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I saw this result mentioned a lot in many references, but it is always stated as a fact or an exercise.

My approach would be to see the oriented grassmannian as the quotient $$\frac{SO(n)}{(SO(k)\times SO(n-k))},$$ but then I'm unsure how fundamental groups behave under quotient. I've proved that it is a $2$-covering of the classical grassmaniann and I think it should represent its orientation cover (because I read that it is orientable), but proving that it is simply connected would be slightly more and would imply (together with the fact that it is a $2$-covering) the two facts.

Can someone provide me a solution, a reference or some hints?

There is a related question here, but the answer didn't provide any detail in the case I'm interested in.

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Luigi M
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  • What's wrong with the paragraph starting with "In the real case, a bit more work needs to be done. For the oriented Grassmanian, it's enough to note that the canonical map $SO(k)\rightarrow SO(n)$ is a surjection on $\pi_1$..." in the answer you link? It gives a complete answer. Do you understand how the homotopy long exact sequence associated to the fiber bundle $H \to G \to G/H$ works? – Najib Idrissi Jan 13 '16 at 13:44
  • @NajibIdrissi I think the author doesn't provide sufficient details to make me able to prove what he claims to be true. In particular, why the inclusion induces a surjection on fundamental groups and how this implies the solution. Yes I'm aware of the l.e.s. associated to a fibration – Luigi M Jan 13 '16 at 13:50

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I'll try to fill in some of the missing details, hopefully this will be enough. Let the unoriented Grassmanian be $X = \widetilde{\mathrm{Gr}}(k, \mathbb{R}^n) \cong SO(n) / (SO(k) \times SO(n-k))$. Assume $0 < k < n$ (otherwise there's not much to prove). There is thus a fiber bundle $SO(n) \to X$, with fiber $SO(k) \times SO(n-k)$. Since $SO(n)$ is path connected, so is $X$. This fiber bundle then induces a homotopy long exact sequence: $$\dots \to \pi_1(SO(k) \times SO(n-k)) \to \pi_1(SO(n)) \to \pi_1(X) \to 1.$$ One must see that $\pi_1(SO(k) \times SO(n-k)) \to \pi_1(SO(n))$ is surjective to prove the claim about $X$. It is sufficient to prove that $\pi_1(SO(k)) \to \pi_1(SO(n))$ is surjective when $2 \le k \le n$ (because if $k = 1$, then $n-k = n-1 \ge 2$).

Let $S^{n-1} \subset \mathbb{R}^{n}$ be the standard $(n-1)$-sphere, and let $e_n = (0,\dots,0,1) \in S^{n-1}$ be the last standard basis vector. The application $SO(n) \to S^{n-1}$, $A \mapsto A \cdot e_n$, is a fiber bundle. Its fiber $F$ over $e_n$ is the subgroup of $SO(n)$ consisting of block matrices of the type $$\begin{pmatrix} A' & 0 \\ 0 & 1 \end{pmatrix}$$ where $A'$ is an $(n-1) \times (n-1)$ matrix. This subgroup is isomorphic to $SO(n-1)$. You thus get a homotopy long exact sequence associated to $SO(n-1) \to SO(n) \to S^{n-1}$. There are two cases:

  • when $n \ge 4$, $S^{n-1}$ is 2-connected, and the LES tells you that $\pi_1(SO(n-1)) \to \pi_1(SO(n))$ is an isomorphism.
  • when $n = 3$, there is a short exact sequence $0 \to \mathbb{Z} \to \pi_1(SO(2)) \to \pi_1(SO(3)) \to 0$. In particular $\pi_1(SO(2)) \to \pi_1(SO(3))$ is surjective. (We do not need this, but this short exact sequence is isomorphic to $0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0$).

By induction, you find that $\pi_1(SO(k)) \to \pi_1(SO(n))$ is surjective for $k \ge 2$ (in fact it's an isomorphism when $k \ge 3$). Going back to the long exact sequence of the beginning, this implies that $\pi_1(SO(k) \times SO(n-k)) \to \pi_1(SO(n))$ is surjective for $1 \le k \le n$ and $n \ge 3$, and this $\pi_1(X) = \pi_1(\widetilde{\mathrm{Gr}}_k(\mathbb{R}^n)) = 0$

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