Consider the tautological bundle $S$ on a Grassmannian $G(r,n)$ of $r$-subspaces in $\mathbb{C}^n$. Is $S$ trivial outside (large degree) hypersurfaces on $G(r,n)$. Morel's theorem seems to confirm when $r\neq 2$.

thanks. but I want the complement to be large degree hypersurface sections.

  • $\begingroup$ Do you mean all hypersurfaces from some degree on or just there exists one? The latter is trivial if you admit reducible hypersurfaces. $\endgroup$ Aug 1, 2016 at 7:04
  • $\begingroup$ yes at least general irreducible smooth hypersurfaces of large degree $\endgroup$
    – john
    Aug 1, 2016 at 7:06
  • $\begingroup$ @john: for an irreducible hypersurface of degree $d > 1$ this is not true, see the second part of my answer. $\endgroup$
    – Sasha
    Aug 1, 2016 at 8:01

1 Answer 1


It is even trivial outside appropriate hyperplane section. Let $V$ be a vector space of dimension $n$ such that the Grassmannian is $Gr(r,V)$, and let $V_0 \subset V$ be a subspace of codimension $r$. Consider the composition $$ f: S \to V \otimes O \to (V/V_0) \otimes O. $$ It is an isomorphism outside of the zero locus of the determinant $$ \det(f): \det(S) \to \det(V/V_0) \otimes O. $$ It remains to note that $\det(S) \cong O(-1)$, so the vanishing locus of $\det(f)$ is a hyperplane section. To be more precise, it is the Schubert cycle, parameterizing subspaces of $V$ that intersect $V_0$ nontrivially.

EDIT. If you want $S$ to be trivial on a complement of a general irreducible hypersurface of degree $d > 1$, then this does not hold. Indeed, if $S$ would be trivial, then $\det(S) \cong O(-1)$ would be trivial as well. But the Picard group of a complement of an irreducible hypersurface of degree $d$ in $Gr(r,n)$ is isomorphic to $\mathbb{Z}/d\mathbb{Z}$, and $O(-1)$ is a nontrivial element there.


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