Question
Let $\gamma$ be the tautological line bundle of the $n$-dimensional real projective space $\mathbb{P}^n$.
Is $\gamma$ is orientable as a vector bundle?
My guess
I only consider in case of $n=1$.
Let $n=1$, and consider the one dimensional projective space.
let $\nu$ be a normal bundle on the two dimensional sphere $\mathbb{S}^1 \subset \mathbb{R}^2$. Then, there is a bundle map $ \nu \to \gamma$. Then the pull back of the cohomology class $u_F$ of $\gamma$ also gives a orientation of the bundle $\nu$. However, the antipodal points of $\mathbb{S}^1$ are sended to the same point of $\mathbb{P}^1$ and thus, such orientation will be given by some same vector in $\mathbb{E}^2$. But it is impossible, because the orientation of antipodal points should be opposite directions. Thus $\gamma$ is not orientable as a vector bundle.
Definition of orientation 1 Each fiber $F$ of $\gamma$ is given an orientation as an vector space and ,in addition, for each $p \in
\mathbb{P}^n$ there are a neighborhood $N$ of $\mathbb{P}^n$ centered
at $p$ and a local section $s$ of $\gamma$ defined on $N$ such that
the value of section $s$ at each $q \in N$ gives the orientation of
the fiber at $q$.Definition of orientation 2 For each fiber, there is a generator $u_F \in H^1(F,F-0,\mathbb{Z})$ such that for each $q \in
\mathbb{P}^n$ there is an element $u \in
H^1(\pi^{-1}(N),\pi^{-1}(N)-0,\mathbb{Z})$ whose restriction to $F$
is coincides with $u_F$.
Best Answer
It is not orientable since it is not trivial a $1$-dimensional vector bundle is orientable if and only if it is trivial. A way to see this, if $E\rightarrow M$ is a $1$-dimensional orientable vector bundle, there exist coordinate change $(U_i,g_{ij}:U_i\cap U_j\rightarrow \mathbb{R}_+^*$ so the bundle has a $\mathbb{R}_+^*$ reduction since this group is contractible, the bundle is trivial.
See also the answer here to see why the tautoligical line bundle is not trivial.
How to distinguish the tautological line bundle and the trivial line bundle on $P^n$?