# Oriented Grassmann is a $2$-sheeted covering space of Grassmann

Let $$G_n(\Bbb R^k)$$ denote the Grassmann manifold (consisting of all $$n$$-planes in $$\Bbb R^k$$), and let $$\tilde{G}_n(\Bbb R^k)$$ denote the oriented Grassmann manifold, consisting of all oriented $$n$$-planes in $$\Bbb R^k$$. Let $$V_n(\Bbb R^k)\subset (\Bbb R^k)^n$$ denote the space consisting of all tuples $$(v_1,\dots,v_n)$$ such that $$\{v_1,\dots,v_n\}$$ is linearly independent. Then there are natural surjections of $$V_n(\Bbb R^k)$$ onto $$G_n(\Bbb R^k)$$ and $$\tilde{G}_n(\Bbb R^k)$$. We topologize $$G_n(\Bbb R^k)$$ and $$\tilde{G}_n(\Bbb R^k)$$ as quotient spaces of $$V_n(\Bbb R^k)$$. Clearly there is a $$2$$-$$1$$ continuous surjection $$p:\tilde{G}_n(\Bbb R^k)\to G_n(\Bbb R^k)$$ (orientation-ignoring map). How do we know that this map is a covering map?

Edit. I've found a relevant comment in https://pi.math.cornell.edu/~hatcher/VBKT/VB.pdf, p.31 (of the book, not the pdf), but I can't understand what "using for example the local trivializations constructed in Lemma 1.15" means. (I've read the proof of Lemma 1.15)

• I don't think you are going to get out of just doing the work of checking the local homeomorphism, but it is worth noting this is a special case of the orientation cover of a manifold, and Hatcher gives a proof that the orientation cover is a covering space. Aug 14, 2020 at 23:50
• @ConnorMalin: I wouldn't call this a special case of the orientation cover as $G_n(\mathbb{R}^k)$ can be orientable. Aug 15, 2020 at 1:29
• @MichaelAlbanese the orientation double cover is defined for any space and it is connected, if and only if, you are non orientable. Aug 15, 2020 at 1:33
• @ConnorMalin: I agree, but for example, $G_1(\mathbb{R}^2) = \mathbb{RP}^1$ is orientable and $\tilde{G}_1(\mathbb{R}^2) = S^1$ which is not the orientation cover of $\mathbb{RP}^1$. Aug 15, 2020 at 1:36
• @MichaelAlbanese Oh yes I see, that's my mistake. Aug 15, 2020 at 1:41

Write \begin{align} \pi &: V^n(\mathbb R^k) \to G_n(\mathbb R^k), \\ \tilde \pi &: V^n(\mathbb R^k) \to \widetilde G_n(\mathbb R^k) \end{align} as the quotient maps which defines the topologies of $$G_n(\mathbb R^k)$$, $$\widetilde G_n(\mathbb R^k)$$ respectively. Then we have $$p \circ \tilde \pi = \pi$$, where
$$p : \widetilde G_n(\mathbb R^k) \to G_n(\mathbb R^k)$$ is the $$2-1$$ continuous surjection.
Now we show $$p$$ is a 2-sheeted covering. It more or less follows from definitions. Let $$\ell \in G_n(\mathbb R^k)$$. Then there is $$(v_1, \cdots, v_n) \in V^n(\mathbb R^k)$$ so that $$\pi (v_1, \cdots, v_n) = \ell$$. Note that $$p^{-1}(\ell) = \{ \ell_+:= \tilde \pi (v_1,v_2, \cdots, v_n),\ell_-:= \tilde \pi (-v_1, v_2, \cdots, v_n)\}.$$
Let $$W = \{w_{n+1}, \cdots, w_k\}$$ be a fixed set of vectors in $$\mathbb R^k$$ so that $$\{ v_1, \cdots, v_n, w_{n+1}, \cdots, w_k\}$$ forms a basis of $$\mathbb R^k$$. Let $$U_+ \subset V^n(\mathbb R^k)$$ be the collection of all $$(\bar v_1, \cdots, \bar v_n)$$ so that $$\tag{2} \frac{\det (\bar v_1 , \cdots, \bar v_n , w_{n+1}, \cdots, w_k)}{\det (v_1 , \cdots, v_n , w_{n+1}, \cdots, w_k)} >0.$$ (in particular $$\{\bar v_1, \cdots, \bar v_n, w_{n+1}, \cdots, w_n\}$$ forms a basis of $$\mathbb R^k$$) Similarly define $$U_-$$ by using $$-v_1$$ instead of $$v_1$$ in (2). Clearly $$U_\pm$$ contains $$(\pm v_1,v_2, \cdots, v_n)$$, is open in $$V_n(\mathbb R^k)$$, $$U_+\cap U_- = \emptyset$$ and $$\tilde \pi^{-1} (\tilde \pi(U_\pm)) = U_\pm, \ \ \pi^{-1} (\pi (U_\pm ))= U_- \cup U_+.$$ In particular, $$V_\pm = \tilde \pi (U_\pm)$$ are open sets in $$\widetilde G_n(\mathbb R^k)$$, $$V = \pi (U_+)$$ is open in $$G_n(\mathbb R^k)$$ and $$p|_{ V_\pm} : V_\pm \to V$$ is a homeomorphism. Since $$\ell$$ is arbitrary, we show that $$p$$ is a 2-sheeted covering.