I was looking at the following paper by Tango:


<https://projecteuclid.org/journals/journal-of-mathematics-of-kyoto-university/volume-14/issue-3/On-n-1-dimensional-projectlve-spaces-contained-in-the-Grassmann/10.1215/kjm/1250523169.full>


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.



Trivial subbundle of universal bundle on the Grassmannian $\mathbb{G}(k,n)$
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 $Z\_p$).


