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...



Orientation form on the blow up of a Kaehler manifold
In case anyone else is interested... 


A recent paper by Haibao Duan and Banghe Li seems to settle this issue (arXiv version [here](http://arxiv.org/abs/0906.4152)--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. 


