Minimal Manifold, $\mathbb{R}^4$

Question

We know that all the minimal surfaces of revolution in $\mathbb{R}^3$ are the Catenoid and the euclidean plane $\mathbb{R}^2$ in $\mathbb{R}^3$.

Do we know what are the minimal surfaces of revolution in $\mathbb{R}^4$?

Answer 1

You might want to look at a related question, Rotation in 4D?, where I show that if we have a rotation about some axis $v \in \Bbb R^4$, i.e., a distance-and-orientation-preserving linear transformation for which $Rv = v$, then there's a linearly-independent vector $w$ with $Rw = w$ as well.

To have a "surface of revolution" in 4-space, we'd need to have some arc $c$ and a circle-worth of rotations fixing some axis $v$. If $R$ is a "small" rotation (i.e., if $\|Ru - u\| << 1$ for all unit vectors $u$), then we can just take the shortest geodesic from the identity through $R$ and extend it to get a circle subgroup $H$ of $SO(4)$. I believe that in this case, you'd find that both $v$ and $w$ are fixed by all elements of $H$, as is the plane they span. Therefore we might as well, by change of coordinates, assume $v = e_3$ and $w = e_4$, at which point the surfaces of revolution look a lot like surfaces of revolution in just $xyz$, with an additional "wobble" in the $w$-direction (albeit one that varies only with $z$, not with $x$ and $y$).

Has someone written down what the minimal-surface equations look like in this case? Probably. But they're much more likely to have done so in odd dimensions, where you can have a circle-subgroup of the rotation group that leaves fixed only a 1-dimensional subspace.

I know this isn't a complete answer, and I'm not going to go search the minimal-surfaces literature to determine whether someone's done it or not. Maybe someone else knows the answer off the top of their head. But maybe thinking about what I've written might help you realize that your question (i.e., the exact meaning of what you're asking, not the answer!) isn't quite as simple as you think.