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

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.

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 $$Z_p$$).