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Let $(X,\omega)$ be a complex Kaehler manifold of (complex) dimension $d$, and let $Y\subset X$ a complex submanifold of dimension $k$. Evidently $[\omega]^d\in H^{2d}(X,{\mathbb{R}})$ is always nonzero, so we may as well take this to be a orientation for $X$, and define the integral on $X$ so that $\int_X\omega^d=1$. Also evident is that the restriction $\omega|_Y$ is a Kaehler form on $Y$, hence we may also take $\left(\omega|_Y\right)^k=\left(\omega\right)^k|_Y$ to be an orientation on $Y$ and define an integral on $Y$ so that $\int_Y\left(\omega|_Y\right)^k=1$ as well.

Now suppose we blow up $X$ along $Y$. Let $\tilde{X}$ be the blow up, $\tilde{Y}\subset \tilde{X}$ the exceptional divisor, and $\tau\in H^2(\tilde{X})$ the Poincar\'e dual of the exceptional divisor. Since $\tilde{X}$ is also a complex manifold of dimension $d$, and the pullback of the blow down map $\pi^*\colon H(X,{\mathbb{R}})\rightarrow H(\tilde{X},{\mathbb{R}})$ is injective, $\pi^*\left(\omega^d\right)$ must also be (cohomologous to) an orientation on $\tilde{X}$, so we can define an integral on $\tilde{X}$ so that $\int_{\tilde{X}}\pi^*\left(\omega^d\right)=1$.

$\textbf{Question:}$ What, if anything can we say about (the sign of) $$ \int_{\tilde{X}}\tau^{d-k}\cdot\pi^*\left(\omega^k\right)=\int_{\tilde{Y}}\left(\tau|_{\tilde{Y}}\right)^{d-k-1}\cdot\pi_0^*\left(\omega|_Y\right)^k? $$

According to Griffiths and Harris, the restriction $\zeta:=\tau|_{\tilde{Y}}\in H^2(\tilde{Y},{\mathbb{R}})$ satisfies the relation $$ \zeta^{d-k}-c_1(\mathcal{N})\zeta^{d-k-1}+\cdots+(-1)^{d-k}c_{d-k}(\mathcal{N})=0 $$ where $c_i(\mathcal{N})$ is the $i^{th}$ Chern class of the normal bundle $\mathcal{N}$ of $Y\subset X$. In particular, I think the top Chern class $c_{d-k}(\mathcal{N})$ is equal to the restriction of the Poincar\'e dual of the submanifold $Y\subset X$, so it seems a reasonable guess that $\tau$ would satisfy a similar relation $$ \left(\dagger\right) \ \ \ \ \ \ \tau^{d-k}-c_1(\mathcal{N})\tau^{d-k-1}+\cdots+(-1)^{d-k}\pi^*\left([Y]\right)=0 $$ where $[Y]\in H^{d-k}(X,{\mathbb{R}})$ is the Poincar\'e dual to $Y\subset X$, which would seem to imply that $$ \int_{\tilde{X}}\tau^{d-k}\cdot\pi^*\left(\omega^k\right)= (-1)^{d-k}. $$

Does the relation $\left(\dagger\right)$ always hold? Does it affect the answer if we assume that $X$ is projective?

$\textbf{Edit:}$

I apologize if my formulation in terms of integrals is confusing or incorrect. Let me rephrase my question without integrals.

The two orientations on $\tilde{X}$, $\pi^*(\omega^d)$ and $\tau^{d-k}\cdot\pi^*(\omega^k)$ define cohomology classes that are constant multiples of each other in $H^{2d}(\tilde{X})$, i.e. $$ \pi^*\left(\left[\omega^d\right]\right) = A\cdot \left[\tau^{d-k}\cdot\pi^*(\omega^k)\right]. $$ What is the sign of $A$?

I would venture to guess that it is always $(-1)^{d-k}$, but if there is a counterexample out there, I would love to hear about it...

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    $\begingroup$ I don't see how you can normalize the integral to be 1 on $X$ and $Y$ simultaneously. For example, take $X$ to be the product of two $\mathbb P^1$s with areas $r, 1/r$, and $Y = \mathbb P^1 \times pt$. $\endgroup$ – Allen Knutson Jul 4 '15 at 11:26
  • $\begingroup$ @Allen I guess I'm thinking of the integral as taking a top form and picking out the coefficient of whatever orientation form I specify. In your example, I would specify the forms $\omega=\omega_{FS}\times\omega_{FS}$ on $X$ and $\omega|_Y=\omega_{FS}\times 1$ on $Y$ which I guess would make $r=1$... $\endgroup$ – Chris McDaniel Jul 4 '15 at 15:40
  • $\begingroup$ I agree that in that example you could pick the form you said. What my example shows is that if you pick a random form and then scale it to be good for $X$, it may not work for $Y$. So if you want it for both (which I'm not convinced you really do -- but it's what you wrote), you need a real argument. $\endgroup$ – Allen Knutson Jul 6 '15 at 15:42
  • $\begingroup$ But I don't have to scale anything before I pick an orientation form to scale against, right? Or perhaps there is a canonical choice of orientation form here...In any event sorry for the confusion. And you are correct: All I really care about are signs, so in what I wrote above, you may as well interpret my "=" as "is a positive multiple of"... $\endgroup$ – Chris McDaniel Jul 7 '15 at 2:53
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In case anyone else is interested...

A recent paper by Haibao Duan and Banghe Li seems to settle this issue (arXiv version here--it does not appear to be published yet). In particular their relation (i) in Theorem 4.1 is exactly my relation $\left(\dagger\right)$ although they have a slightly different notation (theirs $\mapsto$ mine): $t\mapsto -\tau$, $\omega_X\mapsto \pi^*([Y])$, $\gamma_X\mapsto\mathcal{N}$. Moreover their proof seems to be valid in the symplectic case as well.

So to answer my own question: Yes $\left(\dagger\right)$ always holds.

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