Show that the tautological vector bundle over $G_1(\mathbb{R}^2)$ is smoothly isomorphic to the Mobius bundle

Recall that the tautological vector bundle over the Grassmanian $G_k(V)$ is the smooth rank-$k$ vector bundle of the form $T = \{(S,v) \in G_k(V) \times V | v \in S\}$, which has the property that the fibre over a point $S \in G_k(V)$ is isomorphic as a vector space to $S$.

I have a proof attempt below but I think the proof is wrong because it is not using any of the structures or properties specific to the tautological bundle or the Mobius bundle. It seems like the proof can be adapted to any two line bundles with homeomorphic base spaces:

Proof Attempt:

Let $T$ be the tautological bundle of $G_1(\mathbb{R}^2)$ and $E$ be the Mobius bundle. Note that by definition, $G_1(\mathbb{R}^2) = \mathbb{RP}^1$. Let $f$ be the homeomorphism between $\mathbb{RP}^1$ and $S^1$. Let $S \in \mathbb{RP}^1$. Then $\pi^{-1}_T(S) \cong S$ as vector spaces, since $T$ is the tautological bundle of $\mathbb{RP}^1$. Since $S$ is a $1$-dimensional subspace of $\mathbb{R}^2$, there is a single basis vector $s$ which spans $S$. Hence, every $v \in S$ can be written in the form $v = a_vs$, for some $a_v \in \mathbb{R}$. Now $f(S) \in S^1$, and so let $R = \pi^{-1}_E(f(S)$. Then since the Mobius bundle is a line bundle, $R$ is a $1$-dimensional vector space and thus there is a single basis vector $r$ which spans $R$. Hence, any $u \in \pi^{-1}_E(f(S))$ can be written in the form $u = b_ur$, for some $b_u \in \mathbb{R}$. Since $S$ and $R$ are vector spaces with the same dimension, there is an isomorphism between them. Hence, for every $v = a_vs \in S$, there is a unique linear correspondence to a vector $u = b_ur \in R$. Since every point of $T$ lies in the fibre of some point in $\mathbb{RP}^1$, define the following map fibre-wise:

\begin{gather*} F: T \rightarrow E \\ v = a_vs \in \pi^{-1}_T(S) \mapsto u = b_ur \in \pi^{-1}_E(f(S)) \end{gather*}

Due to the isomorphism between $\pi^{-1}_T(S)$ and $\pi^{-1}_E(f(S))$, $F$ is a bijection and is linear when restricted to each fibre. What remains to show is that $F$ is smooth, which shouldn't be hard since $F$ is smooth at least between each fibre and hence should be smooth using the charts of $T$ and $E$.

End Of Proof Attempt

I would like to know whether this proof can be saved, what is going wrong with this proof, and what a better approach to this problem would be.

  • 2
    $\begingroup$ You said it yourself - this proof does not use the characteristics of the specific vector bundles in question. If it were a valid proof, it would mean that any two vector bundles of the same rank are isomorphic, and this is far from being the case. Constructing a bundle isomorphism is mostly a matter of choice. Over any point in the manifold there are many possible isomorphisms of vector spaces, but the hard part is to choose these isomorphisms continuously along the whole base manifold. In your proof you seem to ignore this. You claim that between two lines there is a single isomorphism, $\endgroup$ – Amitai Yuval Nov 1 '18 at 22:29
  • $\begingroup$ which is wrong. $\endgroup$ – Amitai Yuval Nov 1 '18 at 22:29
  • $\begingroup$ I see, yes. This proof would show that $F$ is bijective and linear on each fibre, but what's missing it to show that $F$ is smooth on the entire bundle, not just on the fibres. I'm not sure sure where to progress from here though. What is a better approach to solve this problem? $\endgroup$ – Frederic Chopin Nov 1 '18 at 22:43
  • 1
    $\begingroup$ To globalise the choice of basis vectors $s$ and $t$ in each fibre requires the choice of non-vanishing sections of each bundle, and these do not exist since the bundles are not trivial. $\endgroup$ – Tyrone Nov 2 '18 at 9:32

Let $E\to M$ and $F\to M$ be vector bundles. This answer suggests ways to define bundle morphisms $E\to F$. The "iso" issue is less significant, as a bundle morphism is an isomorphism if and only if it is iso on each fiber, and the latter is usually easy to confirm.

First suggestion - relying on extra structure: As I wrote in my comment, there are many different ways to map a fiber of $E$ into a fiber of $F$. Sometimes, however, the base manifold and the bundles carry extra structure that can be used to specify a morphism. A basic example is using a Riemannian metric to construct an isomorphism between a bundle and its dual over a smooth manifold. Put simply, an inner product on a (real) vector space $V$ gives rise to a "natural" isomorphism between $V$ and $V^*$. Hence, if the bundle $E$ is equipped with a fiber metric, this can be used to identify $E$ with its dual, $E^*$. The fact that the metric changes continuously (or smoothly) along the base manifold guarantees that so does the isomorphism it yields.

Second suggestions - trivializations and transition maps: Let $\{U_\alpha\}_{\alpha\in I}$ be an open cover of $M$ which trivializes both $E$ and $F$. Let $\varphi^E_{\alpha\beta}$ and $\varphi^F_{\alpha\beta}$ denote the transition maps. This means that over both $U_\alpha$ and $U_\beta$ we identify $E$ with the trivial bundle, and we glue $E|_{U_\alpha}$ to $E|_{U_\beta}$ using $\varphi^E_{\alpha\beta}$: for a point $x\in U_\alpha\cap U_\beta$ and a vector $v\in\mathbb{R}^k,\; k$ being the rank of $E$, we declare that $(x,v)$ as a point in $E|_{U_\alpha}$ is identical to $(x,\varphi^E_{\alpha\beta}(x)\cdot v)$ as a point in $E|_{U_\beta}$. The same goes for $F$ and the $\varphi^F_{\alpha\beta}$'s. Then, constructing a bundle morphism $E\to F$ amounts to specifying a linear map $L_\alpha(x):\mathbb{R}^k\to\mathbb{R}^k$ over every point in each of the $U_\alpha$'s and making sure that these maps agree on the intersections. "Agree" means that for $x\in U_\alpha\cap U_\beta$ we have $$\varphi^F_{\beta\alpha}(x)\circ L_\beta(x)\circ\varphi^E_{\alpha\beta}(x)=L_\alpha(x).$$

Now, in your case this shouldn't be so hard to do, as your base manifold is a circle. This means that you can cover it by two open sets over which every vector bundle is trivial. Moreover, you are working with line bundles, and so, all linear maps involved are just multiplication by scalars.


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