Meromorphic Functions on Riemann Surfaces (Pt. II)
Point of Post: This is a continuation of this post.
Let be a Riemann surface and suppose that is meromorphic at . We know then that for each chart at one has that is meromorphic at . Thus, from basic complex analysis we can form the Laurent series at relative to , which is defined on a punctured disc . We then define the order of relative to be . As one would hope, despite the fact that the Laurent series of varies greatly depending on which chart one takes it relative to, the order of does not!
Theorem: Let be a Riemann surface and be meromorphic at . Then, the order of is independent of chart.
Proof: Let and be any two charts at . We first consider the Laurent series of relative , so that on some disc we have that
where is the order of relative (note that this order is finite by the assumption that is meromorphic at ). Now, we consider a disc contained in . We see then that on this disc
Now, since is biholomorphic on a neighborhood of we have that we can extend into it’s Taylor series
Now, note that , and . This second fact is because if [remember that is biholomorphic!) then for all in some neighborhood of , and so , so . Thus, we see that
where are some coefficients in . Now, by the uniqueness of the Laurent series on a punctured disc, this must be the Laurent series of at relative . But, note that and are both nonzero so that is non-zero. It follows easily that the order of at relative is also . Since and were arbitrary the conclusion follows.
Thus, if is a meromorphic function at , we may unabashedly define the order of at to be its order at relatively any chart at . We denote the order of at by or sometimes (although rarely, if at all) by (this is supposed to be suggestive to anyone familiar with some parts of algebra!).
Let’s give an example of a meromorphic function (finally, right?). Consider the extended complex plane and a rational function . It’s trivial that is meromorphic at each point of , and checking the is meromorphic at is equivalent to checking that is meromorphic at –a classical exercise in basic analysis. This should be expected considering our intuition that meromorphic functions are just holomorphic functions to the Riemann sphere, since we have already shown that is a holomorphic function .
Now, having the notion of order allows us to classify singularities. Namely, let be meromorphic at , then we say that is a removable singularity if and that is a pole if . Note that the name ‘removable singularity’ makes sense since evidently if then is bounded in a neighborhood of and so Riemann’s Removable Singularity Theorem allows us to holomorphically extend to . In particular, we see that is holomorphic at if and only if . Along with this, it’s easy to see that has a pole at if and only if .
The other thing to notice about this definition is that the poles of are necessarily discrete in . Indeed, if is meromorphic at then we know that is meromorphic at for any chart at . Thus, we know there exists a neighborhood such that is holomorphic on and thus we see that is a neighborhood of which intersects the poles of only at –and thus discreteness follows. This is comforting since we always had the discreteness of poles for meromorphic functions on subsets of . In fact, not only is this comforting, but it should be expected. For, if meromorphic are really just holomorphic then the poles of should be and fibers are discrete!
This above analysis gives us an equivalent way of defining holomorphic functions on a Riemann surface –by specifying poles. Namely, suppose that is a Riemann surface and is open. Then, we say that a function is a meromorphic function on if it is a holomorphic function on , if is discrete, and if for each . In essence, this says that a meromorphic function on is a holomorphic function with a discrete set of poles.
Now, we leave it to the reader to verify the following simple properties of the order:
Theorem: Let be meromorphic at , then:
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