In my recent studies of curvature, I worked with the Riemann tensor and the equation:(adsbygoogle = window.adsbygoogle || []).push({});

([itex]\delta[/itex]V)^{a}= A^{[itex]\mu[/itex]}B^{[itex]\nu[/itex]}R^{a}_{b[itex]\mu[/itex][itex]\nu[/itex]}V^{b}

Now previously, I worked in 2D with the 2 sphere. While doing so, I learned that if I set my x^{1}coordinate to be θ and my x^{2}coordinate to be ø, then the vectors that serve to be the legs of the loop that I am transporting around would be as follows:

A^{[itex]\mu[/itex]}= <θ, 0>

B^{[itex]\nu[/itex]}= <0, ø>

and then of course the vector that I parallel transport would be as follows:

V^{b}= <θ, ø>

Now this may work for 2 dimensions, but what if I have 3 or more dimensions? With only 2 vectors being the legs of the loop, there wouldn't be enough vectors for me to give each individual coordinate its own leg with every other component being 0 (as shown above with A^{[itex]\mu[/itex]}and B^{[itex]\nu[/itex]}).

How do I deal with this? Is it even a requirement for every coordinate to have its own leg that is reminiscent of a unit vector? Page 5 on the following PDF gave me the impression that it is a requirement:

http://www.physics.ucc.ie/apeer/PY4112/Curvature.pdf

Is it possible for one of the legs of the loop to have more than one type of coordinate in it (like A^{[itex]\mu[/itex]}= <r , θ, 0>) ?

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