The mapping properties of filling radius and packing radius and their applications

Transcription

1 Differential Geometry and its Applications ) The mapping properties of filling radius and packing radius and their applications Luofei Liu Nankai Institute of Mathematics, Nankai University, Tianjin , P.R. China Received 4 July 2003; received in revised form 8 November 2003 Available online 8 September 2004 Communicated by O. Kowalski Abstract Let f : V W be a map between closed, oriented Riemannian n-manifolds. It is shown that FillRadW) dilf ) FillRadV ), if degf ) =1. By this mapping property, we obtain an estimate from below for the filling radius of a closed, oriented, nonpositively curved manifold, or a manifold with sectional curvature bounded above by a positive constant. In addition, a similar mapping property of packing radius and a corollary are also obtained Elsevier B.V. All rights reserved. MSC: 53C23; 53C20 Keywords: Filling radius; Dilatation; Injectivity radius; Packing radius 1. Introduction First recall the precise definition of filling radius from [4]. Let V be a closed, oriented Riemannian n-manifold and L V ) be the Banach space of bounded Borel functions f on V with the norm f =sup x fx). It is easy to see that the map J : V L V ) defined by x f x ) = distx, ) is an isometric embedding in the strong sense: it preserves not only lengths of paths but also distances). Let U ε V ) be the ε-neighborhood of JV) in L V ), andleti ε : JV) U ε V ) be the inclusion map. address: luofei-liu@263.net L. Liu) /$ see front matter 2004 Elsevier B.V. All rights reserved. doi: /j.difgeo

2 70 L. Liu / Differential Geometry and its Applications ) Gromov s filling radius is defined by FillRadV ) = inf { ε>0: H n i ε ) = 0 }, where H n i ε ) : H n V, Z) H n U ε V ), Z) is the homomorphism induced by i ε. Replacing Z by Q, one defines the filling radius with rational coefficients FillRadV, Q) [9]. IfV is nonorientable, one uses the coefficient group Z 2. This invariant gives important information about the systole the length of a shortest non-contractible loop) of an essential manifold V [4]: sys 1 V ) 6FillRadV ). Up till now, very little is known about filling radius. The exact value of this invariant has been computed for very few manifolds [4,8]. However some important results about positively curved manifolds with large filling radius have been obtained by Wilhelm [11]. In this note, we will prove a mapping property of filling radius, which is related to the dilatation of a map. Recall that the dilatation of a map f : X Y between two metric spaces X, d X ) and Y, d Y ) is defined by d Y f x), f x )) dilf ) = sup. x,x X,x x d X x, x ) If f is a Lipschitz map, dilf ) is the minimal Lipschitz constant C satisfying d Y f x), f x )) Cd X x, x ). The dilatation of a non-lipschitz map is infinity. The local dilatation at x is the number: dil x f ) = limf BX ε 0 x,ε)), where B X x, ε) is an open ball of radius ε centered at x of X. IfX, d) is an inner metric space e.g., Riemannian manifold), then dilf ) = sup x X dil x f ). IfX and Y are Riemannian manifolds, and f is differentiable, then dil x f ) = df x where df x : T x X T fx) Y is the differential of f at x. Throughout this paper, V and W are always closed, connected, oriented Riemannian n-manifolds unless specifically stated, and S n is the standard sphere with Riemannian metric of constant curvature +1. The following theorem shows a mapping property of filling radius: Theorem 1.1. Suppose that f : V W has nonzero degree. Then i) if degf ) =1, FillRadW) dilf ) FillRadV ); ii) if degf ) > 1, FillRadW, Q) dilf ) FillRadV, Q). For closed, connected, oriented Riemannian n-manifold V, there are maps V S n of every degree k Z by the Hopf s theorem. We denote by dil{v S n ; deg =±k} the smallest number α such that there is a map V S n with degree k or k and dilatation α. Corollary 1.2. If dil{v S n ; deg =±1} 3/2, then FillRadV)>π/6. Moreover, if K V 1, thenv is a twisted sphere. Proof. Let f : V S n be a map with degree 1 or 1 and dilatation less than 3/2, we have: FillRadV ) 1 dilf ) FillRadSn )

3 L. Liu / Differential Geometry and its Applications ) = 1 dilf ) 1 2 arccos 1 ) from [8, Th. 2] n + 1 > π 2 = π 6. By [11, Th. 2, 4)], we know that V is a twisted sphere if K V 1. For packing radius in the sense of Grove and Markvorsen [5], we have a similar mapping property Proposition 4.1). In the mapping property, the degree 1 is not required. For a closed Riemannian n-manifold, we denote by dil{v S n } the smallest α such that there is a surjective map V S n with dilatation α. Although the mapping property for packing radius is trivial, the following corollary, by Proposition 4.1 and the pack n 2 sphere theorem of Grove and Wilhelm [7, Theorem C], strengthens the result in Corollary 1.2 under weaker assumptions. Corollary 1.3 To be proven as Corollary 4.2). LetV be a closed Riemannian n-manifold. If n 4, K V 1 and dil{v S n } < 2 π arccos 1 n 3 ),thenv is diffeomorphic to Sn. In view of the difficulty in calculating the filling radius of a Riemannian manifold, it is natural to estimate this invariant by various method. Gromov [4] obtains an universal upper bound for the filling radius of a closed connected Riemannian manifold. In addition, Katz [8] has shown that for any closed manifold V, FillRadV ) 1 DiamV ). In[8], Katz also gives a lower bound for the filling radius of 3 CP n, HP n, CaP 2. In this note, we attempt to give a preliminary lower bound for the filling radius of a Riemannian manifold under certain conditions about sectional curvature and the largest injectivity radius. Thanks to the work of Katz [8], the exact value of the filling radius for S n has been obtained: FillRadS n ) = 1 2 arccos 1 ).SoweuseTheorem 1.1 to compare the filling radius of V with that of n+1 S n, and hence obtain a lower bound for the filling radius of V. Firstly the following proposition gives an estimate from below for the filling radius of a nonpositively curved manifold, in which we use the expression Inj max V ) = max{inj x V : x V }, the largest injectivity radius of V. Proposition 1.4 To be proven as Proposition 3.2). For any closed, oriented, nonpositively curved manifold V, we have: FillRadV ) Inj maxv ) π FillRadS n ) Inj maxv ). 4 Remark ) Note that nonpositively curved manifolds are essential [4], Gromov s famous isosystolic inequality, sys 1 V ) 6 FillRadV ), is valid. The injectivity radius of a compact, nonpositively curved manifold V can be related to the systole [1, Th. 3.4]: InjV ) 1 2 sys 1V ). ByProposition 1.4, if Inj max V ) InjV ) k, then: sys 1 V ) 8 FillRadV ). k For nonpositively curved manifold with Inj maxv ) InjV ) than the isosystolic inequality. 4, we obtain a better estimate from above for systole 3

4 72 L. Liu / Differential Geometry and its Applications ) ) For a flat torus T n, the inequality is weak. Indeed, we have that FillRadT n )/Inj max T n ) = FillRadT n )/InjT n ) = 1/3 [4,p.11]. Next, we shall give a lower bound for the filling radius of manifold with sectional curvature bounded above by a positive constant. Proposition 1.6 To be proven as Proposition 3.3). IfK V c, c > 0, and Inj max V ) π 2 c,then FillRadV ) 1 FillRadS n ) 1 c π 4 c. Using Klingenberg s injectivity radius estimate theorem [2, Th. 5.9 and the remark after that], the following corollary is immediate: Corollary 1.7. Let V be a closed, oriented Riemannian n-manifold. Then FillRadV ) 1 c π FillRadS n ) 1 4 in each of the following cases: c i) n is even and 0 <K V c; ii) n 3, V is simply connected and 0 < 1 4 c K V c. Remark ) Removing the simply connected restriction and relaxing the pinching constant in odd dimensional cases, a recent result, due to Fang and Rong [3, Cor. 0.3], says that if 0 <δ c K V c, 0 <δ 1, and π 1 V ) = Z q with prime q 2 n 1)/2 and the second Betti number b 2 V, Z) = 0, then InjV ) in,δ,q), a constant depending only on n, δ and q. Under these topological conditions, if the upper bound c for sectional curvature satisfies in,δ,q), one has the same conclusion as Corollary ) Recall that the sectional curvature of CP n satisfies 1 4 K CP n 1. For CP n also for HP n,cap 2 ), the estimate in Corollary 1.7 is inferior to Katz s [8]: FillRadCP n ) 1 2 arccos 1 ). The result of Proposition 1.6 is merely a general, rough filling radius estimate for manifold with K V c, c>0. 3 3) Combining the corollary with [11, Th. 2 1)] gives that if V is an even-dimensional manifold and 1 K V c, then 1 4 c FillRadSn ) FillRadV ) FillRadS n ). π 2 c 2. Proof of the mapping property of filling radius The inwardness of the following lemma was stated in [4] without detailed proof. To give a detailed proof, it seems to the author that some mild assumption is needed. Lemma 2.1. Let Y be a subset of a metric space X. Then every Lipschitz map ϕ : Y L V ) has a Lipschitz extension ϕ : X L V ) with dil ϕ) = dilϕ) in each of the following cases: i) Y is Lindelöf ;

5 L. Liu / Differential Geometry and its Applications ) ii) Y is separable; iii) Imϕ) C 0 V ) the set of bounded continuous functions on V ). Proof. We define ϕ : X L V ) by ϕx) := f x, f x v) := inf ϕy))v) + α dx x, y) ), where x X, v V and α = dilϕ). Step 1. ϕ is well-defined. We must check that f x0 : V R is a bounded Borel function on V for every fixed X. Fixay 0 Y and assume that ϕy 0 ) L V ) = C 0.For y Y, since sup v V ϕy)v) ϕy 0 )v) = ϕy) ϕy 0 ) α d X y, y 0 ),weget ϕy)v) ϕy 0 ) ) v) α d X y, y 0 ) C 0 α d X y, y 0 ) for v V. Then clearly ϕy)v) + α d X,y) C 0 α d X y, y 0 ) + α d X,y) C 0 α d X,y 0 ) hence f x0 v) C 0 α d X,y 0 ). On the other hand, f x0 v) = inf ϕy))v) + α dx,y) ) ϕy 0 ) ) v) + α d X,y 0 ) C 0 + α d X,y 0 ). Therefore f x0 v) C 0 + α d X,y 0 ) for v V, i.e., f x0 is bounded. The remainder is to prove that {v V : f x0 v)<t} is a Borel subset of V for t R. Indeed, { v V : fx0 v) < t } { = v V :inf ϕy))v) + α dx,y) ) <t} = { v V : ϕy))v) + α dx,y) ) <t } := A y, where A y := {v V : ϕy))v) + α d X,y)<t}={v V : ϕy))v) < t α d X,y)} is a Borel subset of V for every y Y,sinceϕy) is a Borel function on V. We denote the open metric ball of radius 1/n centered at y of Y by By,1/n). In the case i), the open covering {By,1/n): y Y } of Y contains a countable subcovering {By n,i, 1/n): i N}. Checking on the following equality is a routine: A y = A yn,i, n=1 i=1 hence {v V : f x0 v) < t} is a Borel subset of V. The case ii) is similar to the case i). In the case iii), it is easy to prove that every f x0 : V R is continuous.

6 74 L. Liu / Differential Geometry and its Applications ) Step 2. ϕ Y = ϕ. Forafixedy 0 Y and v V we have: f y0 v) ϕy 0 ) ) v) + α d X y 0,y 0 ) = ϕy 0 ) ) v). On the other hand, for y Y we have: ϕy0 ) ϕy) ) v) ϕy0 ) ϕy) = dl V ) ϕy0 ), ϕy) ) 1) and hence α d X y 0,y) f y0 v) = inf ϕy))v) + α dx y 0,y) ) ϕy 0 ) ) v). 2) Then 1) and 2) above give that ϕy 0 ) = f y0 = ϕy 0 ) for y 0 Y. Step 3. dil ϕ) α. Forfixedx i X, i = 1, 2, and v V, n N, one can choose elements y i = y i v, n) in Y so that f xi v) = inf ϕy))v) + α dxi,y) ) ϕy i ) ) v) + α d X x i,y i ) 1 n. Then we have that f x1 v) f x2 v) = inf ϕy))v) + α dx x 1,y) ) inf ϕy))v) + α dx x 2,y) ) [ ϕy 2 ))v) + α d X x 1,y 2 ) ] [ ϕy 2 ))v) + α d X x 2,y 2 ) ] + 1 n = α d X x 1,y 2 ) α d X x 2,y 2 ) + 1 n α d X x 1,x 2 ) + 1 n. Since the inequality above holds for any n N, f x1 v) f x2 v) α d X x 1,x 2 ). Similarly, f x1 v) f x2 v) α d X x 1,x 2 ). The two inequality yield: f x1 v) f x2 v) α d X x 1,x 2 ) for x 1,x 2 X, v V. Then we can deduce that d L V ) ϕx1 ), ϕx 2 ) ) = ϕx1 ) ϕx 2 ) = fx1 f x2 = sup fx1 v) f x2 v) α dx x 1,x 2 ). v V Hence dil ϕ) α, which completes the proof of Lemma 2.1. Now we are in position to prove Theorem 1.1. One may suppose that f is a Lipschitz map if not, the inequality in Theorem 1.1 is obvious). According to the standard isometric embedding J, V, W may be viewed as a subspace of L V ), L W), respectively. From Lemma 2.1, themapf : V W L W) possesses an extension f : L V ) L W) with dil f)= dilf ). Suppose that U ε V ) is any ε-neighborhood of V in L V ), V bounds in U ε V ) in the ordinary algebraic topological meaning i.e., H n i ε ) : H n V, Z) H n U ε V ), Z) is the zero homomorphism).

7 L. Liu / Differential Geometry and its Applications ) First prove the following inclusion: f U ε V ) ) U αε W), where α = dilf ) and U αε W) is the αε-neighborhood of W in L W). Indeed, for any h U ε V ) L V ), choose v 0 V so that d L V )h, v 0 ) close enough to d L V )h, V ),thenwehave: d L W) fh),w ) d L W) fh), fv 0 ) ) α d L V )h, v 0 ) α ε. fv 0 ) = fv 0 ) W) Lemma 2.1: dil f)= α) Hence ) holds. Next, since f is an extension of f, we have the commutative diagram: ) V f W J V J W L V ) f L W) From ), we further have the following commutative diagram: V f W i ε U ε V ) f i αε U αε W) i.e., i αε f = f i ε,wherei ε and i αε are the inclusion maps. Let [V ], [W] denote the fundamental homology class of V,W, respectively. When degf ) =±1, H n f ))[V ]=±[W]. Thenwehave: Hn i αε ) ) [W] ) =±H n i αε ) H n f ) ) [V ] ) =± H n i αε ) H n f ) ) [V ] ) =± H n f) H n i ε ) ) [V ] ) = 0. Hence H n i αε ) = 0, i.e., W bounds in U αε W) in the algebraic topological meaning, therefore FillRadW) α ε. Takingε above close enough to FillRadV ), we obtain the inequality in the case degf ) =1. If degf ) > 1, H n f ))[V ]=degf ) [W], we just deduce that FillRadW, Q) degf ) FillRadV, Q), which completes the proof of Theorem Estimates from below for filling radius For a fixed V,letB Tx0 V 0,r 0 ) denote the ball of radius r 0 centered at the origin of the tangent space T x0 V and B V,r 0 ) the ball of radius r 0 centered at of V.Whenr 0 is not bigger than Inj x0 V ), we have the inverse of the exponential map at : exp 1 : B V,r 0 ) B Tx0 0,r 0 ).

8 76 L. Liu / Differential Geometry and its Applications ) Lemma 3.1. If r 0 Inj x0 V ), and dilexp 1 ) β, there exists a map ϕ : V S n with degree one and dilatation π β/r 0. Proof. Fix a point p S n and consider the composite ϕ: B V,r 0 ) exp 1 BTx0 V 0,r 0 ) f B Tx0 V 0,π) g B Tp S n0,π) h S n in which f is the obvious diffeomorphism with dilatation π/r 0 ; g is the obvious isometry; h has degree 1, dilatation 1, and which maps B Tp S n0,π) onto p, the point of S n diametrically opposed to p. The composite map ϕ sends B V,r 0 ) to p.extendϕ to V S n by setting ϕv \ B V,r 0 )) = p.by what we have seen above, degϕ) = 1anddilϕ) dilh) dilf ) dilexp 1 ) π β/r 0. This completes the proof of Lemma 3.1. We are now ready to give the proof of Proposition 1.4, which depends mainly on Lemma 3.1 and Theorem 1.1. First recall the statement: Proposition 3.2. For any closed, oriented, nonpositively curved manifold V, FillRadV ) Inj maxv ) π FillRadS n ) Inj maxv ). 4 Proof. Since the function Inj : V 0, + ] is continuous [1, Th. 3.3], there exists a point V such that Inj x0 V ) = Inj max V ) := r 0.FromLemma 3.1, there exists a map ϕ x0 : V S n with degree one and dilatation less than π β/r 0.ByTheorem 1.1, wehave 1 FillRadV ) dilϕ x0 ) FillRadSn ) Inj maxv ) FillRadS n ). πβ The remainder is to prove β 1. Since V is nonpositively curved, using Rauch comparison theorem [2, Th. 1.28] and comparing with Euclidean space, we can deduce that V has no conjugate points and dexp x0 ) v X) X, v B Tx0 V 0,r 0 ), X T x0 V, where dexp x0 ) v : T v B Tx0 V 0,r 0 )) T y V, y = exp x0 v), denotes the differential of exp x0 at v and the tangent space T v B Tx0 V 0,r 0 )) is naturally identified with T x0 V.Sinceexp x0 : B Tx0 V 0,r 0 ) B v,r 0 ) is a diffeomorphism, dexp 1 ) y =[dexp x0 ) v ] 1 for v B Tx0 V 0,r 0 ), y = exp x0 v) B V,r 0 ).Let dexp x0 ) v X) = Y,wehave dexp 1 ) y Y ) = X dexpx0 ) v X) = Y. Therefore dil y exp 1 ) = dexp 1 ) y 1, and hence, dil ) exp 1 = sup y B,r 0 ) dil y exp 1 ) 1, i.e., β 1, which completes the proof of the proposition. The following method of estimating from below for the filling radius of a manifold with sectional curvature bounded above by a positive constant is similar to that given above. The only difference is that the estimate for dilexp 1 ) is more difficult here.

9 L. Liu / Differential Geometry and its Applications ) Proposition 3.3. If K V c, c>0, and Inj max V ) π/2 c),then FillRadV ) 1 FillRadS n ) 1 c π 4 c. Proof. Similarly, first take V such that Inj x0 V ) = Inj max V ). Taker 0 = π 2 c and consider exp : B Tx0 V 0,r 0 ) B v,r 0 ).Thenwehave: FillRadV ) r 0 πβ FillRadSn ), where β = dilexp 1 B V,r 0 )). SinceK V c, andr 0 = π 2 c < π c,exp x0 B Tx0 V 0,r 0 ) is nonsingular [2, the remark after 1.29]. Using the geometric Rauch theorem [1, Th.7.3 and the remark 7.1 after that], we have: dexp x0 ) v X) X S c v ) v = sin c v ) c v for v B Tx0 V 0,r 0 ), X T x0 V.Letv = exp 1 y), y B v,r 0 ),anddexp x0 ) v X) = Y.Since dexp 1 ) y =[dexp x0 ) v ] 1, one deduces dexp 1 ) y Y ) c v Y sin c v ), ) dil y exp 1 c v sin c v ). Note that c v / sin c v ) is an increasing function on 0,r 0 ],then dil ) c v c exp 1 sup sin c v ) = r0 sin c r 0 ), v B Tx0 V 0,r 0 ) i.e., β c r 0 / sin c r 0 ). Therefore r 0 πβ r 0 π sin cr0 ) = 1 cr0 π c sin c r 0 ) = 1. c π This completes the proof of the proposition. 4. A mapping property of packing radius and differentiable sphere theorem Recall that the qth q 2) packing radius, pack q X, of a compact metric space X in the sense of Grove and Markvorsen [5] is the largest r > 0 for which X contains q disjoint open r-balls, i.e., pack q X = 1 2 max min d Xx i,x j ), x 1 x q 1 i<j q where the maximum is taken over all configurations of q points in X.

10 78 L. Liu / Differential Geometry and its Applications ) Proposition 4.1. Let f : X, d X ) Y, d Y ) be a map between two compact metric spaces. If f is surjective, then pack q X 1 dilf ) pack qy. Proof. If f is not a Lipschitz map, the inequality is obvious. Hence we may assume that f is a Lipschitz map and dilf ) = α. Letpack q Y = r and B Y y 1,r),...,B Y y q,r)be q disjoint open r-balls centered at y 1,...,y q, respectively. It is easy to see that f 1 B Y y 1,r)),...,f 1 B Y y q,r)) are q disjoint subsets of X. Sincef is surjective, choose x i X such that fx i ) = y i, i = 1,...,q. Consider the open ball B X x i,r/α)of X.For x B X x i,r/α), ) d Y fx),yi = dy fx),fxi ) ) α d X x, x i )<α r α = r. Therefore fx) B Y y i,r), i.e., x f 1 B Y y i,r)), hence B X x i,r/α) f 1 B Y y i,r)). Then B X x 1,r/α),...,B X x q,r/α)are q disjoint open r α -balls of X, hence pack qx r α. Note that if X and Y are Riemannian manifolds, the degree 1 is not required here, and we do not even require that X and Y have the same dimension. If dim X>dim Y, in general the inequality above is weak. Indeed, an early theorem due to Lawson [10, Theorem 1] states that every differentiable map f : S m S n, m>n>0, has dilatation greater than or equal to 2 if f is not nullhomotopic. An inspection of Lawson s proof will convince us that the same conclusion, dilf ) 2, holds only if f : S m S n is a surjective map, on the other hand, pack q S n is independent of n [6, Prop. 2.4]. Corollary 4.2. If n 4, then any closed Riemannian n-manifold with sectional curvature K V 1 and dil{v S n } < 2 π arccos 1 n 3 ) is diffeomorphic to Sn. Proof. Let f : V S n be a surjective map with dilf ) < 2 π arccos 1 n 3 ).Wehave: pack n 2 V 1 dilf ) pack n 2S n = 1 dilf ) 1 2 arccos > π 4. 1 ) n 3 from [6, Prop. 2.4i)] By the pack n 2 sphere theorem [7, Theorem C], we know that V is diffeomorphic to S n. Remark ) In the case n = 3, using the pack n 1 sphere theorem [6, Theorem A] we have that any closed Riemannian 3-manifold with K V 1anddil{V S n } < 2 is diffeomorphic to S 3. 2) The curvature pinching differentiable sphere theorems or other differentiable sphere theorems probably imply the conclusion of Corollary 4.2, though the author does not know how to derive it.

11 L. Liu / Differential Geometry and its Applications ) Acknowledgements It is my pleasure to thank Professor F. Fang for guiding me to this direction, and for his encouragement and many stimulating discussions. I would especially like to thank the referee for suggesting all contents of Section 4,which ledup to Corollary 1.3 given here, and for pointing out a little hole in an earlier proof of Lemma 2.1. References [1] I. Chavel, Riemannian Geometry A Modern Introduction, Cambridge Tracts in Mathematics, vol. 108, Cambridge University Press, Cambridge, [2] J. Cheeger, D. Ebin, Comparison Theorems in Riemannian Geometry, North-Holland, Amsterdam, [3] F. Fang, X. Rong, The second twisted Betti number and the convergence of collapsing Riemannian manifolds, Invent. Math ) [4] M. Gromov, Filling Riemanning manifolds, J. Differential Geom ) [5] K. Grove, S. Markvorsen, New extremal problems for the Riemannian recognition program via Alexandrov geometry, J. Amer. Math. Soc ) [6] K. Grove, F. Wilhelm, Hard and soft packing radius theorems, Ann. of Math ) [7] K. Grove, F. Wilhelm, Metric constraints on exotic spheres via Alexandrov geometry, J. Reine Angew. Math ) [8] M. Katz, The filling radius of two-point homogeneous spaces, J. Differential Geom ) [9] M. Katz, The rational filling radius of complex projective space, Topology Appl ) [10] H.B. Lawson, Stretching phenomena in mappings of spheres, Proc. Amer. Math. Soc ) [11] F. Wilhelm, On the filling radius of positively curved manifolds, Invent. Math )