Let $\nabla$ be a connection on a vector bundle $\pi:E\to M$. The curvature of $(E, \nabla)$ is the tensor $R:\mathcal X(M)\times \mathcal X(M)\times \Gamma(M, E)\to \Gamma(M, E)$ defined as $$R(X, Y, \sigma) = \nabla_X\nabla_Y\sigma - \nabla_Y\nabla_X\sigma - \nabla_{[X, Y]}\sigma$$

(Here $\mathcal X(M)$ is the set of all smooth vector fields on $M$ and $\Gamma(M, E)$ denotes the set of all the smooth sections of $E$).

Now the connection $\nabla$ on $E$ defined a connection $1$-form $\phi$ on the frame bundle $FE\to M$. And the curvature of $\phi$ is defined as the $\mathfrak g\mathfrak l_n(\mathbf R)$-valued $2$-form $\Omega$ on $FE$ as $$\Omega(X, Y) = d\phi(X, Y)+[\phi(X), \phi(Y)]$$ for vector fields $X$ and $Y$ on $FE$.

I am wondering what is the relation (connection?) between the curvature $R$ coming from $\nabla$ and the curvature $\Omega$ coming from $\phi$. Is there a way to get $R$ from $\Omega$? Any further interesting comments relevant to the above scenario are very welcome.


The two are actually the same, once you know how to think of them.

Note that the curvature of $(E,\nabla)$, a vector bundle equipped with a connection, can be thought of as an $\mathrm{End}(E)$-valued $2$-form, where $\mathrm{End}(E)$ is the endomorphism bundle of $E$. On the other hand, if $P$ is a principal $G$-bundle over $M$, and $\phi$ is a connection $1$-form on $P$, then the curvature of $(P,\phi)$ can be thought of as an $\mathrm{adj}(P)$-valued $2$-form on $M$. Here, $\mathrm{adj}(P)$ denotes the adjoint bundle. Now, if $P$ happens to be the frame bundle of $E$, then $\mathrm{adj}(P)$ and $\mathrm{End}(E)$ are the same vector bundle, and both curvatures are the same $2$-form.


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