2
$\begingroup$

Let $V$ be a $k$-vector space with basis $v_1,\dots,v_n$. Let $G(1,V)$ be the Grassmannian of $1$-dimensional subspaces of $V$. In their book on intersection theory, in section 3.2.3, Eisenbud & Harris write that the vector bundle on $G(1,V) \cong \mathbb{P} V $, whose fiber at every point is the corresponding line of $V$, is the one induced by $\mathcal{O}_{\mathbb{P} V}(-1)$. Then they say that the vector bundle on $G(n-1,V) \cong \mathbb{P}V^*$, whose fiber at every point is the quotient of $V$ by the corresponding hyperplane, is $\mathcal{O}_{\mathbb{P} V^*}(1)$.

I have tried to understand these statements by applying the correspondence between locally free sheaves and vector bundles, as described in exercise II.5.18 in Hartshorne, but the role of the twists by $1$ and $-1$ is still unclear to me. For example, in the first case, if we look at the morphism $$\operatorname{Spec} \operatorname{Sym} \mathcal{O}_{\mathbb{P} V}(-1) \longrightarrow \mathbb{P} V$$ locally at the open set $D_+(v_1^*)$ then we get $$\operatorname{Spec} (k[v_1^*,\dots,v_n^*]_{v_1^*})_{\le 0} \rightarrow \operatorname{Spec} k[v_1^*,\dots,v_n^*]_{(v_1^*)}$$ The fiber over a closed point certainly looks like a line (with $k$ algebraically closed), but so it would if we had started with $-1$ replaced by $1$; in that case we would have $k[v_1^*,\dots,v_n^*]_{v_1^*})_{\ge 0}$ instead of $k[v_1^*,\dots,v_n^*]_{v_1^*})_{\le 0}$. So it seems i am missing something basic here and i would be very thankful if someone could kindly shed some light.

$\endgroup$

1 Answer 1

2
$\begingroup$

Question: "So it seems i am missing something basic here and i would be very thankful if someone could kindly shed some light."

Answer: If $V:=k\{e_0,..,e_n\}$ and $V^*:=k\{x_0,..,x_n\}$ let $S:=Sym_k^*(V^*)=k[x_0,..,x_n]$ It follows $P^n:=Proj(S)$ is projective $n$-space. There is a canonical map

$$\pi: P^n \rightarrow Spec(k)$$

and an injection

$$T1.\text{ }\mathcal{O}_{P^n}(-1) \rightarrow \pi^*(V) \cong V\otimes \mathcal{O}_{P^n}.$$

There is a canonical map

$$\phi: V^*\otimes_k k[x_0,..,x_n] \rightarrow k[x_0,..,x_n](1)$$

defined by

$$\phi(f_0,f_1,..,f_n):=f_0x_0+\cdots +f_nx_n\in k[x_0,..,x_n](1).$$

The map $\phi$ defines a surjective map

$$(T1)^*\text{ }V^* \otimes \mathcal{O}_{P^n} \rightarrow \mathcal{O}_{P^n}(1)$$

which is dual to $T1$.

If $x\in P^n(k)$ is a $k$-rational point and you look at the fiber

$$l_x:=\mathcal{O}_{P^n}(-1)(x) \rightarrow V$$

it follows the line $l_x \subseteq V$ is the line corresponding to the point $x$. If $L:=\mathcal{O}_{P^n}(-1)$ and you consider $T:=Sym_{\mathcal{O}}^*(L^*)$ you get the geometric vector bundle $\mathbb{V}(L^*):=Spec(T)$

$$\eta: \mathbb{V}(L^*) \rightarrow P^n$$

corresponding to $L$. You must somehow embed $\mathbb{V}(L^*) \subseteq \mathbb{A}^{n+1}_k \times P^n.$ Dualize $T1$ to get a surjection

$$V^*\otimes \mathcal{O}_{P^n} \rightarrow \mathcal{O}(1):=L^*$$ and take the symmetric algebra

$$ Sym_{\mathcal{O}}^*(V^* \otimes \mathcal{O}_{P^n}) \rightarrow Sym_{\mathcal{O}}^*(L^*)$$

and take relative Spec you get - since taking spec of a surjection gives a closed immersion - a closed subscheme

$$\mathbb{V}(L^*):=Spec(Sym_{\mathcal{O}}^*(L^*)) \subseteq Spec(Sym_{\mathcal{O}}^*(V^* \otimes \mathcal{O}_{P^n})) \cong \mathbb{A}^{n+1}\times_k P^n \rightarrow P^n.$$

Since $V^* \otimes\mathcal{O}_{P^n}$ is a trivial bundle of rank $n+1$ it follows

$$Spec(Sym_{\mathcal{O}}^*(V^*\otimes \mathcal{O}_{P^n}))\cong \mathbb{A}^n_k \times_k P^n.$$

The induced map is the map $\eta$ and you realize $\mathbb{V}(L^*)$ as a sub-scheme of a product. Then you must interpret the embedding $\mathbb{V}(L^*) \subseteq \mathbb{A}^n_k\times_k P^n$ geometrically: The $k$-rational points

$$(v, [l])\in \mathbb{V}(L^*)(k) \subseteq (\mathbb{A}^{n+1}_k \times_k P^n)(k) \cong \mathbb{A}^{n+1}_k(k) \times P^n(k)$$

should correspond to a line $l\in V$ and a vector $v\in l$. The $k$-rational points $P^n(k)$ of $P^n$ correspond 1-1 to lines $l \subseteq V$ - the projective space $P^n$ is a "parameter space" parametrizing lines in $V$. Moreover

$$\mathbb{A}^{n+1}_k:=\mathbb{V}(V^*):=Spec(Sym_k^*(V^*)):=Spec(k[x_0,..,x_n]),$$

hence a $k$-rational point $v\in \mathbb{V}(V^*)$ corresponds 1-1 to a vector $v\in V$: Given a $k$-rational point $v^*\in \mathbb{V}(V^*)(k)$ you get a surjective map of $k$-algebras

$$v^*:k[x_0,..,x_n] \rightarrow k$$

and a corresponding vector $v \in V$

$\endgroup$
2
  • $\begingroup$ Thanks for the answer. Can you please explain how T1 is defined? It would help if you explained this algebraically. Also, in $Sym^*_k(V^*)$ i am not sure what you mean by $Sym^*$, it seems $Sym_k(V^*)$ is ok. $\endgroup$
    – Manos
    Apr 20, 2021 at 6:17
  • $\begingroup$ @Manos - I wrote down a definition of the dual of $T1$ - this is more natural. $\endgroup$
    – hm2020
    Apr 20, 2021 at 9:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .