Unique separation of the vertical and horizontal spaces?

[lichen1983312]

I have trouble understanding how a connection one-form can separate and tangent space${T_u}P$ of a principal bundle uniquely into horizontal and vertical spaces ${H_u}P \oplus {V_u}P$ since from the literature I am learning (mainly Nakahara' book), the connection one-form is a Lie-algebra-valued one-form that satisfies
1, $\omega ({A^\# }) = A$
2, ${R_{g * }}{H_u}P = {H_{ug}}P$
where ${A^\# }$ is the fundamental vector field.
My question is how does it separate ${T_u}P$ uniquely since from the first requirement we only project the vertical space into its corresponding Lie algebra. Should we have some additional requirement like $\omega (X) = 0$ for $X \in {H_u}P$ ? Or this condition can just be derived from the second requirement?

To be more specific, for two connection one-forms ${\omega _1}$ and ${\omega _2}$ that satisfy condition 1 and 2, do they have the same kernel automatically?

 

[lavinia]

I have trouble understanding how a connection one-form can separate and tangent space${T_u}P$ of a principal bundle uniquely into horizontal and vertical spaces ${H_u}P \oplus {V_u}P$ since from the literature I am learning (mainly Nakahara' book), the connection one-form is a Lie-algebra-valued one-form that satisfies
1, $\omega ({A^\# }) = A$
2, ${R_{g * }}{H_u}P = {H_{ug}}P$
where ${A^\# }$ is the fundamental vector field.
My question is how does it separate ${T_u}P$ uniquely since from the first requirement we only project the vertical space into its corresponding Lie algebra. Should we have some additional requirement like $\omega (X) = 0$ for $X \in {H_u}P$ ? Or this condition can just be derived from the second requirement?

To be more specific, for two connection one-forms ${\omega _1}$ and ${\omega _2}$ that satisfy condition 1 and 2, do they have the same kernel automatically?

The horizontal space at a point in the principal bundle is just the kernel of the connection 1-form. The tangent space splits into a direct sum $K⊕V$ where $K$ is the kernel and $V$ is the vertical space. Connection 1-forms differ by their kernels since their restrictions to the vertical spaces are all the same.

In general two linear maps can agree on a subspace but have different kernels. The vector space is always isomorphic to the direct sum of the image and the kernel of the linear map.

 

[lichen1983312]

Thanks very much!