$Gr_2^+(\mathbb R^4) \cong S^2 \times S^2$ Let $Gr_2^+(\mathbb R^4)$ be the oriented Grassmanian of 2-planes in $\mathbb R^4$. How would one go about showing that this is diffeomorphic to $S^2 \times S^2$?
 A: The diffeomorphism can be nicely explicited as follows. Lift the transitive $SO(4)$ action on $\mathbb{Gr}^+_2(\mathbb{H})$ to its universal covering $\mathrm{Spin}(4) = S^3 \times S^3$ via the canonical covering representation
$$
S^3 \times S^3 \to SO(4) \qquad (p,q) \mapsto L_p \circ R_{q^{-1}}
$$
where we identify $\mathbb{R}^4$ with the quaternions $\mathbb{H}$, the sphere $S^3$ is the group of unit norm quaternions, $L_*$ is left multiplication and $R_*$ is right multiplication by the quaternion $*$ (note that taking the diagonal $p=q$ and identifying $\mathbb{R}^3$ with the imaginary quaternions $\mathrm{Im}\mathbb{H}$, this restricts to the canonical covering $S^3 \to SO(3)$).
The $S^3 \times S^3$ action on $\mathbb{Gr}^+_2(\mathbb{H})$ is then
$$
(p,q)\Pi^+ = p \Pi^+ q^{-1}
$$
where $\Pi^+$ is an oriented plane.
To compute the isotropy of the oriented plane $\mathbb{C}^+$ with basis $1, i$, note that
$p \mathbb{C}^+ q^{-1} = \mathbb{C}^+$
implies that $L_p \circ R_{q^{-1}}$ restricts to a rotation in $\mathbb{C}^+$, thus
$$
pzq^{-1} = e^{i\theta} z, \qquad\text{for all } z \in \mathbb{C}
$$
Taking $z = 1$, $z=i$, it follows that $p = e^{i\theta} q$ and
$p i = e^{i\theta} i q$ so that
$$
p i = i ( e^{i\theta} q ) = i p \qquad
e^{i\theta} i q = (p)i = e^{i\theta} q i \Longrightarrow qi = iq
$$
so that both $p, q$ are complex.
Thus the isotropy of $\mathbb{C}^+$ is $S^1 \times S^1$, and the covering transformation descends to the explicit diffeomorphism
$$
S^3/S^1 \times S^3/S^1 \to \mathbb{Gr}^+_2(\mathbb{H}) \qquad (p S^1,q S^1) \mapsto p \mathbb{C}^+ q^{-1}
$$
where $S^3/S^1$ is the 2-sphere, since it identifies with the Riemmann sphere $\mathbb{C}P^1$ via the slope map $S^3 \to \mathbb{C} \cup \infty = \mathbb{C}P^1$, $z + wj \mapsto w/z$, where $z, w \in \mathbb{C}$.
Remark: the same arguments can be used to compute the isotropy of the $S^3 \times S^3$ action on the Grassmanian of non-oriented planes $\mathbb{Gr}_2(\mathbb{H})$ by noting that now $L_p \circ R_{q^{-1}}$ restricts to a rotation or a rotation and reflection in $\mathbb{C}^+$, thus we also have the possibility
$$
pzq^{-1} = e^{i\theta} \bar{z}, \qquad\text{for all } z \in \mathbb{C}
$$
which implies that $p,q$ anti commute with $i$, so that $p, q$ belong to $S^1 j$. The isotropy is thus given by $(S^1 \times S^1) \cup (S^1 \times S^1)j$ and is disconnected, in this case.
A: Not a complete solution, but: $\text{Gr}_2^{+}(\mathbb{R}^4)$ is acted on transitively by $\text{SO}(4)$. $-I$ acts trivially, so this action factors through the quotient, which is isomorphic to $\text{SO}(3) \times \text{SO}(3)$ (this can be seen for example using quaternions). It suffices now to show that the stabilizer of a point is $\text{SO}(2) \times \text{SO}(2)$. 
