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Let $E\to M$ be a smooth oriented real vector bundle of rank 3, so that its structure group can be reduced to $SO(3)$. Suppose $E\to M$ is given a Riemannian metric and a connection $\nabla:\Omega^0(E)\to \Omega^1(E)$ compatible to it.

Consider its endomorphism bundle $\text{End}(E)$. $\nabla$ induces a connection (also denoted by $\nabla$) on $\text{End}(E)$ satisfying $(\nabla \phi)(\sigma)=\nabla(\phi(\sigma))-\phi(\nabla\sigma)$ (cf. exterior covariant derivative of $\operatorname{End}(E)$-valued $p$-form).

Now consider the adjoint bundle $\text{Ad}(E)$ of $E$. The Lie algebra $\mathfrak{g}\subset \text{End}(\Bbb R^3)$ of $SO(3)$ is isomorphic to $\Bbb R^3$, via the isomorphism $\Bbb R^3\to \mathfrak{g}$, $v\mapsto (v\times -)$, where $\times $ is the cross product of $\Bbb R^3$. This gives an isomorphism $E\to \textrm{Ad}(E)$. Also the connection $\nabla$ on $\text{End}(E)$ defined above restricts to a connection on $\text{Ad}(E)$.

So now we have two connections on $E\cong \text{Ad}(E)$ (denoted by the same symbol). But are these two the same? It seems quite natural to be the same, but I'm not sure

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Yes, they are the same.

More formally, the section $c\in\Gamma\mathrm{Hom}(E,\mathrm{Ad}(E))$ defined by $c(\sigma)(\tau)=\sigma\times\tau$ is parallel with respect to the induced connection, which means that for all $\sigma\in\Gamma E$ $$ \nabla \,c(\sigma)=c(\nabla\sigma) $$ This is equivalent to the fact that $$ \nabla(\sigma\times\tau)=\nabla\sigma\,\times\tau+\sigma\times\nabla\tau $$ for all $\sigma,\tau\in\Gamma E$. To show this, fix $p\in M$ and a local positively oriented orthonormal frame $e_0,e_1,e_2$ with ${\nabla e_i}_{|p}=0$ and write $\sigma=\sigma^ie_i$, $\tau=\tau^ie_i$ using the double summation convention. Now consider indices modulo $3$, so that $e_{i}\times e_{i+1}=e_{i+2}$. Then at $p$

$$ \nabla(\sigma\times\tau) =\nabla(\sigma^i\tau^{i+1}e_{i+2}-\sigma^i\tau^{i+2}e_{i+1})\\ =(d\sigma^i)\tau^{i+1}e_{i+2}-(d\sigma^i)\tau^{i+2}e_{i+1}+\sigma^i(d\tau^{i+1})e_{i+2}-\sigma^i(d\tau^{i+2})e_{i+1}\\ =\nabla\sigma\,\times\tau+\sigma\times\nabla\tau $$

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