Tautological vector bundle over $G_1(\mathbb{R^2})$ isomorphic to the Möbius bundle

Question

Let $V$ be a finite dimensional vector space, and let $G_k(V)$ be the Grassmannian of $k$-dimensional subspaces of $V$. Let $T$ be the disjoint union of all these $k$-dimensional subspaces and let $\pi:T\rightarrow G_k(V)$ be the natural map sending each point $x \in S$ to $S$. Then $T$ has a unique smooth manifold structure making it into a smooth rank-$k$ vector bundle over $G_k(V)$, with $\pi$ as a projection and with the vector space structure on each fiber inherited from $V$. $T$ is called the tautological vector bundle over $G_k(V).$

What I want to prove is that tautological vector bundle over $G_1(\mathbb{R^2})$ is isomorphic to the Möbius bundle.

(This is a problem from Introduction to Smooth Manifolds by Lee and Möbius bundle is defined as in Lee's book, page 105. Also I took the definition of the tautological vector bundle over $G_k(V)$ from Lee's book as well.)

What ways do you know of showing two bundles are isomorphic? — Mariano Suárez-Álvarez, Dec 10 '11 at 00:05
@MarianoSuárez-Alvarez: Writing a smooth bundle isomorphism between them. I do not know if it is enough to show that the transition functions are the same. — Karatuğ Ozan Bircan, Dec 10 '11 at 00:51
Can you see that $G_1(\mathbb R^2) \cong \mathbb RP^1 \cong \mathbb S^1$? It's only a step from there, to construct an isomorphism explicitly. — Sam, Dec 10 '11 at 00:54

score 5 · Accepted Answer · answered Dec 10 '11 at 13:07

5

As @Sam pointed out $G_1(\mathbb{R^2}) \cong \mathbb{RP^1} \cong \mathbb{S^1}$. Now, writing a smooth bundle isomorphism is not so hard.

answered Dec 10 '11 at 13:07

Karatuğ Ozan Bircan

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Tautological vector bundle over $G_1(\mathbb{R^2})$ isomorphic to the Möbius bundle

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