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)

  • $\begingroup$ 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. $\endgroup$ Aug 14, 2020 at 23:50
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    $\begingroup$ @ConnorMalin: I wouldn't call this a special case of the orientation cover as $G_n(\mathbb{R}^k)$ can be orientable. $\endgroup$ Aug 15, 2020 at 1:29
  • $\begingroup$ @MichaelAlbanese the orientation double cover is defined for any space and it is connected, if and only if, you are non orientable. $\endgroup$ Aug 15, 2020 at 1:33
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    $\begingroup$ @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 $\mathbb{RP}^1$. $\endgroup$ Aug 15, 2020 at 1:36
  • $\begingroup$ @MichaelAlbanese Oh yes I see, that's my mistake. $\endgroup$ Aug 15, 2020 at 1:41

1 Answer 1


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.


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