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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.

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1 Answer 1

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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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