# Universal quotient bundles of $G(2,4)$ and $\mathbb{G}(1,\mathbb{P}^3)$

Let $V$ be an $n$-dimentional complex vector space, $G=G(k,V)$ the Grassmannian of $k$-planes in $V$, and let $\mathcal{V}:=V \otimes \mathcal{O}_G$ the rank-$n$ trivial vector bundle on $G$. We denote with $\mathcal{S}$ the rank $k$ subbundle of $\mathcal{V}$ whose fiber at a point $[\Lambda] \in G$ is the subspace $\Lambda$ itself. Let $\mathcal{Q}$ the quotient bundle $\mathcal{V}/\mathcal{S}$. If we have $G=G(2,4)$ (the Grassmannian of $2$-planes in $\mathbb{C}^4$)

1) How can I describe the quotient $\mathcal{Q}$? I think that this is a $2$-vector bundle (not trivial).

It is well-know that $G(2,4) \simeq \mathbb{G}(1,\mathbb{P}^3)$.

2) In this situation how can I describe the quotient bundle $\mathcal{Q}$?

We know that if $X$ is a compact manifold every $n$-vector bundle on $X$ can be realized as the pull-back of the tautological bundle $\gamma^n \to G(n,k)$ via the classifying map $f:X \to G(n,k)$. Is there a similar result when $X$ is a scheme? Thank you.

• As for $\mathcal Q$, over $[\Lambda]\in G(2,4)$ there lies $V/\Lambda$.
• Let $$\mathcal L\subset \mathbb G(1,3)\times\mathbb P^3\overset{\pi}{\longrightarrow}\mathbb G(1,3)$$ be the universal line. This means that $\pi^{-1}([\ell])=\ell$. Here, to give $\ell$ is to give $\mathbb P(\Lambda^\vee)$, exactly as to give $\mathbb P^3$ is to give $\mathbb P(V^\vee)$. To say that $\mathcal L$ is the universal line means just that $$\pi^{-1}([\ell])=\mathcal S_{[\ell]}.$$ For the quotient bundle on $\mathbb G(1,3)$, this would translate as $$\mathcal Q_{[\ell]}=\mathbb P((V/\Lambda)^\vee).$$ (The appearance of the dual is due to the fact that in Algebraic Geometry the space of lines in the vector space $V$ is $\mathbb P(V^\vee)$.)
• To give a map $f$ from a scheme $X$ to $G(k,n)$ is to give a rank $k$ vector subbundle of $V\otimes \mathscr O_X$. The correspondence is $$f\longleftrightarrow f^\ast\mathcal S,$$ where $\mathcal S$ is the universal subbundle. It is the universal property of the Grassmannian.