Suppose X and Y are two unit length vector fields on a Riemannian manifold which are orthogonal at each point. Is it true that the lie bracket of X, Y belongs to the span of the vector fields at each point in the manifold? Is there a way to prove or disprove this without going into local coordinate expressions. Thanks
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2 Answers
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Pick any manifold of dimension 3 or higher and any two non-vanishing vector fields that do not satisfy the Frobenius condition. Define a Riemannian metric such that the vector fields are orthonormal.
answered Dec 9, 2014 at 1:24
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$\begingroup$ Thank you. But what would happen if I impose $X= \frac{\nabla \phi}{|\nabla \phi|}$ and $Y = \frac{\nabla \psi}{|\nabla \psi|}$ so that the vector fields have explicit information about metric. Do you suspect that the answer is again negative? $\endgroup$– AliCommented Dec 9, 2014 at 2:17
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$\begingroup$ I don't see why this should be any different. I suggest trying to construct a counterexample with the standard flat metric. $\endgroup$ Commented Dec 9, 2014 at 15:13
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1$\begingroup$ With the standard flat metric, any affine function $\phi$ will have gradient perpendicular to that of a function $\psi$ just when their gradients commute, so you get some examples. However, for Morse $\phi$ you find that there are no possibilities for $\psi$, because there are no integrals of motion of the gradient vector field. $\endgroup$ Commented Dec 9, 2014 at 19:42
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Whether the two fields are orthonormal or not has nothing to do with integrability: After doing Gram-Schmidt orthonormalization you have the same integrability property as before, since $[X,fY] = f[X,Y] + X(f).Y$ etc.
answered Dec 9, 2014 at 14:54