The univalent Bloch-Landau constant, harmonic symmetry and ...
The univalent Bloch-Landau constant, harmonic symmetry and ...
The univalent Bloch-Landau constant, harmonic symmetry and ...
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We consider an (i+1)-face f 1 of Σ 1 , the i-faces that bound it, <strong>and</strong> the<br />
corresponding (i + 1)-face f 2 of Σ 2 . Assuming that T maps each of<br />
these i-faces onto the corresponding i-face of Σ 2 , it follows fromthe key<br />
property of T that the image of the i-faces bounding f 1 has winding<br />
number 1(mod2) about each interior point of f 2 . Consequently, T is<br />
onto f 2 .<br />
□<br />
4.3. Harmonically symmetric arcs don’t grow. It is natural to<br />
ask whether the <strong>harmonic</strong>ally symmetric arcs γ k in <strong>The</strong>orem 3 can be<br />
described by means of a differential equation of Löwner type. As our<br />
final result, we show that this is not possible, even in the case of two<br />
<strong>harmonic</strong>ally symmetric arcs. We note that if the data in <strong>The</strong>orem 3<br />
is symmetric with respect to R, then, by construction, the resulting<br />
<strong>harmonic</strong>ally symmetric curves can be taken symmetric with respect<br />
to R.<br />
<strong>The</strong>orem 4. We suppose that ζ lies on the upper half of the unit circle<br />
<strong>and</strong> that a 1 <strong>and</strong> a 2 lie in (0,1) with a 1 < a 2 . <strong>The</strong>orem 3 is applied twice<br />
to construct two pairs of <strong>harmonic</strong>ally symmetric curves, {γ 1 ,γ 1 } <strong>and</strong><br />
{γ 2 ,γ 2 } respectively, the first from the data a 1 , a 1 , ζ, ζ, <strong>and</strong> the second<br />
from the data a 2 , a 2 , ζ, ζ. <strong>The</strong>n γ 1 ⊈ γ 2 except in the case when ζ = i.<br />
Proof. Let us suppose that γ 1 ⊆ γ 2 . Since D\{γ 1 ∪¯γ 1 } is <strong>harmonic</strong>ally<br />
symmetric in ¯γ 1 with respect to 0, ¯γ 1 is a geodesic arc in D\γ 1 . Since<br />
D\{γ 2 ∪¯γ 2 } is <strong>harmonic</strong>ally symmetric in ¯γ 2 with respect to 0, ¯γ 2 , <strong>and</strong><br />
hence its subarc ¯γ 1 , is a geodesic arc in D \γ 2 . Thus ¯γ 1 is a geodesic<br />
arc with respect to both the domain D\γ 1 <strong>and</strong> the domain D\γ 2 .<br />
We write Γ 1 for the full geodesic in D\γ 1 of which ¯γ 1 is a part <strong>and</strong><br />
write Γ 2 for the full geodesic in D\γ 2 of which ˜γ 1 is a part. Both Γ 1<br />
<strong>and</strong> Γ 2 pass through the origin. We map D\γ 1 onto the unit disk by<br />
a conformal map f 1 so that Γ 1 is mapped onto (−1,1) <strong>and</strong> map D\γ 2<br />
onto the unit disk by a conformal map f 2 so that Γ 2 is mapped onto<br />
(−1,1).<br />
We consider the map g = f 1 ◦ f2 −1 , which maps the unit disk into<br />
itself conformally. Moreover, g is real-valued on f 2 (¯γ 1 ), which itself is<br />
a subinterval of (−1,1). Thus g is real-valued on the entire interval<br />
(−1,1). Since f2 −1 (−1,1) = Γ 2 , it then follows that<br />
f 1 (Γ 2 ) ⊆ (−1,1) = f 1 (Γ 1 ),<br />
so that<br />
Γ 2 ⊆ Γ 1 .<br />
<strong>The</strong>re arenow two possible geometric situations to consider, depending<br />
on where the geodesic Γ 2 might end (both it <strong>and</strong> Γ 1 begin at the<br />
endpoint of ¯γ 1 on the unit circle). Suppose that Γ 2 were to end at a<br />
boundary point of D \ γ 1 . In this case, Γ 2 <strong>and</strong> Γ 1 coincide <strong>and</strong> the<br />
final step is to recall that the <strong>harmonic</strong> measure of the boundary is<br />
split evenly in two along a hyperbolic geodesic. Let E be that part of<br />
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