Monday, January 28, 2013

Complex Analysis: #11 Monodromy Theorem

  • Complex Analysis: #11 The Monodromy Theorem

Why confine our thoughts on analytic continuation to a single continuous path γ? After all, it seems obvious that we can move the path back and forth to some extent — at least when staying within the open discs which we are using — without affecting the arguments of the previous section. This is the idea of the Monodromy Theorem. But first we must define what “moving a path” is supposed to mean.

Definition 9
Let Q = {(x, y) ∈ ℝ2 : 0 ≤ x, y ≤ 1} be the unit square, and let H : Q → ℂ be a continuous mapping such that
  • H(0, y) = H(0, 0), for all y ∈ [0, 1], 
  • H(1, y) = H(1, 0), for all y ∈ [0, 1], 
Let α : [0, 1] → ℂ be the path α(t) = H(t, 0) and let β(t) = H(t, 1). Then we have α(0) = β(0) and α(1) = β(1). The mapping H is said to be a homotopy from α to β. One also says that α and β are homotopic to one another. If G ⊂ ℂ is a region, and H(Q) ⊂ G, then it is a homotopy within the region.

In the more general setting of topology, this idea of homotopy is quite important. But historically, the idea grew out of these applications in complex analysis. If we work with closed paths α, so that α(0) = α(1), then we can define the fundamental group of the topological space. This is dealt with to a greater or lesser degree in all of our textbooks. But I will skip over these things here. Of course, as an additional remark, you should note that the fact that the paths are being parameterized using the unit interval [0, 1] represents no loss of generality. It all works just as well if we use some other interval [t0, t1]. Finally, note that if f : G → ℂ is given as an analytic function and α and β are homotopic to one another within G, then theorem 4 shows that

 Complex Analysis: #11 The Monodromy Theorem equation pic 1



Theorem 23 (Monodromy Theorem)
Let α and β be two homotopic paths in ℂ. Assume there is an open disc B0 centered on α(0) = β(0) and an analytic function f0 : B0 → ℂ. For each τ ∈ [0, 1] let hτ : [0, 1] → ℂ be the path hτ(t) = H(t, τ). (Thus α = h0 and β = h1) Assume that f0 has an analytic continuation along the path hτ for each τ. In particular there exists an open disc B1 centered at α(1) = β(1) such that the analytic continuation along α produces the function f1 : B1 → ℂ and the analytic continuation along β produces the function f1 : B1 → ℂ. Then f1 = f1.

Proof
The proof uses the technique which we have seen in theorem 4. The construction of an analytic continuation along each of the paths hτ involves some finite chain of open discs. So the set of all such discs covers the compact set H(Q). Take a finite sub-covering. Take the inverse images of the sets in this sub-covering. We obtain a finite covering of Q by open sets. Take a subdivision of Q into sub-squares of length 1/n, for n sufficiently large, so that each of the sub-squares is contained in a single one of these open sets covering Q. Then let γj = hj/n, for j = 0, . . . , n. Now the argument in the proof of the previous theorem (theorem 22) shows that the analytic continuation along γj leads to the same function as that along γj+1, for each relevant j. In particular, the function is uniquely defined through the power series representation in each of the sub-squares. Therefore it’s values along one segment of the curve γj determines uniquely it’s values along the corresponding segment of the next curve γj+1. Since the endpoints of all of the curves are identical (that is, γj(1) = α(1) = β(1) for all j), we must have the power series expression at this endpoint for each of the analytic continuations of the original function being the same.

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