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The Grassmann manifold $\widetilde{Gr}(k,\Bbb{R}^n)$ of oriented $k$-planes in $\Bbb{R}^n$ is a double cover of the Grassmann manifold $Gr(k,\Bbb{R}^n)$ of non-oriented $k$-planes. We can give $\widetilde{Gr}(k,\Bbb{R}^n)$ the covering metric making the covering a local isometry.

Is it possible to describe the geodesics in $\widetilde{Gr}(k,\Bbb{R}^n)$ simply in terms of the geodesics in the non-oriented $Gr(k,\Bbb{R}^n)$?

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    $\begingroup$ Yes of course. Not only is the map from the oriented Grassmannian to the unoriented ones a covering map but it's a local isometry. So there's a direct correspondence between the geodesics given by lifting and taking the composite with the covering map, respectively. Does this answer your question? $\endgroup$
    – Ryan Budney
    Dec 27, 2012 at 4:30
  • $\begingroup$ @Ryan: I'm not sure that it does. I'm having trouble picturing the geodesics in the oriented Grassmannian in terms of the geodesics in the unoriented Grassmannian. For example, consider a plane with its two orientations, which corresponds to two distinct points in the oriented Grassmannian. Under the covering map, these points get map to a single point. The geodesic through these two points in the oriented Grassmannian get map presumably to a closed geodesic in the unoriented Grassmannian. But how do we describe the geodesic through these two oriented planes in terms of the closed geodesic? $\endgroup$
    – Oliver Jones
    Dec 27, 2012 at 6:36
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    $\begingroup$ @Oliver: I think the answer to your comment is already in Ryan's comment: lift. In more detail: Imagine your unoriented plane P, moving along a closed geodesic back to its starting position. Now attach to the initial P one of its orientations and, as P moves continuously along the geodesic, keep it oriented in a continuous way; in other words, drag the orientation along with the plane. When P returns to its initial position, the orientation might return to the initial orientation or to the opposite one. The latter possibility is the situation in your comment. $\endgroup$
    – Andreas Blass
    Dec 27, 2012 at 15:43
  • $\begingroup$ @Andreas: You're right, I didn't fully appreciate that covering spaces have the lifting property. Thanks for clarifying. This brings me to a related question. There are two ways in which to define a metric on the Grassmnnian of oriented planes; one is to treat it as a homogeneous space and the other is to pull back the metric from the Grassmannian of unoriented planes under the covering map. Do these metrics coincide? $\endgroup$
    – Oliver Jones
    Dec 28, 2012 at 21:39
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    $\begingroup$ @Oliver: I assume the metric on the unoriented version is obtained by treating it as a homogeneous space. Then the two metrics in your comment should coincide, provided you've normalized them the same way. The reason is that the 2-to-1 projection from the oriented to the unoriented version is invariant under the actions (on both versions) of the orthogonal group, so pulling back an invariant metric from the unoriented version will produce an invariant metric on the oriented version. $\endgroup$
    – Andreas Blass
    Dec 30, 2012 at 2:12

1 Answer 1

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(CW answer just to get this off the unanswered list. See the comments above by Ryan Budney 1 and Andreas Blass 1 2 for more details.)

Yes. Let $f:(M,g) \to (N,h)$ a covering map that is a local isometry, and let $p\in M$. If $\gamma:[0,1]\to N$ is a geodesic such that $\gamma(0) = f(p)$, then we can lift $\gamma$ to a geodesic $\tilde{\gamma}:[0,1]\to M$ a geodesic with $\tilde{\gamma}(0) = p$.

Of course, it is possible that $\gamma$ is a closed geodesic while $\tilde{\gamma}$ is not: consider the simple case of the covering $\mathbb{R}^1\to \mathbb{T}^1$.

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  • $\begingroup$ 'Twas a noble effort, anyway. $\endgroup$
    – LSpice
    Sep 10, 2022 at 20:53

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