# homogeneous surface in $\mathbb{R}^4$

It is well known that the only homogeneous surfaces in $\mathbb{R}^3$ are the spheres, cylinders or planes. My question is about other examples in dimension $4$. Such a surface should have "constant curvature" but it seems that there is no good scalar invariant such as mean curvature or Gauss curvature...except perhaps the square norm of the second fundamental form. In another codimension 2 situation, namely curves in $\mathbb{R}^3$, there are with lines and circles,also the helix. Is there a homogeneous surface in $\mathbb{R}^4$ which is not $S^2$, $S^1\times \mathbb{R}$ or $\mathbb{R}\times \mathbb{R}$?

Edit: and $S^1\times S^1$ but connected

• What is your definition of "homogeneous"? – Igor Rivin Jul 24 '17 at 15:07
• for all $x,y$ in the surface there exists an isometry of $\mathbb{R}^4$ which sends the surface to itself and $x$ to $y$. – Paul Jul 24 '17 at 15:14
• Is a homogenous surface in $\mathbb R^4$ a 3-manifold or a 2-manifold? From your question it sounds like you might want 3-manifolds. – Ryan Budney Jul 24 '17 at 18:37
• it is a 2-manifold – Paul Jul 25 '17 at 8:34

I'm rearranging my answer a little bit because I realized that I overlooked an apparent possibility (that turns out not to occur), and I didn't want my answer to be misleading:

If the surface in Euclidean $\mathbb{R}^4$ has positive Gauss curvature and is homogeneous, it will be complete and hence compact. Hence the group of ambient symmetries will have to preseve its center of mass, which we can take to be $0\in\mathbb{R}^4$. Thus, the group of symmetries will lie in $\mathrm{O}(4)$. However, the identity component of the symmetries of a compact surface of positive curvature must be a quotient of $\mathrm{SU}(2)$, so we are looking at a subgroup of $\mathrm{O}(4)$ whose identity component must be either $\mathrm{SU}(2)$ or $\mathrm{SO}(3)$. The former can only be represented one way, up to conjugacy, in $\mathrm{O}(4)$, and that action has no $2$-dimensional orbits (they are either the origin or $3$-spheres). Thus, it must be $\mathrm{SO}(3)\subset\mathrm{O}(4)$, uniquely up to conjugacy. The two dimensional orbits of $\mathrm{SO}(3)$ are round $2$-spheres lying in $3$-planes, so this is all of the possible homogeneous surfaces with positive curvature. (Note that in $\mathbb{R}^5$, though, there is a homogeneous $\mathbb{RP}^2$ that sits linearly fully in $\mathbb{R}^5$.)

There can't be any homogeneous surface of negative Gauss curvature whose ambient symmetries act as the full isometries of the surface, since the identity component of the symmetry group of its simply-connected cover would be $\mathrm{PSL}(2,\mathbb{R})=\mathrm{SO}(2,1)^\circ$, which has no nontrivial homomorphism into the group of isometries of $\mathbb{R}^4$ (for Lie algebra reasons). Thus, the action of the identity component of the ambient symmetries of the surface would have to be a 2-dimensional subgroup of $\mathrm{SO}(2,1)^\circ$, and these are all conjugate to each other. In particular, the ambient symmetries would preserve a foliation by geodesics and, consequently, an orthonormal coframing on the surface. Now one can set up the structure equations for such an ambient coframing and easily derive that this case cannot occur. Thus, homogeneous negatively curved surfaces do not exist in $\mathbb{R}^4$.

Finally, there remain only the homogeneous surfaces with zero Gauss curvature in $\mathbb{R}^4$ to classify, and a calculation with the structure equations shows that, up to isometry, there are only two types: First, there is the product of a homogeneous curve (i.e., line, circle or circular helix) in $\mathbb{R}^3\subset\mathbb{R}^4$ with an orthogonal line; second, there are the products of two circles in orthogonal 2-planes in $\mathbb{R}^4$.

In higher dimensions, of course, the number of types of homogeneous surfaces increases.

• Thanks for the nice answer. Can you explain why " However, the identity component of the symmetries of a compact surface of positive curvature must be a quotient of SU(2)", in fact a reference will suffice. – Paul Jul 25 '17 at 15:31
• @Paul: There are (up to constant multiples) only two compact surfaces of positive curvature, namely, $S^2$ and $\mathbb{RP}^2$, and the identity component of their group of isometries is $\mathrm{SO}(3)$, which is double-covered by $\mathrm{SU}(2)$, which is simply connected. See any introductory book on Lie groups. – Robert Bryant Jul 25 '17 at 15:53

Obviously, $S^1\times S^1$ is homogeneous in ${\mathbb R}^4$.

If disconnected surfaces are allowed, then there are many more, already curves in ${\mathbb R}^3$ and surfaces in ${\mathbb{R}}^4$.

• Also called the Clifford torus. – Ben McKay Jul 24 '17 at 18:01
• Also, it would be interesting to know all connected homogeneous curves in ${\mathbb R}^4$, curves and surfaces in ${\mathbb R}^5$, and so on. – Wlodek Kuperberg Jul 24 '17 at 18:15
• Hypersurfaces seem more interesting... – Igor Rivin Jul 24 '17 at 18:27
• Yes, I forget it in my question. – Paul Jul 25 '17 at 10:20
• @IgorRivin: In fact, though, the homogeneous hypersurface case in Euclidean space is not very interesting: It's a classical fact that a homogeneous hypersurface in $\mathbb{E}^{n+1}$ is an orthogonal product of the form $S^p\times \mathbb{E}^{n-p}$ for some $0\le p\le n$. (This is far from the case in positive curvature, of course, i.e., in $S^{n+1}$, as there are quite a few interesting homogeneous hypersurfaces in that case. See the extensive work on isoparametric hypersurfaces, for example.) – Robert Bryant Jul 25 '17 at 15:59