Computing $\pi_3(\mathrm{Gr}_2(\mathbb{R}^4))$

Question

How can one go about computing the 3rd homotopy group of the Grassmannian manifold of 2-planes through the origin in $\mathbb{R}^4$?

I don't want to be more general in the question, because:
1) I think this is related to homotopy groups of spheres which is currently intractable.
2) This particular group pops up in my thoughts on vector bundles.

Answer 1

$\text{Gr}_2(\mathbb{R}^4)$ is the homogeneous space $\text{O}(4)/(\text{O}(2) \times \text{O}(2))$ and as such it fits into a fibration

$$\text{O}(2) \times \text{O}(2) \to \text{O}(4) \to \text{Gr}_2(\mathbb{R}^4).$$

The higher homotopy groups of $\text{O}(2) \times \text{O}(2)$ vanish, so the associated long exact sequence shows that $\pi_3(\text{Gr}_2(\mathbb{R}^4)) \cong \pi_3(\text{O}(4))$. To compute the latter, note that

$$\pi_3(\text{O}(4)) \cong \pi_3(\text{SO}(4)) \cong \pi_3(\text{SU}(2) \times \text{SU}(2))$$

since higher homotopy groups are invariant up to taking covering spaces. Note also that $\text{SU}(2) \cong S^3$ satisfies $\pi_3(S^3) \cong \mathbb{Z}$, so the answer is $\mathbb{Z} \times \mathbb{Z}$.

Answer 2

$Gr_2(\mathbb R^4)$ is the quotient $O_4 / O_2 \times O_2$, which is double covered by $SO_4 / SO_2 \times SO_2$. $SO_4$ is double-covered by $S^3 \times S^3$. So $Gr_2(\mathbb R^4)$ has $(S^3 / SO_2)^2$ as a covering space, and this is just $(S^2)^2$. So $\pi_3 Gr_2(\mathbb R^4)$ is the same as $\pi_3 (S^2 \times S^2)$ which is a free abelian group of rank two.