By John Edward Campbell

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**Additional info for A Course of Differential Geometry**

**Example text**

10. Let Ω1 , Ω2 be domains and ρ1 , ρ2 be metrics on these respective domains. If f : Ω1 → Ω2 is a holomorphic isometry (in particular, an onto mapping) of (Ω1 , ρ1 ) to (Ω2 , ρ2 ), then the following three properties hold: (a) If γ : [a, b] → Ω1 is a continuously diﬀerentiable curve, then so is the push-forward f∗ γ ≡ f ◦ γ and ρ1 (γ) = ρ2 (f∗ γ). (b) If P, Q are elements of Ω1 , then dρ1 (P, Q) = dρ2 (f (P ), f (Q)). (c) Part (b) implies that the isometry f is one-to-one. Then f −1 is well deﬁned and f −1 is also an isometry.

Then c will be mapped by ϕ to another circle in A that has Bergman distance τc from ϕ(C). In short, ϕ maps circles to circles. And the same remark applies to ϕ−1 . Now the orthogonal trajectories to the family of circles centered at P will of course be the radii of the annulus A. Therefore (by conformality) these radii will get mapped to radii. Let R be the intersection of the positive real axis with the annulus A. After composition with a rotation, we may assume that ϕ maps R to R. If M has just one circle in it, then that circle C must be ﬁxed.

This is so because we have already selected the curve to have ∂/∂θ length as small as possible; a curve whose tangents have components in the ∂/∂r direction will a fortiori be longer. In more detail, the length of any curve γ(t), 0 ≤ t ≤ 1, is calculated by 1 ρ (γ) = 1 γ (t) 0 ρ,γ(t) dt = 0 γr (t) + γθ (t) ρ,γ(t) dt , 26 Problems for Study and Exploration where γr and γθ are, respectively, the normal and tangential components of γ . Clearly, if we construct a new curve γ by integrating the vector ﬁeld γθ , then the result is a curve that is shorter than γ.