On Deciding Whether a Surface is Parabolic or HyperbolicAuthor(s): John MilnorSource: The American Mathematical Monthly, Vol. 84, No. 1 (Jan., 1977), pp. 43-46Published by: Mathematical Association of AmericaStable URL: http://www.jstor.org/stable/2318308 .

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1977] CLASSROOM NOTES 43

This harvest plan is one of several used in laboratories. Other plans recommend themselves if we are concerned about condition of the organism (determined by position on the growth curve) or if we wish instead to maximize yield per unit of nutrient used. One apparatus used to grow and harvest simple organisms in laboratories is pictured in Figure 2. Each drop of nutrient added to the culture causes one drop of the culture, a mixture of nutrient and organisms, to overflow and be harvested. The size of each harvest is determined by the volume of the drop, the timing by the time between addition of drops. With this apparatus our theoretical solution of harvesting small amounts near the inflection point can be realized.

The author wishes to thank the referees for their helpful suggestions.

DEPARTMENT OF MATHEMATICS, SUFFOLK COUNTY COMMUNITY COLLEGE, SELDEN, NY 11784.

ON DECIDING WHETHER A SURFACE IS PARABOLIC OR HYPERBOLIC

JOHN MILNOR

Let M be a simply connected open surface, differentiable of class say C', with a C' Riemannian metric. It follows from complex function theory that M is conformally diffeomorphic either to the complex plane, or to the open unit disk in the complex plane. (See for example [Springer, p. 225] together with the local argument as given in [Chern] or [Chern, Hartman, Wintner] or [Courant-Hilbert, p. 350] or [Milnor, pp. 1110, 1111].) In the first case one says that M is parabolic, in the second case hyperbolic.

This note will attack the problem of deciding whether M is parabolic or hyperbolic in the very special circumstance that M is complete and rotationally symmetric about a point p, so that the Gaussian curvature K can be expressed as a function of the distance r from p. Let 2irg(r) denote the circumference of the circle of radius r centered at p. Introducing geodesic polar coordinates r, 0 the Riemannian metric takes the form dr2 + g(r)2d02. Computation shows that Gauss curvature is given by

K= -(d 2gldr 2)g

or briefly, K = - g/g. (See for example [18, pp. 136-138] or [16, pp. 3B, 32-37].) Thus, the function g satisfies the linear differential equation g +Kg = 0 with initial conditions g(O) = 0, g(O) = 1.

THEOREM. If K _? - 1/(r2logr) for large r, then the surface is conformally parabolic. But if K ? - (1 + E)/(r2log r) for large r, and if the function g (r) is unbounded, then the surface is hyperbolic.

Here ? > 0 can be any positive constant. Note that the hypothesis that g(r) is unbounded is essential. In fact, if g(r) is bounded, then the

surface is necessarily parabolic. Compare the lemma below. An example is provided by the function g(r) = re7r2, with K(r) = 6- 4r2. This yields a parabolic surface since g is bounded, even though K tends to - oo as r -- oo. (Such an example, with g bounded, cannot have K 0 o for all values of r. For if K 0 O or equivalently g _ 0 for all r, then it follows by integration that g ' 1 and g(r) _ r so that g must be unbounded.)

This theorem surprised the author: It seemed strange that a parabolic surface could be converted into a hyperbolic surface simply by replacing its curvature function K(r) = - 1I(r2log r) for r ' c by (1 + t)K(r) for r-' c.

The proof will be based on the following. (Compare [1].) Choose a > 0.

LEMMA. The surface M is hyperbolic if and only if the integral fJ dr/g(r) is finite.

Proof. Introduce a new coordinate p in place of r by setting p = fa dslg(s). Then the metric takes

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44 JOHN MILNOR [January

the form

dr2+ g2d02 = g2(dp2 + dO2).

(Coordinates p, 0 for which the metric takes the latter form, conformally equivalent to the flat metric dp2 + d02, are known classically as isothermal coordinates.) Now map M conformally into the complex numbers by sending the point with coordinates p, 0 to the complex number exp (p + iG) = epcos 0 + iePsin 0. This conformal mapping is at first not defined at the base point p. However, it is bounded, smooth, and conformal throughout a neighborhood of p, except at the point p itself. Hence the apparent singularity at p is removable, and we obtain a smooth conformal mapping which is defined and one-to-one throughout M.

[Expressing this construction in a different way, we map the point of M with geodesic polar coordinates (r, 0) to the complex number Re" with polar coordinates (R, 0), where R = exp (f, dslg(s)) = exp (f' dslg(s) + constant). This last formula for R is the most general one which yields a radial stretching factor dR/dr precisely equal to the circumferential stretching factor 27R/2rg(r), so that the mapping is conformal.]

Evidently the image of this mapping is either the disk of radius expf - drlg(r) or the entire complex plane according as this integral is finite or infinite. U

REMARK. If the total absolute curvature f!f K dA of a complete simply connected surface is finite, then [Blanc and Fiala] have shown that the surface is parabolic. Compare [8], [6]. (In the rotationally symmetric case we can recover this result from the Lemma as follows. Since the Riemannian area element dA is equal to gdrd6, it follows that

(2r)-1f I K IdA = I K lgdr= I g ldr

is equal to the total variation of g. If this total variation is finite, then g is bounded, hence g(r) Cr, and f?' drig = oo, so that the surface must be parabolic.)

Here is a basic example which illustrates the Lemma. Consider the function go(r) = rlog r for r ?> 2 (choosing say go(r) = r for r ' 1 and interpolating smoothly in the interval 1 ? r -? 2). The correspond- ing surface is parabolic since f2 drl(r log r) = [log log r]2 = co. The associated Gauss curvature Ko is given by

Ko= -go/go= -1I(r2 log r)

for r ' 2. (In this example, the total absolute curvature f f I Ko IdAo is infinite, so that the Blanc-Fiala criterion does not apply.)

Proof of Theorem. Let g(r)> 0 be any smooth function whose associated curvature K = -gg satisfies the inequality K(r) ? Ko(r) for r ' a; where Ko = - 1/(r2 log r) as above. (Note that K may take both positive and negative values.) Multiplying g by a small positive constant if necessary, we may assume that the initial inequalities

(1) g(a) < go(a),

(2) g(a) < go(a)

are satisfied. It follows that g(r) < go(r) for all r ' a. For otherwise, if b were the smallest number greater than a with g(b) = go(b), then integrating the inequality

g = - Kg - - Kogo = go

from a to r and adding (2) we would obtain g(r) < go(r) for a ' r ' b. Integrating this new inequality from a to b and adding (1) we would obtain g(b) < go(b), which contradicts the choice of b. Thus g(r) < go(r) for all r ' a. Hence the integral fJ drlg(r) is also divergent, and the surface is parabolic.

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1977] CLASSROOM NOTES 45

For the proof in the other direction, we use a comparison function of the form

g, (r) = r(log r)'+' for r ' 2.

This yields a hyperbolic surface since the integral

dr/r(log r)"E = [- 1/e (log r)' ]2

is finite. Computation shows that the corresponding Gauss curvature is given by

K, = - (1+)(1 + E/log r)/(r2log r)- -(1 + 2)I(r2log r) for large r.

Now consider a smooth function g(r) > 0 whose associated Gauss curvature K = - g/g satisfies

K(r) c - (1 + 2?)I(r2 log r) - K, (r)

for r ' a. If g(a)>0 then, after multiplying the function g(r) by a large constant if necessary, a completely analogous argument shows that g(r) > g. (r) for all r ' a. Hence the integral fJ dr/g(r) is also convergent, and the surface is hyperbolic.

This argument wilt work whenever g(r) >0 for some sufficiently large r. In particular, it will certainly work whenever the function g(r) is unbounded. On the other hand, if g(r) c 0 for all large r, then the function g(r) is bounded, and hence by the Lemma the surface is parabolic. U

In conclusion, here are two further problems: (1) Does the Theorem proved above remain valid in the more general case of a complete Riemannian

surface which is not rotationally symmetric? The statement would still make sense as long as the base point p has no conjugate points, so that

there exists a global geodesic polar coordinate system with metric of the form dr2 + g(r, G)2do2. Robert Osserman points out to the author that if K(r, 0) ? - 1I(r2 log r) for large r then, using a theorem of Ahlfors [1] in place of the Lemma, one can indeed prove that M is parabolic. Greene and Wu [7] prove an analogous result in higher dimensions, assuming the sharper hypothesis that all sectional curvatures satisfy 0 ? K ? - C/r2+3. For discussion of a related problem in the hyperbolic case, (see [12] ?6.4).

(2) Consider an embedded surface in Euclidean 3-space of the form z = f(x, y). How can one decide effectively whether this surface is parabolic or hyperbolic? This problem has been studied by Osserman [12, 13], H. Huber [9], Jenkins [10], and by Riiedy [14].

References

1. L. Ahlfors, Sur le type d'une surface de Riemann, C. R. Acad. Sci., Paris, 201 (1935) 30-32. (Compare [Sario and Nakai, p. 329].)

2. C. Blanc and F. Fiala, Le type d'une surface et sa courbure totale, Comment. Math. Helv., 14 (1941-42) 230-233.

3. S. S. Chern, An elementary proof of the existence of isothermal parameters on a surface, Proc. Amer. Math. Soc., 6 (1955) 771-782.

4. S. S. Chern, P. Hartman, and A. Wintner, On isothermic coordinates, Comment. Math. Helv., 28 (1954) 301-309.

5. R. Courant and D. Hilbert, Methods of Mathematical Physics, vol. 2, Interscience, Wiley, New York, 1962. 6. R. Finn, On a class of conformal metrics, with application to differential geometry in the large, Comment.

Math. Helv., 40 (1965-66) 1-30. 7. R. E. Greene and H. Wu, Curvature and complex analysis, II, Bull. Amer. Math. Soc., 78 (1972) 866-870. 8. A. Huber, On subharmonic functions and differential geometry in the large, Comment. Math. Helv., 32

(1957) 13-72. 9. H. Huber, Riemannsche Flichen von hyperbolischem Typug im euklidischen Raum, Math. Ann., 139 (1959)

140-146.

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46 F. R. BEYL [January

10. J. A. Jenkins, On hyperbolic surfaces in three-dimensional euclidean space, Michigan Math. J., 8 (1961) 1-5.

11. J. Milnor, A problem in cartography, this MONTHLY, 76 (1969) 1101-1112. 12. R. Osserman, Riemann surfaces of class A, Trans. Amer. Math. Soc., 82 (1956) 217-245. 13. , Hyperbolic surfaces of the form z = f(x, y), Math. Ann., 144 (1961) 77-79. 14. R. Riiedy, Embeddings of open Riemann surfaces, Comment. Math. Helv., 46 (1971) 214-225. 15. L. Sario and M. Nakai, Classification Theory of Riemann Surfaces, Grundlehren 164, Springer-Verlag,

New York, 1970. 16. M. Spivak, A Comprehensive Introduction to Differential Geometry II, Publish or Perish, Inc., Boston,

1970. 17. G. Springer, Introduction to Riemann Surfaces, Addison-Wesley, Reading, Mass., 1957. 18. D. Struik, Lectures on Classical Differential Geometry, Addison-Wesley, Reading, Mass., 1950.

INSTITUTE FOR ADVANCED STUDY, PRINCETON, N.J. NJ 08540.

CYCLIC SUBGROUPS OF THE PRIME RESIDUE GROUP

F. RUDOLF BEYL

The study of metacyclic groups, i.e., groups which are extensions of a cyclic group by a cyclic group, leads immediately to the following problem of elementary number theory: Which prime residues (mod M, say) are mutual powers of each other? Here the term prime residue mod M means a residue class modulo M which can be represented by an integer prime to M

If we shift attention from the prime residue to the cyclic subgroup it generates in the group ZM of all prime residues mod M, the problem reads: Describe the cyclic subgroups of ZM*. The following can be treated in a course on elementary number theory just before primitive elements. Most of the statements can be found in Basmaji [1, ? 3 p. 177] as well as in other papers on metacyclic groups, but the presentations there are not fit for classroom use.

If n is a natural number and p a prime, let #p(n) denote the exact power of p dividing n.

PROPOSITION 1: Let p be a prime, r an integer 1 mod p, rk - 1, let n be a natural number and S = 1 + r + r2 + * + rn-1. If p is odd, then #p(S) = #p(n). If p = 2 and n is odd, then # 2(S)= # 2(n) = 0. If p = 2 and n is even, then #2(S) = #2(n) + #2(r + 1) - 1, so that #2(S) = #2(n) if r 1 mod 4.

PROPOSITION 2: Let p be a prime, m ' 1, and M = pm. The prime residue r lies in the p-Sylow subgroup of Z* if and only if p r - 1. Let p" be the order of such an r.

CASE (i): p odd. Then q = max(m - #p(r - 1),0).

CASE (ii): p = 2. If r- 1 then q = 0, and if r - 1 then q = 1. For m -' 3 any other prime residue has the form r = 4w + 1 or r = 4w -I with 2m2t w; in either case q = m -2-#2(w).

Proof ofiProposition 1, by induction on the number of prime factors in n. The case n = 1 is trivial. Next let n be a prime. If n 7 p then S n mod p, and we are done. If p = n = 2, then S = r + 1, and we are also done. If p = n is odd, then

1 S= (1 +Ap)l p +y2 p*(p-1) Ap-p mod p2

as desired. Now consider composite n. Let n = k * 1 with 1 < k < n, and assume 21 k if 2 In. Then

f18 C ~~~~~~~= tv r8 tv s,i = r r

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