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Let $G(2, 4)$ denote the space of two dimensional planes in $\mathbf R^4$.

I have found that the integral homology is the following: $H_0 = \mathbf Z, H_1 = \mathbf Z / 2 \mathbf Z, H_2 = 0, H_3 = 0$ and $H_4 = \mathbf Z$.

Now, I would like to argue by using the Hurewicz homomorphism that $G(2, 4)$ is not simply connected since $H_1$ would have to be trivial, but it seems that I would need to establish that $G(2, 4)$ is path-connected. Is there a simple way to do that? I would also appreciate it if someone could check my homology computations against the literature (a quick search on my part did not yield anything).

Thank you.

user25784
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2 Answers2

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We can view $G(2, 4)$ as the homogeneous space $SL(4, \mathbb{R}) / P$, where P is the (parabolic) subgroup that stabilizes a reference $2$-plane in $\mathbb{R}^4$. In particular, it is the quotient of path-connected Lie group, and so it is path-connected.

Travis Willse
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Firstly, you have computed the $H_0$; since $H_0 \cong \mathbb{Z}$ this tells you that the space is path connected.

But anyway, you don't even need that from a formal point of view. Either (a) the space is path connected, and thus you can apply Hurewicz, or (b) the space isn't path connected, so in particular it's not simply connected. In both cases, you get the answer.

Najib Idrissi
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