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I would like to see some examples of how to compute Chern classes of blow-ups via the formula

$$c(Y')-f^{*}c(Y)=j_{*}(g^{*}c(X)\cdot \alpha)$$,

which appears on page 300 of Fulton's intersection theory. Here $Y'$ is the blowup of $Y$ along $X$, the f,g, and j are from the usual blow-up diagram and

$$\alpha=\frac{1}{\zeta}\left(\sum_{i=0}^{d}g^{*}c_{d-i}(N)-(1-\zeta)\sum_{i=0}^{d}(1+\zeta)^{i}g^{*}c_{d-i}(N)\right)$$

where $d$ is the rank of the normal bundle $N$ of $X$ in $Y$ and $\zeta=c_{1}(\mathscr{O}(1))$. Thanks.

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DZN -- I've tried to fix the LaTex. –  algori Oct 3 '11 at 18:52
    
Thanks (I don't know why it always gives me problems). –  DZN Oct 3 '11 at 20:35
    
DZN -- Mathjacks, the software MO uses to process LaTeX, doesn't seem to like asterisks very much. I hope I got the brackets right in the second formula. –  algori Oct 3 '11 at 21:10
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1 Answer

Suppose we have finite-dimensional complex vector spaces $U$ and $V$, and we consider the projective spaces $X=P(U)$ and $Y=P(U\oplus V)$. The blowup $Y'$ is then $$ Y' = \{ (L,M) \in P(U\oplus V) \times P(V) : L \leq U\oplus M\}. $$ The map $f:Y'\to Y$ is just $f(L,M)=L$. The exceptional divisor is $$ X' = \{ (L,M) : L\in P(U), M\in P(V) \} = P(U)\times P(V). $$ Let $x$ be the first Chern class of $L$ in the evident sense, and let $y$ be the first Chern class of $M$. Put $n=\dim(U)$ and $m=\dim(V)$. Note that $Y'$ can be regarded as the projective bundle of the tautological bundle plus $U$ over $P(V)$, which determines its cohomology. We have

\begin{align*} H^\ast(X) &= \mathbb{Z}[x]/x^n \\\\ H^\ast(X') &= \mathbb{Z}[x,y]/(x^n,y^m) \\\\ H^\ast(Y) &= \mathbb{Z}[x]/x^{n+m} \\\\ H^\ast(Y') &= \mathbb{Z}[x,y]/(x^n(x-y),y^m) \\\\ \end{align*}

The tangent bundle to $X=P(U)$ is $\text{Hom}(L,U)-1$, and using this we get $c(X)=(1+x)^n$ (or at least $(1\pm x)^{\pm n}$, depending on your conventions). Similarly $c(Y)=(1+x)^{n+m}$ and $c(X')=(1+x)^n(1+y)^m$. If we regard $Y'$ as a bundle over $P(V)$ then the horizontal tangent bundle is $\text{Hom}(M,V)-1$ and the vertical tangent bundle is $\text{Hom}(L,M+U)-1$ which I think gives $c(Y')=(1+x)^n(1+y)^m(1+x-y)$.

This gives most of what you want. I don't have time to write more just now.

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