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Does there exist a compact connected Lie group $G$ acting smoothly as isometries on the standard sphere $S^n$ for some $n\ge 3$, so that no totally geodesic hypersphere $S^{n-1}$ is $G$-invariant, but there exists embedded $G$-invariant diffeomorphic $S^{n-1}$?

By $G$-invariant, I mean the submanifold is a union of orbits. I've searched the literature, and a direct search using some combination of the keywords didn't give anything immediately relevant. My guess is that if the principal orbits are of low enough dimension, then we would have enough degree of freedom to produce an invariant diffeomorphic hypersphere. However, I don't know whether this is the case.

I'm aware that Hsiang and Lawson have classified some low cohomogeneity actions back in the 1970s in their paper Minimal Submanifolds of Low Cohomogeneity. However, a quick look at those classifications doesn't seem to produce an example.

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    $\begingroup$ Did you mean "there exists embedded $G$-invariant diffeomorphic $S^{n - 1}$"at the end of the first paragraph? $\endgroup$ – user44191 Nov 9 '18 at 3:22
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    $\begingroup$ @user44191 Yes, of course. Thank you. $\endgroup$ – Zhenhua Liu Nov 9 '18 at 3:27
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The answer below is by no means complete, but at least it treats the case when $G$ is a torus, and proposes an idea of what might be tried in the general case. The following fact is useful:

Remark. $G$ preserves a geodesic sphere on $S^n$ iff it fixes a point on $S^n$.

Statement 1. If the compact Lie group is a torus, then such an action doesn't exist.

Proof. Indeed, suppose that $\mathbb T^k$ preserves a sphere $S^{n-1}$ in $S^n$. We will deduce, that it fixes a point on $S^n$. By Schoenflies theorem $S^{n-1}$ cuts $S^n$ into two manifolds $B_1$ and $B_2$ both homeomorphic to a $n$-ball, in particular they have Euler characteristic $1$. So $\mathbb T^k$ must fix a point $x_1$ and $x_2$ in both. Indeed, any orbit of $\mathbb T^k$ of dimension $\ge 1$ has zero Euler characteristic, so in case there were no fixed points in $B_1$ and $B_2$ they would both have Euler characteristic zero. QED.

Other cases. I think, that this reasoning has a potential to be extended to other cases. Indeed, recall the following classical statement.

Statement. Let $G$ be a compact Lie group and $H$ be a subgroup. Then the homogeneous space $G/H$ has Euler characteristic $0$ if and only if $G$ and $H$ have the same rank. Moreover, in case of the same rank the Euler characteristic can be explicitly calculate. See the beginning of

http://moroianu.perso.math.cnrs.fr/tex/2014crelle.pdf

So I wonder when is the Euler characteristic of the quotient is equal to $1$ exactly. Probably this doesn't happen very often. On the other hand, if we want this strange action, so that $G$ doesn't fix any point on $S^n$ but fixes a hyper-sphere, it looks like one of the orbits should be of Euler characteristic exactly $1$.

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  • $\begingroup$ This question might be elementary but why does the action fix a point in $B_1$ or $B_2$? They consist of orbits and they're not themselves orbits, so I don't understand why your reasoning about orbits work for them. $\endgroup$ – Zhenhua Liu Dec 2 '18 at 20:53
  • $\begingroup$ I assume that you are asking about the torus case. In such a case, if there is no fixed point, then all orbits in $B_1$ are tori, and tori have zero Euler charateristic. Now, there is a fact, that if you can foliate (even in a singular way) your manifold by submanioldfs of zero Euler characteristic, then you manifold has zero Euler Characteristic as well. But $B_1$ doesn't have zero $\chi$ $\endgroup$ – Dmitri Panov Dec 2 '18 at 21:00
  • $\begingroup$ Thank you for your patient reply. I understand it better now. $\endgroup$ – Zhenhua Liu Dec 2 '18 at 21:20

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