As mentioned here, the algebraically distinct line bundles $\mathcal{O}(n)$ on $\mathbb{RP}^1$ are equivalent as smooth line bundles depending on whether $n$ is even or odd. So I want to work out a particular case: $\mathcal{O}(2) \cong \mathcal{O}(0)$.
In order to prove $\mathcal{O}(2) \cong \mathcal{O}(0)$, it suffices to show their transition functions are equivalent. Let $[x,y]$ be the homogeneous coordinates on $\mathbb{RP}^1$. Let $U_0$ and $U_1$ be the standard affine opens, i.e., $U_0 = \{[x,y]:x \neq 0\}$ and $U_1 = \{[x,y]:y \neq 0\}$. We know $\mathcal{O}(0)$ is the structural sheaf, i.e., the trivial bundle over $\mathbb{RP}^1$, so its transition function is just $1$. The transition functions of $\mathcal{O}(2)$ are given by $$g_{01}: U_0 \cap U_1 \to \mathbb{R}^*, \ \ \ [x,y] \mapsto (y/x)^2;$$ and $$g_{10}: U_1 \cap U_0 \to \mathbb{R}^*, \ \ \ [x,y] \mapsto (x/y)^2.$$ To show $g$ is equivalent to the trivial transition function $1$, we want to have two nowhere vanishing smooth functions $\lambda_0$ and $\lambda_1$ on $U_0$ and $U_1$, respectively, such that over $U_0 \cap U_1$, $$g_{01} = \lambda_0 \lambda_1^{-1}, \quad \text{and} \quad g_{10} = \lambda_1 \lambda_0^{-1}.$$
Is there an easy way to cook up such $\lambda_0$ and $\lambda_1$? If we allow only continuous functions, then we can choose a function $f: \mathbb{R} \to \mathbb{R}$ with $f(t) = t^2$ when $|t| \geq 1 $, and $f(t)=1$ when $t<1$. Then $\lambda_0([x,y])=f(y/x)$ and $\lambda_1([x,y])=f(x/y)$ will work. But I'm not sure how to construct smooth functions $\lambda_0$ and $\lambda_1$ to make things work.
Best Answer
We don't need to work this specifically. Topologically (and smoothly) it's always true that if $L$ is any real line bundle whatsoever, then $L^{\otimes 2}$ is trivial(izable), and we have $\mathcal{O}(2) \cong \mathcal{O}(1)^{\otimes 2}$. In cohomology this is the observation that the first Stiefel-Whitney class $w_1$ takes values in $H^1(-, \mathbb{Z}_2)$ and hence is $2$-torsion, but we can be somewhat more explicit about it, as follows:
Pick a (smooth) Riemannian metric on $L$ (using partitions of unity). This induces a perfect pairing $L \otimes L \to \mathbb{R}$ ($\mathbb{R}$ the trivial line bundle), which is an isomorphism.