3
$\begingroup$

I have trouble understanding how a connection one-form can separate and tangent space $T_u$ 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?

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ Yes. $H_uP$ is defined to be the space of all $X$ such that $\omega(X) = 0$. $\endgroup$
    – Deane Yang
    Commented Sep 16, 2016 at 15:37
  • $\begingroup$ So for two connection one-forms ${\omega _1}$ and ${\omega _2}$ that satisfy condition 1 and 2, do they have the same kernel automatically? $\endgroup$
    – Chen Li
    Commented Sep 16, 2016 at 16:13
  • $\begingroup$ No. The connection is not unique, and different connections have different horizontal subspaces. $\endgroup$
    – Deane Yang
    Commented Sep 16, 2016 at 17:57

1 Answer 1

Reset to default
9
$\begingroup$

Your definition is a sort of hybrid of two standard definitions of connection 1-form. These are:

  1. A connection 1-form is a Lie-algebra valued 1-form $\omega$ which satisfies $\omega(A^\#) = A$ for all $A \in \mathfrak{g}$ and $ad_g(R_g^*\omega) = \omega$.
  2. Let $VP$ denote the vertical bundle of $P$, i.e. the kernel of $d\pi: TP \to TM$. An Ehresmann connection is a complementary subbundle of $TP$, i.e. a subbundle $HP$ of $TP$ such that $TP = VP \oplus HP$ as vector bundles over $P$, which is invariant under the $G$-action: $R_g^* H_u P = H_{ug} P$.

There is a one-to-one correspondence between connection 1-forms and Ehresmann connections as follows. Given a connection 1-form $\omega$, the kernel of $\omega$ is $G$-invariant and complementary to $VP$, so it defines an Ehresmann connection. Conversely, given an Ehresmann connection $TP = HP \oplus VP$, let $q: TP \to VP$ denote the natural projection map (a morphism of vector bundles). Note that the vector fields $A^\#$ define an isomorphism between $VP$ and the trivial bundle $P \times \mathfrak{g} \to P$, so composing $q$ with the projection $VP \cong P \times \mathfrak{g} \to \mathfrak{g}$ gives a map $TP \to \mathfrak{g}$, i.e. a $\mathfrak{g}$-valued 1-form. This is a connection 1-form.

Unwinding these identifications, you could just as well use a third definition:

  1. A connection on a principal bundle is a Lie-algebra valued 1-form $\omega$ which satisfies $\omega(A^\#) = A$ for all $A \in \mathfrak{g}$ and whose kernel $H = \ker \omega$ is $G$-invariant.

This is your definition. From here you can deduce that $H$ complements $VP$ in $TP$, (or that $\omega$ is compatible with the adjoint action of $G$) and hence connections in this sense are the same as the objects defined in the previous two definitions.

Improve this answer
$\endgroup$
10
  • $\begingroup$ Thanks very much for you detailed explanation, that is really helpful, but I still have one question. Let me put it this way: $\endgroup$
    – Chen Li
    Commented Sep 16, 2016 at 20:20
  • $\begingroup$ (Please ignore the previous comment) Thanks very much for you detailed explanation, that is really helpful, but I still have one question. Let's stick to the first definition, assmue there are two connection 1-forms $${\omega _1}$$ and $${\omega _2}$$ which both satisfy $ \omega ({A^\# }) = A$ and $$\ker {\omega _1} = {H_u}P$$. if the kernel of the $${\omega _2}$$ is only a subspace of $$\ker {\omega _1} $$ or $$ \ker {\omega _2} \subset {H_u}P$$, $${\omega _2}$$ still satisfies the second condition,right ? so $${\omega _2}$$ is still a connection one form ? $\endgroup$
    – Chen Li
    Commented Sep 16, 2016 at 20:30
  • $\begingroup$ If $\ker \omega_2$ is a subspace of $\ker \omega_1$ but the two kernels are not equal then the codimension of $\ker \omega_2$ must be strictly larger than the codimension of $\ker \omega_1$ which in turn is the dimension of $VP$. So in that case $\ker \omega_2$ cannot be complementary to $VP$ in $TP$ and hence $\omega_2$ cannot be a connection 1-form. $\endgroup$
    – Paul Siegel
    Commented Sep 16, 2016 at 20:44
  • $\begingroup$ Yes, that is exactly what confuses me, so should we add a third condition, "the kernel of a connection one form is the horizontal space (bundle)?" $\endgroup$
    – Chen Li
    Commented Sep 16, 2016 at 21:37
  • $\begingroup$ Condition 1 forces the kernel to have the right dimension. $\endgroup$
    – Deane Yang
    Commented Sep 16, 2016 at 21:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .