12

We recall the following facts about total spaces of bundles: Let $X$ be a scheme and $\mathcal{E}$ an invertible sheaf on $X$. The total space of $\mathcal{E}$, $\Bbb{V}(\mathcal{E})$ is defined as $\textbf{Spec} \operatorname{Sym}^\bullet (\mathcal{E}^\vee)$ where $\textbf{Spec}$ is the relative spectrum.

Now I have read many times that if $X = \Bbb{P}^1_k$ and $\mathcal{E} = \mathcal{O}(1)$ then $\Bbb{V}(\mathcal{O}(1))$ is the same as the blow-up of $\Bbb{A}^2_k$ at the origin. How can we see this?

My understanding: Let $x_1,x_2$ be the coordinates on $\Bbb{P}^1$ I have tried to see what the projection $\Bbb{V}(\mathcal{O}(1)) \to \Bbb{P}^1$ looks like. It seems to me in the chart $D_+(x_1)$ this is given by $$\operatorname{Spec} \bigoplus_{n \leq 0} k[x_1,x_2]_{x_1}(n) \to \operatorname{Spec} k[x_2/x_1].$$ via the map $ k[x_2/x_1] \to \bigoplus_{n \leq 0} k[x_1,x_2]_{x_1}(n) $ that is division by $x_2$. 

<p>On the other hand, consider $\text{Bl}_{\Bbb{A}^2}(0,0)$. Let $y_1,y_2$ be the coordinates on $\Bbb{A}^2$. I compute that over $D_+(x_1) \subseteq \Bbb{P}^1$ the projection map is given by 
$$\operatorname{Spec} k[y_1, x_2/x_1] \to \operatorname{Spec} k[y_2/y_1].$$</p>

It looks to me that based on these computations they are the same thing (I looked at how open sets glue too). Is there a nice coordinate free way to see they are the same?

  • 6
    I think you're missing a dualisation somewhere. The total space of the tautological line bundle $\mathscr{O} (-1)$ on $\mathbb{P}^1$ is the blowup of affine plane at the origin, and accordingly there are no global sections. But $\mathscr{O} (1)$ has plenty of global sections. – Zhen Lin Apr 09 '14 at 09:19
  • @ZhenLin Ok I forgot to take the dual. Let me edit the post. –  Apr 09 '14 at 09:29
  • @ZhenLin I have edited my post now. Is it ok? –  Apr 09 '14 at 09:32
  • @Zhen Lin: could you briefly explain why there are no sections $s: \mathbb{A^2} \to Bl_0(\mathbb{A^2})$ on the blowup of affine plane at the origin? – user267839 Nov 11 '20 at 22:34

1 Answers1

4

I think you could use the universal property of the Blow-up and the fact that $\mathbb{V}(\mathcal{O}(1))$ represents the functor $(h:T\to X )\mapsto \Gamma(T,h^*\mathcal{O}(1)^\vee)$. I do not take the dual in the definition of $\mathbb{V}$ though.

more details: the zero section $\mathbb{P}^1\to\mathbb{V}(\mathcal{O}(1))$ shows that via the universal property of $B:=\text{Bl}_0(\mathbb{A}^2)$ that there is a unique $\mathbb{A}^2$-morphism $\mathbb{V}(\mathcal{O}(1))\to B$. The blow-up $B$ is via the restriction of the second projection $q$ of $\mathbb{A}^2\times\mathbb{P}^1$ a $\mathbb{P}^1$ scheme. One can show $\Gamma(B,q^*\mathcal{O}(1)^\vee)$ is zero and thus there is a unique $\mathbb{P}^1$ morphism $B\to \mathbb{V}(\mathcal{O}(1))$. You can use this to show that there is a unique $\mathbb{A}^2$ morphism $B\to \mathbb{V}(\mathcal{O}(1))$ and thus $\mathbb{V}(\mathcal{O}(1))$ is isomorphic to the blowup $B$ up to unique isomorphism.

EDIT: as an alternative, you could consider transition functions. The blow up $B$ is given by the $\mathbb{P}^1$-scheme $V_+(T_ix_j=T_jx_i)$ in $\mathbb{P}^1\times \mathbb{A}^2$, $T_i$ being homogeneous coordinates on $\mathbb{P}^1$, $x_i$ coordinates on $\mathbb{A}^2$. Locally on $D_+(T_i)$, $B$ is given by the equation $\frac{T_j}{T_i}x_i=x_j$, so there is an isomorphism $$B_{|D_(T_i)}\cong D_+(T_i)\times \mathbb{A}^1$$ and one checks the transition functions are $ g_{ij}=\frac{T_i}{T_j}$. These are the same as the transition functions of $\mathbb{V}(\mathcal{O}(1))$.

Carsten
  • 673