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Riemann curvature s denoted by $R^a_{bcd}$. If you try to attempt to parallel transport a tangent vector $\psi$ along an infinitesimal parallelogram given by two tangent vectors along non-parallel sides as $\eta$ and $\nu$, the following holds true. $$\psi^a_{;cd} - \psi^a_{;dc} = R^a_{cdb}\psi^b$$ where the $;$ indicates covariant differentiation. Here the $c$ and $d$ on the L.H.S. implies the two directions of the parallelogram. My professor said that the Riemann curvature tensor denotes the extent of the non commutativity of the covariant diferentiation along directions. In flat space, I would expect the Riemann Tensor to be zero. But does this imply commutativty of covariant differentiaton? Because covariant differentiation of a vector along a direction denotes how much the vector deviates during the parallel transport and one would not expect a deviation in flat space. So this should make the left hand side zero independent of the right hand side. What do you think?

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    $\begingroup$ Note that it is not the commutativity of covariant derivatives, but, specifically, the commutativity of the 2nd-order covariant derivatives! $\endgroup$ Commented Dec 30, 2017 at 23:59

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In flat space the covariant derivative reduces to the partial derivative (by definition). The partial derivatives do commute, hence the L.H.S. vanishes and the R.H.S., that is the Riemann tensor, measures zero.
Just a comment on the statement, it should be:
$\psi^a_{;cd} - \psi^a_{;dc} = R^a_{bdc}\psi^b$
As I know the indices of the non-parallel sides of the infinitesimal parallelogram should be the third and the fourth. A possible ambiguity remains in the sign, as you can decide which way of the loop to assume as positive, however it is a matter of convention.

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  • $\begingroup$ If the covariant derivative reduces to the partial derivative, the notion is still that of parallel transporting vector $\psi$. In this case, it might help to think of $\psi$ as a constant vector field $\psi(x)$ over the manifold. Partial differentiating a constant vector field at any point would give zero for each of the terms individually on the L.H.S. Would this be correct? $\endgroup$
    – Matrix23
    Commented Dec 31, 2017 at 22:42
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    $\begingroup$ However in a flat manifold the parallel transport is independent of the path, so after a loop it delivers the same vector. No change in the vector means a zero R.H.S. in any case. $\endgroup$ Commented Jan 1, 2018 at 18:07
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The right hand side in that equation is identically $=0$ since the component of the vector are smooth and the covariant derivative reduces to partial derivative, in fact the $\Gamma$s are zero in a flat space.

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