Metric Distortion and Triangle MapsMario Bonk1 and William Cherry2March 2, 1998AbstractSuppose T is a geodesic triangle with respect to the spherical or thehyperbolic metric. Let f :T ! E be a triangle map onto a euclideantriangle E; and assume that the angle measures of two of the verticesare preserved by f: We prove that the metric distortion of f is extremalon the side of T opposite the angle whose measure is not preserved.As applications we determine the minimum points of the hyperbolicand spherical densities on some symmetric regions.AMS Subject Classi�cation: Primary: 30C20. Secondary: 30F45, 33C991 IntroductionWe identify the Riemann sphere with C = C [ f1g; and we considertwo non-degenerate circular-arc triangles T and E: We think of T andE as closed subsets of the Riemann sphere. A triangle map f :T ! Eis a homeomorphism f from T onto E which is a conformal map fromthe interior of T onto the interior of E and maps the vertices of T ontothe vertices of E: We will assume that T is a geodesic triangle, either onthe Riemann sphere equipped with the spherical metric or in the unit discequipped with the hyperbolic metric. According to these two cases we callT a spherical or a hyperbolic triangle. The triangle E will always be aeuclidean triangle, meaning the edges of E will be straight line segments inthe complex plane.Themetric distortion associated to a triangle map f :T ! E at z 2 Tis the ratio of the euclidean length element jd�j at � = f(z) to the spherical1Research partially supported by TMR fellowship ERBFMBICT 961462 and by aHeisenberg fellowship of the DFG.2Research partially supported by US NSF Grants # DMS 9505041, # DMS 9304580,and by the Alexander von Humboldt Foundation.1

length element 2jdzj=(1 + jzj2) at z; or to the hyperbolic length element2jdzj=(1 � jzj2) at z; depending on whether T is a spherical or hyperbolictriangle. The main result of this paper is the following theorem on metricdistortion.Theorem 1.1 Suppose f :T ! E is a triangle map of a spherical (respec-tively hyperbolic) triangle T onto a euclidean triangle E such that f pre-serves the angle measure at two vertices of T: Any point where the metricdistortion of f attains its minimum (respectively its maximum) over T lieson the side of T that is opposite the angle whose measure is changed by f:The sum of the angle measures of the vertices of a triangle is largerthan �; equal to �; or less than � according to whether the triangle isspherical, euclidean, or hyperbolic, respectively. Under the assumptions ofthe theorem, the triangle map f preserves the angle measure at two verticesof T; which we will call A and B; say. It follows that the angle measureof the third vertex of T; which we will call C; is decreased by f if T isspherical and increased if T is hyperbolic. At C the metric distortion of fbecomes singular, i.e., it tends to in�nity as we approach C if T is sphericaland vanishes at C if T is hyperbolic. One might then expect that the otherextreme would occur on the edge of T opposite C; which is exactly whatTheorem 1.1 says.In general, there is no obvious point on the side opposite C where theextremum of the metric distortion is attained. In special cases, however, thelocation of the extremum can be precisely determined (cf. Corollary 2.9).Using well-known expressions for the triangle functions considered herein terms of hypergeometric functions, the metric distortion can be writtendown more or less explicitly. Thus, at �rst sight, the reduction of the state-ment of Theorem 1.1 to a straightforward calculus problem seems plausible.However, because of the highly involved nature of these explicit formulas,this approach seems not to be feasible, and so we take a di�erent approach.Rather than attempting to use explicit formulas, we use the di�erentialequation for triangle maps together with geometric arguments to prove ourtheorem. The proof runs along the following lines. We consider the functionon T which is the logarithm of the ratio of the two length elements. This isa real-valued function, so to �nd its critical points, we set its z -derivativeequal to zero, where z is a holomorphic coordinate on the geodesic triangleT: We get an equation involving z and �z: This equation can be solvedfor �z in terms of a function H that is meromorphic in T: Surprisingly,the function H has remarkable mapping properties. Namely, the image of2

T under H is also a circular-arc triangle. In fact, the image of T underH; the mapping H post-composed by complex conjugation, is a circular-arctriangle T 0 which is complementary to T in a sense we will precisely describebefore the statement of Theorem 2.5. From these mapping properties of H;Theorem 1.1 will follow as an easy corollary (cf. Corollary 2.8). The factthat H has the stated mapping property is the deepest and most strikingresult of the present paper. The proof, if appropriately organized, can bereduced to a direct computation checking a di�erential equation for H:In section 3 we give two applications of Theorem 1.1. Our �rst appli-cation is a new proof of a theorem of A. Baernstein II ([Ba], [Ba-V]) andH. Montgomery [Mo], which says that if one considers the complex planeminus a symmetric lattice, then the ratio of the hyperbolic to the euclideanlength elements is smallest at the natural points of symmetry that are far-thest from the lattice points. Our second application is to determine sharpupper bounds on the spherical derivative at the origin of a map from theunit disc into a symmetric region in the Riemann sphere, as in our previouspaper [B-C]. In particular we are able to obtain precise upper bounds onthe spherical derivative at the origin of a map from the unit disc into theRiemann sphere minus the vertices of a regular inscribed dodecahedron oricosahedron, two cases we were not able to handle in [B-C].Acknowledgements. Much of the present work was done while the authorswere visiting the University of Michigan, the University of Jyv�askyl�a, andthe Technische Universit�at Berlin. The authors would like to thank theseinstitutions for their kind hospitality. The second author would also like tothank the Institute for Advanced Study, where he was a member.Figure 1:f ��Euclidean Triangle EGeodesic Triangle TeBC eCA B eAz -plane � -plane�� ����� �~

3

2 Maps from Geodesic Triangles onto EuclideanTrianglesSuppose f :T ! E is a triangle map. In this section we make the assump-tions of Theorem 1.1: namely, that T is a hyperbolic or spherical triangle,that E is a euclidean triangle, and that f preserves the angle measure oftwo vertices. We continue to denote the vertices of T by A;B; and C;and we denote the corresponding angle measures by ��; ��; and � : Thevertices of E we will denote by ~A; ~B; and ~C; with corresponding anglemeasures �~�; � ~�; and �~ : Under our assumptions on f; we may assumef(A) = ~A; f(B) = ~B; f(C) = ~C;� = ~�; and � = ~�: With this notation, Cis the vertex of T whose angle measure is changed by f and so 6= ~ : SinceE is a euclidean triangle, we must also have that �+� < 1: We let z denotea complex variable on the domain triangle T; and we let � denote a complexvariable on the image triangle E: The length element in the domain of f isgiven by 2jdzj1� jzj2 ;where here and henceforth, we make the following sign convention: Whenwe use the symbols � and �; the upper sign corresponds to the sphericalcase and the lower sign to the hyperbolic case.The sense preserving isometries of the Riemann sphere with the sphericalmetric or the unit disc with the hyperbolic metric are given by the M�obiustransformations U : ~z 7! z = ei� ~z � ~z01� ~z0~z :Here � 2 [0; 2�]: Moreover, j~z0j < 1 in the hyperbolic case. In the sphericalcase, ~z0 2 C [ f1g: If ~z0 = 1; the de�nition of U needs to be adjustedappropriately, i.e., U : ~z 7! �ei�=~z: Similar adjustments have to be made inthe following to include the case of in�nity in formulas, and we will usuallynot mention this explicitly. The sense preserving isometries of the euclidean� -plane are the maps V : � 7! ~� = ei�� + �0:Here � 2 [0; 2�] and �0 2 C:We use I to denote the mappingI(z) = �1�z :4

Note that I is the map to the antipodal point on the Riemann sphere inthe spherical case or the inversion through the unit circle in the hyperboliccase.We want to �nd the extrema of the metric distortion by the map f: It istechnically a little easier and amounts to the same to consider the logarithmof (twice) this quantity. Explicitly, it is given by�(z) = log(1� jzj2) + 12 log (jf 0(z)j2):Let x and y be the real and imaginary-parts of the holomorphic coor-dinate z = x + iy: If �(z) is a function with values in C; we denote by �the function z 7! �(z); i.e., the composition of complex conjugation and �:For di�erentiable � de�ned on an open subregion of the z -plane, we de�ne@�@z = 12 �@�@x � i@�@y� and @�@�z = 12 �@�@x + i@�@y� :We denote by � the function de�ned by�(z) = (1� jzj2)@�@z (z):Note that both � and � are smooth, i.e., in�nitely di�erentiable, in theinterior of T:The function � is essentially @�=@z: The factor (1� jzj2) ensures that� has certain useful invariance properties that we will make precise momen-tarily. To solve our main extremal problem, we will want to �nd the criticalpoints of �: They are given by the zeros of �:To simplify the appearance of the formulas below, we will use the fol-lowing short-hand notation:Dz� = ddz log�d�dz� = d2�=dz2d�=dzThe following function will play an important role in determining thecritical points of � : H(z) = � Dzf(z)zDzf(z) + 2 :Note that the denominator in the fraction de�ning H is not identically zeroand hence H is a meromorphic in the interior of T: Indeed, if the denomina-tor were identically zero, we would get a di�erential equation for f: It would5

then be straightforward to show that f would have to be a M�obius trans-formation. But, M�obius transformations preserve angle measures, whereasf changes the angle measure at the vertex C of T: This would then be acontradiction.Remark. The idea to consider this function H comes from the work ofRuscheweyh and Wirths [R-W].Solving for Dzf in the de�nition of H; we getDzf(z) = �2H(z)1� zH(z) : (1)The following proposition describes the invariance behavior of the func-tions � and H:Proposition 2.1 Let the notation be as above. Consider the function~f = V � f � U; ~z 7! ~� = V � f � U(~z);and let ~�; ~�; and ~H be de�ned as �; �; and H; respectively, using ~f insteadof f . Then(i) ~� = � � U; (ii) ~� = � � U;(iii) U � I = I � U; (iv) eH = U�1 �H � U:Proof. Equation (i) follows from the invariant de�nition of � as thelogarithm of (twice) the ratio of the in�nitesimal length elements and thefact that U and V are isometries in the appropriate geometries.Equation (ii) follows from the chain rule and the equation(1� j~zj2) ����dUd~z (~z)���� = 1� jU(~z)j2:In the spherical case equation (iii) follows from the fact that rotationsmap antipodal points to antipodal points. Similarly, suppose two points aremapped to each other by inversion with respect to the unit circle. Then theimage points of these points under a M�obius transformation preserving theunit disc have the same property.To prove (iv); �rst note thatD~z ~f = (Dzf � U) � dUd~z +D~zU: (2)6

Moreover, the following identity is true for a; b 2 C; a 6= I(b);�2U(b)1� U(a)U(b) � dUd~z (a) +D~zU(a) = �2b1� ab: (3)Using (2); (1); and (3) with a = ~z and b = U�1 �H � U(~z); we obtainD~z ~f(~z) = �2 (H � U(~z))1� U(~z)(H � U(~z)) � dUd~z (~z) +D~zU(~z)= �2 (U�1 �H � U(~z))1� ~z(U�1 �H � U(~z)) :Note that in this equation the denominator is not identically zero, sinceH 6� I: This equation together with equation (1) where z; f; and H arereplaced by ~z; ~f; and ~H; respectively, implies (iv): 2We now prove that the functions �; �; and H have extensions to theboundary of T .Proposition 2.2 Let the notation be as discussed above. Then(i) � extends continuously to T with values in R = R[f�1;+1g: Thefunction � can be 1 or �1 only at the vertex C; where �(C) =1in the spherical case and �(C) = �1 in the hyperbolic case.(ii) If one of the vertices A or B of T has internal angle measure strictlyless than �=2; then this vertex is a local minimum for � if T is spher-ical and a local maximum for � if T is hyperbolic.(iii) The function � extends continuously to T with values in C: The onlypoint where � =1 is the vertex C: We have � = 0 at the vertices Aand B:Proof. By Schwarz's re ection principle, at all boundary points of T whichare not vertices, f has a unique local extension as a conformal map. Itis then clear that � and � extend smoothly to the boundary of T; exceptpossibly at the vertices. Thus, we need only check what happens there. Let�� denote the angle measure of one of the vertices of T (i.e. � = �; �; or ), and let �~� denote the angle measure of the image vertex (i.e. ~� = �; �;or ~ ).We �rst consider the case that � 6= 0: In other words, we exclude for themoment the case that T is hyperbolic with the vertex in question lying onthe boundary of the unit circle. 7

By the invariance properties proved in Proposition 2.1, we can move thisvertex to z = 0 by an isometry. Moreover, we can assume that one of theedges meeting at this vertex lies along the positive real-axis with the interiorof the triangle lying to the left and that the other edge meeting at this vertexis a straight line segment. Post-composing f by an euclidean isometry, wecan assume that the image vertex is also at the origin, that one of the sidesof the euclidean triangle is along the positive real � -axis, and that the otherside meeting the vertex at � = 0 is a line segment in the upper half-plane.In this case, for z 2 T near the origin, f is of the formf(z) = [h1(z1=�)]~�:Here and in the following, h1; h2; etc. will denote functions holomorphic atthe origin. Here we have h1(0) = 0; h01(0) 6= 0; and h1 positive on thepositive real axis. We will always choose branches of the power functionsu 7! u�; � 2 R; that are positive on the positive real axis. Then,f 0(z) = (~�=�)h1(z1=�)~��1h01(z1=�)z1=��1= z~�=��1h2(z1=�); h2(0) 6= 0; (4)and Dzf(z) = ~�=� � 1z + 1� � h02(z1=�)h2(z1=�) � z1=��1= 1zh3(z1=�); h3(0) = ~�=� � 1: (5)If ~� = �; then � < 1: Hence Dzf(z)! 0 as z ! 0; z 2 T . In this case Dzfcontinuously extends to the origin with value 0 at this point. On the otherhand, if ~� 6= �; then Dzf(z)!1 as z ! 0; z 2 T .The above expansions for f and its derivatives imply the statements ofthe proposition. To see this note that if ~� 6= �; then near 0 we have�(z) = (~�=� � 1) log jzj+O(1):If ~� = �; we have �(z) = log jh2(0)j+O(jzjminf2;1=�g):If in addition � < 1=2; then�(z) = log jh2(0)j � jzj2 + o(jzj2):Statements (i) and (ii) follow. 8

Since �(z) = ��z + 12(1� jzj2)Dzf(z); (6)statement (iii) follows from the above remark about Dzf .It remains to treat the case � = 0: In this case, T is hyperbolic, the ver-tex in question is C; and it lies on the boundary of the unit circle. Withoutloss of generality (again cf. Proposition 2.1), we may assume the interior ofthe triangle is contained in the upper half plane, the vertex C is located atz = 1; and one of the sides meeting at C lies along the positive real-axis.We can also translate and rotate the target triangle E so that the imagevertex lies at the origin and so that the image of the side of T along thepositive real-axis is mapped into the positive real-axis. Note that since T isin the upper half-plane, E will be in the lower half-plane. With this setup,the function f near z = 1 can be written asf(z) = h4 �e� z+1z�1�~�with h4(0) = 0; h04(0) 6= 0; provided we choose � as a positive real numberso that the function z 7! �(z+1)(z�1)�1 maps the angle of T at z = C toa strip of width �: Computation yields the following expansions for z 2 Tnear the point 1 :f 0(z) = 1(z � 1)2 e~�� z+1z�1h5 �e� z+1z�1� ; h5(0) 6= 0;and Dzf(z) = � 2z � 1 + 1(z � 1)2h6 �e� z+1z�1� ; h6(0) = �2~�� 6= 0:Noting that (1 � jzj)=jz � 1j ! 1 as z ! 1; z 2 T; the assertions easilyfollow from these expressions. We leave the details to the reader. 2Proposition 2.3 Let the notation be as discussed above. Then, the functionH extends continuously to T with values in C: Moreover,H(A) = A; H(B) = B; and H(C) = I(C): (7)Proof. The Schwarz re ection principle and formula (1) imply that Hextends meromorphicly and hence continuously to the boundary points ofT di�erent from the vertices. To see what happens at the vertices we canuse the invariance property of H and the invariant nature of the equations9

in (7) to reduce to the same normalizations for T and E as in the proof ofProposition 2.2. Using the same notation as in in that proof, we can assumethat the vertex is located at the origin if � 6= 0. If in addition ~� = �; thenDzf(z)! 0 as z ! 0; z 2 T: This together with the de�nition of H impliesthat H has a continuous extension to the origin with H(0) = 0.If � 6= 0 and ~� 6= �; then it follows from (5) that H has a continuousextension to the origin if we set H(0) =1 = I(0).Finally, if � = 0; we may assume that the vertex (i.e. C ) is located atz = 1: The expansion for Dzf in the second part of the proof of Proposi-tion 2.2 shows that H extends continuously to C if we put H(1) = 1 = I(1):2 We will be interested in the critical points of �; and it turns out theyoccur if and only if �z = H(z):Proposition 2.4 Let the notation be as discussed above. Suppose T isspherical (respectively hyperbolic), and z0 2 T is a point where � has alocal minimum (respectively maximum). Then �(z0) = 0: Moreover, for allz 2 T n fCg; we have �(z) = 0 if and only if �z = H(z):Proof. By the invariance properties of � and �; we may assume z0 6=1:If z0 is in the interior of T; then z0 is a critical point of �; and so �(z0) = 0:By Schwarz's re ection principle, this is also true for points on the edgesof T which are not vertices. Finally, z0 cannot be the vertex C; and wealready know �(A) = �(B) = 0 by statement (iii) in Proposition 2.2.To prove that �(z) = 0 if and only if �z = H(z); note that for z 6= C;1;we have �(z) = 0 if and only if @�=@z = 0: By direct computation,@�@z (z) = ��z1� z�z + 12Dzf(z):Setting this equal to zero and solving for �z; we get�z = � Dzf(z)zDzf(z) + 2 :The expression on the right is just the de�nition of H; and so we are done inthe case that z 6=1: The case z =1 follows from the invariance propertiesof H and �: 2Before coming to the main result of this section, we introduce one lastidea. Given a distinguished side of the triangle T; which in our case will bethe side connecting the vertices A and B; we de�ne the triangle T 0 com-plementary to T with respect to the distinguished side AB as follows.10

Each side of T is a circular-arc on the Riemann sphere, and can thus beextended to form a complete circle on the Riemann sphere. If we so ex-tend the two sides of T other than our distinguished side connecting A toB; i.e., the two sides meeting at C; then we obtain two complete circles.They determine a unique subregion of the Riemann sphere which containsT and is bounded by subarcs of these circles. This region is the union of Tand another circular-arc triangle. We de�ne this second triangle to be thecomplementary triangle T 0: Note that the distinguished side connecting Aand B is a common side of both triangles T and T 0: Hence T 0 shares thevertices A and B with T; and T 0 has angle measures �(1��) and �(1��)at the vertices A and B; respectively. From the fact that T is a geodesictriangle, it follows that the third vertex of T 0 is I(C); and the angle measureof T 0 at I(C) is � : Note that the relation of T 0 to be the complementarytriangle of T is invariant under M�obius transformations. Figure 2 shows acircular-arc triangle T and its complementary triangle T 0: To simplify thedrawing, the triangles in this �gure are not geodesic.Figure 2:I(C)A C BTT 0

An anti-conformal triangle map of a circular-arc triangle T1 onto acircular-arc triangle T2 is a homeomorphism from T1 onto T2 which is ananti-conformal map of the interior of T1 onto the interior of T2 and whichmaps the vertices of T1 onto the vertices of T2 .We now state the main result of this section, which describes the mappingproperties of the meromorphic function H:Theorem 2.5 Under the assumptions of Theorem 1.1, the function H isan anti-conformal triangle map of T onto the complementary triangle T 0of T with respect to the side of T connecting the two vertices whose anglemeasures are preserved by the given triangle map f; i.e. the side connect-11

ing A and B: The correspondence of the vertices of T and T 0 is given byH(A) = A; H(B) = B; H(C) = I(C).Note that we have already proven the statement about the vertex cor-respondence in Proposition 2.3. Before giving the rest of the proof of The-orem 2.5, we outline our strategy. We begin by recalling that we can detectfunctions which map conformally onto the interior of circular-arc trianglesby examining their Schwarzian derivatives. If G is a holomorphic functionof a complex variable t; then we de�ne the Schwarzian derivative of Gwith respect to t; denoted fG; tg; byfG; tg = ddtDtG� 12 (DtG)2 :Schwarzian derivatives are useful because they remain invariant if G is post-composed by an arbitrary M�obius transformation. We recall the followingwell-known facts, the second of which is an implication of the Schwarz-Christo�el formula; see for example [Ne] or [Ca] for details.Theorem 2.6 Let t be a complex variable. A function G meromorphic inthe upper half-plane Im t > 0 maps the upper half-plane conformally ontothe interior of a circular-arc triangle with internal angle measurements �a;�b; and �c; and maps the real-axis onto the boundary such that 0; 1 and 1map to the vertices of measure �a; �b; and �c respectively if and only iffG; tg = 1� a22t2 + 1� b22(t� 1)2 + a2 + b2 � c2 � 12t(t� 1) :Moreover, if the target is a euclidean triangle, thenDtG(t) = a� 1t + b� 1t� 1 :The strategy for proving Theorem 2.5 is then to change variables to theupper half-plane, to compute the Schwarzian derivative of H in order tocheck that it maps to a circular-arc triangle, and then to determine theplacement of that triangle.So, let t be a complex variable, let �(t) be the conformal map fromIm t > 0 onto the interior of the euclidean triangle E such that when ex-tended to the boundary 0; 1; and 1 map to the vertices eA; eB; and eC ofE respectively. Similarly, let z(t) be the conformal map from Im t > 0 ontothe region T in the z -plane such that when extended to the boundary 0; 112

and 1 map to the vertices A;B and C of T respectively. We can useTheorem 2.6 to compute fz; tg and f�; tg; and from this we can computefH�z; tg: Before we do that, we will �rst state some useful identities betweenthe Schwarzian derivatives of the various functions we are considering.Proposition 2.7 Letu = f�; tg � fz; tg; and v = z(dH=dt)1� zH :We then have the following equalities:f�; zg = ff; zg = �2dH=dz(1� zH)2 ; (8)u = �2(dH=dt)(dz=dt)(1� zH)2 ; (9)DtH = ddt log u�Dt� � 2v: (10)Proof. To show (8); we begin with the de�nitionff; zg = ddzDzf � 12 (Dzf)2 :We now use equation (1) to getff; zg = ddz � �2H1� zH �� 12 � �2H1� zH�2 :After taking the derivative of the �rst term, squaring the second term, andcancelling, we are left with the expression in (8):For (9); note that the \chain rule" for Schwarzian derivatives readsf�; tg = f�; zg(dz=dt)2 + fz; tg:Thus, u = f�; tg � fz; tg = f�; zg(dz=dt)2:Combining this with (8); we haveu = �2dH=dz(1� zH)2 � (dz=dt)2 = �2(dH=dt)(dz=dt)(1� zH)2 ;which shows (9): 13

For (10); we begin by computing the t-derivative of log u; which appearson the right in equation (10): Using (9);ddt log u = ddt log ��2(dH=dt)(dz=dt)(1� zH)2 �= DtH +Dtz � 2H(dz=dt) + z(dH=dt)1� zH : (11)On the other hand,Dt� �Dtz = Dz� � dzdt = �2H(dz=dt)1� zH ; (12)where the left equality is the chain rule for the operator D and the rightequality is equation (1): Solving for Dtz in equation (12); substituting thisexpression in for Dtz in equation (11); and then cancelling gives usddt log u = DtH +Dt� � 2z(dH=dt)1� zH :Solving this last equation for DtH results in (10): 2Proof of Theorem 2.5. As we said in our outline, we need to computefH � z; tg and verify that it is in the form of a Schwarzian derivative ofa triangle function. To compute fH; tg; we need to know DtH and itsderivative. We have an expression for DtH by Proposition 2.7. So, we nowdi�erentiate equation (10) :ddtDtH = d2dt2 log u� ddtDt��2[(dz=dt)(dH=dt) + z(d2H=dt2)](1� zH)(1� zH)2�2[H(dz=dt) + z(dH=dt)](z)(dH=dt)(1� zH)2= d2dt2 log u� ddtDt� � 2(dz=dt)(dH=dt)(1� zH)2 � 2z(d2H=dt2)(1� zH)2+2z2H(d2H=dt2)(1� zH)2 � 2z2(dH=dt)2(1� zH)2 :Using equation (9) to replace the term�2(dH=dt)(dz=dt)(1� zH)214

with �u; and using the fact thatd2Hdt2 = DtH � dHdt ;we get ddtDtH = d2dt2 log u� ddtDt� � u�2z(dH=dt)(DtH)(1� zH)2 [1� zH]� 2z2(dH=dt)2(1� zH)2 :Next, we let v be as in Proposition 2.7, and we haveddtDtH = d2dt2 log u� ddtDt� � u� 2vDtH � 2v2:Replacing the occurrence of DtH on the right-hand side in the above equa-tion with the right hand side of equation (10); we getddtDtH = d2dt2 log u� ddtDt� � u� 2v � ddt log u�Dt��+ 2v2:Then, we combine this last equation with equation (10) to compute fH; tg :fH; tg = ddtDtH � 12 (DtH)2= d2dt2 log u� ddtDt� � u� 2v � ddt log u�Dt��+ 2v2�12 � ddt log u�Dt� � 2v�2= d2dt2 log u� ddtDt� � u� 12 � ddt log u�Dt��2 :Now, using Theorem 2.6 to compute f�; tg and fz; tg;u(t) = f�; tg � fz; tg = 2 � ~ 22t(t� 1) :Moreover, since the triangle in the � -plane is euclidean, the second part ofTheorem 2.6 tells us Dt� = �1� �t � 1� �t� 1 :15

Thus, ddt log u�Dt� = ��t � �t� 1 :Di�erentiating this equation with respect to t; we �ndd2dt2 log u� ddtDt� = �t2 + �(t� 1)2 :Combining these last two equalities with our equation for fH; tg; we �ndfH; tg = �t2 + �(t� 1)2 � 2 � ~ 22t(t� 1) � 12 ���t � �t� 1�2= 2�� �22t2 + 2� � �22(t� 1)2 + ~ 2 � 2 � 2��2t(t� 1)= 1� (1� �)22t2 + 1� (1� �)22(t� 1)2 + ~ 2 � 2 � 2��2t(t� 1) :Because E is euclidean, we have ~ = 1� �� �; and so~ 2 � 2 � 2�� = (1� �� �)2 � 2 � 2�� = (1� �)2 + (1� �)2 � 2 � 1:Thus,fH; tg = 1� (1� �)22t2 + 1� (1� �)22(t� 1)2 + (1� �)2 + (1� �)2 � 2 � 12t(t� 1) ;and hence by Theorem 2.6, H maps T conformally onto the interior ofa circular-arc triangle with internal angle measures: �(1 � �); �(1 � �);and � : Therefore, H maps the interior of T anti-conformally to the in-terior of a circular-arc triangle with these same angle measurements, whichare the angle measurements of T 0: The Schwarzian derivative computationwe have done determines H; and hence H; up to post-composition by aM�obius transformation. But, any M�obius transformation is determined bythe image of any three distinct points, for example the vertices of T: ByProposition 2.3, H maps the vertices of T onto the corresponding verticesof T 0: Therefore, H must map T onto T 0: 2The following corollary implies Theorem 1.1 by Proposition 2.4.Corollary 2.8 Let z0 2 T be a point where �(z0) = 0. Then z0 lies on theside of T connecting the two vertices whose angle measures are preserved byf: In particular, if T is spherical (respectively hyperbolic), then any pointz0 2 T where � attains a global minimum (respectively maximum) over Tlies on this side. 16

Proof. By Proposition 2.4, z0 = H(z0): By Theorem 2.5, the image ofT under H is T 0: Therefore, z0 2 T \ T 0: The only common points of Tand T 0 are the points on the edge connecting A and B; except possibly thevertex C if C = I(C): (The possibility C = I(C) can occur only in thehyperbolic case, and then only when = 0; in which case C lies on the unitcircle.) Since z0 6= C by statement (iii) of Proposition 2.2, the corollaryfollows. 2In case that one of the angles preserved by f has measure �=2; we cansay even more.Corollary 2.9 Under the assumptions of Theorem 1.1, and with the addi-tional assumption that � = 1=2; i.e. the angle at A has measure �=2; wehave �(z0) = 0 for z0 2 T if and only if z0 = A or z0 = B: Moreover, if Tis spherical (respectively hyperbolic), then the unique point where � attainsits minimum (respectively maximum) over T is at the vertex B .We need the following geometric statement for the proof of the corollary.We denote by R the re ection across the circle which contains the side ofT opposite to the vertex C .Lemma 2.10 Let the notation be as discussed above. If �; � � 1=2, thenwe have the strict inclusion R(T ) � T 0:Proof. Let D1; D2; D3 , be the unique closed discs in the Riemann spherethat contain T and whose boundary circles contain the sides of T opposite toA;B; and C; respectively. Then R is the re ection across @D3 . Moreover,T = D1 \D2 \D3 and T 0 = D1 \D2 \R(D3):The hypothesis � � 1=2 implies R(D2 \D3) � D2; and similarly, � � 1=2implies R(D1 \D3) � D1: Therefore,R(T ) � R(D1 \D3) \R(D2 \D3) \R(D3)� D1 \D2 \R(D3) = T 0:Note that the angle measures of the triangle R(T ) at the vertices A and Bare �� and ��; respectively. The angle measures of the triangle T 0 at thevertices A and B are (1� �)� and (1 � �)�; respectively. It follows fromthis that the above inclusion is strict if � < 1=2 or � < 1=2: This is alwaysthe case under our assumptions, since �� and �� are angle measures of thenon degenerate euclidean triangle E which implies �+ � < 1: 217

Proof of Corollary 2.9. Assume T is normalized so that the vertex A isat the origin, that one side of T lies along the positive real-axis, and thatone side of T lies along the positive imaginary-axis. The mapping propertiesof H and Lemma 2.10 imply that H (not H!) maps T onto a circular-arctriangle that strictly contains T: This triangle is nothing other than there ection of T 0 across the real-axis.Let H�1 denote the inverse mapping of H; which is de�ned on T: ThenH�1 maps T onto a proper subset of itself so that the edge of T alongthe real axis is mapped onto itself and the edge along the imaginary axis ismapped into itself. Since T is a right triangle with right angle at 0; we canre ect T across the real and imaginary axis and through the origin. Denotethe union of T with the image triangles of T under these involutions byU . The interior of the set U is a simply connected region and contains theedge AB of T apart from the point B: By the Schwarz re ection principle,the map H�1 has a continuous extension to U which is conformal in theinterior of U . This mapping maps the side AB of T onto itself and has the�xed points A and B: This map is not the identity map, since the imageof U is a proper subset of U: By the Schwarz Lemma, a map of a simplyconnected region into itself that is not the identity can have at most oneinternal �xed point. Thus, the vertex A; i.e., the origin, is the only �xedpoint of H�1 in the interior of U: Since the edge AB lies along the real-axis,this implies that the only solutions of �z = H(z) on the edge AB of T areA and B: The �rst part of the corollary now follows from Proposition 2.4and Corollary 2.8.In particular, � has no critical point in T di�erent from the vertices.Since � is invariant under re ection across any side of T; the derivative of� in the direction normal to the side will vanish along each side. Since �has no critical point except at the vertices, this means that the directionalderivative of � in the direction of a side cannot vanish anywhere along theside, except at the vertices. Thus � is strictly monotonic along each side.Suppose T is spherical. As we remarked in proof of the previous lemma,�+ � < 1. Therefore, � < 1=2 and the function � has a local minimum atB by Proposition 2.2 (ii). Since � is strictly monotonic along the side ABof T; we have �(A) > �(B). We have seen in the �rst part of the proof thatA and B are the only points where � can possibly have a global minimumon T: It follows that z0 = B is the unique point where this global minimumis attained.If T is hyperbolic, it follows similarly that z0 = B is the unique pointwhere the global maximum of � on T is attained. 218

3 ApplicationsWe recall that a region � C is called hyperbolic if and only if thereexists a meromorphic universal covering map f from the unit disc onto .This is the case if and only if the complement of in C contains at leastthree points.If � C is hyperbolic, the hyperbolic density � at a point w 2 is de�ned as the ratio of the hyperbolic length element at any preimagepoint of w under f to the euclidean length element at w: This de�nition isindependent of the choice of f and of the preimage point w . Explicitly,�(w) = 2(1� jzj2)jf 0(z)j ; if w = f(z):As our �rst application of the results of the previous section, we give anew proof of the following theorem of A. Baernstein II and H. Montgomery.Theorem 3.1 Let = CnLj; were Lj is one of the following two lattices:L1 = fa+ bi : a; b 2 Zg; L2 = (a+ b 12 + p32 i! : a; b 2 Z) :Then the set of minimum points of the hyperbolic density � of consistsof the points congruent modulo Lj to (1 + i)=2 if j = 1; or to either12 + p36 i or 1 + p33 iif j = 2:Remark. L1; a square lattice, and L2; a hexagonal lattice, are, upto rotation and stretching, the two lattices that have non-trivial rotationalsymmetries. In the case of the second lattice L2; the region is importantbecause of its connection to the conjectural extremal functions for Bloch'sand Landau's constants. See [Ba] for a discussion of this problem, and see[Ba-V] for a proof of the stronger result that in the case of L2; not onlyare the speci�ed points z0 minimum points for �L2 for the �xed lattice L2;but the pairs (L2; z0) (z0 one of the points speci�ed in the statement ofthe theorem) are local minimum points for �L(z) if the lattice L is alsoallowed to vary (given a certain constraint) together with the point z: Thework of H. Montgomery is not phrased in terms of the hyperbolic density.For a discussion of how Montgomery's Theorem implies the hexagonal caseof Theorem 3.1, see [Ba-V]. 19

Figure 3:

Proof. The density � is doubly periodic, so we need only minimizeit over one period parallelogram. From the further symmetries, we mayconsider � restricted to right euclidean triangles Ej for j 2 f1; 2g as shownin Figure 3. The right-hand side of Figure 3 shows the square (above) andhexagonal lattice (below), together with lines of symmetry. In each case, theright triangle over which it su�ces to minimize � is outlined in bold. Theleft-hand side of Figure 3 shows the unit disc as the universal covering spaceof ; with the hyperbolic triangle outlined in bold as one of the inverse imagetriangles of the bold triangle on the right. Using the standard notation ofthe previous section we have ~�1 = 1=2; ~�1 = 1=4; ~ 1 = 1=4; and ~�2 = 1=2;~�2 = 1=3; ~ 2 = 1=6: The universal covering map of the unit disc onto is the analytic continuation of a triangle map fj:Tj ! Ej , where �j = ~�j ;�j = ~�j ; and j = 0 for j 2 f1; 2g: Note that log �(fj(z)) = log 2 � �(z);where � is as in the previous section. Therefore, the minimum of � occursat the maximum of �; which by Corollary 2.9 is at the vertex whose anglemeasure is preserved by f and less than �=2: The location of this vertex,modulo Lj ; is located precisely as stated in the theorem. 2For our second application, we de�ne the spherical density � at apoint w of a hyperbolic region � C as follows. Let f be a meromorphicuniversal covering map from the unit disc onto : Then �(w) is the ratioof the hyperbolic length element at any preimage point of w under f to thespherical length element at w: Again this de�nition is independent of the20

choice of f and of the preimage point w . Explicitly,�(w) = 1 + jwj2(1� jzj2)jf 0(z)j if w = f(z):Since we have �(w)!1 as w ! @; one should expect that the minimumpoints of � are located \as far" from the boundary of as possible. In [B-C]we established methods to prove that the minimum points � in symmetricsituations are located where they should be. The following theorem coverssome cases that could not be treated with the methods in [B-C].We need the following well-known fact about tesselations of the Riemannsphere by spherical triangles, cf. [Ca]. If T is a spherical triangle, and were ect T through its edges, then we obtain new spherical triangles. If wesuccessively repeat this process with the new triangles, then we obtain a�nite tesselation of the Riemann sphere by non-overlapping triangles if andonly if T has angle measures at its vertices A;B;C given by �=k; �=`;�=m; respectively, where k; `;m are integers satisfying1=k + 1=`+ 1=m > 1:In this case, we say T generates a �nite tesselation of the Riemann sphereof type (k; `;m). The re ections along the sides of T generate a �nite groupof isometries of C; the triangle group corresponding to T .Theorem 3.2 With the above conventions let T be a spherical triangle gen-erating a �nite tesselation of the Riemann sphere of type (2; `;m); ` � 3.Suppose is the complement of the orbit of the vertex C of T under thetriangle group � generated by T . Then the the set of minimum points of thespherical density � of is equal to the orbit of the vertex B of T under�. Before proving Theorem 3.2, we state some corollaries and discuss itssigni�cance.Corollary 3.3 Let be the complement (in the Riemann sphere) of then-th roots of unity (n � 3:) Then � is minimal at z = 0 and z =1:Proof. Take for T the circular-arc triangle with vertices at z = 0; z = 1;and z = e2�i=2n: This is the case (2; `;m) = (2; n; 2): 2Corollary 3.4 Let be the complement (in the Riemann sphere) of thevertices of a regular polyhedron P inscribed in the Riemann sphere. Then21

� is minimal at the radial projections to the Riemann sphere of the centersof the faces of the inscribed polyhedron. Put another way, the minimumpoints of � occur at the vertices of the inscribed polyhedron dual to P:Proof. Choose a vertex V; an edge L; and a face F of P such thatL is one of the edges of the face F; and such that L meets V: Take forT the great circle triangle whose vertices are V; the radial projection tothe Riemann sphere of the midpoint of L; and the radial projection tothe Riemann sphere of the center of the face F: Here we are in one ofthe cases: (2; `;m) = (2; 3; 3) (tetrahedron), (2; `;m) = (2; 4; 3) (cube),(2; `;m) = (2; 3; 4) (octahedron), (2; `;m) = (2; 5; 3) (dodecahedron), or(2; `;m) = (2; 3; 5) (icosahedron).The triangle T is illustrated for the tetrahedron in Figure 4. 2Figure 4:By a di�erent method, in an earlier paper [B-C], we were able to proveCorollary 3.3, and also Corollary 3.4 in the case that the inscribed polyhe-dron was a tetrahedron, cube, or octahedron. Thus, Theorem 3.2 shouldperhaps be regarded as providing a new proof for �nding the minimum of �for some of the regions considered in [B-C]. Nonetheless, methods like thoseused in [B-C] do not seem adequate to prove Corollary 3.4 in the case ofthe dodecahedron or icosahedron, and so for these two regions Theorem 3.2provides the �rst proof that the minimum points for � are as expected.We remark, that as discussed in [B-C], knowing the minimum value for� for a region ; allows one to answer the questionQuestion 3.5 Given a hyperbolic region in the Riemann sphere, what issup ff ](0) : f :D! a holomorphic map g?Here D denotes the unit disc, and the spherical derivative f ](z) is de�nedby f ](z) = 2jf 0(z)j1 + jf(z)j222

if f(z) 6=1; and f ](z) = 2j(1=f)0(z)j if f(z) =1: The spherical derivativef ](z) measures how much f distorts length, locally at z; if the length in theimage is measured with respect to the spherical metric, and length in thedomain is measured with respect to the ordinary euclidean metric. Thus,combining Corollary 3.4 with the numerical computations in the last sectionof [B-C], we have that if f is a meromorphic function on the unit discwhose image is contained in the complement of the vertices of a regulardodecahedron or icosahedron inscribed in the Riemann sphere, thenf ](0) � 1:350058in the case of the icosahedron (12 vertices) andf ](0) � 1:079998in the case of the dodecahedron (20 vertices). (See [B-C] for the preciseupper bounds expressed in terms of the �-function.)Figure 5:~hh

FeF ~gg�~ � 1 � 2 ���� ��

Proof of Theorem 3.2. Finding the minimum points for � is equivalentto �nding the minimum points of � = log �: By symmetry, we only need tominimize � over the closure of the triangle T: We let F denote the universalcovering map of ; and we let eF denote its locally de�ned inverse. Weremark that the inverse image under F of the circular-arc triangle T; willbe a union of geodesic triangles in the unit disc. We only need to considerF restricted to the closure of one of these triangles. Rather than considerthe map F from the hyperbolic triangle to the spherical triangle, we will�nd it easier to factor F through a euclidean triangle, so we can apply ourmain theorem in the form of Corollary 2.9. That is, we write F = g � h;23

where h conformally maps the interior of the hyperbolic triangle onto theinterior of a euclidean triangle, and g conformally maps the interior of thateuclidean triangle onto the interior of the spherical triangle T: We choosethe intermediate euclidean triangle so that g and h both preserve the twoangles preserved by F: We denote by ~g and ~h the inverse mappings of gand h; and so we have eF = ~h � ~g: This setup is illustrated in Figure 5.Notice that the measure of the angle that is not preserved is increasedas we move from left to right in Figure 5. Observe further that�(z) = log(1 + jzj2) + log j eF 0(z)j � log(1� j eF (z)j2)= log(1 + jzj2) + log j~g0(z)j + log j~h0(~g(z))j � log(1� j~h(~g(z))j2)= log(1 + jzj2) + log j~g0(z)j � log jh0(~h(~g(z)))j � log(1� j~h(~g(z))j2)= �+(z)� ��(~h(~g(z)));where �+(z) = log(1 + jzj2) + log j~g0(z)j;and ��(w) = log(1� jwj2) + log jh0(w)j:The point is that �+ and �� are very similar sorts of functions. In particu-lar, they are both built out of a conformal map from a circular-arc triangleto a euclidean triangle. In other words, they are precisely the kind of func-tion � we considered in section 2. By Corollary 2.9, �+ attains its minimumat the vertex of T which is interior to and with internal angle measure< �=2: Similarly �� attains its maximum at the vertex of the hyperbolictriangle which is inside the unit disc and which has internal angle measure< �=2: Thus, ���(~h(~g(z))) attains its minimum at exactly the same placewhere �+(z) attains its minimum. Since � is the sum of these two functions,it must be minimal at exactly the same location. 2References[Ba] A. Baernstein II, Landau's constant, and extremal problems involv-ing discrete subsets of C; in V. P. Havin and N. K. Nikolski, eds.,Linear and Complex Analysis Problem Book 3, Part II, LectureNotes in Mathematics 1574, Springer-Verlag, 1994, 404{407.24

[Ba-V] A. Baernstein II and J. P. Vinson, Local minimality results relatedto the Bloch and Landau constants, in P. Duren, et. al., eds., Qua-siconformal Mappings and Analysis, Springer-Verlag, 1998, 55{89.[B-C] M. Bonk and W. Cherry, Bounds on spherical derivatives for mapsinto regions with symmetries, J. Anal. Math. 69 (1996), 249{274.[Ca] C. Carath�eodory, Theory of Functions of a Complex Variable, Vol-ume Two, Chelsea, New York, 1960.[Mo] H. L. Montgomery, Minimal Theta Functions, Glasgow Math. J. 30(1988), 75{85.[Ne] Z. Nehari, Conformal Mapping, Dover, New York, 1952.[R-W] S. Ruscheweyh and K.-J. Wirths, On extreme Bloch functions withprescribed critical points, Math. Z. 180 (1982), 91{105.FB-Mathematik, TU-Berlin, Sekr. MA 8-2, Stra�e des 17. Juni 136, 10623Berlin, Germany.

25