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 $S^2 \times S^2$?
$Gr_2^+(\mathbb R^4) \cong S^2 \times S^2$
4
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differential-geometry
differential-topology
1 Answers
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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 $\text{SO}(2) \times \text{SO}(2)$.
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1@Eric: I think I was thinking of the proof that $\pi:G\rightarrow G/H$ is an $H$-principal bundle, or the more general case that $M\rightarrow M/G$ is a principal bundle when $G$ acts with the same conjugacy class of isotropy group at every point. – 2012-10-23