Is the degree of a del Pezzo surface same as the usual definition of the degree?

Question

In the definition of del Pezzo surface I'm looking and at they stat that the degree $d$ of a del Pezzo surface $X$ is by definition the self intersection number $(K, K)$ of its canonical class $K$. Is this the same as the usual notion of degree?

Answer 1

Yes. First, we should standardize what we mean by degree; for me, the most useful definition is the cardinality of $X \cap H^2$, where $H$ is the class of a generic hyperplane so $H^2$ is the class of a generic codimension $2$ linear space (and of course for a variety of dimension $n$ you would take the $n$-th power of $H$ instead) .

A del Pezzo surface is embedded $i:X \hookrightarrow \mathbb P^d$ by its complete anticanonical system $|-K_X|$: in other words, $i^* \mathcal O_{\mathbb P^d} (1) \cong \omega_X^\vee$. Hence the restriction of $H$ to $X$ is just $-K_X$, and the self-intersection is $(-K_X)^2 = (-1)^2 K_X^2 = K_X^2 = (H|_X)^2 = H^2|_X = \deg(X)$.

So in a nutshell, this comes down to 1) the fact that this is the degree of the anticanonical embedding, and 2) the fact that a divisor class on a surface and its dual (or negative) have the same self-intersection, which itself is just because $(-1)^2 = 1$ rather than any sophisticated geometry $:)$