] is written
[tex]\nabla X = (d X^a + \omega^a{}_b X^b) \otimes e_a[/tex] where [itex]e_a[/itex] is any frame (i.e., a differentiable choice of basis at every point in some open patch). In principle the only restriction on the matrix [itex]\omega^a{}_b[/itex] is that it be invertible; i.e. [itex]\omega^a{}_b \in \mathfrak{gl}(n, \mathbb{R})[/itex]. There may be further restrictions due to global topology of the manifold, to insure that the connection is continuous everywhere. 



#3
Nov311, 12:11 AM

P: 2,045

But, other than those pretty general restrictions, I am free to choose them however I want? In what sense is the connection giving me a "parallel' transport then, if it's so arbitrary?




#4
Nov311, 12:23 AM

Sci Advisor
P: 1,563

How arbitrary are connections?
A connection gives you the definition of "parallel". The point is that you can define "parallel" however you want.
There are other nice properties you can ask for, such as 1. Torsionfree (i.e. vectors do not twist helically around paths) 2. Volumepreserving (i.e., having a symmetric Ricci tensor) 3. Lengthpreserving (i.e., metric compatibility) 



#5
Nov311, 01:03 AM

P: 2,045

I see...thanks. =]



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