3
$\begingroup$

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 $\textrm{Gr}_\mathbb{C}(2,2)$?

$\endgroup$
6
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.