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The geometry of finite topology Bryant surfaces

In this paper we shall establish that properly embedded constant mean curvature one surfaces in H^3 of finite topology are of finite total curvature and each end is regular. In particular, this implies the horosphere is the only simply connected such example, and the catenoid cousins the only annular examples of this nature. In general each annular end of such a surface is asymptotic to an end of a horosphere or an end of a catenoid cousin.


Introduction
In this paper we shall establish that properly embedded constant mean curvature one surfaces in H 3 of finite topology are of finite total curvature and each end is regular. In particular, this implies the horosphere is the only simply connected such example, and the catenoid cousins the only annular examples of this nature. In general each annular end of such a surface is asymptotic to an end of a horosphere or an end of a catenoid cousin.
Robert Bryant discovered a holomorphic parametrization of (simply connected) mean curvature one surfaces in H 3 which can be thought of as a generalization of the Weierstrass representation of minimal surfaces in R 3 [2]. Each (simply connected) minimal surface in R 3 is isometric to a mean curvature one surface in H 3 (and vice versa); R. Bryant calls this the cousin of the minimal surface. This correspondence follows easily from Bonnet's existence theorem for surfaces in the space forms. This may have been R. Bryant's motivation to seek a meromorphic Weierstrass type representation of mean curvature one surfaces in H 3 .

Definition.
A Bryant surface is a surface in H 3 of constant mean curvature one.
The Weierstrass pair of a minimal (local) surface in R 3 is a pair of meromorphic data (g, ω). The cousin in H 3 has more local structure; in particular, one also has the hyperbolic Gauss map G. The surfaces are isometric so the metric is determined by (g, ω): ds = |ω| 1 + |g| 2 . However, the Gauss map G is fundamental to the geometry of the cousin in H 3 (G is also meromorphic on the minimal cousin in R 3 , but this seems never to have been considered).
An annular end of a finite total curvature minimal surface in R 3 is conformally a punctured disk and the Gauss map g extends meromorphically to the puncture. However, the Gauss map of an annular cousin in H 3 may have an essential singularity at the puncture; R. Bryant observed this for Enneper's minimal surface in R 3 and its cousin in H 3 [2]. An annular end in H 3 is called regular if it is conformally a punctured disk and G extends meromorphically to the puncture. This notion was introduced and developed for Bryant surfaces by M. Umehara and K. Yamada [21]. The idea of regular ends originated in the paper of R. Schoen [19], where he introduced and studied regular minimal annular end hypersurfaces in R n .
A properly embedded minimal annular end in R 3 of finite total curvature is asymptotic to an end of a plane or catenoid. In H 3 , a properly embedded Bryant annular end, regular and of finite total curvature, is asymptotic to an end of a horosphere or a catenoid cousin [15].
We will prove that a properly embedded Bryant annular end in H 3 is of finite total curvature and regular. This is the main result of our work and answers affirmatively a conjecture by M. Umehara and K. Yamada: there are no embedded irregular Bryant annular ends of finite total curvature [20].
When the annular end is part of a properly embedded Bryant surface, we prove it is asymptotic to a catenoid cousin end and not a horosphere end (unless M is equal to a horosphere). This is Theorem 12. This is quite different from properly embedded minimal surfaces in R 3 , where an annular end can be asymptotic to a catenoid or planar end, as in Costa's surface.
The analogous theorem for minimal annular ends in R 3 is not true: the helicoid has an annular end of infinite total curvature. The cousin of the helicoid in H 3 is not embedded. In fact, our initial motivation was the search for a properly embedded simply connected Bryant surface in H 3 , other than a horosphere (the cousin of a plane in R 3 ). Now we know there is no such simply connected surface.
It is still unknown if the helicoid and plane are the only properly embedded minimal surfaces in R 3 that are simply connected.
However, the geometry of properly embedded minimal M in R 3 , of finite topology and with at least two ends, is understood: M has finite total curvature; each annular end of M is asymptotic to a plane or catenoid end [3], [13].
There has been much important work done on the geometry of properly embedded annular H-ends in R 3 and in H 3 [9], [10], [12]. In R 3 , for H = 0, they prove such an end is asymptotic to a Delauney end [10]. Also it is proved that if H > 1 for such an end in H 3 then it is also asymptotic to a Delauney end [9]. In fact, the linking number argument of our Theorem 9 is inspired by the linking number argument of [10]. However this argument needs to be adapted to our situation. Essentially, because of the noncompactness of horospheres, we cannot use them directly as barriers. So, we will construct stable surfaces with sufficiently known behavior at infinity and use them as comparison surfaces with horospheres.
There are examples of higher genus, mean curvature one surfaces in H 3 of finite topology. Many such examples have been constructed by W. Rossman, M. Umehara and K. Yamada and computer images indicate many of these surfaces may be embedded [18]. In R 3 , N. Kapouleas has constructed many properly immersed and embedded H-surfaces by desingularizing certain families of touching spheres. We hope that this may be done in H 3 , by desingularizing certain families of touching horospheres. For example, consider the three horospheres intersecting in three points as in Figure 1-a. One should be able to attach catenoid cousin necks near the three singular points and show there is a Bryant surface in a neighborhood of this new surface by Schauder fixed point techniques. This surface would have genus one and three ends, each asymptotic to a catenoid cousin end; see Figure 1-b. The paper is organized as follows. In Section 2, we give a (brief) description of the Bryant representation; the interested reader may consult [2], [15] and [21] for a serious discussion.
In Section 3, we analyze the connected component of the intersection of a Bryant surface in H 3 with its tangent horosphere at a point. There is more structure here than the trace of a minimal surface in R 3 on its tangent plane at a point. We describe here this trace for properly embedded annular ends E with E ∩ H(q) compact, q ∈ E, and ∂E ∩ H(q) = ∅; H(q) is the tangent horosphere at q.
In Section 4 we study properly embedded Bryant annular ends E, which are not dense at infinity. We first prove this end is regular: it is conformally the punctured disk and G extends meromorphically to the puncture (Theorem 1). We then prove the asymptotic boundary of the end E is precisely the limiting value of G at the puncture (Theorem 2).
In Section 5 we continue the study of properly embedded annular ends E assuming E is regular. We prove E then has finite total curvature. This is done by first proving such an end E has finite total curvature if it is on the mean convex side of a catenoid cousin (Theorem 3). Then we prove this end can be placed on the mean convex side of a catenoid cousin (Theorem 4). This last result requires an analytic theorem concerning H = 1 graphs over noncompact domains (Theorem 5).
In Section 6 we prove the nondensity at infinity of finite topology properly embedded Bryant surfaces. Using the trace on horospheres, we show that if M is dense at infinity there is a proper arc γ on M with ∂ ∞ γ two distinct points at infinity. Using M as a barrier we construct stable H = 1 surfaces with boundary γ. Analyzing the asymptotic behavior of stable surfaces we see that such stable surfaces cannot exist.

The Bryant representation
Let L 4 be Minkowski 4-space with the Lorentzian metric of signature (−, +, +, +). Hyperbolic 3-space can be represented as with the metric induced from L 4 . It is useful to identify L 4 with the space of 2 × 2 hermitian matrices: a point (t, Notice that H 3 is the set of such matrices of determinant one, t > 0, and one has Let M be a simply connected Riemann surface and F : M → SL(2, C) a holomorphic immersion satisfying: Then f = F F * : M → H 3 is a conformal immersion of mean curvatureone. If H ∈ SU(2) then f = F 1 F * 1 where F 1 = F H. Conversely any mean curvature one surface in H 3 is given locally by such an F . The reader should consult [2] and [21] for the details.
In the upper half-space model of H 3 , one can express the immersion in terms of F .
The Weierstrass data of the minimal cousin in R 3 are given by: Then one obtains The metric induced on M is ds = |ω| 1 + |g| 2 .

The tangent horosphere
For M an immersed surface in H 3 and q ∈ M , the tangent horosphere H(q) of M at q is the horosphere tangent to M at q whose mean curvature vector at q has the same direction as that of M at q. This horosphere is unique when the mean curvature of M at q is nonzero.
Note that H(q) separates H 3 into two components. We let H(q) + denote the mean convex component bounded by H(q) and we call it the inside of H(q). The surfaces at a constant distance t from H(q) are also horospheres with the same point at infinity as H(q) and they foliate H 3 . We denote this equidistant horosphere by H t (q), and for t > 0, H t (q) will be inside H(q) and outside H(q) for t < 0. Now suppose M is a Bryant surface properly embedded in H 3 . We allow M to have a compact boundary since many of our results concern the ends of such surfaces. We also assume M is not a part of a horosphere.
For q ∈ int(M ), the intersection of M and H(q) is an analytic curve near q with isolated singularities, and at the singularity q, there are 2k + 2 smooth branches meeting at equal angles where k is an integer at least one. In fact, k is the same as the order of contact of the cousin minimal surface in R 3 with its tangent plane, and k − 1 is the order of q as a branch point of the Gauss map G.
When ∂M = ∅, M separates H 3 into two connected components since M is properly embedded. We let W denote the mean convex component bounded by M . When ∂M = ∅ and is compact, we introduce a mean convex component W as follows. It is not hard to see that there is an embedded compact orientable surface Σ such that ∂Σ = ∂M and Σ ∩ int(M ) = ∅ (take a large ball B of We will now derive properties of E ∩H(q) where E is a properly embedded Bryant annular end (homeomorphic to S 1 × [1, +∞[). We henceforth assume ∂E ∩ H(q) = ∅ and E is topologically the unit disk punctured at the origin.
Proof. If this were not so, then E 1 would be compact and ∂E 1 ⊂ H(q). Consider a "large" horosphere H t (q) outside H(q) such that E 1 ⊂ H t (q) + (so t is near −∞ in our notation). Then increase t: since E 1 is compact there will be a largest t 0 such that H t 0 (q) touches E 1 at a point x ∈ int(E 1 ). But then E = H t 0 (q) by the maximum principle, a contradiction; cf.
Proof. Consider a "small" horosphere H t (q) contained in H(q) + − Q 1 (so t is near +∞). When t decreases, there will be a positive t 0 where H t 0 (q) touches Q 1 for the first time. The point x where they touch is in E 1 and by the maximum principle, the mean curvature vector of E 1 at x is the negative of that of H t 0 (q) at x. So this mean curvature vector points into Q 1 and Q 1 is mean convex; see Figure 2 Lemma 3. There is at most one compact component at q of E − H(q), whose boundary is in H(q); by "at q" is meant a connected component of E − H(q) containing q in its closure.
Proof. Suppose this fails. Then at least two components E 1 , E 2 at q are compact and (∂E 1 ∪ ∂E 2 ) ⊂ H(q). By Lemma 1, we know that Since E is a graph over H(q) near q, the mean curvature vectors (which we denote by − → H ) of E 1 and E 2 point into the same connected component C of points to the noncompact component of H(q) + − E 2 ; this contradicts Lemma 2; see Figure 3.
If C is noncompact, then − → H (q) points into C, so along E 1 , − → H points into C as well. But − → H points into Q 1 along E 1 by Lemma 2. This proves Lemma 3.  Proof. Let 2k + 2 denote the number of branches of E ∩ H(q) at q. Let f : D → E be a local parametrization of a neighborhood of q on E by a disk of C so that f (0) = q and lines passing through 0 of slope an integral multiple of π k+1 are sent to E ∩ H(q). Let A i be the sector of D defined by We know that the B i are alternatively in H(q) + and H(q) − around q.
If B 1 and B 3 are not in the same component of E − H(q), then by the observation above, B 2 yields a third component so that Lemma 4 is true.
If B 1 and B 3 are in the same component of E − H(q), we can construct a cycle α 13 on E as follows: let a 1 , a 3 be two points of A 1 , A 3 respectively, and β a path in E − H(q) from f (a 1 ) to f (a 3 ); α 13 = β ∪ β ′ where β ′ is the image by f of the line segment from a 1 to 0, followed by the line segment from 0 to a 3 . Now α 13 meets H(q) exactly at q.
If B 2 and B 4 were in the same component, we could find a cycle α 24 on E which meets α 13 in a single point, which is impossible since the genus of E is zero. Thus we get at least three components in this case as well.
Now we study the case of equality. In this case there is only one component at q in either H(q) + or H(q) − . Then we can assume that all the B i for i odd are in the same global component of E − H(q). If k ≥ 2 this means that we can construct cycles α 15 (with A 1 and A 5 ), and α 35 (with A 3 and A 5 ) exactly as we constructed α 13 . As before, these three cycles separate the components of B 2 , B 4 and B 6 in E − H(q). Hence we obtain at least four components at q in this case. This completes the proof of Lemma 4.
We define Σ(q) to be the connected component of q in E ∩ H(q) and we assume Σ(q) is compact in the rest of this section.
Lemma 5. E − H(q) has exactly three components at q: B, E 1 and E 2 . The first, B, is compact and contains ∂E; E 1 is compact with boundary in H(q); E 2 is noncompact (recall the assumption that ∂E ∩ H(q) = ∅ and Σ(q) is compact).
Proof. First we prove E − H(q) has at most one noncompact component at q. This is immediate if E ∩ H(q) is compact (in fact E ∩ H(q) will always be compact until we arrive at Theorem 7 of this paper. There we will need to work with the weaker hypothesis: Σ(q) is compact). Let F be the noncompact component of E − Σ(q). For ε > 0, ε small, the points of F a distance ε from Σ(q) form a compact curve C disjoint from H(q). Since E is an annulus, C is in fact connected. This Jordan curve C separates Σ(q) from the puncture.
With the notation of Lemma 4, if two B i and B j (i = j) are in noncompact components, then a curve in E − H(q) from q to the puncture starting in B i (and B j ) meets C. So B i and B j are in the same global component.
Then by Lemma 4, there are at least two components at q of E − H(q) that are compact. At most one may contain ∂E, and the others are compact with boundary in H(q) (and there is at least one). But such a component is in H(q) + by Lemma 1 and is unique by Lemma 3. We call E 1 this unique component. Then the other compact component necessarily contains ∂E -we call it B -, and the third component (E 2 , say) is noncompact.
Finally E − H(q) has exactly three components whose closure contains q and by Lemma 4, the order of contact k of E with H(q) is one. Lemma 6. There are two possibilities: figure eight; the union of two Jordan curves C 1 , C 2 on H(q) meeting at q.
Proof. First notice that ∂E 1 contains no cycle c disjoint from q. To see this, let γ be a path in E 2 ∪ B ∪ {q} going from ∂E to the puncture and meeting H(q) exactly at q. The cycle c does not meet γ so c bounds a compact domain D in E, D ∩ ∂E = ∅ and D = E 1 (since q / ∈ c = ∂D). But D would contain a compact domain outside H(q) with boundary in H(q) contradicting Lemma 1. This proves each cycle in ∂E 1 meets q, E 1 is a disk and ∂E 1 − {q} is an embedded curve.
We know that we have locally at q exactly four components Then q is not a double point of ∂E 1 and ∂E 1 = C 1 is a Jordan curve on H(q). On the contrary if B 3 is also in E 1 , then ∂E 1 = C 1 ∪ C 2 ; C 1 and C 2 Jordan curves meeting exactly at q; i.e., ∂E 1 is a figure eight.
Proposition 1. With the notation of Lemma 5: If ∂E 1 = C 1 ∪ C 2 , then every path starting at ∂E, staying in W , and diverging in H 3 must intersect H(q); Q 1 separates W . Moreover ∂E is outside H(q), and C 1 and C 2 are each homologous to ∂E on E.
Proof. The two possibilities are given by Lemma 6. If ∂E 1 = C 1 , then D is the disk of H(q) bounded by C 1 and D ∪ E 1 = ∂Q 1 . At q, − → H (q) must point into Q 1 (since it does so at points of E 1 near q) by Lemma 2 so that E 2 or B is inside Q 1 near q. It cannot be E 2 since E 2 is noncompact (the puncture is in E 2 ) and E 2 is properly embedded. Thus B is inside Q 1 near q. Thus, B ⊂ Q 1 and, in particular, ∂E ⊂ Q 1 . Now any path starting at a point of ∂E and diverging in H 3 , must intersect ∂Q 1 = E 1 ∪ D; see Figure 4. This proves the first assertion of the proposition.
Locally at q, E 1 is a graph over two opposite sectors of H(q) (the projection of the B 1 and B 3 of Lemma 6). By the local structure of E 1 near q, the two complementary sectors are in ∂Q 1 , hence in D 1 . As D 1 is a disk on H(q), at least one projection of B 1 or B 3 , B 3 say, must be in D 1 . Now E 1 is a disk with two points on the boundary identified at q.
Then ∂E 1 − {q} = C 1 ∪ C 2 hence one of the two boundary arcs of B 3 − {q} (near q) is in C 1 and the other in C 2 . But then, points of C 1 are in the interior of D 1 , which is a contradiction.
Then ∂Q 1 = E 1 ∪D 1 ∪D 2 so Q 1 separates ∂E from infinity in W : any path starting at ∂E, in W and diverging in H 3 , must pass through Q 1 . Moreover B and E 2 are outside H(q), in particular ∂E is outside H(q); see Figure 5.
Then Σ(q) is a figure eight, the union of two Jordan curves C 1 , C 2 meeting at q. Figure 5 By the transversality hypothesis and the local structure at q, Σ(q) consists of two analytic curves meeting at equal angles at q. It is then a figure eight. Note that in the first case of Proposition 1, C 2 together with ∂E bounds B (other curves in ∂B would give rise to a compact component outside H(q)).
Corollary 1. Let E be a properly embedded Bryant annular end, q ∈ E with E ∩ H(q) compact and disjoint from ∂E. Then if ∂E ⊂ H(q) + every divergent path starting at x ∈ ∂E must intersect E ∪ H(q) at a point other than x; if ∂E ⊂ H(q) − then every divergent path starting at x ∈ ∂E and staying in W , must intersect H(q).

The regularity and asymptotic boundary
of an annular end not dense at infinity Theorem 1. Let E be a properly embedded Bryant annular end. If ∂ ∞ E = S ∞ , then E is conformally a punctured disk and the hyperbolic Gauss map G extends meromorphically to the puncture (i.e., E is regular ).
Proof. We will now work in the upper half-space model of First we will show that G is bounded on some subend of E. If not, then for some q n ∈ E, diverging on E, we would have |G(q n )| → ∞. Since ∂E is compact, we have x 3 ∂E ≥ δ > 0 for some δ. Choose n sufficiently large so that H(q n ) ∩ B R is below x 3 = δ. This is possible since x 3 (q n ) → 0; cf. Figure 6. However, ∂E ⊂ H(q n ) + and we can find a path from ∂E to G(q n ) which does not intersect E ∪ H(q n ) except at its endpoint (choose a path from a point of ∂E to a point of ∂B R , not meeting E; then choose n big enough so that H(q n ) is below this path, and then continue to G(q n )).
This contradicts Corollary 1, and so G is bounded. To prove Theorem 1, it suffices to prove that E is conformally the punctured disk.
We will prove this by constructing a complete metric on E of the form dσ = λ|dz| where λ is the module of a holomorphic function on E (R. Osserman [14]).
Let E be the universal cover of E, so that F : are the Weierstrass data and , introduced by M. Umehara and K. Yamada [20]. The . This immersion need not define an immersion of E in H 3 but the metric ds # is well-defined and nonsingular since Ψ # is an immersion: In particular, g ′ ω/G ′ is a nonvanishing holomorphic form.
Since |G| is bounded, the metric Let γ be a divergent path on E, which is proper so that γ diverges in H 3 . Now in the Lorentzian model of In particular t # (γ) → ∞ as well, and γ diverges in H 3 on the dual surface. Since ds # is the induced metric on the dual surface from its immersion in H 3 , the ds # length of γ is infinite. This proves Theorem 1.
Remark 2. The metric ds # gives information on values of the Gauss map G. Zu-Huan Yu has proved G is constant if G misses more than four points [22]; he proves more generally that ds # is complete. We have proved G can miss at most three points when M has finite total curvature (and M is not a horosphere) [4].
Theorem 2. Let E be a properly embedded Bryant annular end. If E is conformally the punctured disk D * and G extends meromorphically to the puncture, then ∂ ∞ E = G(0) (the value of G at the puncture).
Proof. As in the proof of Theorem 1, we work in the upper half-space model and assume G(0) is the point at infinity. First observe that Then (as in the proof of Theorem 1) G must be bounded; a contradiction.
Now suppose E accumulates at another point at infinity which we may assume σ = (0, 0, 0). Let C T be the cylinder There are points of E in C T for all T > 0. As q diverges on E, towards σ, G(q) tends to infinity. The geodesic normal to E at q is a half circle meeting x 3 = 0 at two points, one point close to σ (close in the metric dx 2 1 + dx 2 2 ) and the other point G(q) that is "far" from σ. Thus the mean curvature vector − → H (q) of E at q, tends to a vertical vector pointing up. As x 3 (q) decreases this vector becomes more vertical. Now choose T sufficiently small that q ∈ C T implies the angle between − → H (q) and − → e 3 = (0, 0, 1) is less than π/8.
Then for q ∈ E ∩ C T , the vertical segment going down from q to S ∞ does not meet E again, since E bounds a mean convex domain W (this makes sense since C T can be chosen far from ∂E). So E ∩ C T is a vertical graph u over a (possibly disconnected) planar domain. Now we prove that for T sufficiently small, E ∩ C T is a vertical graph over the whole base of C T : Since there can be no points of E below this graph, this contradicts σ ∈ ∂ ∞ E.
We now make useful gradient estimates for this graph u at q ∈ E ∩ C T in the Euclidean metric.
Consider the vertical plane Q containing the unit normal vector − → n to E at q. G(q) is also in this plane and we have Figure 7 in the plane Q. Here O is the center of H(q) and R is the radius of H(q). Then Thus the horizontal component of − → n has length l = 2au a 2 + u 2 , and the vertical Then the auxiliary function v = ln u has bounded gradient.
Thus E ∩ C T is a graph over the base of C T and Theorem 2 is proved.

Theorems 1 and 2 immediately imply:
Corollary 2. Let E be a properly embedded Bryant annular end. If ∂ ∞ E = S ∞ , then E is regular and ∂ ∞ E is the limiting value of G on E.

Finite total curvature of nondense annular ends
Theorem 3. Let E be a properly embedded Bryant annular end. If E is on the mean convex side of a catenoid cousin end, then E has finite total curvature.
Proof. First we make precise "the mean convex side." The ends of the family of catenoid cousins can be written as graphs (in the upper half-space model) over domains at infinity: These ends are asymptotically 1 r α , α > −1 and r 2 = x 2 1 + x 2 2 . Let C α be such a catenoid cousin end and extend C α to an embedded surface with no boundary by attaching the horizontal disk along ∂C α . The mean convex side of C α is then the component to which − → H points along C α (here − → H is pointing up). So our hypothesis on E is that E is contained in this mean convex side of C α . Clearly the catenoidal ends are ordered by α and we can assume α > 0.
Since ∂ ∞ E is the point at infinity, Theorem 1 applies and we know E is conformally a punctured disk D * = {0 < |z| ≤ 1}, and G extends meromorphically to 0. Parametrize so that G(z) = 1 z p for some integer p ≥ 1.
The end E is determined by F : E → SL(2, C), F = A B C D , with C = z ν f , f holomorphic in D * , and similar representations for A, B and D (this is proved in Lemma 7, following the present proof). We know that x 3 = 1 |C| 2 +|D| 2 . Suppose C = z ν f and f has an essential singularity at 0. Then for some sequence z n → 0, we have Let q n be the point on E corresponding to z n . Since E is above the catenoid cousin C α : so by the previous inequality for C(z n ), we conclude r(q n ) ≥ 1 |zn| (p+1) , and for any integer k > 1, and n sufficiently large: That is, the horizontal (Euclidean) distance from the point q n to the point s = (0, 0, x 3 (q n )) is at least 2 |zn| p . Observe that d(q n , G(q n )) is at least d(q n , s) − d(G(q n ), s) where d denotes the horizontal Euclidean distance.
Let l be the horizontal disk of diameter 2 |zn| p , centered at the point p = (G(q n ), x 3 (q n )). Since the horizontal distance from G(q n ) to (0, 0) is 1 |zn| p , the disk l is in the interior of H(q n ) + ; see Figure 8. Now the origin is under one of the boundary points of l. Observe that the catenoid cousin C α is above the segment [p, s] on l, since the height of C α at G(q n ) is asymptotically 1 |G(qn)| α = |z n | pα and x 3 (q n ) ≤ |z n | (p+1)α . Since the graph of C α is monotone decreasing with r, the segment [p, s] is below C α . Also, E is above C α so that [p, s] is disjoint from E.
Moreover let N be a compact embedded surface with boundary the boundary of E so that N ∪ E is an embedded surface. N can be chosen above the union of the catenoid cousin C α and the flat disk capping off C α . Then exactly as in Section 3, N ∪ E separates the ambient space so one can find a path γ from s to ∂E which meets the N ∪ E only at the endpoint (first vertical, then a fixed path). The k of the above inequality can be chosen large enough so that this path, together with the boundary of E, is inside H(q n ). But γ together with [p,s] can be extended to a divergent path disjoint from E ∪ H(q n ), by going down vertically to G(q n ) from p. This divergent path from ∂E ⊂ H(q n ) + does not meet H(q n ) ∪ E again, which contradicts Corollary 1.
Thus A 1 (y + 2πi) = A 1 (y)e iβ2π and similarly for B 1 , C 1 and D 1 . Now define f (y) = e −yβ A 1 (y), so that and f defines a holomorphic map f (z) on D * , by f (z) = f (y), e y = z.
Theorem 4. Let E be a properly embedded Bryant annular end. If ∂ ∞ E is not S ∞ then E has finite total curvature. Corollary 3. Let E be a properly embedded Bryant annular end. If ∂ ∞ E = S ∞ , then E is regular and the total curvature of E is finite. E is asymptotic to a catenoid cousin or horosphere end.
Proof of the corollary. Theorem 4 and Corollary 2 yield the facts that E is regular, ∂ ∞ E is one point and the total curvature of E is finite. Then the theorem of E. Toubiana and R. Sa Earp yields the asymptotic behavior [15].
Proof of the theorem. We know from Theorem 3, that E will have finite total curvature if we can find a catenoid cousin C α with E on the mean convex side of C α ; we will find such a C α to prove Theorem 4. By Theorems 1 and 2 we know that ∂ ∞ E is one point, which we take to be infinity in the upper half-space model of H 3 .
Let B be a ball in H 3 , whose interior contains ∂E and where E is transverse to ∂B. Let E 1 denote the noncompact component of E − B, and let W denote the mean convex domain (along E 1 ) bounded by E 1 and a compact domain on ∂B.
If x 3 ≥ c > 0 on E then C α can be constructed using a catenoid cousin end below height c which is a graph over an exterior domain x 2 1 + x 2 2 > r 2 0 , asymptotic to the plane x 3 = 0 at infinity. So we can assume there is a sequence For q ∈ E 1 , let γ be the minimizing geodesic of H 3 joining q to a point of ∂B. We will be working with q lower than B. Assume B = {x 2 1 + x 2 2 + (x 3 − 4) 2 = 1} for convenience, and x 3 (q) ≤ 1, r(q) > 6. Parametrize γ by arc length so that γ(0) is the highest point of γ (which is not on B by our choice of constants), and γ(t 0 ) = q with t 0 < 0.
Let P (t) be the family of (hyperbolic) planes orthogonal to γ at γ(t). For t very negative, P (t) is disjoint from E 1 since ∂ ∞ E 1 = ∞, and E 1 is proper so that there is a first t 1 ≤ t 0 (as t increases) such that P (t 1 ) touches E 1 at a point q 1 .
We do Alexandrov reflection of E 1 with the planes P (t) as t increases from t 1 to 0. Let S(t) be symmetry of H 3 through P (t), E 1 (t) + the part of E 1 on the side of P (t) not containing B, and E 1 (t) * = S(t) (E 1 (t) + ).
For t slightly larger than t 1 , E 1 (t) + is a graph over (part of) P (t), int (E 1 (t) * ) ⊂ W , and the angle between P (t) and E 1 (t) + is never π/2 along ∂E 1 (t) + . These properties continue to hold until the first t (t 2 say) such that E 1 (t 2 ) * touches ∂B, for if one of these properties failed to hold at some earlier t, P (t) would be a plane of symmetry of E. Then E is part of a properly embedded, mean curvature one, compact surface M , with ∂M = ∅. This is impossible.
Clearly t 2 < 0 since q is lower than B, and so the symmetry of q through some plane P (t), t < 0 meets B. Thus there is some point q ∈ E 1 (t 2 ) + such that S t 2 ( q) ∈ B.
Let δ 1 = dist( q, γ), and q t = S t ( q). Since γ is invariant by S t , we have dist(q t , γ) = δ 1 as well. For t = t 2 , q t is on ∂B, so that dist(q t , γ) ≤ diam(B) = δ. The curve q t joining q to ∂B, as t varies from t 1 to t 2 , is an equidistant curve β whose distance from γ is less than δ, and this equidistant curve is contained in W . We emphasize that this discussion is valid for any q ∈ E 1 with x 3 (q) < 1, r(q) > 6.
In particular, consider the sequence q n ∈ E 1 , satisfying x 3 (q n ) → 0, r(q n ) → ∞. Then a subsequence of the geodesics γ n joining q n to B converges to a vertical geodesic over B and the equidistant curves β n from q n to B are in W and a distance at most δ from γ n . So the equidistant curves β n are in the tubular neighborhood of γ n of radius δ. As n → ∞, the tubular neighborhoods converge to a vertical cone of hyperbolic width δ. Let C(δ) denote this cone; for simplicity we can assume the base of C(δ) is the origin. Now we can prove that E 1 ∩ A is a graph where A = {x 3 < 1, r ≥ 6}. Suppose this were not true. Let N be the Euclidean unit normal to E, N.
− → H > 0 and suppose that N 3 ≤ 0 at some point q ∈ E 1 ∩ A. Then the horosphere tangent to E at q, H(q), is at most of (Euclidean) radius 1 and ∂E ⊂ H(q) − .
Then by Corollary 1, H(q) separates W into three connected components. One is compact and contains part of ∂B. One is noncompact, and contains the points q n , n large. And the third is compact and inside H(q) + . But the equidistant curves β n are in W and disjoint from H(q) for n large; this is impossible since the β n go to ∂B in W . This proves E 1 ∩ A is a graph.
In fact the above argument proves much more: for q ∈ A, H(q) must intersect C(δ); otherwise the equidistant curves β n would be disjoint from H(q) for n large; cf. Figure 9.
For q ∈ E 1 ∩ A, let R be the Euclidean radius of H(q) and let d be the Euclidean distance of q to C(δ). Then (since C(δ) is invariant by homothety from σ and C(δ) ∩ H(q) = ∅) there is a λ > 0 such that 2R ≥ d ≥ 2λr(q), and λ depends only on C(δ). In particular R → ∞ when r(q) → ∞.
Now we shall prove that E is below some horosphere x 3 = constant. We know that E 1 ∩ A is the graph of a function u and in Theorem 2, we derived the formula: Since u ≤ 1 this implies In particular, at the point q n ∈ E 1 , where x 3 (q n ) → 0, r(q n ) → ∞, we obtain for a sequence ε n → 0. Now recall our discussion of Alexandrov reflection by planes orthogonal to the geodesics γ n joining q n to ∂B. We found a point q n in E 1 , associated to the first accident of Alexandrov reflection, and we showed the equidistant curve β n from q n to ∂B was in W . We have |r( q n ) − r(q n )| < 1 by construction, so at q n we also have an estimate for ε n → 0, ε n ∼ 1 r( qn) 1/4 . Then the maximum oscillation of u on the horizontal (Euclidean) disk D of radius x 3 ( qn) εn , centered at q n , is 2ε n x 3 ( q n ). To check this, notice that the most |∇u| can be is r(q) −1/2 , where q is a point of D closest to the origin. Thus, Then r(q) −1/2 ≤ √ 2r( q n ) −1/2 and the oscillation on D is at most Define D n = D + (0, 0, x 3 ( q n )); D n is a horizontal disk above the graph of u over D so that D n ⊂ W and the hyperbolic radius of D n tends to infinity (it is 1/2ε n ). Also the hyperbolic distance between D n and the graph of u over D is bounded by ln (2).
Let t n < 0 denote the first time that S(t n )( q n ) touches ∂B (the first accident when we do Alexandrov reflection with the planes orthogonal to γ n ). We have F n = S(t n )(D n ) ⊂ W and the distance of F n to ∂B is at most ln(2). As n → ∞, F n converges to a horizontal horosphere F which must be in W . Thus E is below F .
Next we observe that E 2 = E ∩ (Ω × R + ) is a vertical graph, where Ω = x 2 1 + x 2 2 > a 2 , for some a > 0. To see this, remark that x 3 (q) ≤ c 0 for some constant c 0 and so if − → H (q) does not point up then q is in the upper hemisphere of its tangent horosphere so x 3 (q) ≥ R = the Euclidean radius of H(q). Hence R ≤ c 0 and H(q) will be disjoint from the cone C(δ) for r(q) larger than some fixed a. As before, this is impossible since the equidistant curves β n , for n large, will not intersect H(q). Now on the domain Ω × R + where the subend E 2 is a graph, we consider the family of catenoid cousin ends C(t) with each C(t) a graph over Ω × R + , tangent to the vertical cylinder ∂Ω × R + and ∂C(t) is at height t on ∂Ω × R + . These surfaces are described in [7].
For t > c 0 , ∂C(t) is above E 2 . If C(t) intersects E 2 , then by Theorem 5, Γ = C(t) ∩ E 2 is compact. Note that Γ is not homologous to zero on E 2 (nor is any subcycle of Γ) since this would yield a compact domain N on E 2 whose boundary is in C(t). Now vary t to obtain a last point of contact of C(t) with N ; then C(t) = E 2 by the maximum principle. It follows that Γ is a Jordan curve on E 2 that generates Π 1 (E 2 ). On C(t), Γ bounds a catenoid cousin end that is below E 2 and Theorem 4 is clear by Theorem 3. Now, we can assume C(t) ∩ E 2 = ∅ for t > c 0 , and then decrease t to 0. There is some largest t where C(t) is disjoint from E 2 and C(s) ∩ E 2 = ∅, for s < t. Since C(t) is vertical along ∂Ω × R + and E 2 is a graph (not vertical) there, ∂C(t) is always above E 2 . Thus we are in the previous situation where C(s) ∩ E 2 = ∅ and ∂C(s) is above E 2 and Theorem 4 is proved.
Let Ω be a noncompact domain in the plane (x 1 , x 2 ) with at least one component of ∂Ω noncompact. Let u 1 , u 2 be defined on Ω with their graphs solutions of the mean curvature equation H = 1 in H 3 . Suppose the following conditions are satisfied: for some positive constants C 1 , C 2 , α (u 2 is the graph of a catenoid cousin), It then follows that u 1 = u 2 on Ω.
Remark 3. In order to apply this theorem to prove Theorem 4, we need to verify that the graph u (= u 1 ) of E 2 in Theorem 4 satisfies the conditions a, b, and c. The conditions a and b are satisfied by construction; the condition c needs some discussion.
In the proof of Theorem 2 we derived the gradient bound for u: where R is the Euclidean radius of the horosphere H(q). Since u ≤ 1, So we need to know R is of order r 2 for the graph u, to satisfy condition c. We see this by considering H(q), q on the graph of u. Let E denote the graph of u (this is the E 2 in the proof of Theorem 4), and let C be the vertical compact cylinder joining ∂E to the plane x 3 = 0. Observe that for q ∈ E, H(q) must intersect C. For if H(q) passes over C, then ∂E ⊂ H(q) − so the figure eight in H(q) ∩ E, would contain a Jordan curve C 1 that is homologous to ∂E on E (Proposition 1). However, u takes its maximum value on ∂E (u has no interior maximum since the graph of u would touch a horizontal horosphere at a local maximum and have the same mean curvature vector). Thus C 1 would be lower than ∂E and link the cylinder C. Hence H q must intersect C. We want to estimate 1/R from above, so that for q ∈ E, we can assume H(q) intersects the vertical segment over the origin at a point p at height x 3 (p) less than some fixed b > 0. Now for x 3 (q) < b, the horosphere H(q) passing through p, intersects the plane x 3 = 0 at the point G(q); see Figure 10. Then . Apply the same (Pythagorean) calculation with p replaced by q to obtain R ≥ τ 2 2b where τ is the horizontal distance from q to G(q). Since τ + t ≥ r, τ or t is at least r 2 so that R ≥ r 2 8b as desired. Before proving Theorem 5, we need some lemmas.
Proof. We have div ∇u Proof. Consider a family of catenoid cousin graphs u τ (x), with u τ strictly above u 2 on ∂Ω and u τ comes down to u 2 as τ → α, with u τ = u 2 for τ = α. Parametrize so that the growth of u τ is 1/r τ , τ < α. As τ → α, one cannot have u τ above u 1 for all τ (otherwise u 1 = u 2 ). Hence the graph of some u β , β < α, intersects the graph of u 1 . As usual, we know the intersection cannot be homologous to zero on the graph (vary τ to get a last point of contact), and the intersection is not one compact cycle (otherwise there is a catenoid cousin below u 1 and Lemma 9 is proved) so that the intersection is not compact and u 1 is above u β on a noncompact domain. Thus M (r) ≥ (α − β) ln r.
Proof of Theorem 5. We study v = ln u 1 − ln u 2 = v 1 − v 2 . Clearly v ≥ 0, v = 0 on ∂Ω and v ≤ γ ln r for some positive γ. We will show that if v is not identically zero, then for some integer k > 1, M (r) grows faster than (ln r) k . This latter growth is impossible and so v ≡ 0.
By Stokes' theorem, where N is the outer conormal along ∂Ω(R). Apply this equation By Lemma 8, and the estimates Since the last integral converges we have constants a 1 , a 2 . Then equation (1) yields Then (2) implies Inequality (3) then implies Now ∂Ω is not compact, v = 0 on ∂C(r) and M (r) is the maximum of v on C(r) so that |∇v| . Next By Lemma 9 we conclude η(r) → ∞, as r → ∞, unless v ≡ 0. Then there is a constant c 3 > 0 and R 0 ≥ 0 such that for r ≥ R 0 , Thus (4) may be replaced by (5) for R 1 ≥ R 0 : Now we will show that for R 1 greater than or equal to some (other) R 0 , we have µ(R 1 ) = a 3 + µ(R 1 ) > 0, for The module of the second integral is at most r . By Lemma 9 and the comparison between η(r) and M (r) we conclude µ(R 1 ) grows at least as fast as ln 3 (R 1 ).
We write equation (5) as: Let ξ be the function defined on the interval J = R 1 , R 1 exp .
On J, ξ satisfies the equation: The connected component of {R ∈ J ∩ [R 1 , R 2 ]; ξ(R) < η(R)} that contains R 1 , is open by construction and closed by equation (6). Thus it is the interval J ∩ [R 1 , R 2 ]. Since ξ(r) → ∞ when r converges (r increasing) to the right end point of J, and η is bounded on [R 1 , R 2 ], we conclude R 2 ∈ J. Thus .
Since A ≤ γ ln(R 2 ), we have for R 0 ≤ R 1 ≤ R 2 . However this contradicts our estimate for the growth of µ(R 1 ) (take R 2 = R 2 1 ). This completes the proof of Theorem 5.

Nondensity at infinity of finite topology surfaces
Let M be a properly embedded Bryant surface with ∂M perhaps not compact. Assume a properly embedded surface Σ exists with ∂Σ = ∂M and Σ ∪ M = ∂W with M mean convex along W . Let P be a (hyperbolic) plane with H 3 −P = P + ∪P − , the connected components of the complement. Assume Σ ⊂ P − . Let M + = M ∩ P + . Theorem 6. There is a constant c > 0 (independent of M ) such that if |K(q)| < c for q ∈ M + , then int (∂ ∞ M + ) = ∅; i.e., M cannot be asymptotic to an open set at infinity in P + . In the half-space model, for q ∈ M + and x 3 (q) sufficiently small, M + is a vertical graph near q, no point of M is below this local graph, and the angle between − → H (q) and − → e 3 is at most π/4. Also notice that C(t) cannot acquire a singularity (i.e., a point where ∇x 3 = 0) as long as C(t) stays compact. For if a singularity occurs at a point q 1 ∈ C(t) then − → H (q 1 ) is vertical. It cannot point up, since then D(s) would contain other parts of M for s > t, s near t, and this is impossible by the previous paragraph; see Figure 11 Thus as long as C(t) stays inside Cyl (2), it is a smooth Jordan curve.
As t decreases to zero, C(t) must leave Cyl(2) since Cap(q) must connect to the rest of M . Thus there are values of t where C(t) traverses ∂Cyl(3/2). Now if x 3 (q) is small and q ∈ Cyl(1/8), there will be points q of C(t) in ∂Cyl(3/2) where − → H ( q). − → e 3 ≤ 0 and x 3 ( q) ≤ 1/8. Then Cap( q) ⊂ Cyl(2)−Cyl(1) and Cap( q) has a local maximum of x 3 near q. So the curves C(t) are not connected before leaving Cyl(2), a contradiction.
It remains to obtain the gradient bound for the graph. For any horosphere of (Euclidean) radius R in H 3 , the part of the horosphere where the mean curvature vector makes an angle greater than π/4 with − → e 3 is of hyperbolic diameter at most 5. So if q ∈ M + and |K| is sufficiently small on M + , then M will be a graph over a geodesic disk in H(q) that contains the northern hemisphere of H(q), if the angle between − → H (q) and − → e 3 is greater than π/4. Now the same argument as before (with Cap(q) and the C(t)) leads to a contradiction. This proves Theorem 6.
Proof. On any subend of E, there must be points q, with 2x 3 (q) < inf (x 3 ∂E ) and − → H (q). − → e 3 ≤ 0; otherwise the subend would be a graph near x 3 = 0, and so could not be dense at infinity.
At such a point q, ∂E ⊂ H(q) − since H(q) is a sphere of Euclidean radius less than 2x 3 (q). If the connected component Σ(q) of q in E ∩ H(q) is not compact then there is an arc γ q on H(q) in Σ(q) joining q to G(q); i.e., γ q is asymptotic to G(q) at infinity. We will show next that such a q can be found so that Σ(q) is not compact.
Suppose Σ(q) is compact. Then Proposition 1 gives E 1 and a figure eight Assume C 1 is the Jordan curve homologous to ∂E in E − E 1 ; cf. Figure 5.
Let E ′ be the subannulus of E bounded by C 1 . We can assume there are points q ′ ∈ E ′ , q ′ = q, with G(q ′ ) = G(q), for we can consider q near q on E; if we could not find q ′ on E ′ with G( q ′ ) = G( q) then G E ′ would miss an open set Ω (the open set being the image by G of an open set about q on E) in S ∞ , Ω a neighborhood of G(q). However E ′ is dense at infinity so there are points y of E ′ converging to Ω with − → H (y). − → e 3 ≤ 0 (otherwise E ′ would be a graph near Ω), and then G(y) ∈ Ω, for x 3 (y) small.
So we can assume there is q ′ ∈ E ′ , q ′ = q and G(q ′ ) = G(q). If Σ(q ′ ) is not compact then the arc γ q ′ joining q ′ to G(q ′ ) exists on H(q ′ ). Thus, we suppose Σ(q ′ ) compact.
There are two possibilities.
• Case 1. ∂E ′ ⊂ H(q ′ ) + . In this case, we have a compact component 1 a compact domain in H(q ′ ) + (Proposition 1 and Figure 4). Also ∂E ′ ⊂ Q ′ 1 . Now Q ′ 1 ∩ H(q) + contains a connected compact component Q with ∂E ′ ⊂ Q; cf. Figure 5. Q is mean convex and E 1 ⊂ Q so that Q 1 ⊂ Q. Also, Q 1 is mean convex along E 1 . Since E 1 and E 2 = (∂Q) ∩ int (H(q) + ) are on E, there must be another component F of E in Q − Q 1 that separates E 1 and E 2 (W is mean convex along E). Then − → H points into the noncompact component of H(q) + − F , along F , and this contradicts Lemma 2.
• Case 2. ∂E ′ ⊂ H(q ′ ) − . In this case, a Jordan curve of H(q ′ ), C ′ 1 say, together with C 2 bounds a compact annulus N ⊂ E. Near C 2 , N is outside H(q), so that N ∩ H(q) − is a compact domain on E with boundary on H(q) and outside H(q). This contradicts Lemma 1. Thus we can construct a proper arc γ q from q to G(q) on E. Now do the same construction at a point q 1 ∈ E with G(q) = G(q 1 ). Join q to q 1 by a path δ on E. Then the arc γ = γ q ∪ γ q 1 ∪ δ works to prove the theorem. Let D 1 ⊂ Q be a least area embedded minimal surface with ∂D 1 = Γ 1 = ∂M 1 (R), and let Q 1 be the compact domain bounded by D 1 ∪ M 1 (R). D 1 is a barrier for the Plateau problem so we can find a least area minimal surface D 2 ⊂ Q − Q 1 with ∂D 2 = Γ 2 = ∂M 2 (R). Let Q 2 be the compact domain bounded by D 2 ∪M 2 (R). We have Q 1 ∪Q 2 ⊂ Q ⊂ W and int(Q 1 )∩int(Q 2 ) = ∅. Now consider domains Q ⊂ Q with ∂ Q = M 1 (R) ∪ Σ, Σ a surface with ∂Σ = ∂M 1 (R) = Γ 1 . The functional on Q, ∂ Q : ( Q, ∂ Q) → area (Σ) + 2Vol Q has a minimum and at such a Q, the smooth points of Σ have mean curvatureone. This is proved in [1] when the mean curvature of ∂Q is strictly greater than one; the only difference is that the minimum may now touch M 1 (R), in which case Σ = M 1 (R) and M 1 (R) is stable in Q.
Observe that Σ 1 ⊂ Q 1 (this is proved in [1]) since, if Q went outside Q 1 , one could remove the part of Q outside of D 1 and reduce the functional.
Notice also that the mean curvature vector of Σ 1 points outside of Q. Otherwise Q would be mean convex so that one could find a least area minimal surface D ⊂ Q, ∂ D = Γ 1 . Then the functional is smaller on the domain bounded by D ∪ M 1 (R); a contradiction. Now working with Γ 2 = ∂M 2 (R) and Q 2 , one finds a mean curvature-one, stable surface Σ 2 ⊂ Q 2 , ∂Σ 2 = Γ 2 and the mean curvature vector of Σ 2 points outside of the domain bounded by Σ 2 ∪ M 2 (R). Thus the domain of W ∩ B bounded by Σ 1 ∪ Σ 2 (and a part of ∂B) is mean convex; cf. Figure 12. Figure 12 For R > r > 0, one has uniform area and curvature bounds of Σ 1 and Σ 2 on balls of radius r a fixed distance from ∂Σ 1 and ∂Σ 2 . Then (as in [1]), one can find a convergent subsequence of Σ 1 and Σ 2 , as R → ∞, which yield the Σ 1 and Σ 2 of Theorem 8.
In the case Σ 1 (or Σ 2 ) ⊂ M then this part of M is stable in W but this easily implies stability in H 3 (look at an unstable domain D corresponding to a first eigenvalue λ 1 < 0). This proves Theorem 8. Proof. Assume the contrary; M is dense at infinity. Then by Corollary 2 for some annular end E of M , ∂ ∞ E = S ∞ , and so Theorem 7 applies: there is a proper arc γ on E and ∂ ∞ γ equals two distinct points p 1 , p 2 . Also, E has genus zero so that γ separates E, hence M as well. Theorem 8 then yields stable surfaces Σ 1 , Σ 2 satisfying the conditions a through d of Theorem 8.
Let T be those points of H 3 whose Euclidean distance to Γ is at most c 1 > 0. Then for c 1 sufficiently small, Σ 1 ∩ T and Σ 2 ∩ T are vertical graphs over domains Ω 1 and Ω 2 ⊂ {x 3 = 0}. We know − → H . − → e 3 > 0 on G = (Σ 1 ∩ T ) ∪ (Σ 2 ∩ T ) and Σ 1 ∪ Σ 2 bounds a mean convex domain R by Theorem 8, so Ω 1 ∩ Ω 2 = ∅. Also we can assume the angle between − → H (q) and − → e 3 is less than π/4 on G. Then for q ∈ G, and t = x 3 (q) sufficiently small, G, near q, is a vertical graph over a horizontal disk D(q), centered at q, of Euclidean radius t.
Let τ > 0 and Γ τ = Γ + τ − → e 3 . Choose τ small so that Γ τ ⊂ T and Γ τ is transverse to Σ. The linking number of Γ τ and γ is one so that Γ τ ∩ Σ 1 consists of an odd number of points. Now, Σ = Σ 1 ∪Σ 2 bounds the mean convex domain R so that there is an arc of Γ τ , which we denote (q 1 , q 2 ), joining a point q 1 ∈ Σ 1 to q 2 ∈ Σ 2 and the interior of the arc is in the interior of R.
For q on the arc (q 1 , q 2 ), let J(q) be the disk D(q) together with the lower hemisphere of the horosphere that contains ∂D(q) and is vertical along ∂D(q). Note that J(q) has a corner along ∂D(q).
For q = q 2 , J(q) ⊂ Ω 2 × R + by our gradient bound on the graph G. Now move q on the arc (q 1 , q 2 ) from q 2 to q 1 . We know that Ω 1 ∩ Ω 2 = ∅ so that J(q 2 ) ∩ Σ 1 = ∅. There will be a first q on the arc where J(q) touches Σ 1 . We will next see that J(q) touches Σ 1 at infinity.
Suppose J(q) first touches Σ 1 at a smooth point p on the horosphere in J(q). The mean curvature vector of the horosphere points up at p, and the mean curvature vector of Σ 1 points up at p too. So the vectors are equal and Σ 1 is a horosphere. This is impossible because the proper arc γ is on Σ 1 and γ has two points at infinity, p 1 and p 2 ; the horosphere has one point at infinity.
Next suppose the first point p where J(q) touches Σ 1 is on ∂D(q). We know that the horizontal segment in D(q), joining p to q (which we call [p, q]) meets Σ 1 only at p. Also this segment does not meet Σ 2 because our gradient bound implies [p, q] ⊂ Ω 1 × R + .
Thus the segment [p, q] is contained in R. The (Euclidean) tangent plane to Σ 1 at p is a support plane of J(q) and − → H (p) points up at p. This contradicts the fact that R is mean convex: − → H (p) points into R, and [p, q] (⊂ R) is on the other side of the tangent plane than − → H (p). Thus there is a point q on the arc where J(q) touches Σ 1 for the first time at a point q ∞ ∈ Γ; see Figure 13. Figure 13 Now consider q ′ on the arc (q, q 2 ) at Euclidean distance less than τ from q, such that the point of Γ below q ′ is not in ∂ ∞ Σ 1 but q ∞ is below D(q ′ ). By Lemma 10, there exists a one-parameter family of vertical graphs C(t), 0 < t ≤ 1, such that C(1) is the original horosphere of J(q ′ ), and C(t) (t < 1) is a catenoid cousin end; each C(t) is vertical along ∂C(t) and ∂C(t) is contained in the vertical cylinder containing ∂D(q ′ ). As t → 0, x 3 C(t) → 0. Since Σ 1 is a graph in this cylinder, C(t) cannot meet Σ 1 for the first time at a point of ∂C(t) (where C(t) is vertical). Also, C(t) cannot touch Σ 1 at an interior point by the maximum principle, nor at infinity. So C(t) never touches Σ 1 and q ∞ cannot be in the asymptotic boundary of Σ 1 . This proves Theorem 9.
Lemma 10. Let C be a circle in {x 3 = 0} with center q ∞ = (0, 0). There is a one-parameter family of catenoid cousin (and horosphere) ends C(t), 0 < t ≤ 1, satisfying: Proof. J.M. Gomes has proved that a family of this nature exists as graphs over the exterior domain of C [7]. To get the C(t) of the lemma, one does inversion of this family through a plane P with ∂ ∞ P = C, followed by a homothety from q ∞ ; cf. Figure 14; the homothety takes B to A. In the appendix we show how these surfaces can be obtained. Theorem 10. Let E be a properly embedded Bryant annular end. Then E is not dense at infinity, has finite total curvature and is regular. Hence (by Corollary 3) E is asymptotic to a catenoid cousin end or to a horosphere end.
Proof. We remark that the proof of Theorem 9 proves Theorem 10 when E is part of a properly embedded surface M as in Theorem 9. Here is the argument in general.
Let Σ be a compact embedded surface such that ∂Σ = ∂E and M = Σ ∪ E is an embedded surface (not necessarily smooth along ∂E). Change the metric of H 3 in a compact neighborhood of Σ so that M has mean curvature greater than 1 near Σ. Now prove Theorem 8 with M in this new metric. The Σ 1 , Σ 2 one obtains will satisfy all the conditions necessary to do the argument of Theorem 9. What matters is the structure of Σ 1 , Σ 2 near infinity. The same argument as in the proof of Theorem 9 then shows E cannot be dense at infinity. Thus, Corollary 3 yields Theorem 10.
Theorem 11. Let M be a properly embedded finite topology, Bryant surface. If M is simply connected (more generally if M has only one end), M is a horosphere. If M has two ends then M is a catenoid cousin. If M has three ends, then M is a bigraph over a plane P ; i.e., M is invariant by symmetry in P and each component of M − P is a geodesic graph over P .
Proof. When M is simply connected, ∂ ∞ M is one point by Theorem 10. Then M. do Carmo and B. Lawson [5] proved M is a horosphere. When M has two ends, ∂ ∞ M is two points and M is invariant by rotations about the geodesic joining the two points [11]. Thus M is a catenoid cousin. When M has three ends, ∂ ∞ M consists of three points so that ∂ ∞ M is contained in a circle of S ∞ . The conclusion is then proved in [11].
Theorem 12. Let M be a properly embedded Bryant surface, M not a horosphere. Then each annular end of M is asymptotic to a catenoid cousin end.
Proof. We know by Theorem 10, that each annular end E is asymptotic to a catenoid end or to a horosphere end. We will assume E is asymptotic to a horosphere end and obtain a contradiction.
We work in the upper half-space model of H 3 , {x 3 > 0}, and assume E is asymptotic to a horosphere x 3 = c > 0. In particular the mean curvature vector of E points up outside of some compact set of E. There are no ends of M above E since their mean curvature vector would also point up (each such end is asymptotic to a horizontal horosphere or a catenoid cousin end whose limiting normal points vertically up) and M separates H 3 into two connected components so that no such end is above E.
Then for ε > 0, the part A of M above c + ε is compact. At the highest point of A (if A is not empty) the mean curvature vector of M points down. But this highest point can be joined by an arc in H 3 − M to a point of E where the mean curvature vector points up. Thus M is completely below x 3 = c.
Let ε > 0 and let C be a small circle in the plane x 3 = c − ε so that C is above M . Just as in the proof of the half-space theorem for properly immersed minimal surfaces in H 3 [16], one can take a family of catenoid cousin ends C(λ), ∂C(1) = C with C(1) above M , where C(λ) converges to the plane x 3 = c − ε as λ → 0. Then some C(λ) touches M at a point q ∈ M and the maximum principle would yield M equals this catenoid cousin. Thus each end of M is asymptotic to a catenoid cousin.
Appendix: The family of graphs of Lemma 10 Consider the family of vertical catenoids in R 3 whose waist circle is of length |λ| and in the {x 3 = 0} plane. Orient by the inner pointing normal. The Weierstrass data on the simply connected covering space C are given by g(z) = e z , ω(z) = |λ|e −z dz, and the metric is ds = |ω| 1 + |g| 2 = 2|λ| cosh(x)|dz|, z = x + iy.
The cousins of these catenoids (as λ varies) have second fundamental form II = II + ds 2 , and II is the second fundamental form of the catenoid in R 3 . The second fundamental form of the catenoid is calculated with respect to the inner pointing normal if λ > 0 and the outer normal for λ < 0.
One can explicitly find the cousins by solving for F in This is done in [21] and [17], and one obtains in the upper half-space model: where α 2 = 1 4 + λ. This is a surface of revolution for λ > − 1 4 , embedded for λ > 0 and immersed for − 1 4 < λ < 0. Let a = 1 2 + α, b = 1 2 − α. The generatrix Γ in the (x 1 , x 3 ) plane of these surfaces of revolution is then x 1 (t) = ab e t + e −t e 2αt b 2 e −t + a 2 e t , x 3 (t) = 2αe 2αt b 2 e −t + a 2 e t .
The points of Γ with vertical tangents are the solutions of ∂x 1 ∂t = 0 and are the solutions of a 2 e 2t − (2ab + 1) + b 2 e −2t = 0.
More precisely, let G α be the graph over (0, A), defined for −∞ < t < τ . We have on G α : ab e t + e −t e 2αt b 2 e −t + a 2 e t , x 3 (t) = A x 1 (τ ) 2αe 2αt b 2 e −t + a 2 e t .