# Degree of a Divisor = Self Intersection?

This has been bugging me for some time now. I know that in certain cases that the self-intersection of divisors and it's degree are the same. Like hyerplanes in projective space. I sometimes read that certain degrees are actually defined as the self-intersection.

Is this always the case, are self-intersection and degree basically the same thing? And how are they related?

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There are curves with negative self-intersection, for example—how would that play with your guess? –  Mariano Suárez-Alvarez Jul 24 '12 at 13:38
Ahhh i guess that would mean it doesnt work –  Mike Jul 24 '12 at 13:49

The answer is that the self-intersection of a divisor $D$ on a complete smooth surface $S$ and the degree of the divisor are unrelated. Here is why:

The self-intersection of the divisor is intrinsically defined: it is a certain number $D\cdot D\in \mathbb Z$, depending only on $S$ and $D$.
In contrast the degree of $D$ is only defined after you have embedded $S$ into some projective space $\mathbb P^n$.
By artificially embedding $S$ in huge projective spaces (say by a Veronese map) you can obtain arbitrarily large degrees for $D$ , but the self intersection $D\cdot D$ will not change by one iota.
And, as Mariano wisely observes, the self-intersection of an effective divisor can be $\lt 0$, whereas its degree will be $\gt0$ for any embedding $S\hookrightarrow P^n$ .

Caveat
I only mentioned divisors on curves.
One can also define self-intersections for divisors on higher dimensional varieties ; however they will no longer be numbers but elements of a Chow ring.

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I think I know what Mike meant to ask. I think the question was: Let $D \subset X$ be a divisor, and embed $X$ into $\mathbb{P}(H^0(X, \mathcal{O}(D))^{\vee})$ by the standard construction.

Is the degree of $X$ inside this projective space equal to the self-intersection $D^d$?

This is what I have always understood the phrase "the degree of $D$", without further context, to mean.

As Georges points out, another reasonable interpretation is that we already have some embedding $\phi: X \to \mathbb{P}^N$ and we ask about the degree of $\phi(D)$; that has no relation to the self intersection of $D$.

Now, even with the interpretation above, one needs some caveats. An arbitrary divisor $D$ may not give an embedding to projective space. (Indeed, for an arbitrary $D$, $H^0(X, \mathcal{O}(D))$ may be zero.) But, if $X$ does embed in $\mathbb{P}(H^0(X, \mathcal{O}(D))^{\vee})$ (i.e. if $D$ is very ample), the answer is yes. Correctly formulated, one could probably weaken "very ample" to "ample", or maybe to one of the weaker adjectives like "big" or "nef", but I'll just stick to the very ample case, where there is literally an embedding.

Proof: Let $d= \dim X$. The degree of $X$ in $\mathbb{P}(H^0(X, \mathcal{O}(D))^{\vee})$ is the intersection of $X$ with $d$ generic hyperplanes; $X \cap H_1 \cap H_2 \cap \cdots \cap H_d$. Each those hyperplanes pulls back to a divisor on $X$ which is linearly equivalent to $D$. So $X \cap \bigcap H_i = \bigcap (X \cap H_i)$ and the latter has size $D^d$.

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Wow, thanks alot for clarifying! I wish I could accept both answers –  Mike Jul 25 '12 at 6:13