Let $(M^2,g)$ be a 2-dimensional Riemannian manifold. For any point $p\in M^2$ can we always find coordinates $(u,v)$ in a neighborhood $U$ of $p$ such that the Gaussian curvature is only a function of $u\pm v$, i.e, $K=K(u\pm v)$ ?
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2$\begingroup$ Near a nondegenerate critical point of Gauss curvature, clearly the answer is no. $\endgroup$– Ben McKayCommented Nov 23, 2021 at 14:04
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2$\begingroup$ Near a regular point of Gauss curvature, clearly the answer is yes. $\endgroup$– Ben McKayCommented Nov 23, 2021 at 14:05
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$\begingroup$ @BenMcKay Thanks. In the affirmative case, do we have an idea about how to construct such a system of coordinates ? $\endgroup$– Daniel CastroCommented Nov 23, 2021 at 14:15
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1$\begingroup$ Near a regular point of Gauss curvature $K$, i.e. where $dK\ne 0$, take any function $v$ with nonzero differential linearly independent of $dK$ at that point. Let $u=K+v$. Near degenerate critical points, I am not sure. $\endgroup$– Ben McKayCommented Nov 23, 2021 at 14:31
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2$\begingroup$ A necessary condition: at any critical point, the Hessian $D^2K$ becomes well defined (independent of coordinates or metric), and must have rank $0$ or $1$. If rank $0$, i..e vanishing Hessian, then $D^3K$ becomes well defined and must have rank $0$ or $1$. If $D^2K$ has rank $1$, then $D^3K$ becomes defined along the null space of $D^2K$, and must vanish there, along with all higher $D^n K$ for all $n\ge 3$. $\endgroup$– Ben McKayCommented Nov 23, 2021 at 14:37
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1 Answer
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Curvature is a smooth function on the surface, and locally, any smooth negative function can serve as a curvature of some surface (M. S. Berger, Riemannian structure of prescribed Gaussian curvature for compact 2-manifolds, J. Differential Geom. 5 (1971), 325-332.)
So the question is whether for an arbitrary smooth function there exists a coordinate system such that the function depends on only one coordinate. The answer is certainly negative, since the function can have critical points.
answered Nov 23, 2021 at 13:43