# 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.

• The only thing that comes to my mind is that one could consider the Segre embedding of $M$ into $\mathbb P^8$ and remark that the diagonal $\Delta$ gets sent to the image of the Veronese embedding $v_2:\mathbb P^2\to \mathbb P^8$ by the quadratic monomials. That said, I hasten to add that I cannot solve your great question. Sep 14 '13 at 14:38

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.
1. Restriction of $\mathcal{L}$ to the diagonal is isomorphic to $\omega_X$, that could be checked directly.
2. $H^i(X\times X, \mathcal{L}) \cong 0$ 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.