I already asked the same question at stack exchange but got no response for quite a while, so I thought I'd ask here. I also know that this has a certain resemblance to this question, but I cannot really make much sense of the answer given there.

Write $\mathbb{P}^n:=\mathbb{P}^n_\Bbbk$ for projective space over some (algebraically closed) 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 $c_2(\tilde Y)c_1^{n-2}(\tilde Y)$. If I could understand the total chern class $c(\tilde Y)$, that would be even better.

My thoughts so far: 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 $c_1(\mathcal{O}_{\tilde{X}}(1))$.

From Fulton's Example 3.2.12, we know that $c_1(X)=(m+1-d)\cdot\xi$ and $c_1(\mathcal{N})=d\cdot\xi$ with $\xi = c_1(\mathcal{O}_X(1))$. Now, I am kinda stuck. I am not sure what the push-forwards and pullbacks really do - in particular, what is $g^\ast(\xi)$ in terms of $\zeta$? What is $f^\ast$, applied to the class of a hyperplane? What does $j_\ast$ do? More importantly, are these the right questions to ask?

Ultimately, I thought (hoped) it would be possible to express $c_2(\tilde Y)$ or even $c(\tilde Y)$ as a sum of intersections of "obvious" cycles in $\tilde Y$, possibly involving only the class of the (strict) transform of a hyperplane and the exceptional divisor.


3 Answers 3


My understanding of this is very unsophisticated, but perhaps that means that what I can explain is precisely what you want.

To understand $f^*[H] \in H^2(\tilde Y)$, where $H \subset Y = \mathbb{P}^n$ is a hyperplane, it may help to think of $H$ as the zero set of section $s$ of the anticanonical bundle $\mathcal{O}_Y(1)$. Then the zero set of $f^*s$ (essentially just the preimage of $H$) is a divisor representing $f^*[H]$. If $H$ contained the blow-up locus $X$, then the resulting divisor is the sum of the proper transform $P$ of $H$ (the blow-up of $H$ at $X$) and the exceptional set $\tilde X$. If $H$ was transverse to $X$, then the divisor is the proper transform of $H$ (which is exactly the pre-image of $H$ in this case), which is the blow-up $\tilde H$ of $H$ at $H \cap X$.

I presume $d = n-m$, the codimension of $X$ in $Y$.

If $D$ is a divisor in $\tilde X$ representing $[D] \in H^2(\tilde X)$, then $j_*[D]$ is the class $[D] \in H^4(\tilde Y)$ that you get by considering $D$ as a cycle in $\tilde Y$. $g^*\xi$ can be represented by a divisor that is the preimage in $\tilde X$ of a hyperplane in $X$. Writing that hyperplane as the intersection of $X$ with a transverse hyperplane $H \subset Y$, we find that $j_*g^*\xi = [\tilde H \cap \tilde X] \in H^4(\tilde Y)$. $\zeta \in H^2(\tilde X)$ corresponds to the conormal bundle of $\tilde X$ in $\tilde Y$, so it is the restriction of $-[\tilde X] \in H^2(\tilde Y)$ to $\tilde X$. Therefore $j_*\zeta = -[\tilde X]^2 \in H^4(\tilde Y)$.

To describe $j_*\zeta$ another way, note that since $\mathcal{N} = \mathcal{O}_X(1)^d$, $\tilde X$ is a trivial bundle $X \times \mathbb{P}^{d-1}$. You can get an explicit trivialisation by picking copy of $\mathbb{P}^{d-1} \cong Z \subset Y$ disjoint from $X$: given points $x \in X$ and $z \in Z$, the line from $x$ to $z$ defines an element in the projectivisation of the fibre of $\mathcal{N}$ over $x$. Let $h$ be the projection $\tilde X \to Z$. Then $\mathcal{O}_{\tilde X}(-1) = g^*\mathcal{O}_X(1) + h^*\mathcal{O}_{Z}(-1)$. $h^*\mathcal{O}_{Z}(1)$ corresponds to a trivial $\mathbb{P}^{d-2}$ subbundle of $\tilde X$. Such a divisor is the intersection of $\tilde X$ with the proper transform $P$ of a hyperplane $H$ containing $X$ and some hyperplane in $Z$. In other words, $-\zeta = g^*\xi -[P \cap \tilde X] \in H^2(\tilde X)$, so $j_*(\zeta + g_*\xi) = [P \cap \tilde X] \in H^4(\tilde Y)$.

Sanity check: $-[\tilde X] + [\tilde H] = [P]$ implies $-[\tilde X]^2 + [\tilde H \cap \tilde X] = [P \cap \tilde X]$, so it adds up.

  • $\begingroup$ First off, thanks for the great reply, +1. I just need to resolve a tiny bit of confusion before I accept - If $\tilde H$ is the blow-up of $H$ at $H\cap X$, it is the proper transform of $H$, by my understanding. I think we have to define $\tilde H$ as $f^{-1}(H)$ since we have $j_\ast g^\ast \xi = [\tilde H\cap\tilde X]=[\tilde H][\tilde X]$ by your first part and $j_\ast g^\ast \xi = [P\cap\tilde X] - j_\ast(\zeta) = [P\cap\tilde X] + [\tilde X]^2 = [\tilde X]([\tilde X]+[P])$ by your second part. Am I right? $\endgroup$ Mar 8, 2012 at 7:20
  • $\begingroup$ I guess what I meant was that when $H$ is transverse to $X$ the proper transform is the same as the pre-image. I've made a small edit to clarify. $\endgroup$ Mar 8, 2012 at 9:36

Consider the differential of $g$, which is a morphism of vector bundles $T_{\tilde Y} \to g^*T_Y$. It is clear that it is an isomorphism out of $\tilde X$, so let us investigate what goes on $\tilde X$. On $\tilde X$ we have two exact sequences: $$ 0 \to T_{\tilde X} \to T_{\tilde Y|\tilde X} \to N_{\tilde X/\tilde Y} \to 0 $$ and $$ 0 \to g^*T_X \to g^*T_{Y|X} \to g^*N_{X/Y} \to 0. $$ Moreover, there is a morphism from the first exact sequence to the second such that the map on the middle terms is the restriction of the map $T_{\tilde Y} \to g^*T_Y$. Since the map $T_{\tilde X} \to g^*T_X$ is an epimorphism, we conclude that the cokernel of the map $T_{\tilde Y|\tilde X} \to g^*T_{Y|X}$ is isomorphic to the cokernel of the map $N_{\tilde X/\tilde Y} \to g^*N_{X/Y}$. Since the latter map is an embedding, we conclude that there is an exact sequence $$ 0 \to T_{\tilde Y} \to g^*T_Y \oplus j_*N_{\tilde X/\tilde Y} \to j_*g^*N_{X/Y} \to 0, $$ It allows to compute what you need --- you compute theChern character of $T_{\tilde Y}$ in terms of those of $T_Y$, $N_{\tilde X/\tilde Y} \cong O_{\tilde X}(\tilde X)$ and of $N_{X/Y}$ (to compute how the Chern character changes under the pushforward $j_*$ you will need the Grothendieck-Riemann-Roch).

  • $\begingroup$ That yields the Formula from Fulton's book, but I am looking for a little more concrete help here. As I said, I would, hands on, really like to know what $c_2(\tilde Y)$ in my special case is, expressed as a sum of intersection products only involving "well-known" cycles in $\tilde Y$. $\endgroup$ Mar 7, 2012 at 8:27

Since you are blowing up $\mathbb{P}^n$ along a linear subvariety, which in turn is a smoooth complete intersection in $\mathbb{P}^n$, you can compute its Chern classes very easily via a formula of Aluffi (Lemma 1.3 in http://www.math.fsu.edu/~aluffi/archive/paper348.pdf). After denoting the first Chern class of a hyperplane in $\mathbb{P}^n$ by $H$, $f:\tilde{\mathbb{P}^n}\to \mathbb{P}^n$ the blowup of $\mathbb{P}^n$ along a linear subvariety say of codimension $k$ and the exceptional divisor by $E$, then

$$ c(\tilde{\mathbb{P}^n})=\frac{(1+E)(1+f^*H-E)^k}{(1+f^*H)^k}f^*c(\mathbb{P}^n). $$

(To get the individual Chern classes, just replace $H$ by $t\cdot H$ and $E$ by $t\cdot E$ in the expression above, expand the expression as a powerseries in $t$, then the coefficient of $t^m$ will be $c_m$.) Then if you want to compute any Chern number, take the product of the corresponding Chern classes and push them forward to $\mathbb{P}^n$, which amounts to just knowing the pushforwards of self-intersections of the exceptional divisor via the projection formula, which can be easily computed via Lemma 2.1 in http://arxiv.org/pdf/1211.6077v1.pdf.

  • $\begingroup$ First of all, thanks for answering this old question of mine. It's been a while since I have dealt with the problem, but in fact, I never managed to completely resolve it: Hence, I am very grateful, because this looks much more promising than the formulas I have used before. I will look into it soon. In the meanwhile, happy holidays! $\endgroup$ Dec 24, 2013 at 12:39
  • $\begingroup$ Cheers same to you! $\endgroup$
    – dezign
    Jan 2, 2014 at 14:04

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