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Is there any way to identify the Grassmann manifold of 2-plane in $R^4$, $G_2(R^2)$ with $S^2\times S^2$?

I tried to give a map, but I have not get it.

Thanks, advance.

Selvakumar A
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    As explained here, the integral homology groups of $G_2(\Bbb R^4)$ have torsion, therefore this cannot be true. https://math.stackexchange.com/questions/1396570/integral-homology-of-real-grassmannian-g2-4 – Arnaud Mortier Feb 20 '18 at 15:49
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    It becomes true when you take the universal (double) cover, which is the Grassmanian of oriented $2$-planes. – Nick L Feb 20 '18 at 15:49
  • Oh. Thanks.@ArnaudMortier and @Nick L – Selvakumar A Feb 20 '18 at 15:54

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As mentioned in the comment the Grassmanian of oriented 2-planes in $\mathbb{R}^{4}$, denoted $\tilde{Gr}(2,4)$ (which is the universal cover of the usual Grassmanian) is diffeomorphic to $S^{2} \times S^2$. There are 2 main steps to show this, which I sketch.

Step 1. Identify $S^{2} \times S^{ 2}$ with the Quadric $Q = \{Z_{0}^{2} + Z_{1}^{2} + Z_{2}^{2} + Z_{3}^{2} = 0\} \subset \mathbb{C} \mathbb{P^{3}}$. (hint: Segre embedding).

Step 2. Show that the map $F: \tilde{Gr}(2,4) \rightarrow \mathbb{CP}^{3}$, given by $F(\Pi) = [a+bi]$ (where $a,b \in \mathbb{R}^{4}$ are an orthonormal,oriented basis of the plane $\Pi$), is 1. well-defined 2. a diffeo to it's image and finally 3. it's image is exactly $Q \subset \mathbb{CP}^{3}$.

Nick L
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  • In general this trick can be used to shown that $\tilde{Gr}(2,k)$ is diffeomorphic to a smooth quadric in $\mathbb{C} \mathbb{P}^{k-1}$. – Nick L Feb 20 '18 at 16:01
  • Thanks, @Nick L – Selvakumar A Feb 20 '18 at 16:04
  • Maybe it's simpler to write the quadric as $Z_0^2Z_3^2 = Z_1^2Z_2^2$ since this is the equation you get from the Segre embedding of $\mathbb P^1 \times \mathbb P^1$ ? – Nicolas Hemelsoet Feb 20 '18 at 21:58
  • The map $F$, defined In step 2 naturally hits the quadric I wrote down (the Fermat Quadric). So I guess we should use Segre plus the fact that all smooth quadrics (of the same dimension) are diffeomorphic . – Nick L Feb 20 '18 at 23:03