Note: All arcs and curves are assumed to be smooth in this entry.
Let
be a complex function defined in a disk
about a point
. In this entry, we shall show how to construct a Riemann surface such that
may be analytically continued to a function on this surface by considering paths in the complex plane.
Let
denote the class of paths on the complex plane having
as an endpoint along which
may be analytically continued. We may define an equivalence relation
on this set ---
if
and
have the same endpoint and there exists a one-parameter family of paths along which
can be analytically continued which includes
and
.
Define
as the quotient of
modulo
. It is possible to extend
to a function on
. If
, let
be the value of the analytic continuation of
at the endpoint of
(not
, of course, but the other endpoint). By the monodromy theorem, if
, then
. Hence,
is well defined on the quotient
.
Also, note that there is a natural projection map
. If
is an equivalence class of paths in
, define
to be the common endpoint of those paths (not
, of course, but the other endpoint).
Next, we shall define a class of subsets of
. If
can be analytically continued from along a path
from
to
then there must exist an open disk
centered about
in which the continuation of
is analytic. Given any
, let
be the concatenation of the path
from
to
and the straight line segment from
to
(which lies inside
). Let
be the set of all such paths.
We will define a topology of
by taking all these sets
as a basis. For this to be legitimate, it must be the case that, if
lies in the intersection of two such sets,
and
there exists a basis element
contained in the intersection of
and
. Since the endpoint of
lies in the intersection of
and
, there must exist a disk
centered about this point which lies in the intersection of
and
. It is easy to see that
.
Note that this topology has the Hausdorff property. Suppose that
and
are distinct elements of
. On the one hand, if
, then one can find disjoint open disks
and
centered about
and
. Then
because
. On the other hand, if
, then let
be the smaller of the disks
and
. Then
.
To complete the proof that
is a Riemann surface, we must exhibit coordinate neighborhoods and homomorphisms. As coordinate neighborhoods, we shall take the neighborhoods
introduced above and as homomorphisms we shall take the restrictions of
to these neighborhoods. By the way that these neighborhoods have been defined, every element of
lies in at least one such neighborhood. When the domains of two of these homomorphisms overlap, the composition of one homomorphism with the inverse of the restriction of the other homomorphism to the overlap region is simply the identity map in the overlap region, which is analytic. Hence,
is a Riemann surface.