This question came up while reading Differential Geometry by Loring W.Tu.
Suppose $\pi: E \to M$ is a $\mathcal{C}^{\infty}$ vector bundle. And $f:N \to M$ a $\mathcal{C}^{\infty}$ map of manifolds. Let $\nabla$ be a connection on $E$ with connection matrix $\omega_e$ relative to a local frame $e$ for $E$. Then there is a unique connection $\nabla'$ on the pullback bundle $f^{*}E \to N$ whose connection matrix relative to the frame $f^{*}e$ is $f^{*}(\omega_e)$.
That makes sense so far.
The author then writes:
The induced curvature form on $f^{*}E$ is therefore
$f^{*} \Omega_e = f^{*} \omega_e + \frac{1}{2} [f^{*} \omega_e, f^{*} \omega_e]$.
Now I don't understand why the induced curvature forms are $f^{*} \Omega_e$ or why this formula holds.
Best Answer
First of all, I think that there is a typo, either in your question or in the book. The right equality should be this $$f^*\Omega_e = d(f^*\omega_e) + \frac{1}{2}[f^*\omega_e,f^*\omega_e]$$ or this $$ f^*\Omega_e = f^*d\omega_e + \frac{1}{2}f^*[\omega_e,\omega_e], $$ which are pretty much the same thing, since the pullback commutes with both the exterior differential $d$ and the bracket $[\cdot,\cdot]$. These equalities are trivial, since by definition, $\Omega_e = d\omega_e + \frac{1}{2}[\omega_e,\omega_e]$ (apply the pullback through $f$ to that equality).
The real question is: Why is this the curvature form of $\nabla'$ on $f^*E$ associated to the frame $f^*e$? The answer is because, as you said, the connection matrix $\omega_{f^*e}$ associated to the frame $f^*e$ is $f^*\omega_e$. Therefore $$\Omega_{f^*e} :=d\omega_{f^*e} + \frac{1}{2}[\omega_{f^*e},\omega_{f^*e}]\underset{\omega_{f^*e}=f^*\omega_e}{=} d(f^*\omega_e)+\frac{1}{2}[f^*\omega_e,f^*\omega_e]$$ which has already been shown above to be equal to $f^*\Omega_e$.