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

In this paper, we give an elementary proof of the result that the minimal volumes of R3 and R4 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 (Rn) = 0 for n ≥ 3.


INTRODUCTION
The definition of minimal volume of a C ∞ manifold M (without boundary) was first introduced by Gromov (Gromov 1982).Denote G(M) 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 G(M).The minimal volume of M is a geometric invariant which is defined as MinVol(M) := inf Vol(M, g) | g ∈ G(M) . (1) For closed surfaces M, by Gauss-Bonnet formula, it's easy to see that Thus the minimal volume of a closed surface is actually a topological invariant.For the two dimensional plane, Gromov (Gromov 1982) obtained the following estimate MinVol(R 2 ) ≤ (2 + 2 √ 2)π.

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JIAQIANG MEI, HONGYU WANG and HAIFENG XU Bavard and Pansu proved that this is an equality (Bavard and Pansu 1986).B.H. Bowditch (Bowditch 1993) gave a different proof by using spherical isoperimetric inequality.Gromov had shown that MinVol(R n ) = 0 without going into details in (Gromov 1982, Appendix 2).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 2003), the minimal volume does depend on the smooth structure of the manifold (also see (Bessières 1998)).J. Cheeger and M. Gromov introduced in (Cheeger and Gromov 1986, Gromov 1982) the concepts of F-structure and Tstructure 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 1982) and the later definition given in (Cheeger and Gromov 1986, Paternain andPetean 2003).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 1982) Gromov pointed out that this result holds for odd dimensional manifolds with T -structures.Paternain and Petean proved in (Paternain and Petean 2003) that the result also holds for the family of manifolds which admit general T -structures and for any dimension greater than 2. 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 1990).
The organization of this paper is as follows: In Section 2, we discuss how to realize smooth gluing of metrics on 2-dimensional surfaces and to construct metrics on Y-pieces (Buser 1992) 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 2 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 MinVol(R 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 2 \ D 2 with a cylinder S 1 × I along the circle (boundary).By viewing S 2 \ D 2 and cylinder as surfaces of revolution, what remains to do is to consider smooth gluing of the images of functions and calculate the Gauss An Acad Bras Cienc (2008) 80 (4) 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 2.1).Using this lemma, we can glue smoothly two functions which are tangent at a point (Lemma 2.2).By a result of Dan Henry, we construct a better cut-off function (see Lemma 2.5).
These results are used to construct metrics on disk D 2 such that the metrics have bounded curvatures and the metrics when restricted in a small neighborhood of ∂ D 2 are product metrics.See Lemma 2.6 and 2.8.LEMMA 2.1.There is a smooth function φ (2) PROOF.Let where 0 < δ ≤ a 4 .Set where j (x) is the mollifier function defined on R by and A is equal to Then by calculation, we get Hence, φ δ (x) (simply denoted by φ(x)) is the required function.
Moreover, we have since Ae > 1.
An Acad Bras Cienc (2008) 80 (4) LEMMA 2.2.Suppose f (x), g(x) are two smooth functions defined on R and satisfy at some point x = c ∈ R. Given any δ > 0, there is a smooth function h δ (x) on R such that PROOF.By the proof of Lemma 2.1, there is a smooth function ϕ δ (x) ∈ [0, 1] such that Then h δ is the required function.
Still consider functions as in the above lemma.Denote Let h δ , 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), g(x) ≥ σ > 0 for all x ∈ [c − 1, c + 1], and let δ ∈ (0, 1/2) be small enough.
Then we have the following lemma.
where K h δ is the Gauss curvature of the surface h δ .Moreover, we have PROOF.By equation ( 5), by Taylor expansion formula and equations ( 3)-( 4) in Lemma 2.1, we have Bras Cienc (2008) 80 (4) where By assumption, additionally, and we set by a similar argument as in the proof of Lemma 2.3, we get h δ ≤ 14(M 3 + N 3 ).
LEMMA 2.5.There is a smooth function φ(x)(0 ≤ φ(x) ≤ b) defined on R, such that the following conditions are satisfied.

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Let us recall some properties of the pseudosphere.Suppose the parametric equations of pseudosphere are We only consider the part of x ≥ 0, and we have A direct calculation implies that LEMMA 2.6.For any given 0 < ε << 1 and 0 < δ << 1, there is a smooth function h ε,δ (x) on [0, +∞) such that Here PROOF.Suppose f (x) is a smooth function which is defined as where t ∈ π 2 , π .It is easy to check that point belongs to the image of function f (x).Suppose line AB is tangent to f (x) at point A (Fig. 2).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.D F is parallel to x-axis.And x F = x B .
Figure 2 It's easy to see that |AB| = 1.
Thus, we have and , Define g(x) as follows That is the broken line AD F. We smooth it by the method in Lemma 2.1 and denote the result smooth function by g(x).Choose By inequality (4), we have the estimate: Next, we observe that f (x 2ε ) = g(x 2ε ) and f (x 2ε ) = g (x 2ε ).According to Lemma 2.2, for any 0 < δ < 1 4 |C E|, there is a smooth function k δ (x) on 0, .
According to Lemma 2.3, we have where Then, we get a global smooth function REMARK 2.7.Let h ε,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: So, we can choose a constant C which is independent of ε and δ, such that LEMMA 2.8.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.
, h ε,δ has following parametric equation: x 2ε and t ε are the same as in Lemma 2.6.Moreover, the sectional curvature of ε,δ is bounded by a constant C which is independent of ε and δ.
REMARK 2.9.If we let δ (which used in Lemma 2.2-2.3)be small enough in Lemma 2.8, the area of the smooth surface above is less than (2 + 2 √ 2)π + 1.If we add two ends to the surface, we will get the required Y-piece as showed in Figure 4.By Lemma 2.8 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)

sequence of compact metric spaces. Suppose d n is the distance function on A n , and d n+1 | A n = d n for all n. Define function d
, 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 diameter(B n ) → ∞ as n → ∞, then ( A, d) is a complete metric space.
Before the proof of Proposition 3.2, 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, Then, we have where 2 2i is a surface with nonempty boundary which consists of three circles.
PROOF.Our goal is to construct a sequence of complete smooth metrics g ε on R 3 such that Vol(R 3 , g ε ) → 0, as ε → 0, and the corresponding curvatures are uniformly bounded by 1, i.e.K g ε ≤ 1.In Lemma 2.8, we have constructed a surface (disk) with nonempty boundary (circle) ∂ satisfying Vol( , g ) < +∞, and, if restricted on a tubular neighborhood of ∂ in , g is product metric.
As has been mentioned in Section 2, it is also easy to construct metric on Y-piece 2 2i satisfying the similar conditions as above.With such metric, 2 2i looks like as in Figure 6.Next, we give the explanation in details as follows.
We assign 2 2i a metric to be a Y-piece (see Fig. 6), and denote it by , which means the lengths of the components of the boundary ∂Y 2i,ε are 2π ε 2 i−1 , 2π ε 2 i , 2π ε 2 i+1 respectively.In summary, we use the following notations: , for i 1. ( simply the last equation as Then the process of the metric ε on R 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 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.S 1 ε means that the circle has length 2πε under the given metric ε 2 dθ 2 .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;(2) × I , and ∂Y i;(3) × I denote the three tubular neighborhoods of the three boundary components of Y i for i ≥ 1.

First, the metrics on ∂Y
are chosen to be the following product metrics: and 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 and The order of gluing is that we exchange the variables α and θ in the (or first) so they can glued directly the first (or second) metric.
At last, what remains to to the metrics on . 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 Vol S 1 By scaling and the fact 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 Therefore, MinVol(R 3 ) = 0.
The proposition is proved.
An Acad Bras Cienc (2008) 80 (4) PROOF.We still use the notations ( 13) and ( 14) in the proof of Proposition 3.2.The metric on Y i (i.e.Y 2i,ε or Y 2i;( ε ) ) is denoted by g 2,ε (means the second kind metric).Let the metric on where and f (t) is defined by Replace the metric g 1, ε 2 i on 2 2i+1 (that is, the surface M i , see ( 13)) by the metric g 1, ε(t) and denote the surface , where ε(t) means that the length of the boundary circle is a function 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 2 i and the metric on boundary S 1 , the direct way is to require that the metric on a small tubular neighborhood is a product metric, and But, in order to make the volume Vol S 1 An Acad Bras Cienc (2008) 80 (4) Fig. 7 -Gluing.
The gluing (ii) is similar to (i).Hence, the metric g 2,ε on Y i (i.e.Y 2i;( ε . 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 G(x, t), see ( 36)) which is used to construct the smooth surface M ( f (t)ε)/2 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 G(x, t).By Lemma 2.5, there is a There exists a constant C > 0, such that An Acad Bras Cienc (2008) 80 (4) In particular, where t ∈ R and x = f (t)x.Let where For simplicity, let Then, we have Then we define the function G(x, t) by Step 2: Calculation of the sectional curvature of surface G(x,t) .Now for every t, let G(x,t) be the surface generated by rotation of the graph of the function G . It can also be used to construct the surface where θ, α ∈ [0, 2π ].Let An Acad Bras Cienc (2008) 80 (4) Then by the structure equations, we have , Step 3: We claim that: PROOF.By Equation (32), we have By Equation (33), we have An Acad Bras Cienc (2008) 80 (4) By Equation (36), we have Hence, by Equations ( 30)-( 36), ( 40)-( 42), we have This implies that f (x)x → +∞, and it is a contradiction since f (x) < 1 x .Hence, there is a positive number σ > 0 which is independent of x, t and ε, such that ϕ(x, t) + f (t) h(x) > σ > 0. Therefore, According to inequalities ( 43), ( 44) and ( 47), curvatures in (38) must be bounded.So the Claim is proved.
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 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.