6
$\begingroup$

When we define a connection on one principal bundle $\pi: P\to M$ with structure group $G$ we define it as an association of one subspace $H_pP\subset T_pP$ for each $p\in P$ such that

  1. $T_pP = H_pP \oplus V_pP$
  2. If $\delta_g : P\to P$ is the right action by $g$, then $(\delta_g)_\ast H_p P = H_{p\cdot g}P$ for each $g\in G$

The intuition behind this as I know is that we want to say how do we move from one fibre to the other. Vertical vectors, which live in the vertical spaces $V_pP$ cannot do this, because they point along the fibres. Because of that we want to select at each point one complement, the horizontal vectors, which will point to another fiber.

Now, one alternative way to define a connection is by means of one $\mathfrak{g}$-valued one-form $\omega$ such that

  1. If $A\in \mathfrak{g}$ and $X^A$ is the associated vector field in $P$ then $\omega(X^A) = A$.
  2. $\delta_g^\ast \omega = \operatorname{Ad}_{g^{-1}}\omega$, that is, for each $\tau \in T_pP$, $(\delta_g^\ast \omega)_p(\tau) = \operatorname{Ad}_{g^-1}(\omega_p(\tau))$

Then the horizontal spaces are defined by $H_p P = \ker \omega_p$.

Now, given the intuition on what a connection should be, how intuitivelly we can understand this? How can one, based on the definition of a connection one-form, understand it is able to do the same thing a connection does?

It seems to me that the connection one-form, given one vector $v\in T_p P$ indicating a direction gives one infinitesimal transformation that in some sense corrects the direction so that is becomes horizontal. Because of that if the vector given to it is already horizontal it gives zero. Is this really what the connection form does? If so, how can one see this in a more precise way?

$\endgroup$
  • 1
    $\begingroup$ Have you done geometry with moving frames and a matrix of connection 1-forms? This is merely an abstraction of that. $\endgroup$ – Ted Shifrin Apr 10 '15 at 2:01
  • 2
    $\begingroup$ @TedShifrin -- I agree with you that it's just an abstraction of part of repère mobile. However, I think what the OP is looking for here is the kind of intuition that you could give to someone with only an understanding of Ehresmann connections and differential forms. In other words, a heuristic rather than experience-based intuition. I don't know of such an explanation, however. $\endgroup$ – Robin Goodfellow Apr 10 '15 at 12:39
  • 1
    $\begingroup$ @TedShifrin, as I said I was trying to understand intuitively this definition based just on the Ehresmann connection and differential forms. On the other hand, I've seem just a little about moving frames on surfaces. But I can't really yet establish a precise relationship between the concepts (although I believe that construction on surfaces is a special case of this one). There the connection one-forms as I know are given by $\omega_{ij}(v) = \nabla_v E_i\cdot E_j(p)$ where $\{E_i\}$ is a frame field, but here there's no frame field. $\endgroup$ – user1620696 Apr 10 '15 at 18:23
2
$\begingroup$

Recall that at every point $p \in P$ the vertical tangent space $V_p P$ can be identified with the Lie algebra of the structure group via $(\Psi_p)': \mathfrak{g} \to T_pP$ where the group action in $P$ is denoted by $\Psi: P \times G \to P$. If $\omega$ is your connection form, then the evaluation $\omega(X) \in \mathfrak{g}$ is nothing else as the vertical part of the tangent vector $X \in T_pP$ via the above identification of $\mathfrak{g}$ with $V_pP$. Hence if you have understood connections in terms of the splitting $TP = HP + VP$, then the connection form is nothing else as the projection on the vertical part.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.