MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\pi_P:P\to M$ be a principal $G$-bundle on a manifold $M$ endowed with a connection $A$ and $\pi_Q:Q\to M$ be a principal $H$-bundle on a manifold $M$. Let $f:P\to Q$ be a morphism of bundles along $\lambda :G\to H$ which means that $$\forall p \in P, \forall g\in G: f(p.g)=f(p)\lambda(g)$$ We can push-forward the connection $A$ to a connection $f_*(A)$ on $Q$. Let $a\in\Omega^1(Q,\mathfrak{h})$ be an $\mathfrak{h}$-valued 1-form on $Q$ then $B=f_*(A)+a$ is also a connection on $Q$. We denote the curvature of $A$ by $F_A$.

can we find the curvature of $B$ in terms of $F_A$, $a$ ,$f$ and $\lambda$?

share|cite|improve this question
If $a=0$ then $F_B=\lambda\cdot F_A$ where $\lambda\cdot\colon\mathfrak g\to \mathfrak h$ is the induced map on Lie algebras. For $a\neq0$ you would in general need also to know $f_*A.$ – Sebastian Jul 25 '13 at 14:38
@Sebastian: I think, in the case $a=0$ your answer need to revision,$f^*F_B=\lambda.F_A$.(Foundations of differential geometry, Kobayashi vol.1 Prop.6.1) – Arash Jul 27 '13 at 13:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.