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Anais da Academia Brasileira de Ciências

versão impressa ISSN 0001-3765versão On-line ISSN 1678-2690

An. Acad. Bras. Ciênc. v.76 n.4 Rio de Janeiro dez. 2004 



A relation between the right triangle and circular tori with constant mean curvature in the unit 3-sphere



Abdênago Barros

Departamento de Matemática-UFC, Bl 914, Campus do Pici, 60455-760 Fortaleza, CE, Brasil





In this note we will show that the inverse image under the stereographic projection of a circular torus of revolution in the 3-dimensional euclidean space has constant mean curvature in the unit 3-sphere if and only if their radii are the catet and the hypotenuse of an appropriate right triangle.

Key words: Flat torus, constant mean curvature, circular tori, stereographic projection.


Neste artigo mostraremos que a imagem inversa pela projeção estereográfica de um toro circular de revolução no espaço euclidiano de dimensão 3 tem curvatura média constante se e somente se os seus raios são o cateto e a hipotenusa de um triângulo retângulo apropriado.

Palavras-chave: Toro plano, Curvatura média constante, Toro circular, Projeção estereográfica.




We will denote by T(r, a) the standard circular torus of revolution in 3 obtained from the circle G in the xzplane centered at (r, 0, 0) with radius a < r, i.e.

Now let r : 3 \ {n} ® 3 be the stereographic projection of the Euclidean sphere 3 = {x Î 4: | x |2 = 1}, where n = (0, 0, 0, 1) is its north pole. The inverse image of a circular torus in 3 under the stereographic projection will be called a circular torus in 3. We would like to know when circular tori in 3 comes from constant mean curvature circular tori in 3 under the stereographic projection. A circular torus in 3 meant that it is obtained from a revolution of a circle in 3 under a rigid motion. A general T(r, a) will not satisfy the above requirement. For instance, it was proved by Montiel and Ros (Montiel and Ros 1981) that a compact embedded surface S with constant mean curvature contained in an open hemisphere of 3 must be a round sphere. Hence for T(r, a) contained inside or outside of the unit ball B(1) Ì 3, r–1(T(r, a)) will be contained in an open hemisphere of 3 and can not have constant mean curvature. Then among all tori T(r, a) which intercept the inside and the outside of the unit ball B(1) we will describe those which have the desired property. We will show that to construct such a torus we take an arbitrary point P(a) = (cos a, 0, sin a) on the unit circle of the xz – plane, 0 < a < p/2, draw its tangent until it meets the x axis at the point (a) = (sec a, 0, 0) which will be the center of the circle G whereas its radius will be a = tan a, i.e. the torus T(sec a, tan a) will satisfy the previous requirement. We note if O denotes the origin of 3 then the triangle O P is a right triangle. This description will yield that the Clifford torus is associated to a right triangle with two equal sides. More precisely, our aim in this note is to present a proof of the following fact:

THEOREM 1. Let T2 Ì 3 be a circular torus of constant mean curvature. Then

T2 = r–1(T(sec a,tan a)) = S1(cos a) × S1(sin a).

Moreover, the mean curvature of T2 is given by = .



For an immersion f : M ® between Riemannian manifolds we will denote by d the induced metric on M by f. Now let Mn, and be Riemannian manifolds, where the superscript denote the dimension of the manifold. Consider y : Mn ® be an immersion, r : ® a conformal mapping and set j = r º y. Let f : M ® be a function verifying d = e2fd. If i and ki denote the principal curvatures of y and j = r ° y, respectively, then we get

where x is a unit normal vector field to y(M), see for instance (Abe 1982) or (Willmore 1982). At first we will recall the following known lemma of which we sketch the proof.

LEMMA 1. Let y = (y1, y2, y3, y4) : M2 ® 3 \ {n} be an immersion of a surface M2, set j = r º y and suppose d = e2fd. Then we get

where g = án, denotes the support function on M2 Ì 3.

PROOF. If we put y = y(u1, u2) then a direct computation gives

where l = (1 – y4)–1 = . So we can write d = e2fd with ef = . Thus if n denotes a unit normal vector field to j(M2) then n = efx, where x stands for a unit normal vector field to y(M2). Hence we have from (1)

as we wished to prove.



PROOF. First we note that the circle G = {(x, 0, z) Î 3 : (x – r)2 + z2 = a2} can be parametrized by the map g : [0, 2p] ® 3 defined by

In fact, it is enough to check that

Representing by Rq a rotation on 3 around the z – axis, we see that Rq(g(t)) is a circular torus T(r, a) if g is a parametrization of the circle G given above. We put now s = , q = ru1/s2 and t = ru2/as. We note that such a choice implies 0 < u1 < (2ps2)/r and 0 < u2 < (2pas) / r. Let us call Rq(g(t)) of j(u1, u2), i.e.

Hence we have

where q(t) = a(s2 – 1) sin t + r(s2 + 1). Now a straightforward computation yields

From that we derive that j is a conformal parametrization of T(r, a) satisfying

Moreover, a unit vector field normal to j is given as follows:

Therefore we conclude that

On the other hand a new computation gives us

From this we have k1 = and k2 = – . Taking into account (5), (7) and (8) we conclude from Lemma 1 that

Now we have that is constant if and only if s2 = 1. Moreover, s2 = 1 yields = (a2 – 1). Since a < r we put a = r sin a, r = sec a and this completes the proof of the theorem.

We point out that = 0 if and only if a = 1 and r = which corresponds to the right triangle with two equal sides.



In this section we will present a simple way to compute òT(r, a) H2dA by using the parametrization of T(a, r) given by (4). We observe that if dA denotes the element of area of T(r, a) then its Willmore measure is given by

Hence, using Gauss-Bonnet theorem, we easily conclude that

Therefore the family of tori T (a, a) , which corresponds to the family of right triangles with two equal sides, yields the minimum for òT(r, a)H2dA among all circular tori. Moreover, from (9) its value is (see also Willmore 1982)

Since a < r, if we choose a such that sin a = , we conclude from (9) the following corollary.

COROLLARY 1. Given a circular torus T(r, a) Ì 3 we have a circular torus T (sec a, tan a) Ì 3 such that òT(r, a)H2dA = òT(sec a, tan a) dAa. In other words, the family of circular tori with constant mean curvature in 3 cover all values of òT(r, a)H2dA.



We point out that Theorem 2 of K. Nomizu and B. Smyth (Nomizu and Smyth 1969) guarantees that a flat torus of constant mean curvature in 3 is isometric to a product of circles. Then r–1T(a, r) is flat if and only if it has constant mean curvature. We notice if we set y = r–1j where j was given by (4) then we have

where q(t) = a(s2 – 1) sin t + r(s2 + 1), (see(5)). Hence by using (3), (5), (6) and putting z = u1 + iu2 we conclude that

According to our theorem the metric d is flat if and only if r–1T(r, a) has constant mean curvature in 3. In this case we have

i.e. r–1T(r, a) is isometric to the product of circles S1 () × S1 (). We note that this yields cos a = and sin a = , i.e. r(S1 (cos a) × S1 (sin a)) = T(sec a, tan a).



This work was partially supported by FINEP-Brazil.



ABE N. 1982. On generalized total curvature and conformal mappings. Hiroshima Math J 12: 203-207.         [ Links ]

MONTIEL S AND ROS A. 1981. Compact hypersurfaces: The Alexandrov theorem for higher order mean curvatures, A symposium in honor of Manfredo do Carmo, Edited by B. LAWSON AND K. TENEBLAT, Pitman Monographs 52: 279-296.         [ Links ]

NOMIZU K AND SMYTH B. 1969. A formula of Simons' type and hypersurfaces with constant mean curvature. J Diff Geom 3: 367-377.         [ Links ]

WILLMORE T. 1982. Total curvature in Riemannian geometry, Ellis Horwood limited, 168 pp.         [ Links ]



Correspondence to
AMS Classification: Primary 53A05, 53A10; Secondary 53A30.

Manuscript received on May 30, 2003; accepted for publication on June 14, 2004; presented by MANFREDO DO CARMO*



*Member Academia Brasileira de Ciências

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