Structure of complex Grassmannian $\textrm{Gr}_\mathbb{C}(2,2)$

I recently asked a question about the topology of real Grassmannian $$\textrm{Gr}_\mathbb{R}(2,2) = \frac{O(4)}{O(2)\times O(2)},$$ see Second homotopy group of real Grassmannians $\textrm{Gr}(n,m)$, special case $n=m=2$ not clear.. It turns out that $S^2 \times S^2$ double covers this Grassmanian, which explains the unusual result for its second homotopy group.

This left me wondering: Can the analogous complex grassmannian $$\textrm{Gr}_\mathbb{C}(2,2) = \frac{U(4)}{U(2)\times U(2)}$$ (which, as opposed to the previous case, is simply connected) be also expressed as a product of two "more simple" manifolds? Or not in any trivial way?

It is easy to see that it cannot be $S^4 \times S^4$ (which has the right dimension and comes to my mind as the first guess) because the homotopy groups $\pi_4[\textrm{Gr}_\mathbb{C}(2,2)] = \mathbb{Z}$ (obtained from long exact sequence) and $\pi_4(S^4\times S^4)=\mathbb{Z}\times\mathbb{Z}$ are different. But perhaps there is a feasible way to factorize $\textrm{Gr}_\mathbb{C}(2,2)$?

I will use the notation $\operatorname{Gr}^{\mathbb{C}}(k, n)$ for the grassmannian of $k$-dimensional complex subspaces of an $n$-dimensional complex vector space.

First recall that $\operatorname{Gr}^{\mathbb{C}}(k, n)$ has real dimension $2k(n-k)$ and $\chi(\operatorname{Gr}^{\mathbb{C}}(k, n)) = \binom{n}{k}$, so $\operatorname{Gr}^{\mathbb{C}}(2, 4)$ is a closed eight-dimensional manifold with Euler characteristic $6$.

If $\operatorname{Gr}^{\mathbb{C}}(2, 4) = M\times N$, then $\dim M + \dim N = 8$ and $\chi(M)\chi(N) = 6$. As odd-dimensional manifolds have Euler characteristic zero, there are two (non-trivial) possibilities:

• one has dimension $2$, and the other has dimension $6$, or
• they both have dimension $4$.

Recall that a manifold of dimension $4k + 2$ has even Euler characteristic, so the product of two such manifolds would have Euler characteristic a multiple of four. Therefore, if $\operatorname{Gr}^{\mathbb{C}}(2, 4)$ is a non-trivial product of manifolds, it must be a product of two four-manifolds.

Suppose now that $\operatorname{Gr}^{\mathbb{C}}(2, 4) = M \times N$ with $\dim M = \dim N = 4$. As $\operatorname{Gr}^{\mathbb{C}}(2, 4)$ is simply connected, so are $M$ and $N$. Now $\chi(M) = 2 + b_2(M)$ and $\chi(N) = 2 + b_2(N)$. As $\chi(M)\chi(N) = 6$ and $\chi(M), \chi(N) \geq 2$, we can assume without loss of generality that $\chi(M) = 2$ and $\chi(N) = 3$. It follows that $M$ is homotopy equivalent to $S^4$, and $N$ is homotopy equivalent to $\mathbb{CP}^2$. However, the cohomology rings of $\operatorname{Gr}^{\mathbb{C}}(2, 4)$ and $S^4\times \mathbb{CP}^2$ are different (the former has a degree two element $\alpha$ with $\alpha^3 \neq 0$, but the latter doesn't).

In conclusion, $\operatorname{Gr}^{\mathbb{C}}(2, 4)$ is not a product of lower-dimensional manifolds.

• This was surprisingly straightforward -- once you know the right tool to approach the problem. Thanks for the brilliant answer @Michael ! – zdus Sep 28 '17 at 1:49