# Why are Klein geometries $G/H$?

The idea behind Klein geometries is simple, clear and beautiful. Simply we have a manifold M and a Lie group G who acts on it and we study the properties that remain invariants under this action.

But then, when I open the books about Klein geometry they start with something like: choose a point $$x \in M$$, take the stabilizer $$H$$ of x, then the Klein geometry is $$G/H$$; instead of $$(M, x)$$ we can study $$(G, H)$$.

Now, even if I "understand" what they say, I cannot really understand what this means and why one chooses this instead of the natural definition. In particular

1. What does it really mean to study $$G/H$$ instead of $$M$$? For example if I want to study properties of curves, I will consider $$\gamma: I \to M$$ and not something like $$\gamma: I \to G/H$$. For example if I consider $$M = \mathbb{R}^2$$, $$x = \vec O$$ and $$G = SO(2)$$ (i.e I do not consider translations) I will have $$H = SO(2)$$ and $$G/H = \{0\}$$: what does it mean that my geometry is "empty"? What is $$G/H$$ in practice?
2. Why do every one define them so? Which are the advantages?

• In your example the action of $G$ under $x$ is trivial, and that is precisely what the quotient group $G/H$ says. Commented Nov 8, 2018 at 15:01
• That's ok, but I cannot understand why I have to care about what happens to x! I mean: if I want to study a curve in that geometry, the action of G is not trivial! Commented Nov 8, 2018 at 15:22
• @MarcoAll-inNervo Why would you want to study curves in this context? Generic curves $I\to M$ don't take the action of $G$ into account at all. Also note that if $G$ is compact then the orbit of $x$ is diffeomorphic to $G/H$, so the substitution makes more sense. Commented Nov 8, 2018 at 15:26
• I think I'm missing something important. I am trying to self-study (with success, really) Cartan's method of moving frames, which, among the other things, let us study curves in arbitrary Klein geometries. For example, in $\mathbb{R}^3$ with $G = SO(3)$, this reduces to the classical Frenet frame. Now we have two invariant: $\kappa (s), \tau (s)$. If two curves are $SO(3)$-equal then they have these invariants equals and if they have these invariants equals they are $SO(3)$-equal . I am asking myself what does the $G/H$ definition mean in this context. Commented Nov 8, 2018 at 15:55
• @MarcoAll-inNervo You usually assume that the action of $G$ on $M$ is transitive. In that case $M$ is diffeomorphic to $G/H$ for some subgroup $H$. Hence $(G,H)$ fully describes $M$. But your case of $SO(3)$ on $\mathbb{R}^3$ is not transitive (assuming standard action). So this context seems to go beyond Klein geometry. Unless you assume that each curve belongs to a single orbit? Commented Nov 8, 2018 at 16:18

A Klein geometry is specified by a Lie group $$G$$ and a closed Lie subgroup $$H < G$$; sometimes we additionally require that the quotient space $$X := G / H$$ be connected. In particular, the action $$G \times X \to X$$, $$g \cdot g' H := (gg')H$$, is transitive.

Conversely, if we have a (perhaps connected) space $$M$$ admitting a transitive Lie group action $$\lambda : G \times M \to M$$ and we fix a point $$m \in M$$, the stabilizer $$H$$ of $$m$$ in $$G$$ is a closed subgroup and we can identify $$G / H \leftrightarrow X$$ via $$g H = g \cdot m$$. (Note that in your example, $$SO(2)$$ does not act transitively on $$\Bbb R^2$$, so this action not define a Klein geometry, at least not on $$\Bbb R^2$$.) Picking a different point, say, $$m' = k \cdot m$$ yields a conjugate stabilizer subgroup, $$H' = k H k^{-1}$$. In particular, the conjugation map $$g \mapsto g h g^{-1}$$ induces an isomorphism $$(G, H) \to (G, H')$$ of Klein geometries. Thus, up to this isomorphism, $$\textbf{the data (G, H) contains the same information as the data (G, M, \lambda, x)} .$$

There are at least two reasons to prefer working with the data $$(G, H)$$. One is that it is manifestly independent of the choice of basepoint $$x \in M$$. The second is that it emphasizes the group action of $$G$$ and whatever geometric structure it preserves rather than the underlying space. Indeed, sometimes a given smooth manifold is the underlying space of several different Klein geometries. For example, $$S^n$$ is the space underlying:

• $$(O(n + 1), O(n))$$: the Riemannian sphere (spherical geometry)
• $$(SL(n + 1), P)$$, where $$P$$ is the stabilizer of a ray: the projective sphere
• $$(SO(n + 1, 1), P')$$, where $$P$$' is the stabilizer of an isotropic ray: the conformal sphere
• $$(Sp(2 m, \Bbb R), P'')$$, $$n = 2 m - 1$$, where $$P''$$ is the stabilizer of a ray: the contact projective sphere
• $$(SU(\ell + 1, 1), P''')$$, $$n = 2 \ell + 1$$, where $$P'''$$ is the stabilizer of an isotropic ray: the CR sphere.
• Thanks for the reply, very helpful! My next question is then: why we care only about transitive actions? Commented Nov 8, 2018 at 16:37
• You're welcome, I'm glad you found it useful. As for transitivity, this is a matter of definition: The model of a Klein geometry $(G, H)$ is $G / H$, and the given action $G \times (G / H) \to G / H$ is transitive: For any element $gH \in G / H$, we have $gH = g \cdot eH$. There are many interesting situations where we have a nontransitive Lie group action on a manifold (for example, the group of isometries of a generic surface of revolution is $O(2)$), but by definition this situation is outside the scope of Klein geometry. Commented Nov 8, 2018 at 18:44
• @AntonioJPan That's a great question: Ultimately you're looking for whatever structure on $M$ has exactly the symmetries $G$, and for the case of Euclidean geometry we know ahead of time that the structure on $M$ is the bilinear form. (More precisely, in this case $G$ preserves and entire homothety class of bilinear forms.) Given $G, H$ it is not always easy to identify the preserved structure, though in practice usually one starts with a geometry and produces $G, H$, and not the other way around. Commented Dec 6, 2020 at 4:49
• @AntonioJPan I'm not sure that there is a general definition of "geometric structure that can be encoded with a Klein geometry" that isn't essentially tautological, formal or otherwise. You're correct that $G$ need not preserve a bilinear form on $G / H$ (and in fact it never does for the important class of Klein geometries where $G$ is semisimple and $H < G$ is a parabolic subgroup, which includes projective geometry, conformal geometry, CR geometry, and many others). Commented Dec 6, 2020 at 8:49
• To give a flavor for how different Klein geometries can be from the Euclidean example, suppose $G=SL(3,\Bbb R)$, and let $H<G$ be the subgroup of upper triangular matrices in $G$. Then the preserved structure is a pair $(E,F)$ of line fields such that the $E \oplus F$ is a contact distribution, and the $H$-action preserves no bilinear form on $T(G/H)$, though it does preserve the bilinear map $E\times F\to T(G/H)/(E\oplus F)$ induced by the Lie bracket. (It turns out that this Klein geometry has a close relationship with the geometry of second-order o.d.e.s modulo so-called point equivalence.) Commented Dec 6, 2020 at 8:56