# Complex Analysis: #11 The Monodromy Theorem

**Deﬁnition 9**

Let Q = {(x, y) ∈ ℝ

^{2}: 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],

*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 [t

_{0}, t

_{1}]. 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

**Theorem 23 (Monodromy Theorem)**Let α and β be two homotopic paths in ℂ. Assume there is an open disc B

_{0}centered on α(0) = β(0) and an analytic function f

_{0}: B

_{0}→ ℂ. For each τ ∈ [0, 1] let h

_{τ}: [0, 1] → ℂ be the path h

_{τ}(t) = H(t, τ). (Thus α = h

_{0}and β = h

_{1}) Assume that f

_{0}has an analytic continuation along the path h

_{τ}for each τ. In particular there exists an open disc B

_{1}centered at α(1) = β(1) such that the analytic continuation along α produces the function f

_{1}: B

_{1}→ ℂ and the analytic continuation along β produces the function f

_{1}: B

_{1}→ ℂ. Then f

_{1}= f

_{1}.

*Proof*

The proof uses the technique which we have seen in theorem 4. The construction of an analytic continuation along each of the paths h

_{τ}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 γ

_{j}= h

_{j/n}, for j = 0, . . . , n. Now the argument in the proof of the previous theorem (theorem 22) shows that the analytic continuation along γ

_{j}leads to the same function as that along γ

_{j+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 γ

_{j}determines uniquely it’s values along the corresponding segment of the next curve γ

_{j+1}. Since the endpoints of all of the curves are identical (that is, γ

_{j}(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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