I am interested to know if there is a way of calculating the fundamental group of the real Grassmannian $\mathbf{G}(k,n)$, I have already proved using van Kampen that the fundamental group only depends on its $2$-skeleton and previously I proved that the fundamental group of $\mathbb{RP}^n$ is $\mathbb{Z}_2 \cong \frac{\mathbb{Z}}{2\mathbb{Z}}$ for $n\geq 2$, therefore, $\pi_1(\mathbf{G}(k,n)) = \mathbb{Z}_2$ for $k=1$ (and $n\geq 3$), so my problem is for $k>1$ because the idea is to use the result that I tested previously to be able to calculate ‘easily’ the fundamental group, so I could also use van Kampen directly but I don't know if I have to use cases, because I understand that in some cases $\mathbf{G}(k,n)$ is simply connected. I am grateful for any help and suggestions.
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1There are fibre bundles $O(n) \to V_n(\mathbb{R}^k) \to \mathbf{G}(k, n)$ which together with the fact that $V_n(\mathbb{R}^k)$ is $(k - n - 1)$-connected you can use to get yourself a good bit further along. – Ben Steffan Jul 24 '24 at 20:21
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1Using another set of fibre sequences, $V_{n - m}(\mathbb{R}^{k - m}) \to V_n(\mathbb{R}^k) \to V_m(\mathbb{R}^k)$ it might be possible to do the whole computation, though I haven't thought it through. I don't see a "low-tech" way on the level of van Kampen to do this; looking at the cell structure doesn't strike me as likely to be a tractable approach. – Ben Steffan Jul 24 '24 at 20:42
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2@Ben: it's a straightforward calculation using the long exact sequence of a fibration applied to $O(k) \times O(n-k) \to O(n) \to \text{Gr}(k, n)$; this is the approach taken in the linked question. The answer is that for $k \ge 1, n-k \ge 1, n \ge 3$ the fundamental group is always $\mathbb{Z}/2$. The double cover is the oriented Grassmannian $SO(n)/(SO(k) \times SO(n-k))$ of oriented $k$-planes in $\mathbb{R}^n$, and the same calculation for this Grassmannian shows (with the same hypotheses on $k, n$) that it's simply connected. – Qiaochu Yuan Jul 24 '24 at 22:29