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I have a problem to understand the section "The Oriented Universal Bundle" in the page 145 of "Characteristic classes" written by J.W. Milnor.

The content in that page is like below,

$G_n(\mathbb{R}^{n+k})$ is an unoriented Grassmann manifold. Let $\tilde{G_n}(\mathbb{R}^{n+k})$ denote the Grassmann manifold consisting of all oriented $n$-planes in $(n+k)$-space. $(G_n = G_n(\mathbb{R}^{\infty})$ and $\tilde{G_n} = \tilde{G_n}(\mathbb{R}^{\infty}))$.

Then the book says that the universal bundle $\gamma^n$ over $G_n$ lifts to an oriented $n$-plane bundle, $\tilde{\gamma}^n$ over $\tilde{G_n}$. Furthermore, for any oriented $n$-plane bundle $\xi$, each bundle map $\xi \rightarrow \gamma^n$ lifts uniquely to an orientation preserving bundle map $\xi \rightarrow \tilde{\gamma}^n$.

But I can not understand how $\gamma^n$ over $G_n$ lifts to an oriented $n$-plane bundle over $\tilde{G_n}$.

In my opinion, $\tilde{\gamma}^n$ is an induced bundle of $\gamma^n$ by a covering map $p : \tilde{G_n} \rightarrow G_n$. Buy why is it orientable?

Can anybody give me a hint to understand this statement?

Thank you.

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Intuitively: the points of $\tilde{G_n}(\mathbb{R}^{n+k})$ are pairs $(p,\varphi)$ of an $n$-plane and an orientation of it. The tautological bundle has fiber $p$ at point $(p,\varphi)$, which can naturally be given the orientation $\varphi$.

A clean proof of orientability is as follows: the oriented Grassmannian is simply connected, as you can compute directly from a homogeneous space description or as follows from the fact that it double covers a space with two-element fundamental group. And every vector bundle over a simply connected manifold is orientable.

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