Consider a projective bundle $\pi:\mathbb P(E)\to X$ that comes from a vector bundle of $E$ on a variety $X$. My question is about what is the line bundle "$\mathcal{O}(1)$" on $\mathbb P(E)$. To my understanding, this line bundle defines $\mathcal{O}(1)$ on each fiber and satisfies some universal properties, and should be unique.
However, in Hartshorne (Cha.2, Lemma 7.9), it says that when twisted by a line bundle $\mathcal{L}$, $\mathbb P(E)\to \mathbb P(E\otimes L)$ is isomorphic canonically, and the isomorphism pullbacks $\mathcal{O}_{\mathbb P(E\otimes L)}(1)$ to $\mathcal{O}_{\mathbb P(E)}(1)\otimes \pi^*L^{-1}$.
This seems to imply that $\mathcal{O}_{\mathbb P(E)}(1)$ is not unique: Let's test this on the Hirzeburch surface. Let's consider $X=\mathbb P^1$ and $E=\mathcal{O}(-1)\oplus \mathcal{O}$, then the divisor $S$ that defines $\mathcal{O}_{\mathbb P(E)}(1)$ is the section at infinity, induced from the constant section on $E$. Moreover $S\cdot S=-1$. (cf. Hartshorne, Cha.5 Example 2.11.3.)
However, if we twist $E$ by $L=\mathcal{O}(1)$, then the Lemma above implies the projective bundle is unchanged $$\mathbb P(\mathcal{O}(-1)\oplus \mathcal{O})\cong \mathbb P(\mathcal{O}\oplus \mathcal{O}(1))$$ but sends $S$ to $S_1+F$, where $F$ is a fiber, and $S_1$ defines $\mathcal{O}(1)$ for $\mathbb P(\mathcal{O}\oplus \mathcal{O}(1))$. Since the map preserves intersection pairing, $S_1$ has self-intersection $-3$. So the divisor that associates to $\mathcal{O}(1)$ seems to be dependent the presentation of the vector bundle. Where is the contradiction?
Best Answer
There's no contradiction here - $\mathcal{O}(1)$ depends on $\mathcal{E}$. In fact, if you're working out of Hartshorne, this is called out several times: lemma II.7.9 shows that if $\mathcal{E}$ is a vector bundle and $\mathcal{L}$ is a line bundle on a (sufficiently nice) scheme $X$, then there is a natural isomorphism $\varphi: P' = \Bbb P(\mathcal{L}\otimes\mathcal{E}) \stackrel{\cong}{\to} \Bbb P(\mathcal{E}) = P$ and $\mathcal{O}_{P'}(1) \cong \varphi^*\mathcal{O}_P(1) \otimes (\pi')^*\mathcal{L}$ for $\pi':P'\to X$; exercise II.7.14 shows that $\mathcal{O}(1)$ need not always be very ample relative to $\pi$; notation V.2.8.1 says that $\mathcal{E}$ for a ruled surface isn't uniquely determined (even when $\mathcal{E}$ is normalized), etc.