$\gamma_n^1 \otimes \gamma_n^1$ is isomorphic to the trivial line bundle over $\mathbb{RP}^1$ I have read that $\gamma_n^1 \otimes \gamma_n^1$ is isomorphic to the trivial line bundle over $\mathbb{RP}^1$, where $\gamma_n^1$ is the canonical line bundle over $\mathbb{RP}^1$, but I am unable to give a vector bundle isomorphism.
I am ok with the fibers being 1-dimensional, but without a trivialization I really don't know how to get a map to $\mathbb{RP}^1 \times \mathbb{R}$.
 A: Note that $\gamma^1$ has total space
$$E(\gamma^1)=\{([x,y],(a,b))\in\mathbb{R}P^1\times\mathbb{R}^2\mid [x,y]=[a,b]\in\mathbb{R}P^2\},$$
and the bundle structure is given by projecting onto the first factor. It follows therefore that
$$E(\gamma^1\otimes\gamma^1)=\{([x,y],(a,b)\otimes(c,d))\in\mathbb{R}P^1\times\mathbb{R}^2\otimes\mathbb{R}^2\mid [x,y]=[a,b]=[c,d]\in\mathbb{R}P^2\}.$$
Since $\gamma^1\otimes\gamma^1$ is $1$-dimensional, to give it a trivialisation it suffices to find a global non-vanishing section $s:\mathbb{R}P^1\rightarrow E(\gamma^1\otimes\gamma^1)$. Now, since $\mathbb{R}P^1\cong S^1/[(a,b)\sim(-a,-b)]$ is a quotient of $S^1$, we can represent any of its points $[x,y]$ by an element $(x,y)\in S^1\subseteq\mathbb{R}^2$. Such a representation is not unique, but there are exactly two elements in each coset. Namely $(x,y)$ and $(-x,-y)$.
Define
$$s[x,y]=([x,y],(x,y)\otimes (x,y))\in E(\gamma^1\otimes \gamma^1).$$
This is well-defined since
$$(-x,-y)\otimes(-x,-y)=(x,y)\otimes (x,y)\in\mathbb{R}^2\otimes\mathbb{R}^2,$$
from which we see that the definition of $s$ does not depend on the choice of reprsentative for the coset $[x,y]\in\mathbb{R}P^1$. Furthermore, $s$ is non-vanishing at each point of $\mathbb{R}P^1$. Thus it gives a global trivialisation of $\gamma^1\otimes\gamma^1$.
