Let $X$ be a smooth scheme of dimension $r$. Given a rank $r$ vector bundle $\pi: E\to X$, we say that $E$ is orientable if there is a line bundle $L$ on $X$ with an isomorphism $L^{\otimes 2} \cong \bigwedge^{\text{top}} E$.
Let $\operatorname{Gr}(2,4)$ denote the Grassmannian of 2-dimensional subspaces of a 4-dimensional vector space. In the paper I am reading, the authors indicate that
The tangent bundle of $\operatorname{Gr}(2,4)$ is not orientable in the above sense.
Does anyone know a quick way (or any way) to see this? Any reference to the literature where this fact is proved would be appreciated too! Thanks!
Samir is correct! There was a typo in the paper. I was reading the arXiv version. This (false) remark does not appear in the updated version in one of the author's webpage.