It is a set $X$ together with a transitive group action of a group $G$. There is only one orbit and so $X$ is isomorphic to the space of cosets $G/H$ being $H$ the stabilizer or isotropy group $Stab_G(x)$ for an arbitrary $x\in X$: the map
$$ \varphi: G/H \rightarrow X, $$being $\varphi(gH)=ghx$ for any $h\in H,$ is bijective.
The choice of $x$ has no influence since the isotropy groups are conjugate.
Basic examples:
1. $X=\mathbb{R}^2$ and $G=E(2)$
2. Any connected topological manifold $X$ is homogeneous with $G=Homeom(X)$, see this answer in Mathstackexchange. The same is true for any connected smooth manifold and its group of diffeomorphisms. But keep an eye: these groups are infinite dimensional.
3. Any finite set $S$ of cardinal $n$ is homogeneous with $G=S_n$ the symmetric group. Indeed, any set is homogeneous with respect to its bijections, I guess.
$\blacksquare$
I think they are the same as Klein geometrys.
When the isotropy group is the identity, $G\equiv X$, and $X$ is called a principal homogeneous space or $G$-torsor. This happens if the group action is also free, as well as transitive, that is, regular.
Proposition. Let $G$ be a connected Lie group and $H$ is a closed subgroup of $G$ such that the homogeneous space $G/H$ is contractible. Then $G/H$ is homeomorphic to a Euclidean space $\mathbb R^n$ for some $n$.
$\blacksquare$
Proof. This question in MO$\blacksquare$
Homogeneous spaces are used, to my knowledge, in two different ways:
Suppose we have a complicated (real-world) object $X$ which we can describe by means of a well known (mathematical-world) object $S$. Examples:
The object $X$ is homogeneous with respect to the group $Bij(S)$ in the sense that:
1. There is a physical linkage, or external linkage, from $S$ to $X$, which is introduced externally. Let's call it $\phi: S\to X$.
2. All the possible descriptions of $X$ are given by $P=\{\phi\circ g: g\in Bij(S)\}$.
3. Suppose we have a distinguished point in $s\in S$ (for example $(0,0)\in \mathbb{R}^2$ or $1\in \{1,\ldots,32\}$). At first it may seem that the point $x=\phi(s)\in X$ is a privileged point, since it is described by the "main point" of $S$. But it turns out that
$$ P=\sqcup_{y\in X} P_y $$being $P_y=\{\phi\circ t:t\in t_y H\}$ (with $H$ the isotropy group of $s$ in $Bij(S)$ and $t_y$ any map sending $s$ to $\phi^{-1}(y)$) all the descriptions of $X$ "centered at" $y$. That is, $P_y$ consist of descriptions of $X$ in which $y$ is described by the distinguished point $s\in S$. It is clear that all $P_y$ are "equal", and so we say that $X$ is homogeneous. For example, no point of the land is special, all of them can be chosen to be the center of the world; and no student is special, all of them can be chosen to be associated with the label 1.
Now, it is interesting to now if we can shrink $Bij(S)$ to a subgroup $G$ in such a way that $X$ still conserve this property. Why do we want to restrict? Because there are descriptions which are more important than others. For example, any "crazy" bijection $g\in Bij(\mathbb{R}^2)$ may not give rise to an interesting description $\phi \circ g:\mathbb{R}^2\to X$. We are interested in bijections that do not clutter the set too much (for example, with our "sense" of continuity). The choice of $G$ is again an external data, which depends on the nature of the problem we are interested in. But the group needs to be transitive, in order to conserve homogeneity.
An element $\phi\circ g\in P$ is a $G$-basis or $G$-description of $X$ in the sense that we can define the coordinates of a point $y\in X$ as
$$ (\phi \circ g)^{-1}(y)\in S $$The group $G$ not only serves to parameterize the "$G$-descriptions" of $X$. It also defines an action on $X$ through $\phi:S\to X$. The action of $g\in G$ on $y\in X$ is $gy=\phi(g(\phi^{-1}(y)))$. If we fix a point $x\in X$ (corresponding to certain $s\in S$ by means of $\phi$) we get a bijection $\Psi:X\to G/H$ (being $H$ the isotropy group of $x$) given by
$$ \Psi:y\mapsto g_y H $$ $$ gx \style{display:inline-block; transform:scale(-1,1);}{\mapsto} gH $$being $g_y$ an element of $G$ carrying $x$ to $y$. So many times we identify $X$ with $G/H$, or even start the construction with $G/H$.
This bijection $\Psi$ let us to express the points of $G/H$ in "coordinates" in $S$. Given $\bar{g} \in G$, the element $\bar{g}H$ have coordinates
$$ (\phi \circ g)^{-1}(\Psi^{-1}(\bar{g}H))=(\phi \circ g)^{-1}(\bar{g}x)= $$ $$ =(\phi \circ g)^{-1}(\phi(\bar{g}(\phi^{-1}(x))))=(g^{-1}\bar{g})\phi^{-1}(x)= $$ $$ =(g^{-1}\bar{g})s $$In particular, if we think in $X=G/H$ described by the set $S=G/H$ by means of the identity and with selected points $s=x=H$, then an abstract point $\bar{g}H \in G/H$ is expressed in a frame $g\in G$ as
$$ g^{-1}\bar{g}H $$Example
$G=\mathbb{R}^2 \rtimes GL(2)$, $H=GL(2)$, $G/H=\mathbb{R}^2$, $x=s=\tau_{(0,0)}H$
An element $g=\tau \circ A\in G$ is a translation $\tau$ following a linear transformation $A$, and sends the canonical reference system $((0,0),(1,0),(0,1))$ to a new one. The coordinates of a point $y\in \mathbb{R}^2$ in the new reference system are computed by means of $A^{-1}\tau^{-1}y$.
$\blacksquare$
In particular $G$ itself is a homogeneous space ($H=1$). Every element $f_1\in G$ is, at the same time, a "point", a "movement" and a "frame". Movements and frames are the same, see general covariance and contravariance. Now, given any other point $f_2\in G$, how it is described from the frame $f_1$? Well, according to this we take
$$ f_1^{-1}f_2 $$See also xournal 190 and maple 071.
Well, in this case we can decompose the space in the orbits $S=\sqcup_i \mathcal{O}_i$. The group acts transitively in every $\mathcal{O}_i$ so we have bijections
$$ \Psi_i: \mathcal{O}_i\to G/H_i $$where $H_i$ is the isotropy group of some point in $\mathcal{O}_i$. Then we have several principal bundles $G\to \mathcal{O}_i$, not only one. I think this has to do with subrepresentations of groups, but I have to think more on this.
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Author of the notes: Antonio J. Pan-Collantes
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