Let $\lambda = (P,\pi,M;G)$ be a smooth principal $G$-bundle (projection $\pi : P \to M)$, $V$ a finite dimensional vector space, and $\rho : G \to GL(V)$ a smooth representation of $G$ in $V$.

We can associate to $\lambda$ a vector bundle on $M$ with total space $P \times_{\rho} V$, wich I note $\lambda \times_{\rho} V$. If $\lambda$ is equipped with a principal connection $H$ whose 1-form is $\omega$, there is a unique connection $\nabla^H$ on $\lambda \times_{\rho} V$.

If $(U_i,G,\Phi_i)$ is a trivialisation of $\lambda$, it induces a trivialisation $(U_i,V,\Psi_i)$ on $\lambda \times_{\rho} V$. Let $s_i$ be the local section of $\lambda$ defined by $s_i(x) = \Phi_i^{-1}(x,e)$, then the local 1-form of $H$ on $M$ is $\omega_i = s_i^* \omega$, and there is a simple relation between the local connection form $\Gamma_i$ of $\nabla^H$ and local 1-form $\omega_i$ : $$ \Gamma_i = \rho_* \omega_i$$

On the other way, if we have a vector bundle $\xi = (E,\pi,M)$ with fiber-type $V$, we associate to $\xi$ its frame bundle, i.e. a $GL(V)$-principal bundle which I note $\lambda_F(\xi)$, whose total space $P$ is the set of couples $(x,F_x)$ where the isomorphism $F_x : V \to E_x$ is a frame of $E_x$.

If there is a connection $\nabla$ on $\xi$, whose horizontal distribution is $H(\nabla)$ we can associate to it a unique connection $H_{\nabla}$ on $\lambda_F(\xi)$ : this horizontal distribution can be defined at a point $p \in P$ as $${H_{\nabla}}_{p} = \{Y \in T_{p}P : \forall v \in V,\, T_{p}Q_{v}(Y) \in H(\nabla)_{[Q_{v}(p)]}\}.$$ where $Q_v : P \to E, \ (x,F_x) \mapsto F_x(v)$.

My question : is there a simple expression relating the given local $\Gamma_i$ of $\nabla$ and the local 1-form of the induced connection $H_{\nabla}$ on $\lambda_F(\xi)$ ?


1 Answer 1


I think I have the answer, and if I am not wrong it's so simple that I even regret to have asked this question : $$ \omega_i = \Gamma_i$$ This comes form the fact that the space of linear connection on $\xi$ and the space of principal connection on $\lambda_F(\xi)$ are affine isomorphic, and the correspondances $$H \mapsto \nabla^H \quad \text{and} \quad \nabla\mapsto H(\nabla)$$ realizes this isomorphism, once we make the canonical identification $$\xi \leadsto \lambda_F(\xi) \leadsto \lambda_F(\xi) \times_{\rho} V \simeq \xi$$ Here the structural group of $\lambda_F(\xi)$ is $GL(V)$ and the representation $\rho$ is the identity. So the induces connection on $\lambda_F(\xi)$ induces a connection on $\lambda_F(\xi) \times_{\rho} V \simeq \xi$ such that in the correspondind trivializations $$\Gamma_i = \rho_* \omega_i = \omega_i$$, and this $\Gamma_i$ should be the same than the connection form of the initial linear connection on $\xi$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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