The Spectrum of the Kinematic DynamoOperator for an Ideally Conducting FluidC. Chicone�, Y. Latushkiny, and S. Montgomery-SmithzDepartment of Mathematics, University of MissouriColumbia, MO 65211, USAApril 25, 1997AbstractThe spectrum of the kinematic dynamo operator for an ideally con-ducting uid and the spectrum of the corresponding group acting inthe space of continuous divergence free vector �elds on a compact Rie-mannian manifold are described. We prove that the spectrum of thekinematic dynamo operator is exactly one vertical strip whose bound-aries can be determined in terms of the Lyapunov-Oseledets exponentswith respect to all ergodic measures for the Eulerian ow. Also, weprove that the spectrum of the corresponding group is obtained fromthe spectrum of its generator by exponentiation. In particular, thegrowth bound for the group coincides with the spectral bound for thegenerator.�carmen@chicone.cs.missouri.edu, supported by the NSF grant DMS-9303767;ymathyl@mizzou1.missouri.edu, supported by the NSF grant DMS-9400518 and by theSummer Research Fellowship of the University of Missouri;zstephen@mont.cs.missouri.edu, supported by the NSF grant DMS-9201357; Key wordsand phrases. Hyperbolic dynamical systems, Mather spectrum, magnetohydrodynam-ics, ideally conducting uid, spectral mapping theorems, weighted composition operators.Mathematics Subject Classi�cation. 76W05, 58F99, 58G25.1

1 IntroductionIn this paper we give a description of the spectrum of the kinematic dynamooperator and of the corresponding group it generates for an ideally conducting uid in the space of continuous divergence free vector �elds.Consider a steady incompressible conducting uid with Eulerian velocityv = v(x) for x 2 R3 and let �t denote the corresponding ow. The kinematicdynamo equations for the induction of a magnetic �eld H by the ow hasthe following form: _H = r� (v �H) + "�H; divH = 0;(1.1)where " = Rem�1, and Rem is the magnetic Reynolds number (see, e.g.,[15, Ch. 6]). The spectral properties of the kinematic dynamo operator L",de�ned by (1.1), have been a subject of intensive study, in particular, inconnection with the famous dynamo problem (see [1, 2, 3, 6, 16, 23] andreferences therein).For the ideally conducting uid, " = 0, these equations become:_H = �(v;r)H+ (H;r)v; H(x; 0) = H0(x); divH = 0:(1.2)The last equation has [15] so-called Alfven solutionsH(x; t) = D�t(��tx)H(��tx; 0);given by the group fetLgt2R with the generator L = L0 that acts by the ruleL : u 7! (u;r)v � (v;r)u:(1.3)In the present paper the kinematic dynamo operator L is considered inthe following well known context. Let v denote a continuous divergence-free vector �eld on a compact Riemannian manifold X without boundary,let �t denote the ow generated by v and let D�t(x) denote its di�erential.Consider the group fetLgt2R of push-forward operators generated by the Liederivative L in the direction v. This group acts on continuous sections of thetangent bundle T X, by the rule�etLu� (x) = D�t(��tx)u(��tx); x 2 X; t 2 R:(1.4) 2

We will consider the group fetLgt2R in the space CND(X; T X) of the contin-uous vector �elds with zero divergence.Operators of the form (1.4) belong to the class of weighted compositionoperators. This class has been widely investigated in connection with hyper-bolic dynamical systems since the celebrated paper by J. Mather [13], seealso [7, 20], the recent papers [4, 9, 10, 22] and the detailed bibliography in[12]. The spectral properties of these operators in spaces of continuous orp-summable vector �elds are by now well understood. However, the investi-gation of their spectral properties in the space of divergence-free vector �eldswas initiated recently by R. de la Llave. His important work [21] inspiredthe present paper.In Section 2 we prove the spectral mapping theorem in CND(X; T X) forthe group fetLgt2R, assuming that the aperiodic trajectories of �t are dense inX and dimX � 3. Theorems of this type for continuous and for Lp-sectionspaces over �nite dimensional manifolds were proved in [5, 8] while similarresults for the in�nite dimensional setting were obtained in [9, 10, 12].The spectral mapping theorem states that the spectrum �(etL) of etLcan be obtained from the spectrum of L by exponentiation. It shows, inparticular, that in the space of divergence-free vector �elds the spectral boundof the generator L coincides with the growth bound of the group, see alsoRemarks 2.9{2.11 below.Our proof of the spectral mapping theorem in Section 2 exploits the fact(cf. [13, 21]) that approximative eigenfunctions of the operator (1.4) can be\localized" along trajectories of the ow. We also show that the spectrumof L is invariant under vertical translations of the complex plane. The sameidea can be used to analyze semigroups of weighted composition operatorswith general cocycles on Banach spaces, see [11].In Section 3, we show that, for dimX � 2, the spectrum �(etL), t 6= 0in CND(X; T X) is exactly one annulus centered at the origin of the complexplane. Our result generalizes a theorem in [21] where this fact was provedunder the restriction that the ow is Anosov with one-dimensional spectralfoliations. The possibility of relaxing the hypotheses of the theorem is dis-cussed as an open problem in [21]. The relatively simple proofs of thesefacts in Section 3 can be read independently of the more di�cult proofs inSection 2.By Mather's theory (see, e.g., [12, 13, 20]), the spectrum �(etL) on thespace C(X; T X) is generally the union of several disjoint annuli centered3

at the origin. Passing to the space CND(X; T X) dramatically changes thespectrum: the gaps, if any, between these annuli are �lled. Also, using theSpectral Mapping Theorem, the spectrum of L on CND(X; T X) is exactlyone vertical strip. Since Lv = 0, this strip always contains iR, and (1.2)does not have an exponential dichotomy. Moreover, the equation (1.2) in thespace of divergence free vector �elds (unlike the situation with Anosov ows,cf. [20]), does not possess a nontrivial uniform exponential dichotomy evenafter \moding out" the direction of the ow, see Remarks 3.7{3.8 below.Using the results in [12], we give a description of the spectrum of L interms of the Lyapunov-Oseledets exponents over all ergodic measures onX, that is, we will determine the boundaries of the spectrum of L via theLyapunov exponents. This description is related to a theorem by M. Vishik[23] that states the \fast" dynamo action is impossible whenever all Lyapunovnumbers are zero.Finally, we remark that we actually prove more general results than thosejust mentioned. In particular, we will only assume the operator L generatesa C0-group of weighted composition operators that preserves the set of di-vergence free vector �elds and has the form�T tu� (x) = �(��tx; t)u(��tx); x 2 X; t 2 R;(1.5)where �(x; t) is a continuous cocycle over �t, that is for x 2 X and t; � 2 Rone has �(x; t + �) = �(�tx; �)�(x; t) and �(x; 0) = I. Throughout thepaper we use M to denote the generator of the group fT tg, T t = etM . Inparticular, if �(x; t) = D�t(x), then (1.5) is the push-forward operator (1.4)and M = L is the Lie derivative.We also note that our technique can be applied to obtain similar resultsfor the space L2.2 The Spectral Mapping TheoremIn this section we will prove the Spectral Mapping Theorem. Throughoutthe section we suppose a smooth vector �eld v to be given on a compactRiemannian manifold X without boundary, that v is divergence free withrespect to the Riemannian volume and that the ow �t of v satis�es thefollowing standing hypotheses: the aperiodic trajectories of �t are dense in Xand n = dimX � 3. LetM be the generator of the group fT tgt2R of weighted4

composition operators, as in (1.5), for some continuous cocycle �(x; t) over�t. We will assume T t is bounded on the space CND(X; T X) of divergencefree vector �elds.There are at least two choices for the space of continuous divergencefree vector �elds depending on whether the divergence is understood in theclassical sense or in the sense of distributions. These spaces are de�ned,respectively, as follows:C0ND(X; T X) = closure ff 2 C1(X; T X) : divf = 0g;(2.6) C1ND(X; T X) = ff 2 C(X; T X) :ZX < f; gradg > d� = 0 8g 2 C1(X; R)g:(2.7)The closure in (2.6) is taken with respect to the sup-norm while the scalarproduct < �; � > and grad in (2.7) are taken with respect to a Riemannianmetric and volume on X. We note that the space C1ND(X; T X) is a closedsubspace of C0ND(X; T X).For a linear operator A in a Banach space E, we will sometimes use�(A;E) to denote the spectrum of A on E and �ap(A;E) to denote itsapproximate point spectrum. Since, as noted above,C1ND(X; T X) � C0ND(X; T X) � C(X; T X)and since an �-eigenfunction for T in C1ND(X; T X) is an �-eigenfunction forT in C0ND(X; T X), one has�ap(T ;C1ND(X; T X)) � �ap(T ;C0ND(X; T X)) � �ap(T ;C(X; T X)):(2.8)Throughout the remainder of the paper the space CND(X; T X) may betaken to be either C0ND(X; T X) or C1ND(X; T X).Theorem 2.1 (Spectral Mapping Theorem) In the space of divergence-free vector �elds CND(X; T X) the spectrum �(M ;CND(X; T X)) is invariantunder vertical translations of the complex plane. Moreover, for each t 6= 0,�(etM ) = exp t�(M):(2.9) 5

Since the proof of Theorem 2.1 is quite technical, we pause to discussour strategy. Using standard facts from the theory of C0-semigroups and byrescaling, we reduce the proof of Theorem 2.1 to the following main assertion(see our Lemma 2.7 below): If 1 2 �ap(T ), T := T 1 then 0 2 �ap(M).Our strategy for the proof of this main assertion develops some ideas ofJ. Mather [13]. The fact that 1 2 �ap(T ) implies the existence of an �-eigenfunction u for T for every � > 0. That is, for every � > 0, there is avector �eld u, with unit norm such that kTu�uk � �. As in [13] and [21], the�-eigenfunctions of the operator T have a nice feature: they can be \localized"along the trajectories of the ow. This means that for every N 2 N thereexist a point x0 2 X, a small neighborhood D of this point, and a vector�eld y with suppy � [Nj=�N�j(D), such that ky � Tyk = O(1=N)kyk. Infact, starting from a given �-eigenfunction u for T , de�ne a \bump"-function� supported in D and let (j) = (N � jjj)=N . The \localization" y isde�ned to vanish outside [Nj=�N�j(D) and by y(x) = (j)�(��jx)(T ju)(x)for x 2 �j(D) and jjj � N , equivalently,y(x) = NXj=�N (j)(T j�u)(x); x 2 X:To prove the main assertion above, our purpose is to construct a vec-tor �eld y with zero divergence such that kMyk = O(1=N)kyk. We startwith a divergence-free approximative eigenfunction u of T . Since the set ofdivergence free vector �elds is not closed under multiplication by \bump"-functions, we can not use Mather's construction directly. Instead, we willconstruct a divergence free vector-�eld w, supported in a small neighbor-hood D of a given point x0 2 X in Lemma 2.3. The main part of thisconstruction takes place in a special neighborhood D of x0, taken to be athin and long \ellipsoid" with the longest axis directed along u(x0). Therequired vector �eld w is constructed in the form w(x) = �(x)u(x0)+wn(x),where � is a \bump"- function, supported in D. The function � is chosen tohave value identically one on a second thin and long \ellipsoid" B containedin D. Some \ uid" leaks from the neighborhood D, but this can be recycledwithin a slightly larger neighborhood.The desired almost-eigenfunction y for M is given by the formulay(x) = Z 1�1 (t)T tw(x) dt; x 2 X;6

where, as above, vanishes outside of [�N;N ]. Direct calculation showsthat (My)(x) = � Z 1�1 0(t)T tw(x) dt; x 2 X:Since suppw � D, the support of the integrand in each integral belongs toft 2 R : jtj � N and�tx 2 Dg. To obtain the desired inequality, we estimatekMyk from above and kyk from below. This requires some estimates of thesojourn time of the trajectory segment f�t(x) : jtj � Ng in D and B. Thisis done in Lemma 2.5.We start, for completeness, from the following simple lemma.Lemma 2.2 If x0 2 X, then there is a coordinate chart at x0, with coordi-nate functions (x1; : : : ; xn), such that the local representation of the volumeelement on X is just the usual volume dx1 ^ � � � ^ dxn on Rn . Moreover, ifz 2 Tx0X, then the coordinates can be chosen so that the local representativeof z is kzk@=@x2.Proof: Let y1; : : : ; yn denote local coordinates at x0. Clearly, there is anonvanishing density function � : Rn ! R such that volume element is givenby �(y1; : : : ; yn)dy1 ^ � � � ^ dyn. We seek new coordinates in the formy1 = f(x1; : : : ; xn); y2 = x2; � � � ; yn = xnwhere the volume element has the form�(f(x1; : : : ; xn); x2; : : : ; xn) @f@x1 (x1; : : : ; xn)dx1 ^ � � � ^ dxn:There is a smooth function f , de�ned in a neighborhood of the origin in Rn ,such that@f@x1 (0; : : : ; 0) 6= 0; @f@x1 (x1; : : : ; xn) = (�(f(x1; : : : ; xn); x2; : : : ; xn))�1:The �rst condition together with the Implicit Function Theorem implies thechange of coordinates is invertible; the second condition ensures the volumeelement in the new coordinates has the desired form.For the second statement of the lemma, note that the volume element isinvariant under a rigid rotation of Euclidean space. 27

A coordinate chart, as in the lemma, is called adapted to the volumeon X and the vector z. Of course, in the adapted coordinates, the Rie-mannian metric will not be the usual one, rather, it will have the form�gij(x1; : : : ; xn)dxi dxj where the components gij form a positive de�nitesymmetric matrix of smooth functions. However, we make the following ob-servation: if u = (u1; : : : ; un) is a vector �eld de�ned in an adapted coordinatechart, then divu = nXi=1 @ui@xi :(2.10)Suppose z 2 Tx0X is a tangent vector and let (x1; : : : ; xn) denote localcoordinates at x0 adapted to the volume on X and the vector z. If theadapted coordinate system is de�ned in a coordinate ball of diameter � > 0and if a; b 2 R are such that 0 < a < b < �=8, we de�neDa;b = f(x1; : : : ; xn) : jxjj � 4b; j = 1; 2; jxjj � a; j = 3; : : : ; ng;Ba;b = f(x1; : : : ; xn) : jxjj � a=2; j = 1; 3; : : : ; n; jx2j � b=2g:Note that the closure of Ba;b is contained in Da;b. We say there is an (a; b)divergence-free extension of the vector z at x0 if there is a smooth bump-function � : Rn ! [0; 1] with �(x) = 1 for x 2 Ba;b and �(x) = 0 for x =2 Da;band a continuously di�erentiable vector �eld wn with support in Da;b suchthati) The vector �eld w(x) := �(x)kzk@=@x2 + wn(x) is divergence free andhas value kzk@=@x2 in Ba;b,ii) There is a number C > 0 independent of a; b such that kwnk � Ca=b.Lemma 2.3 Every tangent vector on X has an (a; b) divergence-free exten-sion.Proof: Let z 2 Tx0X. We will �rst prove the lemma for the case n = 2.To construct a vector �eld W with the required properties in the (x1; x2)coordinate plane, consider the curves given byx41a4 + x42b4 = 1; x41a4 + x42b4 = 2:8

Let � : R ! [0; 1] denote a smooth function such that �(t) = 1 for t � 1,�(t) = 0 for t � 2, and j�0(t)j � 3, t 2 R. Also, de�ne the setsR = f(x1; x2) : jx1j � 21=4a; jx2j � 21=4bg;S+ = f(x1; x2) : (x1 � 21=4a)4a4 + x42b4 � 2; x1 � 21=4ag;S� = f(x1; x2) : (x1 + 21=4a)4a4 + x42b4 � 2; x1 � �21=4agand the functions � : R2 ! R, f : R2 ! R and �� : R ! R by�(x1; x2) = �(x4a4 + x42b4 );f(x1; x2) = �4x32kzkb4 Z x10 �0�s4a4 + x42b4 � ds;�+(�) = �4kzkb4 Z 21=4a0 �0�s4a4 + �b4� ds;��(�) = 4kzkb4 Z 0�21=4a �0�s4a4 + �b4� ds;The vector �eld w2 is de�ned in R by f(x1; x2)@=@x1, in S+ by�+((x1 � 21=4a)4 + x42)x32 @@x1 � �+((x1 � 21=4a)4 + x42)(x1 � 21=4a)3 @@x2 ;in S� by��((x1 + 21=4a)4 + x42)x32 @@x1 � ��((x1 + 21=4a)4 + x42)(x1 + 21=4a)3 @@x2and w2 is de�ned to vanish on the complement of R [ S+ [ S�.We will complete the proof for n = 2 by showing the vector �eldW (x1; x2) := �(x1; x2)kzk @@x2 + w2(x1; x2)is the required extension of z.A direct computation using (2.10) shows divW = 0 in the coordinatechart. Also, using the de�nition of �, we see the support of W is in Da;b. To9

show W is C1, just observe that w2 and each of its �rst partial derivatives iscontinuous on the lines x1 = �21=4a and on the boundary of R [ S+ [ S�.(We remark that additional smoothness can be obtained, if desired, by usingthe function xk1=ak+xk2=bk with k a su�ciently large positive integer in placeof the choice k = 4 used here.)To obtain the required norm bound, let G(x1; : : : ; xn) denote the matrixof the components gij of the Riemannian metric in the adapted coordinates.The square of the norm of a vector V at x = (x1; : : : ; xn) is then given byhG(x)V; V i. Thus, if kGk denotes the supremum of the matrix norms overthe points in the chart, we have kV k � kGkjV j where the single bars denotethe usual norm in Rn . Then, for example, using the usual estimate for theintegral in the de�nition of f , we estimate the norm of w2 in R bykGk sup jf(x1; x2)j � kGk(4kzk23=4b3=b4)3(21=4a) � C1a=bwhere the constant C1 does not depend on a or b. Similarly, we can estimatethe norm of w2 in S�. For example, in S+ we �nd the upper boundkGk sup j�+((x1 � 21=4a)4 + x42)j ((21=4b)6 + (21=4a)6)1=2 � C2a=b:To prove the lemma for the case n � 3, we will show how to extendthe vector �eld W de�ned above to a vector �eld on Rn with the requiredproperties. To do this, let � : R ! [0; 1] denote a smooth \bump"-functionsuch that �(t) = 1 for jtj � a2 and �(t) = 0 for jtj � a, and de�ne(x3; : : : ; xn) = nYj=3�(xj):The required vector �eld w is given byw(x1; : : : ; xn) = (x3; : : : ; xn)W (x1; x2):The fact that the new vector �eld w is continuously di�erentiable, agreeswith kzk@=@x2 in Ba;b, and is supported in Da;b is clear. Also, since the rangeof is the unit interval, the norm bound on wn is the same as the norm boundon w2. To complete the proof we must show divw = 0. But, since the vector�eld w has nonzero components only in the �rst two coordinate directions,divw(x1; : : : ; xn) = (x3; : : : ; xn)divW (x1; x2) = 0;10

as required. 2To estimate the time that trajectories spend in a neighborhood of x0, we�rst need the following observation.Lemma 2.4 Suppose x0 2 X is not a periodic point for the vector �eld vwith ow �t. If N is a positive integer and 0 < s � N , then there is a� > 0 such that every neighborhood D containing x0, with diamD � �, hasthe following property: if x 2 D, then �tx 62 D for s � jtj � N .Proof: Suppose the lemma is false and, for each positive integer k, let Dkdenote the ball centered at x0 with radius 1=k. For each k, there is somexk 2 Dk and some tk in the set J := ft : 0 < s � jtj � Ng with �tk(xk)in Dk. Since J � �D1 for the closure �D1 of D1 is compact, there is a conver-gent subsequence of the pairs (tk; xk). But, by the choice of Dk, the secondcomponent of this sequence converges to x0 and, by the compactness of J ,the limit T of the �rst component satis�es jT j � s > 0. The continuity ofthe ow ensures that �T (x0) = x0, in contradiction to the fact that x0 is notperiodic. 2Consider a vector �eld v on X tangent to the ow �t. For each open setU � X, each non negative integer N and each point x 2 X, de�ne�N;U(x) := ft 2 R : jtj � N; �tx 2 Ug;(2.11) mN;U(x) := mes(�N;U(x)):Lemma 2.5 Suppose X has dimension n � 3. If � > 0, then there is aconstant K > 0 such that for each non periodic point x0 and positive integerN there is a pair of numbers a; b such that � > b > a > 0 and a=b � � togetherwith a pair of open sets B;D at x0 such that B � Ba;b and Da;b � D, withthe following property: for each x 2 X,mN;D(x)mN;B(x0) � K:(2.12)Proof: By Lemma 2.4, there is a neighborhood cD at x0 such that �ty 62 cDwhenever y 2 cD and 1=2 � jtj � 2N . Suppose K; a; b; B;D, with D � cD,are given so that the inequality (2.12) holds for x 2 D. We claim that11

(2.12) holds for all x 2 X. To see this, note �rst that, for y 2 D, we have�N;D(y) = �2N;D(y). Thus, by our de�nition,mN;D(y) = m2N;D(y). If x 2 Xand D\f�tx : jtj � Ng = ;, then mN;D(x) = 0 and (2.12) holds. Otherwise,�x y 2 D \ f�tx : jtj � Ng. Since �N;D(x) � �2N;D(y), we havemN;D(x) � maxy2D m2N;D(y) = maxy2D mN;D(y) � K �mN;D(x0);as required.To complete the proof, we will construct K; a; b; B;D so that (2.12) holdsfor x 2 D. This will require several steps.Step 1. We will work in an adapted coordinate system at x0 with coor-dinate functions (x1; : : : ; xn). We will determine the required sets B;D forappropriate a; b in the formB = (x : x41(a=2)4 + x42(b=2)4 +P4j=3 x4j(a=2)4 � 1) ;D = (x : x41(4nb)4 + x42(4nb)4 +P4j=3 x4j(na)4 � 1) :If a < b, then, clearly, B � Ba;b and Da;b � D. We will show that there isa constant K such that for some choice of a; b and a=b all su�ciently small,the inequality (2.12) is valid for the corresponding set D.We will use the following auxiliary constructions:For each � > 0, let S� denote a section for v at the origin of the coordinatesystem, that is, at x0, such that the Riemannian diameter diamS� < �, andde�ne �� := f�t� : � 2 S�; jtj � �g. Consider the local representation of vgiven by Pni=1 vi(x)@=@xi. By a rigid rotation, if necessary, we can and willarrange the adapted coordinates so that v1(0) = 0. Also, we de�ne V byV (x) := nXi=3 v4i (x); x 2 ��and the numberM� := max8<:maxx2�� jv1(x)j; maxx2�� nXi=3 v4i (x)!1=49=; :12

If V (x0) = 0, then lim�!0M� ! 0. In case V (x0) = 0 and M� 6= 0 forevery �, we will only consider a; b such thata = b � (M�)1=2 :(2.13)Of course, even under the restriction just imposed, a; b, a=b can each bechosen arbitrary small. If V (x0) 6= 0 or if V (x0) = 0 and M� � 0 for allsu�ciently small �, we ignore this restriction.Step 2. For each �, Lemma 2.4 and the de�nition of �� together implythere is an open ball A� � �� at the origin such that, for each x 2 A� andfor each time t with jtj > �, the point �tx is not in ��. If � > 0 is given,choose a; b as required in (2.13) and so small that D is in A�. If x 2 D, let x0denote the point on @D where the segment of the trajectory f�tx : jtj � Ng�rst enters D and let x00 = �tDx0 denote the point of @D where the segmentof the trajectory f�tx : jtj � Ng last exits D. Clearly, mN;D(x) � tD.We use the Mean Value Theorem for integrals on the jth component ofthe vector �eld v to obtain a point �j 2 �� such thatx00j = x0j + tDZ0 vj(�tx) dt = x0j + tDvj(�j):For each j = 1; : : : ; n, let v�j = vj(�j). Also, as an abbreviation, de�ne�j = 4nb for j = 1; 2 and �j = na for j = 3; : : : ; n.Since x0 and x00 both belong to @D, we havenXi=1 x0i�i!4 = 1 and nXi=1 x0i + tDv�i�i !4 = 1:Using a standard inequality for the norm k( i)k = (Pi j ij4)1=4, we �nd1 = Pni=1 x0i + tDv�i�i !4 � Pni=1 �tDv�i�i �4!1=4 � 0@Pni=1 x0i�i !41A1=4= Pni=1 �tDv�i�i �4!1=4 � 1:This computation yields the estimatet4D nXi=1 �v�i�i�4 � 24:(2.14) 13

Step 3. Consider the time tB when the segment of the trajectory f�tx0 :jtj � Ng �rst leaves D. Clearly, mN;B(x0) � tB.Recall that in our local coordinates, x0 resides at the origin and de�ne~x = �tB(x0). As in Step 2, by the the Mean Value Theorem, there is some�j 2 �� such that ~xj = tBZ0 vj(�t(x0)) dt = tBvj(�j):De�ne ~vj := vj(�j), the numbers �j = a=2 for j = 1; 3; : : : ; n and �2 = b=2.Since ~x 2 @B, we have t4B nXj=1 ~vj�j!4 = 1:(2.15)Step 4. In accordance with the previous notation, we de�ne~V = nXi=3 ~v4i ; V � = nXi=3 (v�i )4 :We use (2.14)-(2.15) to obtain the estimate (tD=tB)4 � 216n4 � d whered := ~v41 + �ab ~v2�4 + ~V�ab v�1�4 + �ab v�2�4 + 44V � :We will show that for all su�ciently small � > 0, there are some choicesof a; b such that d < 2. There are several cases.Case 1. If V (x0) 6= 0, then lim�!0 d = 4�4.Case 2. If V (x0) = 0 and M� � 0 for all su�ciently small �, then, sincev2(x0) 6= 0, lim�!0 d = 1.Case 3. Suppose V (x0) = 0 andM� 6= 0. However, note that we still havelim�!0M� ! 0. Also, the restriction we imposed in (2.13) provides thatba~v1=21 � 1; ba ~V 1=8 � 1:In this case, we haved = ba~v1!4 + ~v42 + ba!4 ~V(v�1)4 + (v�2)4 + 44 ba!4 V � � ~v21 + ~v42 + ~V 1=2(v�2)4 :14

Passing to the limit as � ! 0, we see that the last expression converges to1. 2We need the following elementary fact.Lemma 2.6 Suppose A denotes an invertible bounded operator on a Banachspace E and let N 2 Z. If N � 2 and 1 2 �ap(A), then there is a vectoru 2 E with kukE = 1 such that kAkukE � 2 for each integer k with jkj � N .Proof: Set � = 0@ NXk=�N kAkk1A�1. Since 1 2 �ap(A), there is some u 2 E withkukE = 1 such that kAu� uk � �. Also, for 1 � jkj � N , note thatAk � I = 0@k�1Xj=0Aj1A (A� I); k > 0; Ak � I = �0@ kXj=�1Aj1A (A� I); k < 0:Hence, for jkj � N , we have kAku� uk � NXj=�N kAjk � kAu� uk � 1 and, asa result, kAkuk � kAku� uk+ kuk � 2. 2The main result of this section is the following lemma.Lemma 2.7 Suppose etM = T t is the group de�ned in (1.5)and de�ne T :=T 1. If 1 2 �ap(T ;C(X; TX)), then �ap(M ;CND(X; T X)) contains theimaginary axis of the complex plane.In accordance to (2.8) this lemma also shows that 1 2 �ap(T; CND(X; T X))implies 0 2 �ap(M;CND(X; T X)) for both cases:CND(X; T X) = C0ND(X; T X) or CND(X; T X) = C1ND(X; T X):Proof: Let � 2 R and let K be de�ned as in Lemma 2.4. Also, fornotational convenience, de�ne ! := 1=(12K).Since 1 2 �ap(T ), Lemma 2.6 applied to the bounded linear operatorT = T 1 ensures that, for each integer N � 2, there is some vector �eldu 2 C(X; T X) such that kukC(X;TX) = 1;(2.16) kT kukC(X;TX) � 2 for jkj � N:(2.17) 15

Since fT tg is a C0-semigroup, T tu ! u in C(X; T X) as t ! 0. Thus,there is a real number s with 0 < s � 2N such thatkT tu� ukC(X;TX) � ! for jtj � s;(2.18) je�i�t � 1j � ! for jtj � s:(2.19)Also, there is a smooth function : R ! [0; 1] such that: (t) = 0 for jtj � N;(2.20) j 0(t)j � 2N ; for t 2 R;(2.21) (t) = 1 for jtj � s:(2.22)In view of (2.16), and the fact that the non periodic points are dense inX, there is a non periodic point x0 2 X such thatku(x0)k � 12kukC(X;TX) = 12 :(2.23)Use Lemma 2.5 to �nd small a; b with small a=b, and neighborhoodsD � B 3x0, such that (2.12) holds. Moreover, in accordance with Lemma 2.4, we canchoose a; b su�ciently small so that for D := Da;b, for s from (2.18){(2.19)and with c := maxjtj�1 kT tk we have�ty 62 D for any y 2 D provided s � jtj � 2N(2.24)and, for some constant C, the following inequalities:Cab maxjtj�N kT tk! � !;(2.25) maxy2D;jtj�N k�(y; t)u(x0)k � 4c:(2.26)For the last inequality, we use (2.17) to show thatmaxjtj�N k�(x0; t)u(x0)k � maxjtj�N;x2X k�(x; t)u(x)k� maxjtj�N;x2X k�(��tx; t)u(��tx)k = maxjkj�N�1;j� j�1kT k+�uk� c maxjkj�N�1 kT kk � 2c: 16

Since � : (x; t) 7! �(x; t) is uniformly continuous on the compact set X �[�N;N ], we have (2.26) for a su�ciently small neighborhood D of x0.We use Lemma 2.3 with z = u(x0). After a rigid rotation, if necessary, wecan arrange the adapted coordinates so that the component of v(x0) in thedirection of the �rst coordinate vanishes. Then, for this choice of adaptedcoordinates, there is a divergence-free vector �eld of the formw(x) = �(x)u(x0) + wn(x)(2.27)with � and wn supported in D,kwnkC(X;TX) � Cab ;(2.28)and �(x) = 1 for x 2 Ba;b.De�ne the vector �eld y on X byy(x) = Z 1�1 e�i�t (t)T tw(x) dt:We see that y has zero divergence (for this remember T t preserves thedivergence-free vector �elds). By easy computations with My = dd� T �y����=0one has:My = Z 1�1 dd� (e�i�(t��) (t� �))������=0 T twdt = i�y � Z 1�1 0(t)e�i�tT tw dt:To complete the proof, we must show that i� 2 �ap(M ;CND(X; T X)). Thisis an immediate consequence of the following proposition: There is a numberA > 0 that does not depend on the choice of N such that Z 1�1 0(t)e�i�tT tw dt C(X;TX) � AN kykC(X;TX):To prove the proposition, �x x 2 X and note thatkMy(x)� i�y(x)k = Z 1�1 e�i�t 0(t)(T tw)(x) dt � I1 + I2where, by (2.20) and (2.21),I1 = 2N NR�N �(��tx)k�t(��tx; t)u(x0)k dt;I2 = 2N NR�N k�(��tx; t)k kwn(��tx)k dt:17

Since supp� � D and suppwn � D, the integrations I1 and I2 can be re-stricted to �D(x) = �N;D(x), see the notation in (2.11). We use (2.26) toobtain: I1 � 2N Z�D(x) maxy2D;jtj�N k�(y; t)u(x0)k � 2N 4c �mN;D(x):(2.29)We use (2.25) and (2.28) to estimate I2:I2 � 2N Z�D(x) maxy2D;jyj�N k�(y; t)kkwnkC(X;TX) dt(2.30) � 2N maxjtj�N kT tk � Cab �mN;D(x) � 2N !mN;D(x):We obtain the desired upper estimate from (2.29) and (2.30), namely,kMy � i�ykC(X;TX) � A1N maxx2X mN;D(x):(2.31)To determine the lower bound, we de�neJ1 = Z 1�1 e�i�t (t)�(��tx0)u(x0) dt ;J2 = Z 1�1 e�i�t (t)�(��tx0) h(T tu)(x0)� u(x0)i dt ;J3 = Z 1�1 e�i�t (t)(T twn)(x0) and note that kykC(X;TX) � ky(x0)k � J1 � J2 � J3:Again, each integral is equal to its restriction to �D(x0) = �N;D(x0).As in (2.30), we use (2.28) and (2.25) to estimate J3 from above:J3 � Z�D(x0) k(T twn)(x0)k dt � !mN;D(x0):(2.32)Next, we use (2.18) to estimate J2 from above. For this, note that from(2.24) if t 2 �D(x0), then jtj < s. Thus, we haveJ2 � Z�D(x0) kT tu� ukC(X;TX) dt � !mN;D(x0):(2.33) 18

Finally, we estimate J1 from below:J1 = ku(x0)k��� Z�D(x0) e�i�t (t)�(��tx0) dt��� � J11 � J12;(2.34)where J11 = j Z�D(x0) (t)�(��tx0) dtj � ku(x0)k;J12 = j Z�D(x0)(e�i�t � 1) (t)�(��tx0) dtj � ku(x0)k:Since, by (2.24), �tx0 62 D for s � jtj � 2N , equation (2.22) gives (t) = 1for t 2 �D(x0). As �(x) = 1 for x 2 B, we use (2.23) to compute theestimate: J11 � 12 Z�B(x0) �(��tx0) dt � 12mN;B(x0):(2.35)Since ku(x0)k � 1 and jtj � s for t 2 �D(x0), the inequality (2.19)implies: J12 � Z�D(x0) ���e�i�t � 1����(��tx0) dt � !mN;D(x0):(2.36)The estimates (2.35), (2.36), (2.33), (2.32), and (2.12) together with theour choice of ! = 1=(12K) give the following:kykC(X;TX) � 12mN;B(x0)� 3!mN;D(x0)� 12K maxx2X mN;D(x)� 3!maxx2X mN;D(x)= 14K maxx2X mN;D(x):By combining this estimate with (2.31), we have the desired result. 2We are now in the position to prove Theorem 2.1.Proof: It is well-known (see, e.g., [19]) that the Spectral Inclusion Theo-rem �(etM) � exp t�(M); t 6= 0(2.37) 19

holds for any C0-semigroup. Also, the spectral mapping theorem is true forthe point and residual spectrum. Therefore, to prove (2.9) one needs to showthat, in CND(X; T X),�ap(etM ) � exp t�ap(M); t 6= 0:Fix � = j�jei� 2 �ap(etM ;CND(X; T X)). Then � = et� for � = 1t ln j�j+i�t . Consider the cocycle ~�(x; t) = e�t��(x; t), and the group f ~T tg, ~T t =et ~M , de�ned by ~�(x; t) as in (1.5). Then � 2 �ap(etM ; CND(X; T X)) im-plies that 1 2 �ap(et ~M ;CND(X; T X)). By (2.8) and Lemma 2.7, we have0 2 �ap( ~M ;CND(X; T X)). But, since ~M = M � �, this implies � 2�ap(M ;CND(X; T X)).To prove that �(M) is invariant under the translations along the imagi-nary axis, we �x � 2 �ap(M) and � 2 R. By the Spectral Inclusion Theoremfor t = 1 we have 1 2 �ap(e ~M), also, ~M = M � �. By Lemma 2.7, one hasi� 2 �ap( ~M) and, as a result, �+ i� 2 �ap(M). 2We will use notationss(A) := supfRez : z 2 �(A)g and !(A) := limt!1 t�1 ln ketAkfor the spectral bound of a generator A and the growth bound of a C0-semigroup fetAg, respectively. Note [19], that for an arbitrary C0-semigroupfetAg one has s(A) � !(A), but, generally, s(A) 6= !(A). The SpectralMapping Theorem, however, gives for the group of weighted compositionoperators the following fact.Corollary 2.8 In the space of continuous divergence free vector �elds thespectral bound and the growth bound are equal: s(M) = !(M).Remark 2.9. Consider the kinematic dynamo operator L" = L + "� with" > 0 (see (1.1)). This is an elliptic operator, it generates an analytic semi-group, and the spectral mapping theorem is valid [19] for this semigroup.Hence, s(L") = !(L") for " > 0. For L0 = L, Corollary 2.8 shows that thisequality is also valid for " = 0.Remark 2.10. M. Vishik [23] has shown that lim sup"!0 !(L") � !(L0). Inview of Remark 2.9, this theorem can be reformulated as lim sup"!0 s(L") �20

s(L0). We stress that the last assertion does not involve the construction ofthe group fetL"g; it is given in the terms of generators only. See Remark 3.9below for the connection of this assertion to the \fast"-dynamo problem.Our formulation suggests that the validity of the assertion lim sup"!0 s(L") =s(L0) can be approached as a problem from the theory of singular perturba-tions for the generators of C0-semigroups.Remark 2.11. The Spectral Mapping Theorem for semigroups of weightedcomposition operators does not hold without the assumption that aperiodictrajectories are dense in X (see [5, 12] for examples). However, in the spaceC(X; T X) (and L2, see [5, 12]), this assumption is not required to prove thefollowing Annular Hull Theorem:exp t�(M) � �(etM) � H (exp t�(M)) ; t 6= 0;(2.38)where H(�) is the union of the circles centered at origin, that intersect the set(�). We conjecture the assertion (2.38) is valid in CND(X; T X). Note thatthe equality s(M) = !(M) is an immediate consequence of (2.38).3 Description of the SpectrumIn this section we will describe the spectrum �(etM ) in the space CND(X; T X)under the assumptions of the previous section: X is a compact Riemannianmanifold with dimX � 3 and the divergence-free vector �eld v on X gener-ates the ow �t whose aperiodic points are dense in X. However, in fact, allthe results of this section are valid provided dimX � 2.By Theorem 2.1 it su�ces to determine �(etM) for a single value of t, say,for t = 1. As a notational convenience, we de�ne the bounded operator T inC(X; T X) by T = eM . For example, if M = L is the Lie derivative in thedirection v, then, for x 2 X,(Tu)(x) = D�(��1x)u(��1x):Also, we let TND := T jCND(X; T X) denote the restriction of T to thesubspace CND(X; T X).We will prove that �(TND; CND(X; T X)) is exactly one annulus, cen-tered at the origin whose inner and outer boundaries are the boundaries of�(T; C(X; T X)). The proof is based on the following simple idea. We will21

show that both spectra are rotationally invariant and the approximate pointspectra of T and TND coincide. Under the assumption that in the space ofdivergence-free vector �elds �(TND; CND(X; T X)) has a gap, we will extendthe Riesz projection for TND from CND(X; T X) to C(X; T X). To constructthis extension, we will approximate a continuous vector �eld by a linearcombination of locally supported divergence-free vector �elds. The Rieszprojection for TND can be applied to each such divergence-free vector �eldand this extension turns out to be a Riesz projection for T in C(X; T X). ByMather's theory this Riesz projection will be an operator of multiplicationby a continuous matrix-valued function. This multiplication must preserveCND(X; T X), a contradiction.To approximate a continuous vector �eld by a linear combination of lo-cally supported divergence-free vector �elds, we will need the following re-stricted form of Lemma 2.3:Lemma 3.1 If x0 2 X is a non periodic point of v and if z 2 Tx0X, then,for each pair B;D of su�ciently small neighborhoods with D � B 3 x0, thereis a coordinate chart with coordinate functions (x1; : : : ; xn) at x0 containingD and a vector �eld f 2 CND(X; T X) with suppf � D such that the localrepresentative of f in B is given by the constant vector �eld nXi=1 zi @@xi whosecomponents, zi, are the components of the local representative of the vectorz. By a theorem of Mather [13] (see also [5, 12]), the spectrum �ap(T ) inC(X; T X) is invariant with respect to rotations about the origin in the com-plex plane. As a corollary of Theorem 2.1 we have the following two asser-tions. We note, that these two assertions were also proved in [21].Corollary 3.2 �ap(TND;CND(X; T X)) is rotationally invariant.Corollary 3.3 �ap(T ;C(X; T X)) = �ap(TND;CND(X; T X)):Proof: In view of (2.8) and the fact that the spectra TND and T are rotation-ally invariant, the assertion will be proved as soon as we show the followingproposition: If 1 2 �ap(T ;C(X; T X)), then 1 2 �ap(TND;CND(X; T X)).By the Spectral Inclusion Theorem (2.37), to prove this proposition it isenough to show that 0 2 �ap(MND;CND(X; T X)) provided 1 2 �ap(T ) inC(X; T X)). This is done in Lemma 2.7. 222

In accordance with [13] (see also [5, 12]), the set �(T ;C(X; T X)) generallyconsists of several disjoint annuli centered at the origin. Let r� (resp., r+) de-note the radius of the inner most (resp., outer most) circle in �(T ;C(X; T X)).By Corollary 3.3, we have �(TND; CND(X; T X)) � fz : r� � jzj � r+g. Wewill show the set �(TND; CND(X; T X)) is exactly this annulus. This is thecontent of the next theorem.Theorem 3.4 The spectrum �(TND; CND(X; T X)) of T in CND(X; T X) isthe annulus fz : r� � jzj � r+g.Proof: Suppose the theorem is not true, then there is a gap in the spectrum�(TND) in CND(X; T X). Without loss of generality, we can assume there isan annulus fz : r1 � jzj � r2g in the resolvent set of TND containing the unitcircle T and r� � r1 < 1 < r2 � r+. In this case, there is a Riesz projectionP = PND for the operator TND in CND(X; T X) corresponding to the part of�(TND; CND(X; T X)) that lies inside of the unit disc D . In addition, thereare positive constants C1; C2 such thatImP = ff 2 CND(X; T X) :kT nfkC(X;TX) � C1rn1kfkC(X;TX); n 2 Ng;(3.39) Im(I � P ) = ff 2 CND(X; T X) :kT�nfkC(X;TX) � C2r�n2 kfkC(X;TX); n 2 Ng:(3.40)We will construct a projection P in C(X; T X) that commutes with Tand has the following additional properties:�(T jImP; C(X; T X)) � D ; �([T jIm(I � P)]�1; C(X; T X)) � D :In other words, the operator T is hyperbolic in C(X; T X), that is,�(T; C(X; T X)) \ T = ;;and P is the Riesz projection for T in C(X; T X). It follows, see [13] and[12], that the projection P has a form Pf(x) = PC(x)f(x), where PC : X !proj(TxX) is a continuous projection-valued function.Note, that P = PND on CND(X; T X). Hence, P maps CND(X; T X)into itself. We claim that this implies P is either the identity or the zerooperator, in contradiction to the fact that T is hyperbolic. To prove the23

claim, consider local (adapted) coordinates (x1; : : : ; xn) so that the divergenceoperator is given as in (2.10). The projection P is represented by a matrixvalued function with components Pij(x). For each divergence-free vector �eldu, we then have 0 = div(Pu) =Xi;j @Pij@xi uj + Pij @uj@xi :(3.41)For each point x in the coordinate chart and each index pair i; j with i 6= j,there is a divergence-free vector �eld u such that u(x) = 0 and @uj(x)=@xi =�ij where �ij is Kronecker's delta. With this choice of u, (3.41) shows Pij = 0for i 6= j. Using that fact that P is a projection, we have P(P � I) = 0 andit follows that each diagonal element, Pii(x), is either zero or one. Since Ppreserves all divergence-free vector �elds, it is easy to see that all diagonalelements must then be equal and P is as required in the coordinate chart.The desired result follows by continuity and the connectivity of X.We will construct the required projection P in the space C(X; T X).Step 1. We introduce \step-functions" in C(X; T X).Since X is compact, there is a partition of unity f�kgKk=1 with K < 1,that is, for each integer 0 < k � K, the function �k : X ! [0; 1] is continuous,and, for each x 2 X, PKk=1 �k(x) = 1;(3.42) supp�k n ([`=2ksupp�`) 6= ;:(3.43)In particular, there is some xk 2 supp�k n [`=2ksupp�` such that �k(xk) = 1.For each set of vectors u1; : : : ; uK, the vector �eldg(x) = KXk=1 �k(x)ukis continuous, and kgkC(X;TX) = supk kukk. Indeed, if kuk0k = supk kukk; wecan use (3.43) to choose x0 such that �k0(x0) = 1 for some x0 2 supp�k0n[k=2k0supp�k. Then,kgk � kg(x0)k = kXk �k(x0)ukk = k�k0(x0)uk0k = kuk0k:24

On the other hand, using (3.42),kgk = maxx2X kXk �k(x)ukk � maxx2X Xk �k(x)kukk � supk kukk:It is also easy to see that the set G of all such \step-functions" g = Pk �kukis dense in C(X; T X).Step 2. We will de�ne P for g 2 G.Suppose g = Pk �kuk 2 G. Without loss of generality we can assume thepartition of unity is so �ne that Lemma 3.1 is applicable for each k. By thislemma, for each k, there is a section fk 2 CND(X; T X) such that fk(x) = ukfor x 2 supp�k and kfkkC = kukk. Then, for each x 2 X,g(x) =Xk �k(x)fk(x); fk 2 CND(X; T X):(3.44)We de�ne Pg and Qg as follows:Pg(x) =Xk �k(x)(Pfk)(x); Qg(x) =Xk �k(x)[(I � P )fk](x);(3.45)where P = PND is the Riesz projection for TND in CND(X; T X). Formally,the decomposition g = Pg+Qg; depends upon the choice of fk. However, wewill show that, in fact, the de�nition (3.45) does not depend on this choiceand that P is a bounded linear operator on G. Once this is proved, theunique bounded linear extension of P to C(X; T X) is the desired projection.Step 3. De�neF+ := ff 2 C(X; T X) : limn!1 kT nfk = 0g;F� := ff 2 C(X; T X) : limn!1 kT�nfk = 0g:We will show that F+ \ F� = ;.Clearly, F� are linear (not necessarily closed) subspaces in C(X; T X).Assume f 2 F+ \ F� and f 6= 0. We havelimn!�1 kT nfk = limn!�1maxx k�(x; n)f(x)k = 0:In particular, there is some x0 2 X such that f(x0) 6= 0 andsupn2Zk�(x0; n)f(x0)k <1:25

This implies (see [5] or [14]) that T � �ap(T; C(X; T X)). Thus, by Corol-lary 3.3, we have T � �ap(TND; CND(X; T X)). But, this contradicts ourassumption that TND is hyperbolic in CND(X; T X).Step 4. We show that P and Q are well-de�ned on G.Suppose g = P �kuk 2 G and, for each k, the section fk is chosen as in(3.44). Also, de�ne g+ := Pg and g� := Qg. We will show g� 2 F�.Indeed, as Pfk 2 ImP , using the inequality (3.39), we havekT ng+k = kXk �k � �nT nPfkkC(X;TX)� supk kT nPfkkC(X;T X)maxx Xk �k � �n� C1rn1 supk kPfkkC(X;TX) � C1rn1kPk supk kfkkC(X;TX) == C1rn1kPk supk kukk = C1rn1kPk � kgkC(X;TX):In particular, limn!1 kT ng+k = 0 and g+ 2 F+. Similarly, using (3.40), wehave kT�ng�kC(X;T X) = kXk �k � ��nT�n(I � P )fkkC(X;TX)� maxx Xk �k � ��n(x) supk kT�n(I � P )fkkC(X;TX)� C2r�n2 kI � Pk supk kfkkC(X;T X) = C2kI � Pkr�n2 kgkC(X;TX):This implies kT�ng�k ! 0 as n!1 and g� 2 F�.By Step 3, we have F+ \ F� = f0g. Hence, g�, in the decompositiong = g+ + g� with g� 2 F�, are uniquely de�ned. If particular, the de�nitionof Pg and Qg in (3.45) does not depend on the choice of fk.Step 5. We extend P and Q from G to C(X; T X).From the calculations in Step 4 with ~C1 := C1kPk and ~C2 := C2kI �Pk,we have, for n 2 N, thatkT ng+kC � ~C1rn1kgk; kT�ng�kC � ~C2r�n2 kgk:These inequalities, for n = 0, show that P and Q are bounded on G. Tocomplete the proof we will show these operators are linear on G.26

Indeed, for g = i0Xi=1 �iui and ~g = j0Xj=1 ~�j~ujthere are fi; ~fj 2 CND(X; T X) such that g = P �ifi and ~g = P ~�j ~fj. Wede�ne fij = fi and ~fij = ~fj for i = 1; : : : ; i0, j = 1; : : : ; j0, and use (3.42) toobtain: g =Xi;j �i~�jfij; ~g =Xi;j �i~�j ~fij:Then, (3.45) gives:P(g + ~g) =Xi;j �i~�j(Pfij + P ~fij) = Pg + P~g;as required. 2From this theorem and Theorem 2.1 we conclude, that the spectrum�(M) in the space CND(X; T X) of divergence-free vector �elds is exactlyone vertical strip:Corollary 3.5 �(M ;CND(X; T X)) = fz : ln r� � Re z � ln r+g.Our next goal is to characterize the spectra �(TND;CND(X; T X)) and�(MND;CND(X; T X)) via the exact Lyapunov exponents for the cocycle�(x; t) with respect to the set of �t-ergodic measures � 2 E . Recall that,by the Multiplicative Ergodic Theorem [17], for each ergodic measure � 2 E ,there exists a set X� � X with �(X�) = 1 such that for each x 2 X andu 2 TxX there exist exact Lyapunov exponents��(x; u) = limt!�1 1t ln k�(x; t)uk:(3.46)For each �, there may exist n0 = n0(�) � n di�erent Lyapunov exponents;we will denote them by �1� > �2� > : : : > �n0� .Corollary 3.6 The boundary circles of the spectrum �(TND;CND(X; T X))and the spectrum �(MND;CND(X; T X)) are given byln r+ = supf�1� : � 2 Eg; ln r� = inff�n0� : � 2 Eg:There exist measures �+ and �� and exact Lyapunov exponents ��+(x+; u+)and ���(x�; u�), such that the sup and inf above are attained.27

Proof: For the boundaries r� of the spectrum �(T ;C(X; T X)) in C(X; T X)these formulas were obtained in [12] (see also [18]). 2Remark 3.7. The absence of nontrivial spectral components of �(L) for thespace CND(X; T X) leads to the following observation. Consider a situationwhen L acts in the space C(X; T X). After some inessential modi�cations,we can obtain a dichotomic (no spectrum on iR) operator L. For example,starting with an Anosov ow, as usual in Mather's theory, such an operatorcan be obtained by \moding out" the direction of the ow.We note that \moding out" the direction of the ow does not change �(L).The reason is that the spectrum of L on the direct sum of the quotient spaceCND(X; T X=[v]) and the space of sections generated by v is the union of therespective spectra. An element of the second space must be a divergence-freevector �eld of the form �v where � is a function on the manifold. This impliesgrad� = 0 so that � is constant along the trajectories of v. But, then L�v = 0and the spectrum of L on this subspace is f0g. Since �(LND;CND(X; T X))is invariant with respect to vertical translations in the complex plane, theentire imaginary axis must be in �(LND; CND(X; T X)). All the points exceptthe origin must then be in the spectrum of L restricted to the quotient space.But the spectrum of L is closed, thus the origin is already in the spectrumon the quotient. In particular, the spectrum of the quotient is the same asthe spectrum on the original space.Going back to the kinematic dynamo equations (1.1), we make the fol-lowing concluding remark.Remark 3.8. Recall (see, e.g., [1, 2, 3]) that kinematic dynamo is called\fast" provided lim sup"!1 !(L") is positive. M. Vishik [23] gave the fol-lowing su�cient condition for the non-existence of a fast kinematic dynamo:De�ne the Lyapunov numbers��(x; u) = lim supt!1 1t ln kD�t(x)uk:If supf��(x; u) : x 2 X; u 2 T Xg � 0;(3.47)then there is no fast kinematic dynamo. The fact that the spectral bound!(L) is less than or equal to the supremum in (3.47), see Remark 2.10, isused in [23]. Therefore, in view of [23], our Corollary 3.6 gives an alternateform of the su�cient condition for no fast kinematic dynamo.28

4 AcknowledgementWe thank Misha Vishik for several very helpful conversations.References[1] V. I. Arnold, Some remarks on the antidynamo theorem, Moscow Uni-versity Mathem. Bull., 6 (1982) 50{57.[2] V. I. Arnold, Ya. B. Zel'dovich, A. A. Rasumaikin, and D. D. Sokolov,Magnetic �eld in a stationary ow with stretching in Riemannian space,Sov. Phys. JETP, 54 (6) (1981) 1083{1086.[3] B. J. Bayly and S. Childress, Fast-dynamo action in unsteady ows andmaps in three dimensions, Phy. Rev. Let. 59 (14) (1987) 1573{1576.[4] C. Chicone, and Y. Latushkin, Quadratic Lyapunov functionsfor linear skew-product ows and weighted composition operators,J. Int. Di�. Eqns., to appear.[5] C. Chicone and R. Swanson, Spectral theory for linearization of dynam-ical systems, J. Di�. Eqns., 40 (1981) 155{167.[6] S. Friedlander and M. Vishik, Dynamo theory methods for hydrodynamicstability, J. Math. Pures Appl. 72 (1993) 145{180.[7] M. Hirsch, C. Pugh, and M. Shub, Invariant Manifolds, Lect. NotesMath. 583 (1977).[8] R. Johnson, Analyticity of spectral subbundles, J. Di�. Eqns., 35 (1980)366{387.[9] Y. Latushkin and S. Montgomery-Smith, Evolutionary semigroups andLyapunov theorems in Banach spaces, J. Funct. Anal., to appear.[10] Y. Latushkin and S. Montgomery-Smith, Lyapunov theorems for Banachspaces, Bull. AMS 31 (1) (1994) 44{49.29

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