An elementary proof of MinVol(R n ) = 0 for n 3

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1 Anais da Academia Brasileira de Ciências : Annals of the Brazilian Academy of Sciences ISSN An elementary proof of MinVolR n = 0 for n 3 JIAQIANG MEI 1, HONGYU WANG and HAIFENG XU 1 Department of Mathematics, Nanjing University, Nanjing, Jiangsu, P.R. China School of Mathematical Science, Yangzhou University, 5009 Yangzhou, Jiangsu, P.R. China Manuscript received on January 30, 008; accepted for publication on May 7, 008; presented by ARON SIMIS ABSTRACT In this paper, we give an elementary proof of the result that the minimal volumes of R 3 and R 4 are zero. The approach is to construct a sequence of explicit complete metrics on them such that the sectional curvatures are bounded in absolute value by 1 and the volumes tend to zero. As a direct consequence, we get that MinVol R n = 0 for n 3. Key words: minimal volume, smooth gluing, bounded geometry. 1 INTRODUCTION The definition of minimal volume of a C manifold M without boundary was first introduced by Gromov Gromov 198. Denote GM the set of all complete smooth Riemannian metrics on M such that the corresponding sectional curvatures are bounded in absolute value by 1. We say that M, g has bounded geometry Cheeger and Gromov 1985 if its metric belongs to GM. The minimal volume of M is a geometric invariant which is defined as MinVolM := inf { VolM, g g GM }. 1 For closed surfaces M, by Gauss-Bonnet formula, it s easy to see that MinVolM = π χm. Thus the minimal volume of a closed surface is actually a topological invariant. For the two dimensional plane, Gromov Gromov 198 obtained the following estimate AMS Classification: 53C0 Correspondence to: Haifeng Xu beard_xu@yahoo.com.cn MinVolR + π.

2 598 JIAQIANG MEI, HONGYU WANG and HAIFENG XU Bavard and Pansu proved that this is an equality Bavard and Pansu B.H. Bowditch Bowditch 1993 gave a different proof by using spherical isoperimetric inequality. Gromov had shown that MinVolR n = 0 without going into details in Gromov 198, Appendix. Cheeger and Gromov Cheeger and Gromov 1985, Example 1.6 showed that R n n 4 has a similar solid torus decomposition as R 3. Thus R n admits a family of metrics such that the sectional curvatures are bounded and the volumes tend to zero. In this paper, we give a detailed proof of the result in another direct way by constructing the explicit metrics which are different from those in Cheeger and Gromov 1985, Example 1.6 on the higher dimensional Euclidean spaces. We state a few results about minimal volume. As stated in Paternain and Petean 003, the minimal volume does depend on the smooth structure of the manifold also see Bessières J. Cheeger and M. Gromov introduced in Cheeger and Gromov 1986, Gromov 198 the concepts of F-structure and T - structure and obtained some results about F-structure and minimal volume. They proved that if M admits a polarized F-structure then the minimal volume of M vanishes. Notice that there is a little difference between the original definition of T -structure given in Gromov 198 and the later definition given in Cheeger and Gromov 1986, Paternain and Petean 003. The graph manifold is a 3-manifold which admit a polarized T -structure. So graph manifold is a special T -manifold. Thus the minimal volume of graph manifold is zero. Furthermore, T. Soma proved in Soma 1981 that the connected sum of two graph manifold is still a graph manifold. In Gromov 198 Gromov pointed out that this result holds for odd dimensional manifolds with T -structures. Paternain and Petean proved in Paternain and Petean 003 that the result also holds for the family of manifolds which admit general T -structures and for any dimension greater than. The minimal volume is closely related to the collapsing theory in Riemannian geometry. Cheeger, Fukaya and Gromov have developed collapsing theory, and they obtained many important results Cheeger and Gromov 1986, 1990, Fukaya The organization of this paper is as follows: In Section, we discuss how to realize smooth gluing of metrics on -dimensional surfaces and to construct metrics on Y-pieces Buser 199 by a simple method. This method is intuitive even without using uniformization theorem. The Y-piece also called pair of pants means a compact topological surface obtained from a dimensional sphere by cutting away the interior of 3 disjoint closed topological disks. In Section 3, we give an explicit construction of a sequence of complete metrics on R 3 with bounded curvatures such that the corresponding volumes tend to zero. We just take product metric on Y-piece S 1 and equip the metric on each piece. So we didn t change the topology of R 3. We also apply the similar construction to R 4. As an immediate corollary, we have MinVolR n = 0, for n 3. CONSTRUCTION OF METRICS ON Y-PIECES Our goal is going to construct metrics on Y-piece with uniformly bounded curvatures which are independent of the lengthes of the boundary. In order to realize smooth gluing of metrics, we simply require that the metrics on a small tubular neighborhood of the boundary of such Y-piece are product metrics. To do this, it is sufficient to construct metrics on a disk. A simple way is to glue S \ D with a cylinder S 1 I along the circle boundary. By viewing S \ D and cylinder as surfaces of revolution, what remains to do is to consider smooth gluing of the images of functions and calculate the Gauss

3 AN ELEMENTARY PROOF OF MinVolR n = 0 FOR n curvature of surface of revolution. But here the radius of S 1 must be very small e.g. ε for our purpose. Hence, to maintain the curvatures of the surface in [ 1, +1], we must insert a good surface. Here we choose a part of pseudosphere. Then, we prove that the surface after gluing is still smooth, the curvature is uniformly bounded i.e. independent of ε, and the volume changes a little. We first use the mollifier to construct a cut-off function Lemma.1. Using this lemma, we can glue smoothly two functions which are tangent at a point Lemma.. By a result of Dan Henry, we construct a better cut-off function see Lemma.5. These results are used to construct metrics on disk D such that the metrics have bounded curvatures and the metrics when restricted in a small neighborhood of D are product metrics. See Lemma.6 and.8. LEMMA.1. There is a smooth function φx [0, b], b > 0 on R, such that { b, x 0, φx = 0, x a. and PROOF. Let where 0 < δ a 4. Set φ x b a. 3 b, x δ, φ δ x = b ba δ x + a δ a δ, δ < x < a δ, 0, x a δ, φ δ x = R j δ x t φ δ tdt = R 1 δ j x t where j x is the mollifier function defined on R by 1 j x = A e 1 x 1, x < 1, 0, x 1, and A is equal to 1 1 e 1 x 1 dx, so that R Then by calculation, we get j xdx = 1. For any δ > 0, let. j δ x = 1 δ j x δ φ δ x b a δ b a. Hence, φ δ x simply denoted by φx is the required function. Moreover, we have since Ae > 1. φ δ x b Aa δδe δ φ δ tdt, b a δδ b aδ. 4

4 600 JIAQIANG MEI, HONGYU WANG and HAIFENG XU LEMMA.. Suppose f x, gx are two smooth functions defined on R and satisfy f c = gc and f c = g c at some point x = c R. Given any δ > 0, there is a smooth function h δ x on R such that h δ, c δ] = f, h δ [c+δ, + = g. PROOF. By the proof of Lemma.1, there is a smooth function ϕ δ x [0, 1] such that { 1, x, c δ], ϕ δ x = 0, x [c + δ, +. Let Then h δ is the required function. h δ x = f xϕ δ x + gx1 ϕ δ x. 5 Still consider functions as in the above lemma. Denote { f x, x c, hx = gx, x > c. Let hδ, h be the surfaces in R 3 generated by the rotation around the x-axis of the graphs of the functions h δ and h respectively. Assume that f x, gx σ > 0 for all x [c 1, c + 1], and let δ 0, 1/ be small enough. σ is independent of δ. Let Then we have the following lemma. M = sup f x, N = sup g x. x [c 1, c+1] x [c 1, c+1] LEMMA.3. There is a constant Cσ, M, N > 0, such that K hδ Cσ, M, N, 6 where K hδ is the Gauss curvature of the surface hδ. Moreover, we have Vol hδ Vol h 0, as δ 0, 7 PROOF. By equation 5, h δ x σ. For x [c δ, c + δ], by Taylor expansion formula and equations 3 4 in Lemma.1, we have h δ = f ϕ δ + g 1 ϕ δ + f g ϕ δ + f gϕ f + g + f ξ 1 g ξ δ 4M + N, δ 1 δ + 1 f ξ 3 g ξ 4 δ 1 δ

5 AN ELEMENTARY PROOF OF MinVolR n = 0 FOR n where ξ i c δ, c + δ, i = 1,, 3, 4. Hence, By assumption, Cσ, M, N is independent of δ. Since h δ h as δ 0, we have h δ 4M + N K hδ = Cσ, M, N. h δ σ Vol hδ Vol h 0, as δ 0. In fact, if x b, and if x b, h δ f = g f ψ 0, as δ 0; h δ f = f gϕ 0, as δ 0. REMARK.4. In Lemma., if we suppose f c = g c additionally, and we set M 3 = sup x [c 1, c+1] f x, N 3 = sup x [c 1, c+1] g x, by a similar argument as in the proof of Lemma.3, we get h δ 14M 3 + N 3. LEMMA.5. conditions are satisfied. There is a smooth function φx0 φx b defined on R, such that the following 1. φx = b, for x 0, φx = 0 for x a;. φ k 0 = φ k a = 0, for any k Z + ; [ 3. max max φ i x ] ! b + max { 3 3! b, 3b/a }. i=1,,3 x [a,b] PROOF. Let b, x 0, φ 1 x = b 1 e x 1, 0 < x ηa; 8 { 1 be x a, a ηa x < a, φ x = 0, x a. Here we select 0 < η < 1 such that φ 1 ηa > φ a ηa. By a result of Dan Henry Henry 1994 which is available on the net at the address: we have [ ] max x 3 3! b, j = 1,. max i=1,,3 φ i j x [a,b] 9 Let p 1 x, p x be two polynomial functions as in Figure 1 which join φ 1 x, φ x to a line l C -smoothly respectively.

6 60 JIAQIANG MEI, HONGYU WANG and HAIFENG XU Figure 1 Let C i j = max p i j x [a/3,a/3] x, and C j a, b := max { } C 1 j, C j, C 3 j, j = 1,. It is clear that C j a, b is in a small neighborhood of max { 3 3! b, b/1 ηa }. Remark.4, we have max i=1,,3 [ max φ i x ] ! b + max { 3 3! b, 3b/a }. x [a,b] Therefore by Let us recall some properties of the pseudosphere. Suppose the parametric equations of pseudosphere are x = ± ln tan t + cos t, y = sin t cos θ, z = sin t sin θ, where t [ π, π, θ [0, π]. We only consider the part of x 0, and we have 1 x = cosh 1 1 y y. A direct calculation implies that 10 dx dt = cos t sin t, dx 1 y dy =, y x = tan t, y x y = sin t cos 4 t. 11 LEMMA.6. For any given 0 < ε << 1 and 0 < δ << 1, there is a smooth function h ε,δ x on [0, + such that 1. For x [0, x ε 1 4ε ], the parametric equation of h ε,δ x is { x = ln tan t + cos t, y = sin t, where t [ π, t ε], tε satisfies the equation ln tan t ε + cos t ε = xε 1 4ε ;. For x [ x ε + 1 4ε, +, h ε,δ x = ε.

7 AN ELEMENTARY PROOF OF MinVolR n = 0 FOR n Here x ε = ln ε 1 4ε. ε PROOF. Suppose f x is a smooth function which is defined as { x = ln tan t + cos t, y = sin t, where t [ π, π. It is easy to check that point A = ln ε 1 4ε, ε ε belongs to the image of function f x. Suppose line AB is tangent to f x at point A Fig.. B is the intersection point of line AB and x-axis. AC is perpendicular to x-axis. D, E are the midpoints of AB and BC respectively. DF is parallel to x-axis. And x F = x B. Figure It s easy to see that AB = 1. Thus, we have BC = 1 4ε, DE = ε, C E = 1 1 4ε. and x C = x ε := ln ε 1 4ε, ε x E = x C + C E = x ε ε, x B = x C + BC = x ε + 1 4ε. Define gx as follows ε gx = x + ε x ε + 1 4ε, x [0, x E ], 1 4ε 1 4ε ε, x [x E, +.

8 604 JIAQIANG MEI, HONGYU WANG and HAIFENG XU That is the broken line ÃDF. We smooth it by the method in Lemma.1 and denote the result smooth function by gx. Choose By inequality 4, we have the estimate: g x δ 1 = 1 4 C E = ε. ε 1 1 4ε ε = 3ε 1 4ε. Next, we observe that f x ε = gx ε and f x ε = g x ε. According to Lemma., for any 0 < δ < 1 4 C E, there is a smooth function k δx on [ 0, x ε + 1 C E ], such that k δ x [0,xε δ] = f x, k δx [x ε +δ,x ε + 1 C E ] = gx. Choose point P = x P, y P Im f x and Q = x Q, y Q Imgx such that According to Lemma.3, we have where M := Since f x = sin t see 11, f x cos 4 ε = t about x for x 0, 1, ε < y P 4ε, x ε < x Q x ε + 1 C E. 4 k δ x 4M + N, sup f x, N := sup g x = 0. x [x P,x Q ] x [x P,x Q ] M = f x P ε 1 4ε. Because x is a increasing function 1 x 4ε 1 16ε. Then, we get a global smooth function h ε,δ x on [0, + such that h ε,δ x [0,x ε 1 4ε = f x, ] h ε,δ x [x ε + 1 4ε = ε.,+ since k δ x = gx, for x x ε + 14 C E, x ε + 34 C E. REMARK.7. Let hε,δ be the surface in R 3 generated by the rotation around the x-axis of the graph of the function h ε,δ. Then we have the estimate about the sectional curvature: 1. For x [ 0, x ε 1 4ε ], g x K hε,δ gx 3ε 1 4ε ε = 3 1 4ε.

9 AN ELEMENTARY PROOF OF MinVolR n = 0 FOR n For x [ x ε + 1 4ε, +, 4M + N K hε,δ = 4M ε ε 4 4ε 1 16ε ε = ε. So, we can choose a constant C which is independent of ε and δ, such that K hε,δ C. Figure 3 LEMMA.8. For any given 0 < ε << 1 and 0 < δ << 1, there is a smooth surface ε,δ which is generated by rotation of the graph of a smooth function h ε,δ x satisfying the following conditions along x-axis. 1. h ε,δ x = 1 x, for x. For x [ [ 1, δ ]. + δ, x ε 1 4ε + ln 1 + where t 3π 4 + δ, t ε. [ 3. h ε,δ x = ε, for x { x = ln tan t + cos t + ln y = sin t, x ε + 1 4ε + ln 1 + ], h ε,δ has following parametric equation: 1 +,, xε + 1 4ε + ln 1 + ] + 1. x ε and t ε are the same as in Lemma.6. Moreover, the sectional curvature of ε,δ is bounded by a constant C which is independent of ε and δ. PROOF. By Lemma.,.3,.6 and Remark.7. REMARK.9. If we let δ which used in Lemma..3 be small enough in Lemma.8, the area of the smooth surface above is less than + π + 1. If we add two ends to the surface, we will get the required Y-piece as showed in Figure 4.

10 606 JIAQIANG MEI, HONGYU WANG and HAIFENG XU Figure 4 3 MINIMAL VOLUME OF R 3 By Lemma.8 and torus decomposition of R 3 in Cheeger and Gromov 1985, we can construct a sequence of complete smooth metrics on R 3 with bounded curvatures and the corresponding volumes tend to zero. So the minimal volume of R 3 is zero. The completeness of every metric we constructed is followed from Proposition 3.1. Note that the diameters of the Y-pieces we constructed have a positive lower bound. By Hopf-Rinow theorem Chavel 1993, R 3, g ε is also geodesic complete. PROPOSITION 3.1. Let A 1 A A n be a sequence of compact metric spaces. Suppose d n is the distance function on A n, and d n+1 An = d n for all n. Define function d : A = n=1 A n R by dx, y = d m x, y, where m is the positive integer such that x, y A m. Then A, d is a metric space. If additionally we suppose that there is a sequence of concentric balls {B n } such that B n A n and diameterb n as n, then A, d is a complete metric space. Before the proof of Proposition 3., it maybe useful to keep Figure 5 in mind which describes the construction of metrics on R 3. As in Cheeger and Gromov 1985, Example 1.4, we decompose R 3 into a sequence of solid toruses, C 1 C C n, such that R 3 = i=1 C i, where C i = Di S 1, {Di } are closed disks. Every solid torus is contractible in the next. Let C 1 = 1 S1 = D1 S1. Let A i denote the axis i.e. {0} S 1 Di S 1 of C i. The tubular neighborhood of A i+1 denotes by i+1 S1. Then, we have C i+1 \ C i = i S1 i+1 S1, where i is a surface with nonempty boundary which consists of three circles. PROPOSITION 3.. MinVolR 3 = 0.

11 AN ELEMENTARY PROOF OF MinVolR n = 0 FOR n Fig. 5 Metrics on R 3. PROOF. Our goal is to construct a sequence of complete smooth metrics g ε on R 3 such that VolR 3, g ε 0, as ε 0, and the corresponding curvatures are uniformly bounded by 1, i.e. Kgε 1. In Lemma.8, we have constructed a surface disk with nonempty boundary circle satisfying Vol, g < +, Kg C, Length = πε, 1 and, if restricted on a tubular neighborhood of in, g is product metric. As has been mentioned in Section, it is also easy to construct metric on Y-piece i satisfying the similar conditions as above. With such metric, i looks like as in Figure 6. Next, we give the explanation in details as follows. For 1, we assign it a metric g 1, ε as in Lemma.8. Denote the geometric surface by M ε,δ, or simply by M ε, where the ε means the length of M ε is πε under the metric given. By Remark.9, when δ is small enough. For i+1, i 1, define Vol M ε + π + 1, M ε i := i+1, g 1, εi, for i 1.

12 608 JIAQIANG MEI, HONGYU WANG and HAIFENG XU Fig. 6 Metric on Y-piece. We assign i a metric to be a Y-piece see Fig. 6, and denote it by Y i = Y i,ε = Y i; ε i 1, ε i, ε, i+1 which means the lengths of the components of the boundary Y i,ε are πε, πε, πε respectively. i 1 i i+1 In summary, we use the following notations: M 0 = M ε := 1, g 1, ε, M i = M ε i := Y i = Y i; ε i 1, ε i, ε 1 i+1, g 1, εi, for i 1, := i+1 i, g ; ε i 1, ε i, ε i+1, for i We simply write the last equation as Y i = Y i,ε := i, g,ε. 14 Then the process of constructing the metric g ε on R 3 is as follows. According to the torus decomposition of R 3, to get a global smooth metric on R 3, we have to glue the metrics on the pieces M 0 Sε 1, S1 ε M 1, S 1 ε M, S 1 ε M 3, Y 1 S 1 ε, Y S 1 ε, Y 3 S 1 ε, along boundary circles or the product factors S 1 in certain pairs and certain order. The metrics on such product manifolds are chosen simply as product metrics. Sε 1 means that the circle has length πε under the given metric ε dθ. The relation between the lengths of the boundary circles must be related to ε. The order and the relation of gluing is stated as follows. Let Y i;1 I, Y i; I, and Y i;3 I denote the three tubular neighborhoods of the three boundary components of Y i for i 1.

13 AN ELEMENTARY PROOF OF MinVolR n = 0 FOR n First, the metrics on Y i;3 I S 1 ε and Y i+1;1 I S 1 ε are chosen to be the following i i+1 product metrics: and ε i+1 dθ + dt + ε i dα, 16 ε ε dθ + dt + dα. 17 i i+1 The order of gluing here is that we direct glue the first term of 16 and the third term of 17; and direct glue the third term of 16 and the first term of 17. Second, the metrics on Y i; I S 1 ε and S 1 ε I M i are chosen to be i i ε ε dθ + dt + dα, 18 i i and ε ε dα + dt + dθ. 19 i i The order of gluing here is that we exchange the variables α and θ in the second or first metric, so that they can be glued directly with the first or second metric. At last, what remains to do is to glue the metrics on M 0 I S 1 ε and Y 1;1 I S 1 ε. Thus, we get a sequence of complete metrics g ε,δ simply denote by g ε on R 3 such that where C is independent of ε and δ; and we have and By scaling and the fact Kgε C, Vol Y i S 1 ε 3 + π πε, 0 i i Vol S 1 ε M i πε + π i i K λg = 1 λ K g, VolM, λg = λ n/ VolM, g, where n = dim M, we get our metrics still denote by g ε such that the curvature is uniformly bounded by 1, while the volumes of the surfaces are finite, since ε + ε + ε 4 + ε + = ε. 3 8 Hence, Therefore, The proposition is proved. VolR 3, g ε 0, as ε 0. 4 MinVolR 3 = 0.

14 610 JIAQIANG MEI, HONGYU WANG and HAIFENG XU 4 MINIMAL VOLUMES OF R n FOR n 3 PROPOSITION 4.1. MinVolR 4 = 0. PROOF. We still use the notations 13 and 14 in the proof of Proposition 3.. The metric on Y i i.e. Y i,ε or Y i; ε i 1, ε i, ε i+1 is denoted by g,ε means the second kind metric. where Let the metric on Y i S 1 ε i R be and f t is defined by g,ε = g,ε + f tg S 1 ε + dt, i = 1,, 3,..., 5 i ε g S 1 = dθ, θ [0, π], t R, 6 ε i i f t = { 1 e 4 t, t = 0, 1, t = 0. It is easy to prove that f t is a smooth function. Moreover, f t satisfies f t 1, f t f t 3 f t 6, + f tdt = 4 6π <, f t > 0, for all t R. Replace the metric g 1, ε on i i+1 that is, the surface M i, see 13 by the metric g 1, εt and denote i the surface i+1, g 1, εt by M εt, where εt means that the length of the boundary circle is a function i i of t. If we want to realize the smooth gluing i as in Figure 7, that is to glue the metric on factor S 1 ε and i the metric on boundary S 1 εt, the direct way is to require that the metric on a small tubular neighborhood of M i εt is a product metric, and i Thus But, in order to make the volume Vol has the following form S 1 ε i M εt Length M εt i = π εt i = f t π ε i. εt = f tε. R i 7 8 finite, the metric on S 1 ε i M f tε i R should f t ε i dθ + g 1, f tε i + dt. 9

15 AN ELEMENTARY PROOF OF MinVolR n = 0 FOR n Fig. 7 Gluing. The gluing ii is similar to i. Hence, the metric g,ε on Y i i.e. Y i; g, f tε. That is, becomes to Y i; ε i 1, ε i, ε i+1 Y i; f tε f tε. i 1, i, f tε i+1 ε i 1, ε i, ε i+1 should be changed to To realize the gluing i and ii while keeping the curvature bounded, we first construct an ε related smooth function G ε x, t simply denoted by Gx, t, see 36 which is used to construct the smooth surface M f tε/ i. After that, we calculate the sectional curvatures and claim that the sectional curvatures are bounded. Then, by calculating the volumes, we complete the proof. For state clearly, we divided the proof into several steps: Step 1: Construction of function Gx, t. By Lemma.5, there is a C function φ ˉx 0 φ ˉx 1 on [ 1, + such that 1, 1 ˉx 0, 1 e 1 ˉx, 0 < ˉx 1 3, φ ˉx = e 1 30 ˉx 1, ˉx < 1, 3 0, ˉx 1. There exists a constant C > 0, such that φ ˉx, φ ˉx, φ ˉx C. 31

16 61 JIAQIANG MEI, HONGYU WANG and HAIFENG XU In particular, Let 1 φ ˉx = e ˉx ˉx, 3 for ˉx φ ˉx = 1 ˉx 1 3 e ˉx 1, for ˉx 0, 1, 6 5 6, 1. { 1, 1 x 0, ϕx, t = φ f tx, 0 < x < +, 3 where t R and ˉx = f tx. Let Fx, t = ε1 f thx, 33 where hx satisfies hx = 1, for x [ 1, 0], h h h, h C, + 0 hxdx <, 0 < hx 1, for x [ 1, +. For simplicity, let hx = { 1 e 4 x, x > 0, 1, x [ 1, 0]. 34 Then, we have hx, h x, h x Then we define the function Gx, t by Gx, t = Fx, tϕx, t + ε f t, x [ 1, +, t R. 36 Step : Calculation of the sectional curvature of surface Gx,t. Now for every t, let Gx,t be the surface generated by rotation of the graph of the function Gx, t. Gx,t is just a part of surface M f tε i. It can also be used to construct the surface Y i; f tε On S 1 ε i Gx,t R, the metric is given by where θ, α [0, π]. Let f tε i 1, i, f tε i+1. f t ε i dθ G x x, t dx + G x, tdα + dt. 37 ω 1 = ε i f tdθ, ω = 1 + G x dx, ω3 = Gx, tdα, ω 4 = dt.

17 AN ELEMENTARY PROOF OF MinVolR n = 0 FOR n Then by the structure equations, we have ω 1 = 0, ω3 1 = 0, ω4 1 = ε i f tdθ, ω 3 = G x 1 + G x dα, ω 4 = G x G xt dx, 1 + G x ω 4 3 = G t dα. So the curvatures are K 1 = f t f t G x G xt 1 + G x K 13 = f t G t f t G, K 14 = f t f t, K 3 = G xx G1 + G x, + G x G t G xt G1 + G K 4 = G xt + G x G xtt 1 + G x, 1 + G x x, 38 K 34 = G tt G. Step 3: We claim that: Ki j C, for i, j = 1,, 3, PROOF. By Equation 3, we have ϕ x = φ ˉx f t, ϕ t = φ ˉx f t f t ˉx, ϕ xx = φ ˉx f t, f ϕ tt = φ t ˉx ˉx + φ ˉx f t ϕ xt = φ ˉx f t ˉx + φ ˉx f t, xtt = φ f t ˉx f t ϕ f t f t ˉx, ˉx + φ f t ˉx ˉx + φ ˉx f t f t ˉx + φ ˉx f t. 40 By Equation 33, we have F x = ε1 f th x, F t = ε f thx, F xx = ε1 f th x, F tt = ε f thx, F xt = ε f th x, F xtt = ε f th x. 41

18 614 JIAQIANG MEI, HONGYU WANG and HAIFENG XU By Equation 36, we have G x = F x ϕ + Fϕ x, G t = F t ϕ + Fϕ t + ε f t, G xx = F xx ϕ + F x ϕ x + Fϕ xx, G tt = F tt ϕ + F t ϕ t + Fϕ tt + ε f t, G xt = F xt ϕ + F x ϕ t + F t ϕ x + Fϕ xt, G xtt = F xtt ϕ + F xt ϕ t + F x ϕ tt + F tt ϕ x + F t ϕ xt + Fϕ xtt. 4 Hence, by Equations 30 36, 40 4, we have G x ε + εc, G xt ε + 4Cε, G xtt ε + 14Cε. 43 G G xx h ϕ hϕ C. h φ ˉx f t + h φ ˉx f t, f 1 f f 1 f 44 G t G f t + 1 f thx ϕ t 1 f thxϕx, t + f t f t f t f t f t + hx ϕ + t hxϕx, t + f t 1 f t hx ϕ + t hxϕx, t + f t C ϕx, t + f t hx. 45 Similarly, we have G Then, if and only if That is tt G + C + C. 46 ϕx, t + f t hx ϕx, t + f t f t = φ f tx + hx f x 0 φ f tx 0, and f tx 1, and f t f x 0. f tx f xx 0.

19 AN ELEMENTARY PROOF OF MinVolR n = 0 FOR n This implies that f xx +, and it is a contradiction since f x < 1. Hence, there is a positive x number σ > 0 which is independent of x, t and ε, such that Therefore, ϕx, t + f t hx > σ > 0. G t G < C, G tt < C. 47 G According to inequalities 43, 44 and 47, curvatures in 38 must be bounded. proved. So the Claim is Step 4: Conclusion. Therefore, for every piece considered above, we have constructed a metric on it. By scaling if necessary, we can make the metric satisfying the condition that the curvature is bounded by 1. Moreover, we have and Vol S 1 ε M i f tε R πε 4 6π + π + 1, 48 i i Vol Y i, f tε S 1 ε R 3 + π πε 4 6π. 49 i i Hence, we have construct a sequence of complete metrics g ε with bounded curvatures on R 3 R, and Vol R 3 R, g ε 0, as ε 0. Therefore, MinVol R 4 = 0. THEOREM 4.. MinVolR n = 0 for n 3. PROOF. Every positive integer n can be written in one of the forms: Then R n n 3 can be written as 3k, 3k + 1 = 3k 1 + 4, 3k +. R } 3 {{ R } 3, or } R3 {{ R } 3 R4, or R } 3 {{ R } 3 R. Take product metric on it, then MinVolR n = 0 since MinVolR 3 = MinVolR 4 = 0 and MinVolR = + π.

20 616 JIAQIANG MEI, HONGYU WANG and HAIFENG XU ACKNOWLEDGMENTS This work was partially supported by the National Key Basic Research Fund China G , NSFC Grant and NSFC Grant We wish to thank Professor Xifang Cao for his suggestions. The third author would like to thank Professor Detang Zhou for his significant suggestions and discussions about the completeness of the metrics constructed. Also we would like to thank Professor Qing Zhou and Professor Ying Zhang for their important suggestions on metric construction on Y-pieces. We would also like to express our sincere gratitude to the referees for their interest, careful examination and valuable suggestions. RESUMO Neste artigo fornecemos uma demonstração elementar do resultado de que os volumes minimais de R 3 e R 4 são ambos iguais a zero. A abordagem consiste na construção de uma seqüência de métricas completas explícitas nesses espaços cujas curvaturas seccionais são limitadas em valor absoluto por 1 e os volumes tendem a zero. Como conseqüência direta, estabelecemos que MinVolR n = 0 para n 3. Palavras-chave: volume mínimo, colagem diferenciável, geometria. REFERENCES BAVARD C AND PANSU P Sur le volume minimal de R. Ann Scient de l ÉNS 19: BESSIÈRES L Un théorèm de rigidité différentielle. Comment Math Helv 73: BOWDITCH BH The minimal volume of the plane. J Austral Math Soc Series A 55: BUSER P Geometry and Spectra of Compact Riemann Surfaces. Birkhäuser, Boston. CHAVEL I Riemannian Geometry: A Modern Introduction. Cambridge University Press. CHEEGER J AND GROMOV M On the Characteristic Numbers of Complete Manifolds of Bounded Curvature and Finite Volume. H.E. Rauch Memorial Chavel and Farkas Eds. I: CHEEGER J AND GROMOV M Collapsing Riemannian manifolds while keeping their curvature bounded. J Differential Geom 3: CHEEGER J AND GROMOV M Collapsing Riemannian manifolds while keeping their curvature bounded II. J Differential Geom 3: FUKAYA K Hausdorff convergence of Riemannian manifolds and its applications. Recent topics in differential and analytic geometry. Adv Stud Pure Math 18-I: GROMOV M Volume and Bounded Cohomology. IHÉS 56: HENRY D Useless facts about the derivatives of e 1/x, x > 0 manuscript on Feb. PATERNAIN GP AND PETEAN J Minimal entropy and collapsing with curvature bounded from below. Invent Math 151: SOMA T The Gromov invariant of links. Invent Math 64: