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I am really stuck on this one. Let $Y=\mathbb{P}^n$ be the complex projective space and let $\tilde Y$ be the blow-up of $Y$ along a linear subvariety $X$ of codimension $d$. We get the following blow-up diagram: $$\begin{matrix} E & \xrightarrow{\;j\;} & \tilde{Y} \\ \hphantom{\scriptstyle g}\downarrow {\scriptstyle g} && \hphantom{\scriptstyle f}\downarrow {\scriptstyle f} \\ X &\xrightarrow{\;i\;} & Y \end{matrix}$$ Denote by $P$ the proper transform, under $f$, of a hyperplane in $Y$.

I am trying to calculate $c_2(\tilde Y)$, and with the help of a post here at MO, I thought that I had figured it out. However, I wanted to do a quick check if nothing went wrong, but something did go wrong. Assume $n=4$ and $d=2$, then we can use the formula from Fulton's book in Example 15.4.3 to get $$c_2(\tilde Y)=f^\ast c_2(Y) - j_\ast g^\ast c_1(X) - E^2.$$ By the answer to my question by Johannes Nordström, we can write $j_\ast g^\ast c_1(X)=3(E^2 + EP)$ and this yields $$c_2(\tilde Y)=10P^2 - 3EP - 4E^2.$$ Now, we also know from the same example in Fulton's book that $c_1(\tilde Y)=f^\ast c_1(Y) - E=5P-E$. Since the blow-up map $f$ is finite of degree one, the degrees of $c_1^2(\tilde Y)c_2(\tilde Y)$ and $c_1^2(Y)c_2(Y)$ should coincide. Because I was unsure of the calculation, I asked a second question and obtained the answer that $$P^{n-b} E^b = (-1)^{b-1+n-d} \cdot \binom{b-1}{n-d}$$ Now, we can put this all together and obtain $$\begin{align*} c_1^2(\tilde Y)c_2(\tilde Y) &= (5P-E)^2(10P^2 - 3EP - 4E^2) \\&= 250P^4 - 175P^3E - 60P^2E + 37PE^3 - 4E^4 \\&= 250 + 37 + 12 = 299, \end{align*}$$ but $c_1^2(Y)c_2(Y)=(4+1)^2\cdot\frac{4(4+1)}{2} = 250$.

I do not know where the mistake is, since I find both of the answers I received very convincing, but I cannot find a flaw in my calculation either, nor do I doubt Fulton.

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up vote 3 down vote accepted

Note that $P^2 = 0$, since we blow up the self-intersection of a hyperplane.

The pull-back of the hyperplane class is $P+E$, so $c_1(\tilde Y) = 5P + 4E$, and $c_2(\tilde Y) = 10(P+E)^2 - 3EP - 4E^2 = 17EP + 6E^2$.

This still does not yield $c_1(\tilde Y)^2 c_2(\tilde Y) = 250$ (I think it's $512 - 3\cdot 96 = 224$), but I don't see why this Chern number should be preserved by the blow-up.

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The fact that the degree of a zero-cycle is preserved should follow (in my opinion) from example 1.7.4 in Fulton. Note that a finite surjective morphism between nonsingular varieties is always flat. By Corollary 6.7.2 in Fulton, since a generic hyperplane intersects $X$ in dimension $n-1-d$, I concluded that the pull-back of a hyperplane clas is its strict transform. Where is my error? –  Jesko Hüttenhain Mar 13 '12 at 23:52
    
Also, what exactly is your reason for $P^2=0$? –  Jesko Hüttenhain Mar 13 '12 at 23:59
    
$f^*(c_1(Y)^2c_2(Y))$ has the same degree as $c_1(Y)^2c_2(Y)$, but if $f^*c_i(Y)$ equaled $c_i(\tilde Y)$ the calculation would be trivial. Certainly $c_1(Y)^4$ is not equal to $c_1(\tilde Y)^4$. The formula you use for the intersections of $P$ and $E$ requires that $P$ be the proper transform not of a generic hyperplane, but of one that contains $X$. Two such hyperplanes intersect along $X$, so when you blow up $X$ their proper transforms are disjoint. –  Johannes Nordström Mar 14 '12 at 8:48
    
It is beginning to become more clear to me, slowly. I have one more question, though: if $H$ is a hyperplane containing $X$ and $H'$ is a generic one, intersecting $X$ transversally, then we should have $[H]=[H']$ as classes in the Chow ring. Denoting by $P$ and $P'$ the proper transforms, this would imply that $[P']^2=0$ as well. That would mean that all proper transforms of hyperplanes suddenly have self-intersection zero. That sounds weird to me. –  Jesko Hüttenhain Mar 16 '12 at 12:16
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But $[H]=[H']$ does not imply $[P] = [P']$. –  Johannes Nordström Mar 16 '12 at 22:00
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