This might not be research level but I've tried more than once to ask about this in MSE and it got nowhere. So I thought It's fair to at least try.

At the risk of repeating well known stuff I tried to make the question as precise as possible.

Let $\pi: P \to M$ be a $G$-bundle with connection form $\omega \in \Omega^1(P ;\mathfrak{g})$. Let $\rho : G \to Gl(V)$ be a linear representation and $E = P \times_{\rho} V$ the corresponding associated bundle.

Denote by $\Omega^{\bullet}_{\rho}(P ; V) \subset \Omega^{\bullet}(P;V)$ the space of $V$-valued forms satisfying:

  • Vertical: $i_X \eta = 0$ for all $X \in VP = \ker \pi_*$.
  • $R^*_g \eta = \rho(g^{-1}) \cdot \eta$ for all $g \in G$.

There is a one to one between forms in $\Omega^{\bullet}_{\rho}(P ; V)$ and sections of the bundle $\Omega^{\bullet}(M) \otimes E \to M$.

We have an isomorphism $\varphi : \mathfrak{g} \to VP$ sending $X \to (p \mapsto \frac{d}{dt}(e^{tX}\cdot p))$. Therefore, our connection $\omega \in \Omega^1(P ;\mathfrak{g})$ gives rise to a retraction $\varphi^{-1} \circ \omega :TP \to VP$ and thus to a projection onto the horizontal bundle $h= I-\varphi^{-1} \circ \omega$. This determines a derivation on the complex $\Omega^{\bullet}(P;V)$ called the exterior covariant derivative:

$$D: \eta \mapsto d\eta \circ h$$

This derivation descends to a derivation $\Omega^{\bullet}_{\rho}(P ; V)$.

Now to the question:

In the appendix of the book "Dirac operator on riemannian geometry" there's a nice formula for how the exterior covariant derivative acts on $\Omega^{\bullet}_{\rho}(P ; V)$. Let $\eta \in \Omega^r_{\rho}(P,V)$. Here is the formula:

$$D \eta = d\eta + \rho_*(\omega)\wedge \eta$$

Where the wedge of a matrix and vector calued forms is defined as:

$$\rho_*(\omega)\wedge \eta (v_0,...,v_r) = \sum^r_{j=0} (-1)^j \rho_*(\omega(v_j))\cdot \eta(v_0,...,\hat{v_j},...,v_r))$$

First part of question:

  1. How can this formula be neatly derived? It gets messy so quickly for me I don't manage to get very far...

  2. Is there a deeper meaning to wedging a matrix and a vector or is it just a notational conveniance? On the face of it it seems to be a very awkward thing to do.

In the case where $G=GL(V)$ the representation is the identity and $P\cong F(E)$ we can get to a connection on the associated bundle $E$ by simply pulling back along a frame. Indeed if $u: U \to P$ is a local section then the covariant derivative is $\nabla = u^*(\rho_*(\omega)) \in \Omega^1(U ; End(V))$.

This procedure doesn't seem to generalize directly.

  1. How do I get from $\rho_*(\omega)$ to a covariant derivative $\nabla \in \Omega^{\bullet}(M;End(E))$ in the general case?

And finally:

What's a good source to read about transferring connection information from principal bundles to asscoiated bundles and back?

I can't stress enough the extent at which i feel this is glossed over in the familiar literature.

  • $\begingroup$ See Section 3.3.3. of these lectures www3.nd.edu/~lnicolae/Lectures.pdf $\endgroup$ Jan 25, 2016 at 15:48
  • $\begingroup$ @LiviuNicolaescu Thanks, I didn't find anything there about principal bundles though... $\endgroup$ Jan 25, 2016 at 15:53
  • $\begingroup$ Sec. 8.1.1 of the lectures deals with principal bundles. However, the result being true for any connection on a vector bundle, it will be true for a connection induced from a connection on a principal bundle via a representation of the structure group. The key formulas are (3.3.12)-(3.3.14) $\endgroup$ Jan 25, 2016 at 16:01
  • $\begingroup$ @LiviuNicolaescu I tried computing again now and I'm still getting stuck. My problem is there must be a way of understanding what's happening here without computing like hell. Let's simplify and suppose that $P$ is the frame bundle of the vector bundle $E$. Here if i pullback $\rho_*(\omega)$ via a frame I get the usual covariant derivative right? But $P$ here is not necessarily the frame bundle. How can I cook the covariant derivative from $\rho_*(\omega)$? How do i get to the formula above? does it look exactly the same for sections over the base? what is $\rho_*(\omega)$ then? $\endgroup$ Jan 25, 2016 at 16:31
  • $\begingroup$ See 19.8 - 19.15 of mat.univie.ac.at/~michor/dgbook.pdf $\endgroup$ Jan 25, 2016 at 18:43

1 Answer 1


Ad 1.: Since every vector can be decomposed in its horizontal and vertical part. Thus it is enough to consider the case a) where all vectors are horizontal (this is trivial) and b) where at least one is vertical and the rest is horizontal (this has to be calculated explicitly but is relatively simple since one can use the fact that the vertical vector is a Killing field).

Ad 2: As soon as one has a bilinear form $A \times B \to C$ one can form a wedge product of forms with values in $A$ and forms with values in $B$ to get forms with values in $C$ (by the above formula). For example the wedge product of Lie algebra valued forms is defined in the same way using the Lie commutator as the underlying bilinear form.

Ad 3: Use 1 and the identification of sections of $E$ with equivariant maps $P \to V$ (under which $\nabla$ corresponds to $D$)

  • $\begingroup$ Thanks, I finally managed to carry out the computation. Your comments wer'e very helpful! Just one last thing, it looks to me like the same equation should hold if I pull these forms back via a frame and it will look like $d^{\nabla} \alpha = d \alpha + \nabla \wedge \alpha$ where the $\nabla$ here is treated as an $End(E)$ valued form on the base. Is that right? $\endgroup$ Jan 29, 2016 at 21:37

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