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It is trivial that there are a lot of minimal surfaces in the flat $R^3$: for example, for any point, any 2-plane containing this point, and any two othogonal vectors in this plane, and any negative number $K$ there exists a (in fact, infinitely many) minimal suface tangent to the plane at that point such that main curvature vectors are that vectors and the gauss curvature is that number.

Is the statement remains true if our $R^3$ is equipped by a not necessary flat metric?

On the level of equations, the system is of course very underdetermined so one expects a similar freedom to that in the standard flat situation.

The motivation is due to a certain question asked after my talk on projective equivalence of metrics (whether it is reasonable to replace geodesics by minimal surfaces in the definition of projective equivalence); would the answer to the question be positive, there is no sense of doing it since two different metrics in $R^3$ such that each minimal surface of the first is a minimal surface of the second will be proportional with a constant coefficient.

Of course the natural generalisation(s) of this question is also interesting in higher dimensions.

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  • $\begingroup$ It is interesting to fix a metric, take a minimal surface through each point tangent to each 2-plane, and then try to change metrics while keeping that family of surfaces minimal. Robert Bryant has results on this question for the standard metric on Euclidean space. $\endgroup$
    – Ben McKay
    Commented Dec 16, 2015 at 13:42
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    $\begingroup$ What Ben is referring to is the fact (originally proved by Bekkar, though I have a different proof) that the space of metrics on $\mathbb{R}^3$ for which all the planes are minimal surfaces has dimension 20. This is interesting because the space of metrics defined on a neighborhood of $0\in\mathbb{R}^3$ that are projectively equivalent to the standard flat metric is only $9$ dimensional. $\endgroup$
    – Robert Bryant
    Commented Dec 16, 2015 at 14:11

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The existence certainly remains true if the ambient metric is real-analytic, and this follows from the Cartan-Kähler Theorem since all you are asking for is local minimal surfaces.

In the case of a smooth metric, because the minimal surface equation is determined elliptic, the existence of a local solution attaining a specified $k$-jet satisfying the equation follows from standard elliptic theory (and you are only asking to specify the $2$-jet of the minimal surface at a point). (For example, see the discussion on elliptic equations in the appendix to Besse's Einstein Manifolds.

While Igor Rivin's answer does suggest that there ought to be many solutions attaining a given $k$-jet, this does not follow immediately from the existence of solutions to the boundary value problem. At least, I do not see how to prove it without invoking elliptic local solvability, in which case you might as well start with that.

The higher dimensional case is not significantly different: In the smooth case, one can specify the $k$-jet of a minimal surface at a point for any $k\ge 1$ as long as the specified $k$-jet satisfies the constraints of the minimal surface equation. (Again, this follows from the Cartan-Kähler Theorem for real-analytic metrics and standard elliptic theory in the smooth case, see Besse, as above.)

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It is a theorem of Morrey, that there is a minimal surface bounded by an arbitrary Jordan curve in a Riemannian manifold (see this paper of Heinz and Hildebrandt for more), so the answer is "a lot".

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  • $\begingroup$ Thanks a lot, Igor. I did know the result (at least locally) since the proof of Douglas for the existence of the minimal surface, or at least the version of the proof I know survives if the target space is not flat; but the precise reference is very important $\endgroup$
    – Vladimir S Matveev
    Commented Dec 18, 2015 at 11:12
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    $\begingroup$ After some thinking I understood that existence result for minimal surface bounded by an arbitrary Jordan curve is sufficient for my goals (so I do not need arbitrary jet to prove what I want.) Thanks! $\endgroup$
    – Vladimir S Matveev
    Commented Dec 24, 2015 at 14:36

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