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I think that the answer is 'yes' if $U$ is simply-connected, because there is a way to construct a candidate 'approximate surface' from 'approximate solutions' of Gauss and Codazzi, but a more useful …

There exist many algebraic surfaces of constant breadth with no continuous symmetries, even ones with no symmetries at all. To see this, consider the properties of the support parametrization: The su …

I don't know where the orbit types in this case were first explicitly classified, but it is done in my paper Second order families of special Lagrangian 3-folds, Perspectives in Riemannian geometry, 6 …

Yes, $L_\delta$ is algebraic. You can find its equations by elimination theory as follows: Let $L$ be defined by the polynomial equation $F(x,y) = 0$. Now consider the polynomial equations $$ F(x,y …

Note: I'm revising my answer to make the argument/construction more transparent. In the previous version, I stated an existence result about flat surfaces, but didn't indicate a proof (because, at th …

This may be more appropriately regarded as a comment than as an answer. I realize that this question has been around for a while, probably because it's not completely clear what some parts of the ques …

The OP doesn't say what is meant by a 'geometric object', so it's hard to give a definitive answer. However, if one assumes that the geometric object is a smooth manifold $M^7$ and that the action is …

The formulation of Laguerre geometry in terms of dual numbers is a decidedly 'synthetic' one, meant to exhibit how this set of transformations can be regarded as a different 'real form' of the well-kn …

In the course of writing an answer to a related MO question, I realized that there is a surface with a complete Riemannian metric of non-constant negative curvature for which one can write down the di …

You may want to have a look at the article Foliation by constant mean curvature spheres, by Rugang Ye, Pacific Journal of Mathematics, 147 (1991), 381–396. In this article, the author shows, given a R …

This may not reallly be an answer that you like, but I think that, maybe you misunderstood what Ben McKay was trying to describe. Here is a more explicit, extrinsic description that may help: Suppose …

The answer is 'no' for $n=2$ (and hence for all higher $n$). Here is how one can see this. First, when $n=2$, recall that, by the Gauss Lemma, a metric $g$ in geodesic normal coordinates $(x,y)$ cent …

See the following Wikipedia page: https://en.wikipedia.org/wiki/Liouville%27s_theorem_(conformal_mappings)

Your $M_n$ is (two copies of) the Riemannian symmetric space $\mathrm{SO}(2n)/\mathrm{U}(n)$ (which is $DI\!I\!I$ in Cartan's nomenclature). Its topology is well-studied from that point of view.

I have looked at Goldberg's paper referenced by J. J. Castro in his excellent answer. It turns out that there is a simpler (and more general way) to generate Goldberg's non-circular solutions, so I t …