Precourse Worksheet 2 Compact Riemann surfaces

I need only the basic pictures and methods of thinking and calculating.
The topic as a whole would form an entire MA4xx module. Everything here
is treated in much more detail in Simon Donaldson's textbook Riemann
Surfaces, OUP, 2011.

A Riemann surface S is a 1-dimensional complex manifold. This means any
point P in S has a neighbourhood U_P and a local coordinate z_P that
identifies U_P with the disc |z| <= 1 in CC. Hence all the material on
holomorphic functions, Cauchy's theorem on path integrals, and
meromorphic functions from basic complex analysis applies to all small
enough open sets of S. (Please look up the definition of manifold if
necessary: when two neighbourhoods U_P and U_Q intersect, their
coordinates define the same local complex analytic structure.)

We work only with Riemann surfaces that are connected and compact.
https://en.wikipedia.org/wiki/Surface_(topology) treats the topology
of surfaces, and for most of you should be a 5-minute read.

Exercise: Riemann surfaces are naturally oriented. Show that the
Cauchy-Riemann equations imply the Jacobian determinant
   det | partial d(u,v)/d(x,y) |
is positive. As webloc cit. explains, PP^2_RR (the Boy surface) has
a problem with their orientation, as does any surface containing the
one-sided Moebius strip.

In our case, our Riemann surface S has the topology of a sphere with
g handles (the invariant g is the genus). The early pages of Allen
Hatcher's book Algebraic Topology CUP 2011
   https://pi.math.cornell.edu/~hatcher/AT/AT.pdf
features this picture several times.

The case g = 0 is the extended complex plane or Riemann
sphere S^2 (with coordinate at infinity w = 1/z, as in the discussion
of Liouville's theorem).

The case g = 1 is the ordinary torus; it is the complex plane CC divided
by a translation lattice CC/La, where La = Lambda is the lattice
generated by 1 and tau (that is, La = ZZ.1 + ZZ.tau) with tau a point
in the upper half-plane. The fundamental domain of translation by La
is then the parallelogram 0, 1, 1+tau, tau. This is the starting point
for Weierstrass' theory of doubly-periodic functions (esp. the \wp
Weierstrass p function and its derivative \wp') featured in MA426
Elliptic curves.

The topology of a sphere with g >= 2 handles is represented by examples
of algebraic curves. (See for example [UAG, picture at the top of
p.51].) The fundamental result is that there always exist nonconstant
meromorphic functions on any compact Riemann surface. (From this point
of view, this is the essential content of the Riemann-Roch theorem.)
Although conjectured in the 19th century, it depends on results in
functional analysis that only came online in the 1920s See [Donaldson,
Riemann surfaces, Part III, Deeper theory].

The topology of a compact oriented Riemann surface S of genus g is more
elementary. The homology groups H_0, H_1, H_2 have resp. ranks 1, 2g, 1:
   H_0(S, ZZ) = ZZ (class of a point)
   H_1(S, ZZ) = ZZ^{\oplus 2*g} (transverse paths around the g handles)
   H_2(S, ZZ) = ZZ (the whole surface as a homology class)
(and the same for any coefficient ring or field). In particular, S has
Euler characteristic 2-2*g.

Exercise. The Euler characteristic E(S) = v-e+f of a Riemann surface
with a subdivision as a simplicial complex with v vertices, e edges
and f faces is invariant under barycentric subdivision. Deduce that
two subdivisions having a common simplicial subdivision give the
same value of v-e+f. [Of course, E(S) is known to be a topological
invariant.]

Exercise(^*). Construct a simplicial complex whose topological space is
an oriented surface of genus g. This proves that a compact Riemann
surface can be triangulated. (The star indicates this may not be too
easy without hints.)
[Hint: You could do worse than start from the ``spectacles" construction
of [UAG, Figure 2.11, p.50] for a Riemann surface of g = 1; adding
another marked point in the middle of the northern sheet, and linking to
the 4 points where the spectacle bars join the frame. This suggests how
to subdivide the northern sheet into 6 triangles; ditto for the southern
sheet, but be aware that the 4 frame vertices and 4 edges are shared. If
I've got it right, that makes v = 6 vertices, e = 18 edges, f = 12 faces,
so Euler characteristic E(S) = 2-2*g = 0. The case g >= 2 provides more
entertainment.]

Application of contour integrals.
Theorem: A global meromorphic function on a compact Riemann S has equal
number of zeros and poles (counted with multiplicities).

We can use the Cauchy integral function on any small open piece of
a Riemann surface. We integrate a differential around a closed path (or
contour). In basic complex analysis, the differential is often f*dz,
and you get the ritual notification that the integral depends not purely
on the function f, but on f*dz, that is, a combination of f and the
coordinate function z.

Now let S be a Riemann surface, with a given triangulation. Given a
global meromorphic function f, assume that the finitely many zeros and
poles of f are not vertices or on the edges of the triangulation. Assume
the triangulation is compatible with the orientation, so that clockwise
around each triangle gives opposite directions when two triangles meet
along any edge. (This gives the sum of the triangles the boundary
zero, so that it is a cycle, and a generator of H_2(S, ZZ).)

For a global meromorphic function f, consider the exact meromorphic
differential df/f = (d log f/dz)dz. 
Exercise. Show that (d log f/dz)dz is well-defined, independently of the
local coordinate function z.
Exercise. Show that if f has a zero of order d at a point P, the residue
Res_P df/f equals (2*pi*i)*d. If f is regular and invertible at P, the
residue is zero.
Show that if f has a pole of order d at a point P, the residue
Res_P df/f equals (2*pi*i)*(-d).

It follows that the contour integral of the exact meromorphic
differential df/f around a triangle of the triangulation is (2*pi*i)
times the sum of the degree of zeros - sum of the degree of poles.
Now if we sum that over all triangles of the triangulation, we
deduce first that the sum of the contour integrals equals
   number of zeros - number of poles (counted with multiplicities)
and then that this number is zero. This holds because, as explained
above, the contour went along any edge twice in opposite directions.

Exercise. Relate the theorem I stated above to slightly more general
statement in [D, Prop 16, p. 76].