Let $X$ be a nonsingular projective variety over $\mathbb{C}$, and let $\widetilde{X}$ be the blow-up of X at a point $p\in X$. What relationships exist between the degrees of the Chern classes of $X$ (i.e. of the tangent bundle of $X$) and the degrees of the Chern classes of $\widetilde{X}$?



3 Answers 3


For the first Chern class you get the simple formula
$$c_1(\tilde X)=p^*c_1(X)- (n-1)E$$ where $p:\tilde X \to X$ is the projection and $E$ the exceptional divisor.

In general the formula is more complicated and I'll refer you to Fulton's Intersection Theory, where the formula you require is given in Theorem 15.4.

In particular cases the relation may be quite simple: for example if $X$ is of dimension 3, it is just $c_2(\tilde X)=p^*(c_2(X)$ for the second Chern class, as proved in Griffiths-Harris's Principles of Algebraic Geometry, page 609.


Like Georges says, 15.4 of Fulton's Intersection Theory deals with the general theory. For this special case it's not too hard to work out the Chern classes by hand though.

Let $f : \widetilde X \to X$ be the projection and $E \cong \mathbb{C}P^{n-1}$ the exceptional divisor. $H^*(\widetilde X) \cong f^*H^*(X) \oplus \langle \textrm{Poincare duals of planes } P_k \textrm{ in } E $ $\textrm{of dimension }k = 1,\ldots, n-1\rangle$. Note that $[P_{n-i}][P_{n-j}] = -[P_{n-i-j}]$, while $(f^*\alpha) [P_k] = 0$ for any $\alpha \in H^i(X)$ ($i, k > 0$).

$f^* c_i(X)$ and $c_i(\widetilde X)$ are equal outside the exceptional divisor, so their difference is Poincare dual to something in $E$. On the other hand the restriction of $f^*c_i(X)$ to $E$ is 0 (for $i > 0$), while the restriction of $c_i(E)$ is $c_i(\mathcal{O}(1)^n \oplus \mathcal{O}(-1)) = \left({n\choose i} - {n \choose i-1}\right)H^i$, where $H \in H^2(E)$ is the hyperplane class. For $0 < i < n$ we deduce that $c_i(\widetilde X) = f^*c_i(X) - \left({n\choose i} - {n \choose i-1}\right)[P_{n-i}]$ by comparing the evaluations on $P_i$.

  • $\begingroup$ I am interested in when $X$ is a 3-fold and when the hyperplane section $H_{\widetilde{X}}$ of $\widetilde{X}$ is $p^{\ast}(H_{X})-E$ ($E$ the exceptional divisor). By Georges Elencwajg's answer I see $c_2(\widetilde{X})=p^{\ast}(c_2(X))$ and hence $\deg(c_2(\widetilde{X}))=c_2(\widetilde{X})\cdot (p^{\ast}(H_{X})-E)=c_2(X)\cdot H_{X}-p^{\ast}(c_2(X))\cdot E$, but but why you say that $p^{\ast}(c_2(X))\cdot E=0$? $\endgroup$
    – gio
    Commented Mar 30, 2012 at 11:41
  • $\begingroup$ $p^*(c_2(X))\cdot E$ is the integral of $p^*(c_2(X))$ over $E$. But $p$ restricted to $E$ is the constant map. $\endgroup$ Commented Mar 30, 2012 at 11:54

Assume $X$ is smooth compact of dimension $n$ and $x_0\in X$ is the point where we perform the blowup. Set $ X_* := X \setminus x_0 $, $ \tilde{X}_* := \tilde{X} \setminus E$. Denote by $N$ a tubular neighborhood of $E$ in $\tilde{X}_* $. By Mayer-Vietoris, the Chern classes of $ \tilde{X} $ are determined once we know their restrictions to $ X_* $ and $ N $.

We identify $ \tilde{X}_* $ with $ X_* $ via the blowdown map $p:\tilde{X}_* \to X_* $. The restriction of $c_k( \tilde{X}) $ to $X_*$ is equal to the restriction of $c_k(X)$. The restriction of $c_k(\tilde{X})$ to $N$ is easy to determine since

$$TN \cong \pi^* T\mathbb{CP}^{n-1} \oplus \pi^* H^*, $$

where $\pi: N\to E= \mathbb{CP}^{n-1}$ is the natural projection and $H\to \mathbb{CP}^{n-1}$ is the hyperplane line bundle. Thus,

$$ c_k(\tilde{X})|_N = c_k( N ) = \pi^*c_k(\mathbb{CP}^{n-1} ) +\pi^* c_{k-1}(\mathbb{CP}^{n-1} ) \pi^* c_1(H^*) $$

$$ = \pi^*c_k(\mathbb{CP}^{n-1} ) - \pi^* c_{k-1}(\mathbb{CP}^{n-1} )\cup \pi^*[H]. $$

Things can be simplified a bit if we introduce the notation $h=\pi^*[H]\in H^2(N,\mathbb{Z})$ and we observe that for some integers $\nu_k$ and $\nu_{k-1}$

$$ \pi^*c_k(\mathbb{CP}^{n-1} ) =\nu_k h^k, $$

$$ \pi^* c_{k-1}(\mathbb{CP}^{n-1} )=\nu_{k-1} h^{k-1}. $$


$$ c_k(N) = ( \nu_k -\nu_{k-1} ) h^k. $$

As for the integers $\nu_k$ they are determined from the equality

$$ 1+c_1( \mathbb{CP}^{n-1} )+\cdots + c_{n-1}( \mathbb{CP}^{n-1} )= (1+H)^n - H^n $$

$$= \sum_{k=0}^{n-1}\binom{n}{k} H^k. $$


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