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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 $S^2 \times S^2$?

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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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    Here I am making implicit use of a fact which I believe to be true but which I admittedly don't know enough differential geometry to prove: let $M$ be a smooth manifold on which a compact Lie group $G$ acts smoothly and transitively with stabilizer $H$. Then $G/H$ has a natural smooth structure and the map $G/H \to M$ given by $g \mapsto gm$ (for $m \in M$ such that $\text{Stab}(m) = H$) is a diffeomorphism.2012-10-23
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    For what's its worth, you belief is true. In fact, your map is an equivariant diffeomorphism, so is a little bit better. The only proofs I know of all this involve picking an invariant metric (doable because $G$ is compact) and using some Riemannian geometry (geodesics and normal bundles of orbits).2012-10-23
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    @Jason: Cool. Is it still true when $G$ is noncompact or is there an easy counterexample?2012-10-23
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    Unless I'm missing something, $G/H \simeq M$ is a fairly simple thing to prove that doesn't invoke any Riemannian geometry nor compactness of $G$. This is done in chapter 9 of Lee's "Smooth Manifolds," for example.2012-10-23
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    @Eric: thanks for the reference! (For anyone else's reference, this is Theorem 9.24 in Lee.)2012-10-23
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    @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