Why is the standard flop a flop?

Question

I have seen at least two ways to define flops (and similarly flips). We start with $Y \to X$, a surjective birational morphism, contracting a locus of codimension at least 2, such that $K_Y$ is relatively trivial.

1. We start with $\pi\colon Y \to X$ and take $W=\text{Proj}(\bigoplus \pi_*\mathcal{O}(mH))$ for $H$ relatively ample or anti-ample.

2. It is the unique $W \to X$ satisfying the same assumptions as $Y \to X$ but such that $W$ is not isomorphic to $Y$.

Why are these two definitions equivalent? (for flips one has instead $K_Y$ is relatively anti-ample and $K_W$ ample)

There is also a special class of flops, sometimes called standard.

3. Let $C$ be a $(-1,-1)$ curve in $Y$. Blow it up. The exceptional locus is $P^1 \times P^1$. Contract in the other direction and obtain $W$.

Why is this equivalent to other two definitions?

EDIT: I've edited the question following SK's answer. I guess what I really would like to understand is the line in SK's answer which says: "The fact that there are only these two possible maps follows from the fact that one of them is $\text{Proj}(\bigoplus \pi_* \mathcal{O}_Y(−mH))$ and the other is $\text{Proj}(\bigoplus \pi_*\mathcal{O}_Y(mH))$ and one proves that any such small morphism has to be one of them." How do you prove that?

Answer 1

It looks like you are mixing up flips and flops. 1) looks more like a flip, though not really the usual definition. 2) is indeed a flop, but this is also not the usual definition. For certainty look up the definition in Kollár-Mori 98. (Also, flips go from $K$ being anti-ample to ample, not the other way around. This is actually important!!)

The essence of flips and flops is the following: As you say start with a small projective birational morphism $\pi:Y\to X$ and let $-H$ be a $\pi$-ample line bundle on $Y$. Then the $H$-flip of $\pi$ is another small projective birational morphism (isomorphic over the same open set as $\pi$) $\pi^+:Y^+\to X$ such that if $H^+$ denotes the strict transform of $H$ on $Y^+$, then $H^+$ is $\pi^+$-ample.

Now traditionally a $K_Y$-flip is called a flip and an $H$-flip for some $H$ when $K_Y$ is $\pi$-trivial is called a flop.

The fact that there are only these two possible maps follows from the fact that one of them is $\mathrm{Proj}(\oplus \pi_*\mathscr O_Y(-mH))$ and the other is $\mathrm{Proj}(\oplus \pi_*\mathscr O_Y(mH))$. Actually, it is probably even better to say that if $H^+$ is $\pi^+$-ample, then $X^+$ is isomorphic to $\mathrm{Proj}(\oplus \pi_*\mathscr O_Y(mH))$ regardless any other condition. This proves that the $H$-flip is unique. (However, this does not mean that these actually obviously exist, because you need these rings to be finitely generated).

To get that there are no other small morphisms to compete for being the flip/flop of $\pi$ one needs to assume that the relative Picard number of $\pi$ is $1$. In practice, this is usually a given from the way these small morphisms arise. With that assumption, if $\pi$ is $K$-trivial, then no matter what $H$ one picks, the corresponding $H$-flip (i.e., a flop of $\pi$) will be the same.

For number 3, the point is that that $C$ is contractible and so is the curve that $\mathbb P^1\times \mathbb P^1$ contracts to on $W$ and those contractions form a flop, not the blow up, that's just a way to construct it. I suppose it is called standard, because this construction was known before flips and flops were invented and no one really cared about contracting $C$, because it is much nicer to blow it up. However, this method doesn't always work.