Let $X$ be a normal projective variety, and let $\text{Div}(X)$ be the group of Cartier divisors on $X$. I know that $\text{Div}(X)$ is a free abelian group, hence I can consider the associated real vector space $\text{Div}(X)_{\mathbb{R}}$ - which will be of infinite dimension in general -. Moreover, I can consider the set $E(X)$ of effective real Cartier divisors.
There is a natural map $\pi: \text{Div}(X)_{\mathbb{R}}\to \text{N}^1(X)_{\mathbb{R}}$, sending every divisor to its numerical class. Inside $\text{N}^1(X)_{\mathbb{R}}$, which is now a finite dimensional vector space, I can consider the cone of effective divisors $\text{Eff}(X)$, which in general is not closed.
Consider the preimage $\pi^{-1}(\text{Eff}(X))$: is it still a cone? Does it coincide to $E(X)$?
I've thought about it for a while, these are my ideas:
- obviously $E(X)$ is not a vector space, since it is not closed by -negative- scalar multiplication
- I strongly believe $E(X)$ is a cone, since it is closed by addition and -positive- scalar multiplication
- I still don't know though if $E(X)=\pi^{-1}(\text{Eff}(X))$: I think so, but it confuses me if there may exist a non-effective divisor $D$ whose numerical class belong to the effective cone.