§ Ehrsmann connection
Here's my current understanding of how the Ehrsmann connection works.
We have a -bundle .
Let's first think of it as globally trivial so . Now at each point
, we have the fiber over . We now consider the kernel
of the map . What are the elements here?
We know that where
is the Lie algebra of the lie group . So we know that maps .
Thus the kernel of is going to be .
This is called as the "vertical subspace" .
Now, we have a choice in how we pick for each point such that .
This choice of is the connection. We claim that this choice is equally well encoded by a lie-algebra
valued one form, . That is, ,
which is . Intuitively, this tells us how much of the component
along is not covered by the .
The idea is that since , given any vector , I can compute
. Then I will have since I've killed the component in .
However, I know that is the same as . Thus, I spit out the value , treated as an
element of . This tells me how much of is not stolen away by in the decomposition.
So we have the map given by . The kernel
of this map is .
§ Generalizing to Non-trivial bundles
If we have a non-trivial bundle, then I need some way to link with without splitting the bundle
as I did here. The idea is that element of are basically curves . We use the curves
to build derivations. For each lie algebra element , I can build the curve
given by . That is, the curve I get by pushing
the point along . Note that all the points in the curve lie on the same point
in the base manifold, because the group only moves within fibers. So we have that for all .
This means that when we push forward the curve , it represents the constant curve, which has zero derivative!
Thus, we have that all these curves are in the kernel , and hence .
To show the other inclusion, pick some element . The group must be non-trivial,
otherwise the bundle will also be trivial. Let for some . Note that since the bundle is a principal
bundle, we have that the fiber is a -torsor. I guess this is ismorphic as a group to . Now, the
tangent space is the tangent space the group , which has the same dimension as the lie algebra .
Hence, the function we defined above must be surjective.
NOTE TO SELF: there should be a more direct proof that uses the fact that the fiber is -torsor!