Let $L\to M$ be a real line bundle over a manifold $M$, and let us denote by $\Gamma(L)$ its space of sections. I am trying to find a product in $\Gamma(L)$ to make it into an algebra. The naive product defined on every local trivialization $(U,\phi)$ as
$(s_{1}\cdot s_{2})|_{U}(p) =(p, \phi(s_{1}(p))\phi(s_{2}(p)))\, , \qquad p\in M\, ,$
gives as a result a section of the square line bundle, $s_{1}\cdot s_{2}\in \Gamma(L^{2})$, so it doesn't work.
Thanks.
Proof. A map between line bundles which is everywhere nonzero is everywhere an isomorphism, and hence an isomorphism, so $L \otimes L \cong L$. Tensoring by the inverse line bundle $L^{-1} \cong L^{\ast}$, it follows that
$$L \cong L^{-1} \otimes L \otimes L \cong L^{-1} \otimes L \cong 1.$$
Very explicitly, at each point $x \in X$, choose a nonzero vector $v_x \in L_x$. Then $m_x$ acts by
$$m_x : L_x \otimes L_x \ni v_x \otimes v_x \mapsto c_x v_x \in L_x.$$
for some constant $c_x$, depending both on $x$ and on the choice of vector $v_x$. By hypothesis, $m_x$ is nonzero, hence $c_x$ is nonzero, so we can write down the inverse map
$$m_x^{-1} : L_x \ni v_x \mapsto c_x^{-1} v_x \otimes v_x \in L_x \otimes L_x.$$
The corresponding nonzero section of $L$ is, at each point, the map in $L_x^{\ast} \otimes L_x \otimes L_x$ corresponding to $m_x^{-1}$, so explicitly it has value $c_x^{-1} v_x$ at each point $x$, and you can check for yourself that this doesn't depend on any choices.