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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.

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Something in your argument bothers me . It souns like you say every isometry between two oriented inner products spaces must also preserve the orientation. But if we take $-Id:V \mapsto V$, then it is clearly an isometry but it reverses orintation if the dimension isn odd. – Asaf Shachar Jun 7 '15 at 13:28
@AsafShachar I agree that $-I$ defines an isometry on an odd-dimensional inner product space $V$ which does not preserve the orientation. However, the first paragraph is makes no claim that every isometry preserves volume. Really, I'm just trying to get across the idea of the orientation tell us which way is up given a co-dimension 1 object (like a plane in three dimensions) and that we can either take a frame (a set of $n$-ordered vectors) or an $n$-form to describe the orientation... – James S. Cook Jun 7 '15 at 18:31
@AsafShachar For the second paragraph, notice we are given in the problem statement (and this is perhaps a subtle point) that the manifold is oriented. So, obtaining $-I$ in the transport from one tangent space to another is forbidden as that would imply the orientation form changed sign somewhere, and as it is continuous, it follows it is somewhere zero. This contradicts the assumed fact that the manifold is oriented. It follows that the determinant of the isometries implicit within the transport are one (since otherwise we'd spoil the orientation) – James S. Cook Jun 7 '15 at 18:40

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