What happens to the cohomology ring after a “flip-flop”?

I've been trying to understand what happens to the cohomology ring (say with coefficients in $\mathbb{R}$) of a smooth complex projective manifold after blowing up along a smooth complex submanifold. I would first just like to double check my facts, and then I have some other general, perhaps naive, questions about blowing down.

For blowing up, I would just first like to know if I've got my story straight. The following is what I've gathered mostly from Griffiths' and Harris' book (pgs 605-608):

Let $Y\subset X$ be a complex submanifold of codimension $n$ embedded in a complex projective manifold of dimension $d$, let $\tilde{Y}\subset\tilde{X}$ be the exceptional divisor sitting inside the blow up manifold, and let $\beta\colon\tilde{X}\rightarrow X$ be the blow down map.

Then $\tilde{Y}$ is isomorphic to $\mathbb{P}\left(\mathcal{N}_{Y/X}\right)$, the projectivization of the normal bundle of $Y\subset X$, and the restriction of blow down map $\beta_0\colon\tilde{Y}\rightarrow Y$ is the usual projection map $\mathbb{P}\left(\mathcal{N}_{Y/X}\right) \rightarrow Y$. In particular, the fibers of $\beta_0$ are copies of $\mathbb{CP}^{n-1}$, and there is a class $z\in H^2(\tilde{Y})$ that restricts to each fiber to give minus the hyperplane class $H^2(\mathbb{CP}^{n-1})$--this can be taken as the first Chern class of the normal bundle to $\tilde{Y}\subset\tilde{X}$. Thus by the Leray-Hirsch theorem we get an $H(Y)$-module decomposition $$H(\tilde{Y})\cong H(Y)\oplus z H(Y)\oplus\cdots z^{n-1}H(Y).$$

If we take the Poincare dual of the fundamental class of the exceptional divisor, say $\tau\in H^2(\tilde{X})$, then we get an $H(X)$-module decomposition $$H(\tilde{X})\cong H(X)\oplus \tau H(Y)\oplus\cdots\tau^{n-1} H(Y)$$ and the restriction of $\tau$ to $\tilde{Y}$ is equal to $z$.

Question 0: Do I have my story straight?

Also,

Question 1: $\tilde{X}$ is smooth and projective too, right?

Now for blow downs: I believe that in some special cases you can "blow down $\tilde{X}$ along $\tilde{Y}$ in the horizontal direction" to get another complex manifold $W$, different from $X$, and I think that this means that there is some complex submanifold $Z\subset W$ for which the blow up of $W$ along $Z$ is also equal to $\tilde{X}$ with the same exceptional divisor $\tilde{Y}$ realized now as the projectivized normal bundle of $Z\subset W$.

I guess I'm most used to thinking about polytopes and toric varieties. So for example take $\tilde{X}$ to be the toric variety corresponding to a truncated square pyramid, where moving one pair of opposite sides would result in one triangular prism (corresponding to a toric variety $X$), and moving the other pair would result in a different prism (corresponding to a different variety $W$). Here $Y\subset X$ and $Z\subset W$ would be the $\mathbb{CP}^1$'s corresponding to the top edge of their respective prisms and $\tilde{Y}$ would be the product $\mathbb{CP}^1\times\mathbb{CP}^1$.

Question 2: Are there general conditions on $Y\subset X$ that guarantee that we can blow up $X$ along $Y$, then blow down $\tilde{X}$ along $\tilde{Y}$ "in the horizontal direction" to get a new manifold $W$ as in the toric variety example?

From the example above, it seems like under such conditions we should always have:

1. $\tilde{Y}\cong\mathbb{CP}^{m-1}\times\mathbb{CP}^{n-1}$ (where $m$ is the codimension of $Z\subset W$),

2. and $H(\tilde{Y})\cong H(\mathbb{CP}^{m-1})\otimes_{\mathbb{R}} H(\mathbb{CP}^{n-1})\cong \mathbb{R}[x,y]/\langle x^m,y^n\rangle$ (where $x\in H^2(\mathbb{CP}^{m-1})$ and $y\in H^2(\mathbb{CP}^{n-1})$ are the hyperplane classes).

Question 3: In this case, can we explicitly identify the Chern class $z$ as an element of $\mathbb{R}[x,y]/\langle x^m,y^n\rangle$?

The condition that $z$ restricts to every fiber to give minus the hyperplane class seems to indicate that $z=-x-y$? Is that correct or am I missing something?

Finally I have one more general question: I think it's the case that every smooth toric variety can be obtained from a complex projective space by a finite sequence of blow ups and blow downs.

Question 4: What other smooth projective manifolds can be obtained from a projective space this way? Do such things have a name?

References and/or partial answers (and of course corrections) are welcome.

There are many questions here. Let me focus on Question 4.

The answer is all smooth rational varieties. This is a consequence of the following

Weak Factorization Theorem. A birational map between complete nonsingular varieties over an algebraically closed field $\mathbb{K}$ of characteristic zero is a composite of blowings up and blowings down with smooth centers.

References are:

1 D. Abramovich, K. Karu, K. Matsuki and J. Wlodarczyk: Torification and factorization of birational maps, J. Amer. Math. Soc. 15 (2002), 531-572

2 J. Wlodarczyk: Toroidal varieties and the weak factorization theorem, Invent. Math. 154, Issue 2 (2003), 223-331.

• Yes, thanks! This is exactly what I was looking for in Question 4. One question, and please pardon my ignorance: does "blowings up and blowings down with smooth centers" mean that the same exceptional divisor "created" in blowing up is also "destroyed" in blowing down, as I was attempting to describe between Questions 1 and 2? Or does it mean something totally different? – Chris McDaniel Feb 27 '15 at 0:49
• I'm afraid I do not completely understand your question. What do you mean by "distroyed"? In general, I guess that anything can happen. By the way, "weak" factorization means that blow-ups and blow-downs are mixed together, i.e. in general you cannot perform all the blow-ups first and all the blow-downs later. – Francesco Polizzi Feb 27 '15 at 9:18
• By "destroyed" I meant "gets contracted in the blowing down"...sorry I'm rather uninitiated with the jargon so I was trying to guess what "blowings up and blowings down with smooth centers" means, but now I'm guessing that it just means that all blowing ups are taken along smooth subvarieties...again please pardon my ignorance...and thanks again for your answer! – Chris McDaniel Feb 27 '15 at 16:11