How to understand the projective compactification of a vector bundle? The following question confused me a bit:
Given a rank $n$ vector bundle or locally free sheaf $\mathcal{E}$ on $X$, each fiber of this vector bundle is a vector space of dimension $n$. Therefore, we can compactify it into a projective space $\mathbb{P}^n$. And if we do this compatibly for every fiber we get a $\mathbb{P}^n$-projective bundle.
Now my question is whether this projective bundle is $\mathbb{P}(\mathcal{E}\oplus\mathcal{O}_X)$ or $\mathbb{P}(\mathcal{E}^{\vee}\oplus\mathcal{O}_X)$. Here by $\mathbb{P}(\mathcal{F})$ for some locally free sheaf $\mathcal{F}$ I mean the projective bundle of hyperplanes (instead of the projective bundle of lines).
At first I thought the answer should be $\mathbb{P}(\mathcal{E}\oplus\mathcal{O}_X)$. Then something seems to be wrong. If I consider the zero section of the vector bundle $
\mathcal{E}$, it can also be viewed as a section of the projective bundle $\mathbb{P}(\mathcal{E}\oplus\mathcal{O}_X)$. Then it should corresponds to the surjection $\mathcal{E}\oplus\mathcal{O}_X \rightarrow \mathcal{O}_X\rightarrow 0$ by sending $\mathcal{E}$ to $0$ because what else it can be.
However if I consider a concrete example, namely the blowing up of $\mathbb{P}^2$ at a point, this is not the case. The blow-up can be viewed as a $\mathbb{P}^1$-projective bundle (i.e. ruled surface) over $\mathbb{P}^1$. Let $X=\mathbb{P}^1$. Then the resulting space of the blow-up is $\mathbb{P}(\mathcal{O}_X(1)\oplus\mathcal{O}_X)$. Therefore, we can also see it as the projective compactification of the line bundle $\mathcal{O}_X(1)$. But then the surjection
$$
\mathcal{O}_X(1)\oplus\mathcal{O}_X\rightarrow \mathcal{O}_X\rightarrow 0
$$
actually determines the section at infinity, which is the exceptional divisor of the blow-up. Instead, for any section $s\in H^0(X,\mathcal{O}_X(1))$, there is a natural map
$$
\mathcal{O}_X\rightarrow \mathcal{O}_X(1)
$$
given by multiplication of $s$. Therefore, when considered as a section of $\mathbb{P}(\mathcal{O}_X(1)\oplus\mathcal{O}_X)$, $s$ is determined by the induced surjection
$$
\mathcal{O}_X(1)\oplus\mathcal{O}_X\rightarrow \mathcal{O}_X(1)\rightarrow 0.
$$
This example makes me believe that the projective compactification of $\mathcal{E}$ is actually $\mathbb{P}(\mathcal{E}^{\vee}\oplus\mathcal{O}_X)$. Now everything seems to be consistent. Let $s\in H^0(X,\mathcal{E})$. Then we have a map
$$
\mathcal{O}_X\rightarrow\mathcal{E} 
$$
given by multiplication of $s$. Take the dual we have a map
$$
\mathcal{E}^{\vee}\rightarrow \mathcal{O}_X.
$$
And the induced surjection
$$
\mathcal{E}^{\vee}\oplus\mathcal{O}_X\rightarrow \mathcal{O}_X \rightarrow 0
$$
defines the section $s$.
It seems a little bit weird that the projective compactification of $\mathcal{E}$ might be $\mathbb{P}(\mathcal{E}^{\vee}\oplus\mathcal{O}_X)$ instead of $\mathbb{P}(\mathcal{E}\oplus\mathcal{O}_X)$. Is there any explanation for this (if I am not completely talking about nonsense)? My guess is that it should have something to do with the fact I am considering projective bundle of hyperplanes. According to my previous experience, if some results involving projective bundle differs by a dual, it is caused by the two different conventions whether it is projective bundle of lines or hyperplanes.
 A: First let me end the suspense: the  projective closure of the vector bundle $\mathbb V(\mathcal E)$ associated to a locally free sheaf $\mathcal E$ on the scheme $X$ is $$\widehat {\mathbb V(\mathcal E)}=\mathbb P(\mathcal E\oplus \mathcal O_X)\quad (\bigstar)$$ (EGA II,Proposition (8.4.2), page 168).
And now here are some explanations relating to this confusing subject:
Given a quasicoherent sheaf $\mathcal E$ on a scheme $X$, Grothendieck associates to it an $X$-scheme $\pi:\mathbb V(\mathcal E)\to X$ by defining $\mathbb V(\mathcal E)=Spec(\mathbb S(\mathcal E))$, where $\mathbb S(\mathcal E)$ is the Symmetric Algebra sheaf associated to the sheaf of $\mathcal O_X$-Modules  $\mathcal E$.
As with all morphisms of schemes one can associate to $\pi$ the $\mathcal O_X$-sheaf $\mathcal S$  of its sections, with  $\mathcal S(U)$ consisting of the morphisms $s:U\to \mathbb V(E)$ such that $\pi \circ s=Id_U$.
The unfortunate result of these definitions however is that $\mathcal S=\mathcal {E}^\vee$, the source of much confusion !
So if you have a geometric vector bundle $E$ on $X$ with associated sheaf of sections $\mathcal E$, beware that $E=\mathbb V(\mathcal E^\vee)$.
And the projective closure of your  $X$-scheme $ E$ is, according to formula $(\bigstar)$, the $X$-scheme  $$\widehat {E}=\widehat {\mathbb V(\mathcal E^\vee)}=\mathbb P(\mathcal E^\vee \oplus \mathcal O_X)$$
