CONFORMAL GEOMETRY AND DYNAMICS An Electronic Journal of the American Mathematical Society Volume 12, Pages 1–9 (January 22, 2008) S 1088-4173(08)00173-2
NAGATA DIMENSION AND QUASI-MOBIUS¨ MAPS
XIANGDONG XIE
Abstract. We show that quasi-M¨obius maps preserve the Nagata dimension of metric spaces, generalizing a result of U. Lang and T. Schlichenmaier (Int. Math. Res. Not. 2005, no. 58, 3625–3655).
1. Introduction In this note we show that a certain covering dimension of metric spaces – the Nagata dimension – is invariant under quasi-M¨obius maps. Nagata dimension was introduced and studied by Assouad ([A]). It captures both macroscopic and microscopic structures of metric spaces. Hence Nagata dimension has potential applications in both large scale geometry of metric spaces and analysis on metric spaces. S. Buyalo and V. Schroeder ([BS]) have used Nagata dimension to prove that any Gromov hyperbolic group whose Gromov boundary has topological dimension n admits a quasi-isometric embedding into the product of (n+1) trivalent trees. On the other hand, U. Lang and T. Schlichenmaier ([LS]) showed that Nagata dimension is preserved by quasisymmetric maps. Notice that it is clear from the definition (see Section 2) that Nagata dimension is invariant under bi- Lipschitz maps. The purpose of this note is to extend the result of U. Lang and T. Schlichenmaier to quasi-M¨obius maps:
Theorem 1.1. Let f :(X1,d1) → (X2,d2) be a quasi-M¨obius map. Then (X1,d1) and (X2,d2) have the same Nagata dimension. The proof of Theorem 1.1 is not difficult. We first observe that each quasi-M¨obius map can be written as a composition of a quasisymmetric map and at most two metric inversions (see Section 3 for the definition). Hence by the above mentioned result of U. Lang and T. Schlichenmaier it suffices to show that metric inversions preserve Nagata dimension. This statement is proved by modifying the argument of U. Lang and T. Schlichenmaier (see the proof of Theorem 1.2 in [LS]). As an immediate application of Theorem 1.1, consider X = {1/n : n =1, 2, ···} ⊂ R2 and Y = {(m, n) ∈ R2 : m, n ∈ Z}⊂R2 with the metric induced from R2. Since X has Nagada dimension 1 and Y has Nagata dimension 2, they are not quasi-M¨obius equivalent. The author is not aware of a direct proof of this fact. There are two other dimensions closely related to Nagata dimension: capac- ity dimension ([B]) and asymptotic dimension of linear type. Capacity dimen- sion concerns the microscopic structure of a metric space and is also invariant
Received by the editors January 23, 2007. 2000 Mathematics Subject Classification. Primary 54F45, 30C65. Key words and phrases. Nagata dimension, quasi-M¨obius maps, metric inversion.
c 2008 American Mathematical Society Reverts to public domain 28 years from publication 1 2 XIANGDONG XIE under quasisymmetric maps. Asymptotic dimension of linear type captures the large scale structure of a metric space and is invariant under quasi-isometries. No- tice that capacity dimension is not invariant under quasi-M¨obius maps: consider X = {1, 1/2, ··· , 1/n, ···} ⊂ R2 and Y = {1, 2, ··· ,n,···} ⊂ R2 with the metric induced from R2. The inversion about the unit circle in R2 sends X to Y ,soX and Y are M¨obius equivalent; in particular, they are quasi-M¨obius equivalent. However, the capacity dimension of X is 1 while that of Y is 0.
2. Nagata dimension In this section we recall some basic facts about Nagata dimension (see [LS], [BL] and [BDLM] for more details). Let (X, d)beametricspaceandB = {Bi}i∈I a family of subsets of X.The family B is called D-bounded, D ≥ 0, if diam(Bi):=sup{d(x, x ):x, x ∈ Bi}≤D for all i ∈ I.Fors>0, the s-multiplicity of B is the infimum of all integers n such that every subset of X with diameter ≤ s meets at most n members of the family. Definition 2.1. Let (X, d) be a metric space. The Nagata dimension of X, denoted by dimN (X, d), is the infimum of all integers n with the following property: There exists a constant c ≥ 1 such that for all s>0, X has a cs-bounded covering with s-multiplicity at most n +1.
Recall that two metric spaces (X1,d1)and(X2,d2)arebi-Lipschitz equivalent if there is a bijection f :(X1,d1) → (X2,d2)andanL ≥ 1 such that d1(x, y)/L ≤ d2(f(x),f(y)) ≤ Ld1(x, y)holdsforallx, y ∈ X1. It is clear from the definitions that two bi-Lipschitz equivalent metric spaces have the same Nagata dimension. Suppose s>0andB is a covering of X.IfB can be written as a union B n B B = k=0 k such that each k has s-multiplicity at most 1, then it is clear that B has s-multiplicity at most n +1. If X = A ∪ B,thendimN (X, d)=max{dimN (A, d), dimN (B,d)} (Theorem 2.7 of [LS]). In particular, if X contains at least two points and p ∈ X,then dimN (X, d)=dimN (X\{p},d) (since the Nagata dimension of a singleton is 0). A key ingredient in the proof of Theorem 1.1 is the following existence result of a sequence of coverings. There are similar results for asymptotic dimension ([D], Proposition 1) and capacity dimension ([B], Proposition 4.4). Notice that the existence of a sequence of coverings as in Proposition 2.2 implies dimN (X, d) ≤ n. We shall only use properties (i)–(iii). Proposition 2.2 (Proposition 4.1 of [LS]). Suppose (X, d) is a metric space with dimN (X, d) ≤ n<∞. Then there is a constant c ≥ 1 such that for all sufficiently large r>1, there exists a sequence of coverings Bj of X, j ∈ Z, with the following properties: ∈ Z Bj n Bj Bj j (i) For every j , we have = k=0 k where each k is a cr -bounded family with rj-multiplicity at most 1. (ii) For all j ∈ Z and x ∈ X,thereexistsaC ∈Bj that contains the closed ball B(x, rj). ∈Bi ∈Bj ⊂ (iii) Whenever B k and C k for some k and i 3. Quasi-Mobius¨ maps and metric inversions In this Section we recall the notions of quasisymmetric maps, quasi-M¨obius maps and metric inversions, and discuss their relations (see [BHX], [V], [H] and [V2] for more details). Let η :[0, ∞) → [0, ∞) be a homeomorphism. A homeomorphism between metric spaces f :(X, dX ) → (Y,dY )isη-quasisymmetric if for all pairwise distinct points x, y, z ∈ X,wehave d (f(x),f(y)) d (x, y) Y ≤ η X . dY (f(x),f(z)) dX (x, z) A homeomorphism f :(X, dX )→(Y,dY ) is quasisymmetric if it is η-quasisymmetric for some η. Quasisymmetric maps send bounded metric spaces to bounded metric spaces (Proposition 10.8 of [H]). Let Q =(x1,x2,x3,x4) be a quadruple of pairwise distinct points in (X, d). The cross ratio of Q with respect to the metric d is: d(x ,x )d(x ,x ) cr(Q, d)= 1 3 2 4 . d(x1,x4)d(x2,x3) Let η :[0, ∞) → [0, ∞) be a homeomorphism. A homeomorphism between metric spaces f :(X, dX ) → (Y,dY )isanη-quasi-M¨obius map if cr(f(Q),dY ) ≤ η(cr(Q, dX )) for all quadruple Q of distinct points in X,wheref(Q)=(f(x1),f(x2),f(x3),f(x4)). A homeomorphism f :(X, dX ) → (Y,dY )isquasi-M¨obius if it is η-quasi-M¨obius for some η. The inverse of a quasi-M¨obius map is quasi-M¨obius, and the composition of two quasi-M¨obius maps is also quasi-M¨obius (see [V]). Quasisymmetric maps are quasi-M¨obius. In general quasi-M¨obius maps are not quasisymmetric; for example, the inversions about spheres in Euclidean spaces are M¨obius (hence quasi-M¨obius) but are not quasisymmetric. However, quasi-M¨obius maps between bounded metric spaces are quasisymmetric (see [V]). Below we will find further connection between quasisymmetric maps and quasi-M¨obius maps (Proposition 3.6). We first recall the notion of metric inversion, which is a general- ization of inversions about unit spheres in Euclidean spaces. Let (X, d)beametricspaceandp ∈ X.SetIp(X)=X\{p} if X is bounded and Ip(X)=(X\{p})∪{∞} if X is unbounded, where ∞ is a point not in X.Then there is a metric dp on Ip(X) such that the following holds for all x1,x2 ∈ X\{p}: d(x1,x2) d(x1,x2) (3.1) ≤ dp(x1,x2) ≤ . 4 d(x1,p)d(x2,p) d(x1,p)d(x2,p) When X is unbounded, the following also holds for all x ∈ X\{p}: 1 1 ≤ d (x, ∞) ≤ . 4 d(x, p) p d(x, p) Furthermore, the identity map id : (X\{p},d) → (X\{p},dp)isanη-quasi-M¨obius homeomorphism with η(t)=16t (see [BHX] for a proof of the above statements). For any metric space (X, d)andp ∈ X,wecalltheidentitymapid:(X\{p},d) → (X\{p},dp)ametric inversion. 4 XIANGDONG XIE Lemma 3.2. Let (X, d) be a metric space, p ∈ X and A ⊂ X\{p} asubset. Suppose there are positive numbers R>r>0 such that r ≤ d(x, p) ≤ R for all x ∈ A. Then the identity map id :(A, d) → (A, dp) is a bi-Lipschitz map. Proof. Let x, y ∈ A. By the assumption r ≤ d(x, p),d(y, p) ≤ R. Hence dp(x, y) ≤ d(x,y) ≤ 1 ≥ d(x,y) ≥ 1 d(x,p)d(y,p) r2 d(x, y)anddp(x, y) 4 d(x,p)d(y,p) 4 R2 d(x, y). The next two lemmas say that the composition of two suitable metric inversions is bi-Lipschitz. Let (X, d) be an unbounded metric space and p ∈ X.Thendp is a metric on Ip(X)=(X\{p}) ∪{∞}.SetY = Ip(X)andρ = dp.Thenρ∞ is a metric on Y \{∞} = X\{p}. Lemma 3.3 (Proposition 3.7 of [BHX]). Let (X, d) and ρ∞ be as above. Then the identity map id :(X\{p},d) → (X\{p},ρ∞) is 16-bi-Lipschitz. We next describe a procedure which is analogous to obtaining the chordal metric on S2 from the standard metric on R2 via inverse stereographic projection. Let (Y,ρ) be an unbounded metric space and q ∈ Y . It is shown in [BK] (Lemma q 2.2) that there is a metric ρ on Y such that for all y1,y2 ∈ Y , ρ(y1,y2) q ρ(y1,y2) (3.4) ≤ ρ (y1,y2) ≤ . 4(1 + ρ(y1,q))(1 + ρ(y2,q)) (1 + ρ(y1,q))(1 + ρ(y2,q)) The metric ρq can be described as follows. Let Z = Y ∪{q} be the disjoint union of Y and a point q. Define a metric ρ on Z by: ρ(q,q)=0,ρ(y, q)=1+ρ(y, q) ∈ ∈ q for y Y and ρ (y1,y2)=ρ(y1,y2)fory1,y2 Y .Setρ = ρq . Then, by the paragraph preceding Lemma 3.2, ρq satisfies (3.4), and the identity map (Y,ρ) → q \{ } → \{ } (Y,ρ ) is precisely the metric inversion (Z q ,ρ) (Z q ,ρq ). The second inequality in (3.4) also implies diam(Y,ρq) ≤ 1. In particular, (Y,ρq) is bounded. Now let (X, d) be a bounded metric space, p ∈ X a non-isolated point, and q ∈ X\{p}. Then the metric space (X\{p},dp) is unbounded, and so we may form q the bounded metric space (X\{p}, (dp) ). Lemma 3.5. Let (X, d) be a bounded metric space, p ∈ X anon-isolatedpoint,and q q ∈ X\{p}. Then the identity map id :(X\{p},d) → (X\{p}, (dp) ) is bi-Lipschitz. { 1 } { 1 } ∈ Proof. Set a =min 1, 4 d(q,p) , b =max 1, d(q,p) and c =diam(X, d). Let x1,x2 X\{p}.Then q dp(x1,x2) (dp) (x1,x2) ≤ (1 + dp(x1,q))(1 + dp(x2,q)) − − d(x ,x ) d(x ,q) 1 d(x ,q) 1 ≤ 1 2 1+ 1 1+ 2 d(x1,p)d(x2,p) 4d(x1,p)d(q, p) 4d(x2,p)d(q, p) − − d(x ,q) 1 d(x ,q) 1 = d(x ,x ) d(x ,p)+ 1 d(x ,p)+ 2 1 2 1 4d(q, p) 2 4d(q, p) −2 −1 −1 ≤ d(x1,x2) · a (d(x1,p)+d(x1,q)) (d(x2,p)+d(x2,q)) 1 ≤ d(x ,x ). a2d2(p, q) 1 2 NAGATA DIMENSION AND QUASI-MOBIUS¨ MAPS 5 Similarly we have q ≥ 1 (dp) (x1,x2) 2 d(x1,x2) 16b (d(x1,p)+d(x1,q))(d(x2,p)+d(x2,q)) 1 ≥ d(x ,x ) 16b2 · 2c · 2c 1 2 1 = d(x ,x ). 64b2c2 1 2 Proposition 3.6. Let f :(X1,d1) → (X2,d2) be a quasi-M¨obius homeomorphism. −1 ◦ ◦ Then f can be written as f = f2 f f1,wheref is a quasisymmetric map, and fi (i =1, 2) is either a metric inversion or the identity map on the metric space (Xi,di). Proof. Let i ∈{1, 2}.If(Xi,di) is bounded, we let ρi = di and fi =id:(Xi,di) → (Xi,ρi) be the identity map on the metric space (Xi,di). If (Xi,di) is unbounded, q then we pick qi ∈ Xi and set ρi =(di) i and let fi :(Xi,di) → (Xi,ρi)be the identity map (of the set Xi). In this case, fi is a metric inversion (on an auxiliary space) by our discussion in the second paragraph after Lemma 3.3. Set ◦ ◦ −1 → −1 ◦ ◦ f := f2 f f1 :(X1,ρ1) (X2,ρ2). Then f = f2 f f1.Sincef1, f2 and f are quasi-M¨obius maps, f is also quasi-M¨obius. The fact that (X1,ρ1)and (X2,ρ2) are bounded metric spaces now implies that f is actually a quasisymmetric map. 4. Proof of the main theorem In this Section we prove Theorem 1.1. By Proposition 3.6, Theorem 1.1 follows from the following two results. Theorem 4.1 (Theorem 1.2 of [LS]). Suppose (X1,d1) and (X2,d2) are quasisym- metric, then dimN (X1,d1)=dimN (X2,d2). Theorem 4.2. Let (X, d) be a metric space and p ∈ X.ThendimN (X\{p},d)= dimN (X\{p},dp). To prove Theorem 4.2, we need: Proposition 4.3. Let (X, d) be a metric space and p ∈ X.ThendimN (X\{p},d) ≥ dimN (X\{p},dp). Assuming Proposition 4.3, we first finish the proof of Theorem 4.2. Proof of Theorem 4.2. We first suppose that (X, d) is unbounded. Pick p ∈ X. Then dp is a metric on Ip(X)=(X\{p}) ∪{∞}.SetY = Ip(X)andρ = dp. Then ρ∞ is a metric on Y \{∞} = X\{p}. Proposition 4.3 applied to (X\{p},d) → (X\{p},dp)and(X\{p},dp)=(Y \{∞},ρ) → (Y \{∞},ρ∞)=(X\{p},ρ∞)im- plies dimN (X\{p},d) ≥ dimN (X\{p},dp) ≥ dimN (X\{p},ρ∞). On the other hand, Lemma 3.3 implies dimN (X\{p},d)=dimN (X\{p},ρ∞). The theorem fol- lows in this case. Next we suppose that (X, d) is bounded. Pick p = q ∈ X.Ifp is isolated, then Lemma 3.2 implies that the metric inversion (X\{p},d) → (X\{p},dp)isbi- Lipschitz and hence dimN (X\{p},d)=dimN (X\{p},dp). So we assume that p is a non-isolated point. Then the metric space (X\{p},dp) is unbounded, and so we 6 XIANGDONG XIE q may form the bounded metric space (X\{p}, (dp) ). By our discussion in the second q paragraph after Lemma 3.3, the identity map (X\{p},dp) → (X\{p}, (dp) )isa metric inversion on an auxiliary space. Proposition 4.3 implies dimN (X\{p},d) ≥ q dimN (X\{p},dp)≥dimN (X\{p}, (dp) ). Now the theorem follows from Lemma 3.5. The rest of the section is devoted to the proof of Proposition 4.3. We may assume that X contains at least two points and n := dimN (X\{p},d) < ∞. Apply Proposition 2.2. Then there exist c ≥ 1, r>1, and a sequence of cover- ings Bj of (X\{p},d), j ∈ Z, with properties (i)–(iii) as stated in Proposition 2.2. Setc ¯ =16cr and c = 640¯c(¯c + 1). We shall prove that for each s>0, (X\{p},dp) has a cs-bounded covering with s-multiplicity at most n +1. Set a =inf{d(x, p):x ∈ X\{p}} and b =sup{d(x, p):x ∈ X\{p}}.Then 0 ≤ a ≤ b ≤∞. We may assume that either a =0orb = ∞ holds, for otherwise Lemma 3.2 implies that id : (X\{p},d) → (X\{p},dp) is bi-Lipschitz and hence dimN (X\{p},dp)=dimN (X\{p},d)=n. If a>0, b = ∞,ands ≥ (320¯c(¯c +1)a)−1,then d(x, y) 1 1 2 d (x, y) ≤ ≤ + ≤ p d(x, p)d(y, p) d(x, p) d(y, p) a for all x, y ∈ X\{p}. Hence diam(X\{p},dp) ≤ 2/a ≤ c s. In this case the family {X\{p}} consisting of one element is a c s-bounded covering of (X\{p},dp). From now on we assume s satisfies 0 Lemma 4.4. Let x ∈ As and set j j j(x):=sup{j ∈ Z : there is C ∈B with B(x, r ) ⊂ C and diam(C, dp) ≤ cs¯ }. Then j(x) < ∞. Proof. First suppose a = 0. In this case there is a sequence xi ∈ X\{p} with j j d(p, xi) → 0. Let j ∈ Z satisfy r >d(x, p). Then xi ∈ B(x, r ) for all sufficiently d(x,xi) j large i.Sincedp(x, xi) ≥ →∞as i →∞,wehavediam(B(x, r ),dp)= 4d(x,p)d(xi,p) ∞. It follows that j(x)satisfiesrj(x) ≤ d(x, p). In particular, j(x) < ∞. Now assume b = ∞. In this case, there is a sequence of points xi ∈ X with d(x,xi0 ) d(xi,p) →∞. Fix some index i0 such that ≥ 1/2. Let j ∈ Z satisfy d(xi0 ,p) j ∈ j r >d(x, xi0 ). Then xi0 B(x, r )andso d(x, x ) 1 j ≥ ≥ i0 ≥ ≥ diam(B(x, r ),dp) dp(x, xi0 ) 160¯c(¯c +1)s/8 > cs.¯ 4d(x, p)d(xi0 ,p) 8d(x, p) j(x) ≤ ∞ It follows that j(x)satisfiesr d(x, xi0 ). In particular, j(x) < . j(x) By Lemma 4.4, for each x ∈ As,wecanfixsomeCx ∈B that satisfies j(x) B(x, r ) ⊂ Cx and diam(Cx,dp) ≤ cs¯ . Consider the subfamily of B: C := {Cx ∈B: x ∈ As}. NAGATA DIMENSION AND QUASI-MOBIUS¨ MAPS 7 Lemma 4.5. For every member C ∈C, there exists a maximal element C of C with respect to inclusion, such that C ⊂ C. Proof. It suffices to show that there is no infinite strictly increasing sequence in C. { }∞ ⊂C Suppose there is a sequence Cxi i=1 such that Cxi is strictly contained in Cxi+1 ≥ C⊂B B n B for all i 1. Since and = k=0 k, after passing to a subsequence, we may ∈{ ··· } ∈B assume that there is some k 0, 1, ,n such that Cxi k for all i. Property (i) of Proposition 2.2 implies that j(xi) = j(xl) whenever i = l. After passing to a further subsequence, we may assume that j(xi) → +∞ or j(xi) →−∞as i →∞. Bj j →−∞ { } Recall that is cr -bounded. If j(xi) ,thenCxi = xi and xi = xl for all i, l, contradicting the assumption that Cxi = Cxi+1 . Hence we must have → ∞ ∞ \{ } j(xi) + .Sinceallxi lie in the bounded set As we have i=1 Cxi = X p .By ≤ \{ } ≤ the definition of j(x), diam(Cxi ,dp) cs¯ . Hence diam(X p ,dp) cs¯ .Ifa =0, then (X\{p},dp) is unbounded. Hence we must have b = ∞.Pickyl ∈ X with d(yl,p) →∞. Then for any x ∈ As,wehave d(x, yl) 1 dp(x, yl) ≥ → ≥ 160¯c(¯c +1)s/4 > cs,¯ 4 d(x, p)d(yl,p) 4 d(x, p) contradicting diam(X\{p},dp) ≤ cs¯ . Now let D be the subfamily of C consisting of the maximal elements of C with D B n B respect to inclusion. Then is a covering of As.Since = k=0 k,wehave D n D D D∩B = k=0 k,where k = k. Lemma 4.6. Let Cx = Cy ∈Dk for some k.Thendp(x ,y ) >sfor every pair x ∈ Cx, y ∈ Cy. Proof. Suppose there are x ∈ Cx, y ∈ Cy such that dp(x ,y ) ≤ s. We may assume j(x) ≤ j(y). Since Cx and Cy are both maximal elements of C, properties (i) and j(x) (iii) of Proposition 2.2 imply that d(w1,w2) ≥ r for all w1 ∈ Cx and w2 ∈ Cy. j(x) In particular, d(x ,y ) ≥ r . By the definition of j(x)wehavediam(Cx,dp) ≤ cs¯ . ∈ ∪{ } ≤ ≥ d(x,z) Let z Cx y .Thendp(x, z) (¯c +1)s.Sincedp(x, z) 4d(x,p)d(z,p) and ∈ ≤ 1 ≤ x As,weobtaind(x, z) 40¯c d(z,p). It follows that d(z,p) 2d(x, p)andso ≤ 1 d(x, z) 20¯c d(x, p). The triangle inequality implies 1 d(x,p) ≤ 1+ d(x, p) ≤ 11d(x, p)/10, 20¯c ≤ 1 d(x ,y ) 10¯c d(x, p)and d(y,p) ≤ d(y,x)+d(x,p) ≤ 12d(x, p)/10. By property (ii) of Proposition 2.2, there is some C ∈Bj(x)+1 with B(x, rj(x)+1) ⊂ C.Letw ∈ C.Then cr d(x, w) ≤ crj(x)+1 = cr · rj(x) ≤ cr · d(x,y) ≤ d(x, p) ≤ d(x, p)/10, 10¯c 9 ≤ ≤ 11 and so 10 d(x, p) d(w, p) 10 d(x, p). It follows from property (i) of Proposi- tion 2.2 that d(x, w) crj(x)+1 cr · d(x,y) d (x, w) ≤ ≤ ≤ . p d(x, p)d(w, p) 9d2(x, p)/10 9d2(x, p)/10 8 XIANGDONG XIE Using d(x,y) ≤ d (x,y) ≤ s, 4d(x,p)d(y,p) p d(x ,p) ≤ 11d(x, p)/10 and d(y ,p) ≤ 12d(x, p)/10, we obtain dp(x, w) ≤ 8crs for all w ∈ C . It follows that diam(C ,dp) ≤ 16crs =¯cs, contradicting the definition of j(x). Lemma 4.6 implies that each Dk has s-multiplicity at most 1. Let Bs be the complement of As in X\{p}.IfBs = ∅, then the proof of Proposition 4.3 is complete since D iscs ¯ -bounded. Suppose Bs = ∅.Using d(y, z) 1 1 d (y, z) ≤ ≤ + , p d(y, p)d(z,p) d(y, p) d(z,p) we see that diam(Bs,dp) ≤ 2 · 160¯c(¯c +1)s = 320¯c(¯c +1)s. D { ∈D } D { ∈D ≤ } Let 0 = C 0 : dp(C, Bs) >s, 0 = C 0 : dp(C, Bs) s ,andset E = Bs ∪ ∈D C.Then C 0 diam(E,dp) ≤ cs¯ + s + 320¯c(¯c +1)s + s +¯cs ≤ 322¯c(¯c +1)s References [A] P. Assouad, Sur la distance de Nagata, C. R. Acad. Sci. Paris S´er. I Math. 294 (1982), no. 1, 31–34. 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