Suppose we have a compact connected Lie group $G$ acting as isometries on a compact manifold $M^n.$ Then is it necessarily true that the Hausdorff dimension of the union of singular and exceptional orbits are no larger than the Hausdorff dimension of individual principal orbits? If not so, what about the union of singular orbits only?

It's known back in the 1950s through the works of Montgomery and Yang that in some cases, $\dim M^n\setminus M_{\text{principal}}\le n-2,$ the dimension of the complement of the union of principal orbits is at most $n-2.$ However, I'm not sure if they refer to Hausdorff dimension here. A search in the literature seems to indiate that there is no direct result on controlling $\dim M^n\setminus M_{\text{principal}}$ by the dimension of principal orbits.

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    $\begingroup$ I would believe that in your situation ($G$ and $M$ compact and smooth), the union of orbits of one particular type is always a submanifold, and that there are only finitely many orbit types. If this is the case, you can consider the (maximum of the) submanifold dimension instead of the Hausdorff dimension. $\endgroup$ – Sebastian Goette Nov 10 '18 at 10:18
  • $\begingroup$ @SebastianGoette. Yes, I think you are right. Thank you. $\endgroup$ – Zhenhua Liu Nov 10 '18 at 17:52

According to the paper "ORBITS OF HIGHEST DIMENSION" by Montgomery and Yang, by "singular orbits" one means orbits that have dimension less than the generic (or principle) ones.

For this reason, the answer to your question is rather negative. Indeed, take the round $S^n$ and consider on it the action of $S^1$ that fixes pointwise a geodesic $S^{n-2}$. Then the dimension of a generic orbit is $1$, but the union of singular orbits is the fixed $S^{n-2}$, i.e., it has dimension $n-2$.

PS. There is also a way to use this example to get a similar cohomogenity $4$ counter-example. Namely consider $S^4\times S^n$ and take the group $S^1\times SO(n+1)$ that acts diagonally. The first $S^1$-factor fixes $S^2\subset S^4$ (as previously) and the second acts transitively on $S^n$. Then a generic orbit has dimension $n+1$, but $S^2\times S^n$ is composed of obits of dimension $n$.

  • $\begingroup$ Thank you for your reply. Do you think we can have any control if the principal orbits have low cohomogeneity? $\endgroup$ – Zhenhua Liu Nov 11 '18 at 20:12
  • $\begingroup$ Zhenhua, I am not sure if I understood your comment correctly. I guessed that you want the principal orbits to be of low codimension? In such a case a counter-example can be constructed again, and I added it to the answer. Let me know if this what you were asking about. $\endgroup$ – Dmitri Panov Nov 11 '18 at 23:25
  • $\begingroup$ Thank you so much! Your brilliant construction really shows that there is no hope for a generic control by the dimension of principal orbits. $\endgroup$ – Zhenhua Liu Nov 12 '18 at 1:51

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