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.)

  • $\begingroup$ What ways do you know of showing two bundles are isomorphic? $\endgroup$ Commented Dec 10, 2011 at 0:05
  • $\begingroup$ @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. $\endgroup$ Commented Dec 10, 2011 at 0:51
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    $\begingroup$ 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. $\endgroup$
    – Sam
    Commented Dec 10, 2011 at 0:54
  • $\begingroup$ @Sam: Now I see what I need to show. Thanks! $\endgroup$ Commented Dec 10, 2011 at 11:47

1 Answer 1


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.


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