Isoperimetric inequalities and Dirichlet functions of Riemann surfaces

Publicacions Matemàtiques, Vol 38 (1994), 243-253 . ISOPER ;IMETRIC INEQUALITIES AN D DIRICHLET FUNCTIONS O F RIEMANN SURFACE S JOSÉ M . RODRÍGUE Z A,bstract We prove that if a Riemann surface has a linear isoperimetric inequality and verifies an extra condition of regularity, then ther e exists a nonconstant harmonic function with finite Dirichlet integral in the surface . We prove too, by an example, that the implication is not tru e without the condition of regularity . 1 . Introduction . In this paper we study the relationship between linear isoperimetri c inequalities and the existence of harmonic functions with finite Dirichle t integral on Riemann surfaces . By S we denote a Riemann surface (whose universal covering spac e is the unit disk q ) endowed with its Poincaré metric, i . e . the metri c obtained by projecting the Poincaré metric of the unit disk : ds 2(1 — 1With this metric, S is a complete Riemannian manifold wit h constant curvature -1 . The only Riemann surfaces which are left ou t are the sphere, the plane, the punctured plane and the tori . We shall say that a Riemann surface s satisfies a "linear isoperimetri c inequality" ( LII ) if there exists a finite constant h(S) so that far ever y relatively compact open set G with smooth boundary we hav e A(G) h(S) L(aG) . Here and from now on, A, L, d and B refer to Poincaré area, length , distance and open ball of s . There are connections between LII and some conformal invariants o n Riemann surfaces : the bottom of the spectrum of the Laplace-Beltram i operator, b(s) , and the exponentof convergence S (S) : 244  J . M . RODRÍGUE Z Theorem A . ([Ch], [B, p . 228], [FR]) . A Riemann surface S satis - fies a linear isoperimetric inequality if and only if b(S) > O . In fact , 4 < b(S) h(S) 2  and  b(S) h(S) < 2 The next result is a well known theorem of Elstrodt-Patterson-Sulli - van : Theorem B . [S, p . 333] . A Riemann surface S satisfies a linea r isoperimetric inequality if and only if S ( S ) < 1 . In fact ,  ŏ 1  if oCS ( S ) < 1 '  4  _  - 2 ' b(S) =  1 S ( S ) (1 — 6(S)), if 2 — ç «S) ç 1 . A theorem of Myrberg [T, p . 522] states that if S ( S ) C 1 (if S satisfies a LII ) then S has a Green ' s function (S O G in the language o f classification theory) . If S is a plane domain (in fact, if S is a surfac e of almost finite genus [SN, p . 193D, S has Green's function if and onl y if S has a nonconstant harmonic function with finite Dirichlet integra l [SN, p . 194] (s 1 0H D in the language of classification theory) . One would like to understand the relationship between the classe s 0 xD and ,l3 (the Riemann surfaces which do nat satisfy a LII ) . As w e have said above, in the case of surfaces of almost finite genus, OG = V HD C B . The inclusion is strict, as it is shown by the example Sa = q \ (U 1 {2_ k } U {O}) : So O G because it is a plane domain whos e boundary has positive logarithmic capacity [T, p . 81] ; So E B becaus e U =1 {2 —k } U {o} is a discrete set with an accumulation point in q [FR , Theorem 4j . The inclusion 0HD C B is true, in general, with an extra hypothesis : Theorem 1 . Let S be a Riemann surface which satisfies a linea r isoperimetric inequality . lf there exists in S a set of disjoint simple close d curves { y~ }T_ 1 , such that S \ contains n connected components o f infinite area S 1 ,.. . , S n , then dim HD(S) ~ n . This inequalityis the best possible . Here ~-I D (S) denotes the (real) linear space o f harmonic functions i n S with finite Dirichlet integral . The inclusion D HD C B is not true in the general case, even wit h the extra customary hypothesis of bounded geometry [K], which in our ISOPERIMETRIC INEQUALITIES AND HD-FUNCTIONS  24 5 contextmeans that the injectivity radius t(S) is positive . t(S) is define d as t(S) = inf {L(p) : p E S} , where t(p) is the injectivity radius of the geodesic exponential map centered at p . Theorem 2 . There exists a Riemarin surface 7Z E VH D , with c(7Z) ~ O, which satis fies a linear isoperimetric inequality . Acknowledgements . I would like to thank J . L . Fernández for man y useful conversations about these results, and J . Llorente for his carefu l reading of the manuscript and for some helpful suggestions . 2 . Proof of Theorem 1 . Without loss of generality, we can assume that {y i } 7 3 1 1 are simpl e closed geodesics . If this is not the case, we can substitute each curve b y the geodesic in its same free homotopy class . Let S k be a component of infinite area of S \ U j r y 3 , and let S Z be th e Schottky double of [Jk (see [AS, p . 26] for the definition) . Claim . S I *, satisfies a linear isoperimetric inequality . If the claim is true, the theorem of Myrberg [T, p . 522] states that S ~ has Gree n ' s function . This implies that the Royden's harmonic boundar y of S I *, is not empty [SN, p . 1661 . S' k ' is symmetrical with respect to 9 S k , a compact set which separate s S I *, in two connected components . Then the Royden's harmonic boundary of S~ is also symmetrical with respect to O5 k , and contains at leas t two points, one of them corresponding to Sk (one of them is in the closur e of S k in the Royden's compactification of S ; :) . This is true for k = 1, . . . , n . Therefore, the Royde n ' s harmonic bound - ary ofS contains at least n points [SN, p . 191] . This is equivalent [SN , p . 166] to dim HD (S) ~ n . This inequality is the best possible : Let 7Z be the Riemann surface given by Theorem 2 (7Z will be constructed without any mention to Theorem 1) and consider R n , a ncovering of 7Z based in a closed simple geodesic y C R . 7Z, satisfie s the hypothesis of this theorem and also dim .HD (7Z n, ) = n (the Royden' s harmonic boundary of 7Z 7z consists of n points, because the Royden' s harmonic boundary of 7Z consists of one point (7Z E 0 HD / 0 G ) [SN, p . 166]} . 246  J . M . RODRÍGUE Z To finish the proof of Theorem 1, we only need to prove the Claim : By a geodesic domain in a Riemann surface we mean a connecte d domain G with finite area, such that aG consists of finitely many close d simple geodesics . G does not have to be relatively compact since it ma y "surround " finitely many punctures . The following lemma will be very useful : Lemma . [FR, p . 168] . A Riemann surface satisfies a LII if an d only if it satis fies LII for geodesic dornazns . Moreover, if h and h g are , respectively, the usual and geodesic isoperimetric constants, the n h g çhç2+h g . Therefore, we must verify LII only for geodesic domains of SZ . By th e symmetry of S~ and the LII of S, we just need to check this for geodesi c domains which are symmetrical with respect to OS k . Then, we mus t verify A(G) c c LAG ) for geodesic domains G of S k , such that áGf1áS k ~ 0, where óoG mean s aoc ac~as k . Consider the open sets C t = {p E S k : d(p, c7S k ) < t} for positive t . Let G t be the geodesic domain "correspondin g " to C t (each puncture o r boundary curve of G t is freely homotopic to a boundary curve of Ce) . I f G t is empty for all positive t, then S k is a doubly connected domain ( a funnel), S~ is an annulus, and the claim is true with constant 1 . Then , we can assume that G t is connected and not empty for t ~ to . G t i s non decreasingin t, and if t i < t 2 are such that A(Gtx ) < A(G t2 ) , th e constantcurvature -1 and the GaussBonne t theorem give A( G t , )+ 2ir Ç A(G t2 ) . This implies that there exists a positive number T such that G t = G T forallt~T,orA(G t ) ---+ ooast —> oo . The first possibility is easy : there are only a finite number of geodesi c domains . Without loss of generality, we can asssume that A(G t ) —> o 0 ast -~oo . Case 1 . A(G) > 2 h(S) Q, with P = E7_ 1 L(ryj ) . In this case , 2 h(S) ~ < A(G) < h(S) L(aG) h(S) (L(D 0 C) + Q ) and e < L(aoc) . 15OPERIMETRIC INEQUALITIES AND HD-FUNCTIONS  24 7 Therefore, A(G) < 2 h(S) L(aoG) . Case 2 . A(G) < 2 h(S) e . Let S2 be a geodesic domain in S k such tha t ase c ase  and  A(S2) > 2 h(S) Q . We can choose 12, for example, as the first geodesic domainG t satisfyin g A(G t ) > 2 h(S) .e . We defin e a min {L('y) : yclosed simple geodesic, y C a} , b max {L('y) : y closed simple geodesic, y c 5 0 9} . Since A(f2) ~ A(G) and fZ n G ~ sá, one of the two next possibilitie s holds : Case 2 .1 . There exists a closed simple geodesic y C S2 n aoG . The n L(áoG) > L(ry) > a . Case 2 .2 . There exists a closed simple geodesic 'o in aoG, which meet s someyC a ofZ . Then, the Collar Lemma [R] says that L ( r /) ~ 4do, where do (th e width of the greater collar of 'y) satisfies b cosh do > coth L 2 ry) coth 2 , and do ~ . ~ D arc cosh coth 2 Randol [R] states the Collar Lemma if the surface is compact, but th e same proof, without any change, works for a general Riemann surface . Therefore, L(8oG) > L('q) > 4D . In both cases (2 .1 and 2 .2) LAG) > min{a, 4D} co . The n A(G) < h(S) (LAG) + P) < h(S) (L(c) + Q L(8oG) 1 c o J 248  J . M . RODRÍGUE Z and A(G) < h(S) (1 + ~ I L(aoG) . Obviously, .e > a ~ co and 1 + . e/ co > 2 . Therefore, in any case , A(G) < h(S) ( 1+ ~ ) LAG) . Consequently, h(Sk)<2+h y (Sk)<2+h(S)I1+ ~ I , l and the proof of Theorem 1 is now complete . ■ 3 . Proof of Theorem 2 . The desired Riemann surface 7Z will be obtained with the help of a graph G . We will construct this graph in three steps . In the set of vertices of any connected graph we can define a natura l distance : d(p, q) = inf {length of the paths from p to q } This will be " the distance " in all graphs of this section . First, let T be the infinitecomplete binary tree with root ro . Secondly , let Vn be the subset of 2n vertices of T at distance n of ro . We ca n construct new graphs Gn (n > 1) with vertices Vn . In G 1 there is on e edge between the two vertices of V 1 . The edges of G n (n > 2) are chose n as follows : 2' — 1 vertices of Vn are connected by a completebinary tre e with 2 n—1 leaves and with root r n (in any way); we add another edg e between r n and the last vertex v n of V n . In this way, the degree of th e vertices of Gn is one (if the vertex is a leave) or three (if the vertex i s not a leave) . The leaves are at distance n — 1 of rn, except for vn whic h is at distance 1 . Hence, the diameter of Gn is 2n — 2, if n > 2 . Finally, we are ready to construct the graph G . The vertices of G ar e the vertices of T . The edges ofG are the union of the edges of T and the edges of G,, , , for all n ~ 1 . The root ro of G has degree two . Th e other vertices of G have degree four or six . To build up our Riemann surface 7Z, modelled upon the graphG, w e will need the so called Lóbell Y -pieces, which are a standard tool fo r constructing Riemann surfaces . A clear description of these Y -piece s and their use is given in [C, Chapter X .3] . ISOPERIMETRIC INEQUALITIES AND HD-FUNCTIONS  24 9 A LQbell Y-piece is a three-holed sphere, endowed with a metric of constant negative curvature -1, so that the boundary curves are geodesics . We also require that the lengths of the boundary curves are the same , say 2cx, and the distance between any two of these boundary curves is [3 , say . Then cx and ,3 are related b y sinh (2) sinh ~ 2  2 This is the unique restriction on a and  See [C, p . 248] far details . Fix cx and ,Q satisfying the aboye relation . A X-piece (*piece) is a four-holed (six-holed) sphere, endowed wit h a metric of curvature -1, so that the boundary curves are geodesics o f the same length 2a . We can construct these pieces, for example, joinin g two (four) Y -pieces, by identifying corresponding boundary curves . If we nowput together these pieces following the combinatorial desig n of G, with the X-pieces (*pieces) in the place of the vertices of degre e four (six), we obtain a complete surface of constant negative curvatur e 1 . The only non-standard vertex is ro, which has degree two . Therei s not problem if we forget ro and consider that the two vertices of V I , ar e connected by a double edge . Since we have used only two distinct pieces to build up TZ, it is trivia l to see that t(7Z) > O . First of all, we will prove that 7Z E aHD . Let u be a harmonic functio n in R . with finite Dirichlet integral . Without loss of generality we ca n assume that u is a bounded function [AS, p . 203] [SN, p . 1781 . W e want to verify that u is constant . If u has limit at infinity, there is a point p in 7Z such that u(p) is th e maximum or the minimum of u in 7Z . The maximum principie implie s that u is constant . If u is nonconstant and has not limit at infinity, we can assume tha t u is positive an d lim sup u (z ) > 4  and  lim inf u(z) C 1 . z - .00  c o The maximum (minimum) principie implies that each connected com - ponent of the set {u > 4} ({u < 1}) is not a relatively compact set o f R . This implies that, for each n ~ no, there exist points pn, q n in th e pieces of 7Z corresponding to Gn, such tha t u(p n, ) > 4  and  u(q n, ) < 1 . 250  J . M . Ro D RÍGUE Z Since u is a positive harmonic function, the Harnack's Theorem say s that there exists a positive number ~ < Le7Z), independent of n, suc h that, u(z) > 3, for all z E B(p n , e) , u(z) c 2, for all z E B(q n , e) . Let the manifold with boundary 7Zn be the union of the pieces in 7 Z corresponding to the vertices V n of G n . We need a geodesic y n betwee n p n and gn, completely contained in 7Z n , which minimizes distance insid e 7Z n . To prove the existence of such geodesic, consider the Rieman n surface S~ {zEC : 1<Izi Cv 2 } , where the constant v is chosen so that the geodesic {IzI = v} has lengt h 2a, the length of each boundary curve of  If we join a copy of SZo - z {1 < ç v} in each boundary curve of 1?,n , we obtain a new Rieman n surface Te n . Since 7Z° is complete,there is a geodesic y, z between p n and q n such that the length L n of y n is equal to the distance between p n and q n . y n is "completely contained in 72 . n " , because if 'y enters in some cop y of S2o, it lies there forever . Consider now the Fermi coordinates (r, t) [C, p . 247], where r E [0, L n ] describes the curve y , z , and t E [—E, el describes the orthogonal geodesic s to yn . Observe that if we choos e E < 1 arc cosh N /cosh L(TZ) C t(Z ) 2 , (o, Ln) x (—E, e) corresponds injectively to a region A n c 7Z . Let u s denote by 7r this correspondence . Assume that there exists two points (r i,t 1 ) (r 2 ,t 2 ) in (0,L) x ( — E, E), corresponding to the same p EA n . By the definition of t(7Z), i t is not possible that (r 2 , t2) E B (( r 1 , 0), ¿(7? .)) . This implies Ir1 - r 2 l > 2E , because if d = d((r 2 , t 2 ), (r l , 0)), hyperbolic trigonometry [F, p . 92] give s cosh(r 2 — r 1 ) cosh t 2 - = cosh d ~ cosh t(Z) cosh 2 (2E) , and we hav e cosh(r 2 — r 1 ) cosh ~ > cosh(r 2 — 71 ) cosh t 2 ~ cosh (2E) cosh ~ . Then I r 2 — r 1 ~ > 2E, IS4PERIMETRIC INEQUALITIES AND HD-FUNCTIoNS  25 1 and d(7r( r i, O), 7r(r2, O)) < d (7r( r i, O), 7r( r i, t i)) + d err( rz , t z), 7r(r 2 , O ) ) =t 1 +t 2 <2e . But I r l - - r 2 l > 2E and d(7r(r 1 , o), 7r(r 2 , o)} C 2E contradict that -y n minimizes length between pn and gn . Therefore, (o, L n ) x (—e, corresponds injectively to a region A n c R . It is easy to see that for all t E (—E, E), if -yn  {ir(r,t) : o Ç r Ç Ln } , 1 < u(7r(0, t)) — u(7r(Ln , t)) = fvud s n f  1/ 2 IVuIds  J l(L()) 2 . f ;~  7r i But the metric in Fermi coordinates is expressed b y ds 2 = cosh 2 t dr 2 + dt 2 , and so f L n L() =  cosh t dr = Ln cosh t < Ln cosh ~ Ç 2 D n cos h o if D is the maximum of the diameters of the X-pieces and the * -pieces , because the diameter of Gn is 2n — 2 . This give s 1 1Vu1ds ~ f ;', 2 Dn cosh ' ~ Án, 1Vu1 2 — Dncosh e Therefore  q u I 2 > - 2 - 1 E   = no , ~  n ~ no cosh E n _ n a and so u 1 HD (7Z) . This proves that 7Z E GHD . ■ To prove that 7Z has a LII we need to precise the metric relationshi p between G and 7Z . Following Kanai's terminology [K], we say that a n application not necessarily continuous, between two metric space s 99 : ( M i, di) --> (M 2 ,d 2 ) and