Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
add comment

3 Answers

up vote 6 down vote accepted

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.

share|improve this answer
    
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? –  Jesko Hüttenhain Mar 8 '12 at 7:20
    
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. –  Johannes Nordström Mar 8 '12 at 9:36
add comment

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).

share|improve this answer
    
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 $\tilde Y$. –  Jesko Hüttenhain Mar 7 '12 at 8:27
add comment

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.

share|improve this answer
    
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! –  Jesko Hüttenhain Dec 24 '13 at 12:39
    
Cheers same to you! –  James Fullwood Jan 2 at 14:04
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.