# homology class of a rational curve

Let $X\subseteq\mathbb{C}P^n$ be s smooth variety. Let $C\subseteq X$ be an algebraic rational curve [i.e. an algebraic curve which is birational to $\mathbb{C}P^1$].

In what follows, we consider $H_2(X;\mathbb{Q})$.

I think that I was told that the following statement is true:

There exist rational numbers $r_1,\ldots,r_k$ and holomorphic curves $u_1,\ldots,u_k:\mathbb{C}P^1\longrightarrow X$ so that $$[C]=r_1(u_1)_*[\mathbb{C}P^1]+\ldots+r_k(u_k)_*[\mathbb{C}P^1].$$

Why?

• Perhaps I'm misreading your question, but could you not simply take $u_1 \colon \mathbb CP^1 \to C \to X$ to be the normalization of $C$, so that $[C] = (u_1)_*[\mathbb CP^1]$? Commented May 23, 2013 at 22:21
• Thank you. You are correct. So your argument is: let $\si:X\longrightarrow C$ be a normalization of $C$. Since $C$ is birational to $\mathbb{C}P^1$, $X$ is birational to $\mathbb{C}P^1$. Since the genus is invariant under birational transformations, $X$ has genus $0$ so it is (biholomorphic to) $\mathbb{C}P^1$, so we are indeed done. Is there any direct way of proving that $X$ is (biholomorphic to) $\mathbb{C}P^1$? [i.e. without invoking the invariance of the genus under birational transformations] Commented May 24, 2013 at 1:23
• $X$ is not birational to $\mathbb P^1$ Commented May 24, 2013 at 8:04
• Why not? Isn't the normalization a birational transformation? If so, then $X$ is birational to $C$. Furthermore, birational equivalence is an equivalence relation and hence transitive. Where am I making a mistake? Commented May 24, 2013 at 18:46
• if $X$ were birational to a curve, it would already be a curve! Commented May 26, 2013 at 22:08