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Let $S^n$ be the $n$-dimensional round sphere (i.e. with Riemannian metric of constant curvature +1). Is there any classification result of totally geodesic embedded submanifolds? Are they all round spheres? How about general case when the curvature only assumes to be positive?

edit: To make the second question more precise: Let $S^n$ be the standard sphere with Riemannian metric such that the sectional curvature $\ge 1$. Let $N\in M$ be a totally geodesic connected submanifold. (assume $N$ is not a point). What is the possible topology of $N$?

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In the round case, the answer is: yes, they are all round spheres. In the positively curved case, I am not sure what the question is, exactly. –  Igor Rivin Aug 20 '12 at 14:36
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The constant-curvature version seems straightforward to me. Parallels don't exist on $S^n$. That means your submanifold is either a single geodesic circle (whose curvature you can't measure intrinsically) or it contains two or more intersecting geodesics. In the latter case, the intersection of the geodesics guarantees that the submanifold is connected. All simply connected spaces of constant curvature are isomorphic to a sphere of constant curvature, so the answer to the constant-curvature version of your question is yes. –  Ben Crowell Aug 20 '12 at 14:40
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Regarding the case where $S^n$ is assumed with some metric of positive curvature, if you relax "embedded" to "immersed" submanifolds, then it is clear that a totally geodesic immersed submanifold might not be diffeomorphic to a sphere, since, e.g., there are many non-periodic geodesics on ellipsoids. –  Renato G Bettiol Aug 20 '12 at 15:32
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It's not clear how to interpret the last question, in particular, whether the underlying manifold is still assumed to be $S^n$. For example, with the Fubini-Study metric on $\mathbb{CP}^2$ (which has positive sectional curvature), there are at least two kinds of (embedded) totally geodesic $2$-dimensional submanifolds: the complex lines and the real projective planes $\mathbb{RP}^2$. –  Robert Bryant Aug 20 '12 at 19:28
    
Note that you probably want to exclude trivial cases where the submanifold has dimension 0 (and may be disconnected). The case with a submanifold of dimension 1 is also not very interesting, since you can't apply the constant-curvature requirement. –  Ben Crowell Aug 20 '12 at 21:02

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up vote 5 down vote accepted

Contrary to another comment/answer, totally-geodesic $1$-manifolds in $S^3$ are not trivial because they can be disconnected. Great circles do not have to intersect, as can be seen by taking generic planes through the origin in $\mathbb{R}^4$.

Geodesics sometimes can't be isotoped to fibers of the same Hopf fibration. Great circle links were the subject of G. Walsh's thesis. One chapter classified great circle links up to $5$ components. There are $1,1,2,3,7$, respectively.

In higher dimensions, there are disconnected geodesic submanifolds of $S^n$ of dimension up to $(n-1)/2.$ The only place where you can have interesting linking is with totally geodesic $d$-spheres in $S^{2d+1}$, and I think the link theory is always nontrivial in that dimension.

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Thanks for the correction. I'll delete my incorrect answer. –  Ben Crowell Aug 20 '12 at 23:22

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