Tensor product of two Mobius bundles

Question

Consider the Möbius bundle $p : M \longrightarrow S^1$ which is the canonical line bundle over $\mathbb{R}\text{P}^1 \simeq S^1$. I've been able to show that $M \oplus M$ is the trivial bundle and now want to determine the bundle $M \otimes M \longrightarrow S^1$. I was thinking of using the universal property of the tensor product, that is, since we have $M \oplus M \longrightarrow S^1$, we know that this map factors through $$M \oplus M \longrightarrow M \otimes M \longrightarrow S^1$$ But I haven't been able to get any further, even after noting that the map $M \oplus M \longrightarrow S^1$ is defined by $(x,x) \longmapsto x$.

Answer 1

Let $p: M \longrightarrow \mathbb{R}\text{P}^1$ denote the canonical line bundle on $\mathbb{R}\text{P}^1 \simeq S^1$ and consider the map $\varphi : M \oplus M^{\perp} \longrightarrow \mathbb{R}\text{P}^1 \times \mathbb{R}^2$ defined by $$(\ell, x,y) \longmapsto (\ell, x+y),$$ for $\ell \in \mathbb{R}\text{P}^2$ such that $y \perp \ell$. This map gives us an isomorphism between the bundles $M \oplus M^{\perp}$ and $\mathbb{R}\text{P}^1 \times \mathbb{R}^2 \cong S^1 \times \mathbb{R}^2$.

Now let $\psi : M \longrightarrow M^{\perp}$ denote the isomorphism defined simply by rotation by $\pi/2$. Then by consider both $\varphi$ and $\psi$, we have that $$M \oplus M \cong M \oplus M^{\perp} \cong \mathbb{R}\text{P}^1 \times \mathbb{R}^2 \cong S^1 \times \mathbb{R}^2,$$ and so $$M \oplus M \cong S^1 \times \mathbb{R}^2,$$ illuminating the fact that $M \oplus M$ is the trivial bundle.

Now consider that the Mobius bundle is isomorphic to the line bundle over $\mathbb{R}\text{P}^1 \cong S^1$. Therefore we can simply speak about line bundles, rather than Mobius bundles. The set $\text{Vect}^1(B)$ of isomorphism classes of one-dimensional vector bundles over $B$ forms an abelian with respect to the tensor product. Inverses of line bundles are given by replacing the gluing matrices $g_{\beta \alpha}(x) \in GL_1(\mathbb{R})$ with their inverses and the cocyle condition is maintained since $1 \times 1$ dimensional matrices obviously commute. With an inner product on the line bundle, we may rescale local trivializations to be isometries, taking vectors in fibers of the line bundle to vectors in $\mathbb{R}$ of the same scale. Hence the gluing functions $g_{\beta \alpha}$ are simply $\pm 1$. The gluing functions of the tensor product $M \otimes M$ are then the squares of these $g_{\beta \alpha}$ and are subsequently $1$. We therefore see that $M \otimes M$ is the trivial bundle. $\Box$

Answer 2

Given a line $l \in \Bbb P^1(\Bbb R)$, there are two elements $u,v \in l$ that also are on the unit circle, and in $l \otimes l$, $u \otimes u = (-v) \otimes (-v) = -(-(v \otimes v)) = v \otimes v$.

Since this result doesn't depend on how you choose $u$ and $v$, the map $l \mapsto u \otimes u \in l \otimes l$ is continuous because you can cover $\Bbb P^1(\Bbb R)$ with small open sets where you can choose $u$ continuously.

This gives you a nontrivial nonvanishing section of $M \otimes M$, and so $M$ is the trivial bundle.