Let $M$ be a smooth manifold, and $L$ be a smooth real (non-trivial) line bundles over $M$, is it then true that $L\oplus L$ admits a non vanishing section?
Intuitively, I feel like the answer should be yes, since I think I should be able to take a global section $s$ of $L$, and then obtain a new section $t$ by "pushing the zeroes of $s$ around" so that $s$ and $t$ don't vanish at the same point. The section $(s,t)$ would then not vanish. I am having trouble making this precise; I also suspect that I might actually need compactness or something so that I can make sure the zeros of $s$ are isolated and can actually push them off.
My motivation for this, is that I'd like to treat $L\oplus L$ as a trivial complex line bundle (I am pretty sure I can equip $L\oplus L$ with a complex structure since it's structure group should be $GL_1(\mathbb R)\times GL_1(\mathbb R)$ which I believe embeds in $GL_1(\mathbb C)$, but maybe I am also messing that up).
$L \oplus L$ need not admit a non-vanishing section. To see this, note that the Stiefel-Whitney classes of $L \oplus L$ are given by \begin{align} \omega_1(L \oplus L) &= 2\omega_1(L) \omega_0(L) = 0 \\ \omega_2(L \oplus L) &= \omega_1(L)^2 \end{align} with all higher classes zero for degree reasons. Now if $L \oplus L$ has a nonvanishing section, it splits as $L \oplus L \cong L' \oplus \epsilon^1$ where $L'$ is a line bundle, and consequently we compute \begin{align} \omega_1(L \oplus L) = \omega_1(L' \oplus \epsilon^1) &= \omega_1(L') \\ \omega_2(L \oplus L) = \omega_2(L' \oplus \epsilon^1) &= 0 \end{align} so it suffices to find a line bundle $L$ over some manifold $M$ such that $\omega_1(L)^2 \neq 0$, and for this pulling back the universal bundle $\gamma^1$ over $\mathbb{R}\mathrm{P}^\infty$ to some $\mathbb{R}\mathrm{P}^n$, $n \geq 2$ will do.