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Let $X$ be a compact Kähler manifold and $D$ be a snc divisor on it. Then on $X\setminus D$ Stokes theorem holds true

$\int_{X\setminus D}\Delta\alpha=0\; \; \; ?$

In this case $X\setminus D$ is non-compact, but we know of course $\int_{X }\Delta\alpha=0$

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No, in fact, the integral has no reason to even converge.

Take $X=\mathbb{CP}^1$, D a point, and so $X-D=\mathbb{C}$. Then $\alpha(z):=|z|^2$ is a smooth function on $\mathbb{C}$, whose laplacian is a constant.

EDIT: I am interpreting the question by assuming that $\alpha$ is a smooth function on $X-D$, but the last phrase of the question suggests that this interpretation is maybe incorrect (if $\alpha$ is a smooth function on $X$, then the integrals on $X$ or $X-D$ are obviously the same because $D$ is a subset of measure zero).

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  • $\begingroup$ how do you know that a snc divisor has masure zero, also $\alpha$ is a k-form here $\endgroup$
    – Chneg chui
    Commented Jan 2, 2018 at 12:24
  • $\begingroup$ By definition, a snc divisor in a $n$-dimensional complex manifold is locally isomorphic to an union of coordinates hyperplanes in $\mathbb{C}^n$. But it is obvious that an hyperplane in $\mathbb{C}^n$ has measure zero. $\endgroup$
    – user25309
    Commented Jan 2, 2018 at 12:29
  • $\begingroup$ about your first part of answer, $\Delta|\omega|^2=2(|\omega|\Delta |\omega|+|\Delta|\omega||^2)$ $\endgroup$
    – Chneg chui
    Commented Jan 3, 2018 at 3:22

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