What are the first 8 homotopy groups of the complex Grassmannian $G_\mathbb{C}(2,4)$? I was trying to find any information about the first couple of homotopy groups of the complex Grassmannian $G_\mathbb{C} (2,4)$ of complex planes in complex $4$-space. I need the first seven or eight homotopy groups.
Best Regards,
Stephan
 A: This is more of a comment than an answer, but it's not too hard to compute the first four homotopy groups.  In particular, there is a well-known fiber bundle
$$
U(2) \;\longrightarrow\; V_{\mathbb{C}}(2,4) \;\longrightarrow\; G_{\mathbb{C}}(2,4),
$$
where
$$
V_{\mathbb{C}}(2,4) \,=\, \bigl\{(v,w)\in\mathbb{C}^4\times\mathbb{C}^4 \;\bigl|\; \|v\|=\|w\|=1\text{ and }\langle v,w\rangle = 0\bigr\}
$$
is the corresponding Stiefel manifold.  Using the octonians, it is easy to show that
$$
V_{\mathbb{C}}(2,4) \,\approx\, S^7 \times S^5.
$$
In paricular $\pi_n\bigl(V_{\mathbb{C}}(2,4)\bigr) = 0$ for $n\leq 4$.  Using the  long exact sequence for a fibration, it follows that
$$
\pi_n\bigl(G_{\mathbb{C}}(2,4)\bigr) \,\cong\, \pi_{n-1}\bigl(U(2)\bigr)
$$
for $n\leq 4$.  The homotopy groups for $U(2)$ can be computed using the fiber bundle $S^1\to U(2)\to S^3$; in particular $\pi_1\bigl(U(2)\bigr)\cong\mathbb{Z}$ and $\pi_n\bigl(U(2)\bigr)\cong \pi_n(S^3)$ for all $n\geq 2$.  Therefore,
$$
\pi_4\bigl(G_{\mathbb{C}}(2,4)\bigr)\cong \mathbb{Z},\quad 
\pi_3\bigl(G_{\mathbb{C}}(2,4)\bigr)= 0,\quad \pi_2\bigl(G_{\mathbb{C}}(2,4)\bigr)\cong \mathbb{Z},\quad\text{and}\quad \pi_1\bigl(G_{\mathbb{C}}(2,4)\bigr)=0.
$$
The long exact sequence doesn't seem sufficient to compute the homotopy groups for $n\geq 5$, so I'm not sure how to push this any further.  For example, the fifth homotopy group fits into a short exact sequence $\mathbb{Z} \to \pi_5\bigl(G_{\mathbb{C}}(2,4)\bigr) \to \mathbb{Z}_2$, but I'm not sure how to tell whether it's $\mathbb{Z}$ or $\mathbb{Z}\times\mathbb{Z}_2$.
