Is every geometry contained in a maximal geometry?

Question

Following the classical definition (made prominent by Thurston, I believe, but correct me if I am mistaken) we define a geometry to be a pair $(M,G)$, where

1) $M$ is a smooth and $1$-connected (i.e, connected and simply-connected) manifold,

2) $G \subset Diff(M)$ (endowed with the compact-open topology), such that the induced action of $G$ on $M$ is transitive and for some (then any) $x \in M$, the stabilizer $G_x \subset G$ is a compact subgroup.

3) There is a subgroup $H \subseteq G$, acting cocompactly, freely and properly discontiuously on $M$. This is equivalent to saying that the action of $H$ on $M$ is that of a covering group and the quotient $H / M$ is compact.

We say that two geometries $(M,G)$ and $(M',G')$ are equivalent is there is a diffeomorphism $\phi: M \to M'$, such that the obvious induced homeomorphism $\phi_*: Diff(M) \to Diff(M')$, maps $G$ homeomorphically onto $G'$.

Under this relation, the collection of equivalence classes of geometries form a set, equipped with a natural order $\subseteq$, so that $[(M,G)] \subseteq [(M',G')]$ if there are respective representatives $(M,G)$ and $(M',G')$ and a diffeomorphism $\psi: M \to M'$, such that $\psi_*(G) \subseteq G'$. With this in mind, define a model geometry (or maximal geometry) to be a class of geometries that is not properly contained in any other class of geometries with respect to the above define order.

My question is now the following, as stated in the title: Is every geometry contained in a maximal geometry ?

At least to me, this situation doesn't seem suitable for applying Zorn's Lemma: There is neither an obvious upper bound for a chain of geometries, nor is it clear whether the order is even antisymmetric (although it clearly does satisfy both reflexivity and transitivity). However, without the help of Zorn's Lemma, i cannot see a way to give an affirmative answer to this question. Moreover, all geometries I know are contained in a maximal geometry. Does anyone have a (positive, negative) answer to this question ? If it is negative, I am very interested in an explicit example of a such a geometry.

Answer 1

Yes, there is always a maximal geometry. Here are two key facts which one can use to prove the claim:

1. $G_x$ acts faithfully on $T_xM$, hence, there is a natural embedding $G_x\to GL(n, {\mathbb R})$, $n=dim(M)$. (Hint: Put a $G_x$-invariant Riemannian metric on $M$.)

2. $GL(n, {\mathbb R})$ contains a (unique up to conjugation) compact subgroup $O(n)$ containing (up to conjugation) any other compact subgroup.

Answer 2

Here is a complete answer, based on the hints given by Moishe Cohen and expanded from the comments below his answer: Let $(M,G)$ be a geometry. Use Zorn to find a maximal nested set $\{(M,G_i)\}_{i \in I}$ of geometries containing $(M,G)$. This means that:

a) Each $(M,G_i)$ contains $(M,G)$.

b) For every pair $i,j \in I$, there is some $k \in I$ such that $(M,G_k)$ contains both $(M,G_i)$ and $(M,G_j)$.

c) $\{(M,G_i)\}_{i \in I}$ is not properly contained in other set of geometries with properties $a)$ and $b)$.

For fixed $x \in M$, denote by $K \subset G$ and $K_i \subset G_i$ for $i \in I$ the respective compact stabilizers of $x$. By considering the induced action on $T_xM$, this means that each $K_i$ naturally embeds as a compact subgroup into $GL(n,\mathbb R)$. Since each of them contains $K$ and $K$ is contained in a unique maximal compact subgroup $L$ (in fact, $L$ is a conjugate of $O(n)$), it follows that each $K_i$ is also contained in $L$. Therefore, we can define an inner product $\langle \; , \; \rangle_x$ on $T_xM$ invariant under the action of $L$, and thus invariant under the action of each $K_i$.

Now let $y \in M$ be arbitrary. Define an inner product $\langle \; , \; \rangle_y$ on $T_yM$ as the push-forward of $\langle \; , \; \rangle_x$ via some $\alpha \in G_i$ sending $x$ to $y$. To see that this definition doesn't depend on the choice of $\alpha$, note that for some other $\beta \in G_j$ sending $x$ to $y$, we have that $\beta^{-1} \alpha \in K_l \subset G_l$, where $(M,G_l)$ is a geometry containing both $(M,G_i)$ and $(M,G_j)$ (whose existence is guaranteed by b) ).

The collection of those inner products define a Riemannian metric $g$ on $M$. Set $H := Isom(M,g)$. Then $(M,H)$ is a geometry containing all $(M,G_i)$, as $g$ is $G_i$-invariant for each $i \in I$, by definition. Finally, to see that $(M,H)$ is maximal, assume by contradiction that $(M,H)$ is properly contained in some geometry $(M,H')$. Then, the set $\{(M,G_i)\}_{i \in I} \cup \{(M,H')\}$ would be a nested set properly containing $\{(M,G_i)\}_{i \in I}$, a contradiction to property $c)$.

Note that this does not show that the maximal geometry containing a given $(M,G)$ is unique. Indeed, this seems to be true only in some cases, such as when $dim(M) \leq 2$ and $dim(M) = 3$ with $\partial M = \emptyset$. I don't know how the situation is in the general case.