Can the diagonal of a manifold be expressed as the zero set of a section of a vector bundle? Let $\mathbb{C} \mathbb{P}^2$ be the two dimensional complex 
projective space and
$$M:= \mathbb{C} \mathbb{P}^2 \times \mathbb{C} \mathbb{P}^2.$$
Let $ \gamma \rightarrow \mathbb{C} \mathbb{P}^2 $ be 
the tautological line bundle over $\mathbb{C} \mathbb{P}^2$. Denote 
$\Delta_{M}$ to be the diagonal of $M$, i.e., 
$$ \Delta_{M} := \{ (p_1,p_2) \in \mathbb{C} \mathbb{P}^2 \times \mathbb{C} \mathbb{P}^2: p_1 = p_2  \}. $$ 
My question is the following: is there some obvious rank two vector bundle 
$V \rightarrow M$ over $M$ and a section 
$s: M \rightarrow V$ of this vector bundle such that 
$$ s^{-1}(0) = \Delta_{M} $$
?
Moreover $V$ restricted to $\Delta_M$ should be the normal bundle of 
$\Delta_M$ inside $M$. 
It seems to me $V$ should be something obvious that is made up using 
$\pi_1^{*} \gamma$ and $\pi_2^*\gamma $ (and their duals). The section 
$s$ should essentially be 
$$ s(p_1, p_2) = p_1 - p_2 $$ 
Of course $p_1-p_2$ for the time being makes no sense. 
Note: $\pi_1, \pi_2: M \rightarrow \mathbb{C} \mathbb{P}^2$ are the 
two projection maps.  
 A: Let me denote $X=\mathbb{P}^2$, $\mathcal{I}$ ideal of the diagonal, we have a short exact sequence
$$
0 \to \mathcal{I} \to \mathcal{O}_{X \times X} \to \mathcal{O}_\Delta \to 0.
$$
Let $\mathcal{L}=\pi_1^*(\mathcal{O}(-1)) \otimes \pi_2^*(\mathcal{O}(1) \otimes \omega_X)$ be a line budle. This line bundle has two properties that we need for construction.


*

*Restriction of $\mathcal{L}$ to the diagonal is isomorphic to $\omega_X$, that could be checked directly.

*$H^i(X\times X, \mathcal{L}) \cong 0$ for any $i$, by Serre duality and Kunneth formula.
Let apply finctor $Hom(\bullet, \mathcal{L})$ to the short exact sequence, this gives long exact sequence
$$
\ldots \to H^{n-1}(X \times X, \mathcal{L}) \to Ext^{n-1}(\mathcal{I}, \mathcal{L}) \to Ext^n(\mathcal{O}_\Delta, \mathcal{L}) \to H^{n-1}(X \times X, \mathcal{L}) \to \ldots
$$
thus we get an isomorphism $Ext^{1}(\mathcal{I}, \mathcal{L}) \cong Ext^2(\mathcal{O}_\Delta, \mathcal{L})$, and $Ext^2(\mathcal{O}_\Delta, \mathcal{L}) \cong H^2(X, \omega_X) \cong \mathbb{C}$, so lets take an extension that corresponds to the generator. Then by Serre construction we get 
$$
0\to \mathcal{L} \to V^* \to \mathcal{I} \to 0.
$$
where $V^*$ is locally free and its dual is the bundle you are looking for, moreover last exact sequence is exactly Koszul resolution.
