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This is a follow-up to my previous question about chern numbers. Write $\mathbb{P}^n:=\mathbb{P}^n_\Bbbk$ for projective space over some field $\Bbbk$ and assume that $X\subseteq\mathbb{P}^n$ is a linear subvariety, $\mathbb{P}^m\cong X$, say. I now consider the blow-up of $Y:=\mathbb{P}^n$ in $X$, yielding a blow-up diagram $$\begin{matrix} \tilde{X} & \xrightarrow{\; j\;} & \tilde{Y} \\ \hphantom{\scriptstyle g}\downarrow {\scriptstyle g} && \hphantom{\scriptstyle f}\downarrow {\scriptstyle f} \\ X &\xrightarrow{\;i\;} & Y \end{matrix}$$ My question is, what is the second chern class $c_2(\tilde Y):=c_2(\mathcal{T}_{\tilde{Y}})$ of the tangent sheaf of $\tilde{Y}$?

Remark: I am ultimately interested in the degree of $c_2(\tilde Y)c_1^{n-2}(\tilde Y)$.

My thoughts so far: As you can see from my first question, the chern classes of $Y$ (and $X$) have well-known representation, and there is a formula for computing the chern classes of blown-up varieties in Fulton's book Intersection Theory, namely Theorem 15.4. For brevity, I will quote his Example 15.4.3, which gives a formula for $c_2$:

$$c_2(\tilde Y) = f^\ast(c_2(Y)) - j_\ast\left( (d-1) g^\ast(c_1(X)) + \tfrac{d(d-3)}{2} \zeta + (d-2) g^\ast(c_1(\mathcal{N})) \right)$$

Here, $\mathcal{N}=\mathcal{N}_{X/Y}$ is the normal bundle of $X$ in $Y$ and $\zeta$ denotes $c_1(\mathcal{O}_{\tilde{X}}(1))$.

Of course, $c_1(\mathcal{N})=c_1(i^\ast\mathcal{T}_Y)/c_1(\mathcal{T}_X)$ by the well-known exact sequence $$0\to\mathcal{T}_X\to i^\ast\mathcal{T}_Y\to\mathcal{N}_{X/Y}\to 0,$$ but I am not 100% sure what effect the pullbacks and push-forwards have. In the end, I would like to express $c_2(\tilde Y)$ as the sum of intersections of well-known divisors in $\tilde Y$, like the strict transform of a hyperplane and the exceptional divisor.

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I think I was able to solve this with the help of Johannes Nordström on mathoverflow.

Let $\theta$ be the class of the strict transform of a hyperplane in $\tilde Y$ and let $\varepsilon$ be the class of the exceptional divisor. Then, I claim that

$$c_2(\tilde Y) = \binom{n+1}{2} \cdot \theta^2 - \frac{d(d-3)}{2}\cdot\varepsilon^2 + (n+1-dn)(\varepsilon^2+\varepsilon\theta).$$

For the proof, let $\eta:=c_1(\mathcal{O}_X(1))$. For Fulton's Example 15.4.3, we have the formula $c_2(\tilde Y)=f^\ast(c_2(Y))-j_\ast(\delta)$ with

$$\delta=(d-1)g^\ast(c_1(X)) + \frac{d(d-3)}{2}\zeta + (d-2)g^\ast(c_1(\mathcal{N}))$$

Since $X\cong\mathbb{P}^{n-d}$ and $Y=\mathbb{P}^n$, we know from Example 3.2.12 in Fulton's book that $c_1(X)=(n+1-d)\eta$ and $c_1(\mathcal{N})=d\eta$. Thus,

$$\begin{align*} \delta &= (d-1)(s+1-d)\cdot g^\ast(\eta) + d(d-3)/2\cdot\zeta + d(d-2)\cdot g^\ast(\eta)\\ &= (dn-n-1)\cdot g^\ast(\eta) + d(d-3)/2\cdot \zeta. \end{align*}$$

Since $g:\tilde X\to X$ is just the projection map $X\times\mathbb{P}^{d-1}\to X$, we know that $g^\ast(\eta)$ is represented by a divisor that is the preimage of a hyperplane in $X$. Writing that hyperplane as the intersection of $X$ with a transverse hyperplane $H\subseteq Y$, we get $j_\ast g^\ast (\eta) = [\tilde X\cap f^{-1}(H)] = \varepsilon^2 + \varepsilon\theta$. Furthermore, $\zeta$ corresponds to the conormal bundle of $\tilde X$ in $\tilde Y$, so it is the restriction of $-\varepsilon$ to $\tilde X$, in other words $j_\ast(\zeta)=-\varepsilon^2$. This yields

$$j_\ast(\delta) = (dn-n-1)(\varepsilon^2 + \varepsilon\theta) + \frac{d(d-3)}{2}\cdot\varepsilon^2.$$

On the other hand, $f^\ast([H])=\theta$ by Corollary 6.7.2 in Fulton. By Example 3.2.11 (or my previously answered question), we know that $c_2(Y)=\binom{n+1}{2}\cdot [H]^2$. Putting it all together, we get the above formula.

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PS: I'll award the bounty to anyone who can point out a mistake in my reply. That doesn't mean there is one, it's just to make you proof-read it. – Jesko Hüttenhain Mar 9 '12 at 21:30

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