I was looking at the following paper by Tango:


In Lemma 2.4 , at the end, he says that if $Y \subset \mathbb{G}(k,n)$ is a subvariety of the Grassmannian of $k-$planes in $\mathbb{P}^n=\mathbb{P}(V)$ such that for every $y \in Y$ the corresponding $k-$plane $L_y \subset \mathbb{P}^n$ passes through a fixed point $p \in \mathbb{P}^n$ then the restriction of the universal bundle $$S=\{(x,L_x)|\:x \in \mathbb{G}(k,n)\} \subset V \otimes \mathcal{O}_{\mathbb{G}(k,n)}$$ to $Y$, i.e. $S_{|Y}$, has a trivial subbundle as a direct summand.

I understand that the fact that every $L_y$ passes through a fixed point $p$ gives me a section of $S_{|Y}$ simply by associating to every $y \in Y$ the point $p \in L_y$. This gives me the exact sequence $$0 \rightarrow \mathcal{O}_Y \rightarrow S_{|Y} \rightarrow S' \rightarrow 0$$ where $S'$ is a vector bundle with $rank(S')=rank(S_{|Y})-1$.

But I'm not able to see why this implies that $S_{|Y}=\mathcal{O}_Y \oplus S''$ for some $S''$.

Am I missing something? Thanks in advance for the help.


1 Answer 1


Let $Z_p \subset \operatorname{Gr}(k,n)$ be the subscheme parameterizing all subspaces parameterizing all $k$-planes containing $p$. Then $Z_p \cong \operatorname{Gr}(k-1,n-1)$ and the restriction of the tautological bundle to $Z_p$ splits as the sum of $\mathcal{O}$ and the tautological bundle $S'$ of $\operatorname{Gr}(k-1,n-1)$.

Indeed, let $V_1 \subset V$ be the 1-dimensional subspace corresponding to the point $p$. Let $V = V_1 \oplus V'$ be a direct sum decomposition. Then for each $k$-dimensional subspace $U' \subset U$ the sum $V_1 \oplus U'$ is a $(k+1)$-dimensional subspace of $V$, hence the corresponding subbundle $$ V_1 \otimes \mathcal{O} \oplus S' \subset V_1 \otimes \mathcal{O} \oplus V' \otimes \mathcal{O} = V \otimes \mathcal{O} $$ induces a morphism $\operatorname{Gr}(k-1,n-1) \to \operatorname{Gr}(k,n)$ which is an isomorphism onto $Z_p$ and such that the pullback of the tautological bundle is $V_1 \otimes \mathcal{O} \oplus S'$.

Now, it remains to note that $Y \subset Z_p$, hence the restriction of $S$ to $Y$ splits (because it already splits on $Z_p$).

  • $\begingroup$ Thank you very much for the fast response! $\endgroup$
    – gigi
    Jun 16, 2021 at 12:44

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