I am reading the book Complex Geometry - An Introduction by Huybrechts. In proving Lemma 3.2.3 that $\partial$ and $\partial^*$ are formal adjoints to each other, he mention that the following integral vanishes due to Stokes' theorem. \begin{align} \int_{X} d(\alpha \wedge *\bar{\beta}) \end{align}
where $X$ is a compact hermitian manifold, $\alpha$ is of bidegree $(p-1,q)$ and $\beta$ is of bidegree $(p,q)$.
But shouldn't this equal the integration on the boundary?
A friend of me suggested that maybe the manifold has no boundary. I also found a link which says the term "compact manifold" often implies "manifold without boundary". But I am uncomfortable about this idea because I also found the word closed manifold which specifies a compact manifold without boundary.
Also, in case that the manifold has not boundary, I think we can't use Stokes theorem to reason that the integration is zero right? Roughly, I can think of restricting the integration on a smaller region which has a boundary. I would get a non-vanishing value on this smaller region. Then I can grow the region to approximate the value on the whole manifold. It is intuitive unacceptable for me that the integration finally grows to zero.
Any ideas?
In Huybrechts' usage, a manifold does not have boundary. To see this, note that in Appendix $A$ he defines a chart on an $m$-dimensional manifold to be a pair consisting of an open subset of the manifold, together with a homeomorphism from that open subset to an open subset of $\mathbb{R}^m$. For a manifold with boundary, the homeomorphism maps to an open subset of upper half space, not $\mathbb{R}^m$.
By Stokes' Theorem, we have
$$\int_Xd(\alpha\wedge\ast\bar{\beta}) = \int_{\partial X}\alpha\wedge\ast\bar{\beta} = \int_{\emptyset}\alpha\wedge\ast\bar{\beta} = 0.$$
In general, if $X$ is an $n$-dimensional compact manifold (without boundary), then for any $(n-1)$-form $\alpha$,
$$\int_X d\alpha = 0.$$