In a paper I’m reading, the situation is as follows: Let $M$ be a 2-dimensional compact smooth manifold with boundary $\partial M$ and genus $g$ and let $\bar{M}$ be the compact Riemannian surface obtained by attaching a conformal disk at any connected component of $\partial M$. Then the author references p. 261 from „Principles of Algebraic Geometry“ by Griffiths and Harris and concludes:
There exists $\bar{\phi}:\bar{M}\longrightarrow\mathbb{S}^{2}$ non-constant and holomorphic, such that $degree(\bar{\phi})\leq 1+ \lfloor\frac{g+1}{2}\rfloor$, where $\lfloor x\rfloor$ is the lowest integer less than or equal to $x$. Can someone explain this existence, especially the upper bound for the degree?
Further, let $\phi$ be the restriction of $\bar{\phi}$ to $M$. Then $\int_{M}|\nabla\phi|^{2}\leq 8\pi\,degree(\bar{\phi})$. Why is this true?
The only complex structure on $\mathbb S^2$ is that of $\mathbb P^1$. Any non-constant holomorphic function to $\mathbb P^1$ is a branched covering (as $\overline M$ is compact). The statement on p. 261 directly asserts the existence of such branched coverings.
I think that in the second statement they take the volume form $dx\wedge dy$ on $\mathbb P^1$, the integral of which over the whole sphere is $4\pi$:
$$\int_{\mathbb S^2}dx\wedge dy = 4\pi,$$
and then essentially apply the substitution theorem: outside the branch points (which are isolated points), $\overline\phi$ is a holomorphic covering, say $U\subset\mathbb S^2$ is an open set on which it is a covering, and let $V\subset\overline M$ be any component of $\overline\phi^{-1}(U)$, then
$$Area(U) = \int_U dx\wedge dy = \int_{V}\overline\phi^\ast(dx\wedge dy).$$
If you write $\overline\phi = u + iv$ and you work it out, you see that
$$\phi^\ast(dx\wedge dy) = (u_x^2 + v_x^2)dx\wedge dy = \left(\frac{d\overline\phi}{dz}\right)^2dx\wedge dy.$$
The number of such copies of $V$ is the degree of $\overline\phi$.
Now from the fact that
$$dz\wedge d\overline z = -2idx\wedge dy$$
and under the assumption that $\int_M |\nabla\phi|^2$ means $\int_M \left|\frac{d\overline\phi}{dz}\right|^2 dz\wedge d\overline z$, the result can be pieced together.