# parallel transport preserves orientation

In my text its written that parallel transport on a Riemannian manifold preserves orientation. Can someone clarify what does that mean? I am confused about this notion.

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Orientation is defined by a non-vanishing volume form on a manifold. What this means is that given a set of $n$-tangent vectors you either get a positive value and you can say the vectors are positive oriented, or you get a negative value and the set of $n$-vectors is negatively oriented. For example, on $\mathbb{R}^3$ the volume form $dx \wedge dy \wedge dz$ acting on $(u,v,w)$ is given by $$(dx \wedge dy \wedge dz)(u,v,w) = det(u|v|w)$$ This is positive if the vectors are a right-handed triple. This means that the ordered set of vectors $\{ u,v,w \}$ are given such that the third vector is on the same side of the plane spanned by $u,v$ as the cross-product $u \times v$.
For a Riemannian manifold, parallel transport is defined such that the length and angles between vectors is constant along the transport. Hence, given a $n$-tuple of $n$-vectors the original volume defined by the set of vectors will be the same as the volume determined by the transported set. Moreover, their relative angular separations will likewise be preserved. It follows that the orientation of the vectors is preserved. Alternatively, you can prove that the volume form as defined in terms of the metric is constant since the metric is constant under parallel transport.