1. Jan 10, 2017

### hideelo

Suppose I had a principle G-bundle (P,π,M) and I wanted to define for each p in P the vertical and horizontal subspaces of $T_pP$. If I considered any point p, I can consider $\pi_* : T_p P \rightarrow T_pM$. Why cant I divide $T_pP$ into the kernel of $\pi_*$ which I will call my vertical and the rest will be my horizontal? Why do I need connection 1 forms?

2. Jan 10, 2017

### lavinia

The direct sum decomposition of a tangent space at a point in the principal bundle is not determined by the differential of the bundle projection map. Think about what you really mean when you say "the rest".

3. Jan 11, 2017

### hideelo

I guess I wasnt thing that in order to define orthogonal, I need some inner product structure, just being a vector space isnt enough.

Thanks

4. Jan 11, 2017

### lavinia

You do not need an inner product, just a direct sum decomposition of the tangent space to the principal bundle into the tangent space to the fiber and a subspace that projects isomorphically onto the tangent space to the manifold. For a connection, one also requires the horizontal spaces to be invariant under the action of the structure group. This is why a G-invariant 1 form with values in the Lie algebra of the structure group is used. It gives you all of these properties in one fell swoop. The horizontal space at a point is its kernel and G-invariance of the horizontal spaces follows from the G-invariance of the 1-form. In fact, the two, the 1-form and the horizontal spaces - are equivalent. That is: the one determines the other.

Last edited: Jan 11, 2017