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Hi all,

This is my first post on Math Overflow! I've been stuck on the following question and was wondering if anyone might have any insight on it. Here it is:

Let $n \geq 5$. Let $G = SO(n)$ or $SO(1,n-1)$, and let $V$ be the irreducible $G$-module corresponding to algebraic Weyl tensors. What is the "most degenerate" (nonzero) $G$-orbit in $V$ (or the projectivization $\mathbb{P}V$)? More precisely, I want to find the maximum dimension of the stabilizer of a nonzero Weyl tensor. More hopefully, I would like to know what a representative element in this minimal orbit (is it unique?) and its stabilizer look like.

The annoying issue here is that we're working over $\mathbb{R}$. Over $\mathbb{C}$, if $G = SO(n,\mathbb{C})$, then $V$ has highest weight $2\lambda_2$, and the orbit through the highest weight line is the most degenerate orbit. The stabilizer of a highest weight line is (in the standard representation) the parabolic subgroup which is the stabilizer of a null 2-plane. These of course don't exist in Riemannian or Lorentzian signature.

Any tips would be greatly appreciated! Thanks!

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    $\begingroup$ I don't yet know the answer, but there's a reasonable guess: For $G=\mathrm{SO}(n)$, it might be the most degenerate nonzero Weyl tensor in $4$-space (which is the one for $\mathbb{CP}^2$, with stabilizer $\mathrm{U}(2)\subset\mathrm{SO}(4)$) regarded as being in $n$-space, with stabilizer $H = \mathrm{U}(2)\times \mathrm{SO}(n{-}4)$. For $G=\mathrm{SO}(1,n{-}1)$, it might be the same thing with stabilizer $H = \mathrm{U}(2)\times\mathrm{SO}(1,n{-}5)$. The other obvious competitor when $n=2m$, the Weyl curvature of $\mathbb{CP}^m$, has stabilizer $\mathrm{U}(m)$ which is much smaller. $\endgroup$ Aug 31, 2012 at 0:12
  • $\begingroup$ I'm sure Robert will answer this question soon, but I see that you are at ANU, where Michael Eastwood is a faculty member. Have you tried asking him? $\endgroup$
    – Deane Yang
    Aug 31, 2012 at 1:16
  • $\begingroup$ Yes, I've asked Mike (he's my postdoc supervisor), but he wasn't sure. $\endgroup$
    – Dennis
    Aug 31, 2012 at 21:22

1 Answer 1

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Re-Amended Answer:

My guess for $\mathrm{SO}(5)$ appears to have been correct in one sense, but not in another. It's true that a nonzero Weyl tensor in this case has to have stabilizer of dimension at most $4$, but there are two distinct candidates with this property.

First, there is the obvious one, which is the Weyl curvature of $\mathbb{CP}^2\times \mathbb{R}$, which has $\mathrm{U}(2)\subset \mathrm{SO}(4)\subset\mathrm{SO(5)}$ as its stabilizer. However, it turns out that $H = \mathrm{SO}(2)\times\mathrm{SO(3)}\subset \mathrm{SO}(5)$ (which is also of dimension $4$) fixes a nonzero Weyl curvature as well, and this yields a completely different $6$-dimensional orbit of $\mathrm{SO(5)}$ in the vector space of Weyl curvatures. Up to multiples, these two are the only $6$-dimensional orbits.

For $n=5$, there are no other nontrivial orbits of dimension $6$ or less. There is a $7$-dimensional orbit, which is the Weyl curvature of the $5$-dimensional symmetric space $\mathrm{SU}(3)/\mathrm{SO}(3)$. Its stabilizer is the subgroup of $\mathrm{SO}(5)$ that is the irreducibly-acting subgroup isomorphic to the holonomy of this space, i.e., $\mathrm{SO}(3)$. Up to multiples, there are no other orbits of dimension $7$ or less than what I have listed.

When you 'bootstrap' the $\mathrm{SO}(2)\times\mathrm{SO(3)}$-stabilizer example to dimensions higher than $5$, you get an element with stabilizer $\mathrm{SO}(2)\times\mathrm{SO(3)}\times\mathrm{SO}(n{-}5)$, which, when $n>5$, has lower dimension than $\mathrm{U}(2)\times\mathrm{SO}(n{-}4)$, so this gives a higher dimensional orbit when $n>5$ than the Weyl curvature of $\mathbb{CP}^2\times\mathbb{R}^{n-4}$.

In low dimensions (but higher than $n=5$) the Weyl curvature of $\mathrm{CP}^2\times\mathbb{R}^{n-4}$ is not optimal. When $n=2m$, there is the Weyl curvature of $\mathbb{CP}^m$, which has $\mathrm{U}(m)\subset\mathrm{SO}(2m)$ as stabilizer, and this group is bigger than $\mathrm{U}(2)\times\mathrm{SO}(2m{-}4)$ for $2 < m < 7$. However, as soon as $m>7$, the group $\mathrm{U}(m)$ has dimension less than the dimension of $\mathrm{U}(2)\times\mathrm{SO}(2m{-}4)$.

Last night, I thought that Weyl curvature of $\mathrm{CP}^2\times\mathbb{R}^{n-4}$ wins for large enough $n$, but this morning, I had another idea and realized that there is an even better candidate in the 'stable' range: I turns out that, for $n\ge 5$, under the subgroup $\mathrm{SO}(2)\times\mathrm{SO}(n{-}2)\subset \mathrm{SO}(n)$, the space of Weyl tensors in dimension $n$ has a trivial summand, and hence there is a nonzero Weyl tensor (unique up to multiples) whose stabilizer contains $\mathrm{SO}(2)\times\mathrm{SO}(n{-}2)$. Since there is no connected Lie group between this subgroup and $\mathrm{SO}(n)$, it follows that the identity component of the stabilizer of this tensor is $\mathrm{SO}(2)\times\mathrm{SO}(n{-}2)$. Thus, this orbit has dimension $2n{-}4$. I now think that, for sufficiently large $n$ (maybe even $n\ge 9$), this might be the lowest dimensional orbit. It might not be unique, though. For example, when $n=8$, there is an orbit of type $\mathrm{SO}(8)/\mathrm{U}(4)$ and one of type $\mathrm{SO}(8)/\bigl(\mathrm{SO}(2)\times\mathrm{SO}(6)\bigr)$. Both of these orbits have dimension $12$. (NB: Even though these two spaces are isomorphic because of triality, the two Weyl orbits are, of course, quite different.)

There aren't many subgroups of $\mathrm{SO}(n)$ that are larger than $\mathrm{SO}(2)\times\mathrm{SO}(n{-}2)$ when $n$ is sufficiently large. The one obvious exception, $\mathrm{SO}(n{-}1)$, does not fix a Weyl tensor. That's why I'm thinking that this one will win for sufficiently large $n$.

I haven't thought seriously about $\mathrm{SO}(1,n{-}1)$ yet, but, probably, there are at least two orbits of minimal dimension, namely $\mathrm{SO}(1,n{-}1)/\bigl(\mathrm{SO}(1,1)\times\mathrm{SO}(n{-}2)\bigr)$ and $\mathrm{SO}(1,n{-}1)/\bigl(\mathrm{SO}(1,n{-}3)\times\mathrm{SO}(2)\bigr)$, and there might be another 'degenerate' one that fills in the gap between these two.

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    $\begingroup$ Thanks for such a detailed reply, Robert! A result of Egorov is that any non-conformally flat Riemannian space has at most $(n-1)(n-2)/2 +3$ dimensional symmetries. ($n$ suff. large) It is claimed to be sharp, but no explicit maximal model is given. Subtracting the dim of the manifold gives $(n-2)(n-3)/2 + 1$ which is the same dim as $SO(2) \times SO(n-2)$. So your suggestion appears to be correct (for large $n$), though there is still the question of uniqueness. However, independent of Egorov, it would still be nice to have a rep-theoretic argument explaining maximality of your subgroup. $\endgroup$
    – Dennis
    Aug 31, 2012 at 22:24
  • $\begingroup$ @Dennis: You're welcome. It rambles a little bit, I'm afraid; I wandered around trying to get to the point. As for the representation theoretic fact, Dynkin classified the maximal (connected) subgroups of the simple Lie groups (up to conjugacy, of course), and it should be a simple matter to use this to list all of the subgroups of $\mathrm{SO}(n)$ with codimension less than $2n{-}4$ and then check, by hand, which of those fix an element in the Weyl representation. I don't have Dynkin's list handy, but, I expect that, for $n$ large enough (maybe $n\ge 9$?), $\mathrm{SO}(n{-}1)$ is it. $\endgroup$ Sep 1, 2012 at 13:28

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