Most degenerate Weyl tensors in Riemannian and Lorentzian signature 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!
 A: 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.
