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Are the gradients of coordinates, like $\operatorname{grad} u$, $\operatorname{grad} v$, in a non-orthogonal coordinate system of a surface like $(u,v,u^2 +3uv)$, still equal to the dual vectors? (I can get the duals easily using the inverse metric, the gradients are tougher to calculate); it appears the duals in the above aren't even in the same direction as $u$, $v$ (or $dR/du$ and $dR/dv$) which means the idea that $du(v)$ is a "one form" breaks down, because the dual isn't in the direction of the gradient anymore and $du(v)$ needs to be in the direction of gradient.

ronno
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robert prince
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  • I did not know that the dual of a vector is another vector that needs to have the same direction. Perhaps this helps ? – Kurt G. Oct 22 '22 at 04:13

1 Answers1

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It took me a while to realize that it is the V parametric curve which has a constant U and so grad of U is perpendicular to that V curve so its the level sets of v curves(u constant) that dR/du goes through to make calculation.So it works in nonorthogonal coordinates.

robert prince
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