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!

  • 1
    $\begingroup$ The canonical bundle of this Grassmannian is $\mathcal{O}(-4)$, so this claim seems suspicious to me. $\endgroup$ Oct 27, 2018 at 0:46
  • $\begingroup$ @SamirCanning: Thanks! You are correct. $\endgroup$
    – Prism
    Oct 27, 2018 at 17:38

1 Answer 1


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.


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