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The Spectrum of the Kinematic Dynamo Operator for an Ideally Conducting Fluid C. Chicone, Y. Latushkiny, and S. Montgomery-Smithz Department of Mathematics, University of Missouri Columbia, MO 65211, USA April 25, 1997 Abstract

The spectrum of the kinematic dynamo operator for an ideally conducting uid and the spectrum of the corresponding group acting in the space of continuous divergence free vector elds on a compact Riemannian manifold are described. We prove that the spectrum of the kinematic dynamo operator is exactly one vertical strip whose boundaries can be determined in terms of the Lyapunov-Oseledets exponents with respect to all ergodic measures for the Eulerian ow. Also, we prove that the spectrum of the corresponding group is obtained from the spectrum of its generator by exponentiation. In particular, the growth bound for the group coincides with the spectral bound for the generator. [email protected], supported by the NSF grant DMS-9303767; [email protected], supported by the NSF grant DMS-9400518 and by the Summer Research Fellowship of the University of Missouri; z [email protected], supported by the NSF grant DMS-9201357; Key words and phrases. Hyperbolic dynamical systems, Mather spectrum, magnetohydrodynamics, ideally conducting uid, spectral mapping theorems, weighted composition operators. Mathematics Subject Classi cation. 76W05, 58F99, 58G25.  y

1

1 Introduction In this paper we give a description of the spectrum of the kinematic dynamo operator 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 velocity v = v(x) for x 2 R3 and let t denote the corresponding ow. The kinematic dynamo equations for the induction of a magnetic eld H by the ow has the following form: (1.1) H_ = r  (v  H) + "H; divH = 0; 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, in connection with the famous dynamo problem (see [1, 2, 3, 6, 16, 23] and references therein). For the ideally conducting uid, " = 0, these equations become: (1.2) H_ = ?(v; r)H + (H; r)v; H(x; 0) = H0 (x); divH = 0: The last equation has [15] so-called Alfven solutions

H(x; t) = Dt(?tx)H(?tx; 0); given by the group fetL gt2R with the generator L = L0 that acts by the rule (1.3) L : u 7! (u; r)v ? (v; r)u: In the present paper the kinematic dynamo operator L is considered in the following well known context. Let v denote a continuous divergencefree vector eld on a compact Riemannian manifold X without boundary, let t denote the ow generated by v and let Dt(x) denote its di erential. Consider the group fetL gt2R of push-forward operators generated by the Lie derivative L in the direction v. This group acts on continuous sections of the tangent bundle T X , by the rule (1.4)



 etL u (x) = Dt(?tx)u(?tx); 2

x 2 X; t 2 R:

We will consider the group fetL gt2R in the space CND (X; T X ) of the continuous vector elds with zero divergence. Operators of the form (1.4) belong to the class of weighted composition operators. This class has been widely investigated in connection with hyperbolic dynamical systems since the celebrated paper by J. Mather [13], see also [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 or p-summable vector elds are by now well understood. However, the investigation of their spectral properties in the space of divergence-free vector elds was initiated recently by R. de la Llave. His important work [21] inspired the present paper. In Section 2 we prove the spectral mapping theorem in CND (X; T X ) for the group fetLgt2R, assuming that the aperiodic trajectories of t are dense in X and dim X  3. Theorems of this type for continuous and for Lp-section spaces over nite dimensional manifolds were proved in [5, 8] while similar results for the in nite dimensional setting were obtained in [9, 10, 12]. The spectral mapping theorem states that the spectrum (etL ) of etL can be obtained from the spectrum of L by exponentiation. It shows, in particular, that in the space of divergence-free vector elds the spectral bound of the generator L coincides with the growth bound of the group, see also Remarks 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 spectrum of L is invariant under vertical translations of the complex plane. The same idea can be used to analyze semigroups of weighted composition operators with general cocycles on Banach spaces, see [11]. In Section 3, we show that, for dim X  2, the spectrum (etL), t 6= 0 in CND (X; T X ) is exactly one annulus centered at the origin of the complex plane. Our result generalizes a theorem in [21] where this fact was proved under the restriction that the ow is Anosov with one-dimensional spectral foliations. The possibility of relaxing the hypotheses of the theorem is discussed as an open problem in [21]. The relatively simple proofs of these facts in Section 3 can be read independently of the more dicult proofs in Section 2. By Mather's theory (see, e.g., [12, 13, 20]), the spectrum (etL ) on the space C (X; T X ) is generally the union of several disjoint annuli centered 3

at the origin. Passing to the space CND (X; T X ) dramatically changes the spectrum: the gaps, if any, between these annuli are lled. Also, using the Spectral Mapping Theorem, the spectrum of L on CND (X; T X ) is exactly one 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 the space of divergence free vector elds (unlike the situation with Anosov ows, cf. [20]), does not possess a nontrivial uniform exponential dichotomy even after \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 in terms of the Lyapunov-Oseledets exponents over all ergodic measures on X , that is, we will determine the boundaries of the spectrum of L via the Lyapunov exponents. This description is related to a theorem by M. Vishik [23] that states the \fast" dynamo action is impossible whenever all Lyapunov numbers are zero. Finally, we remark that we actually prove more general results than those just mentioned. In particular, we will only assume the operator L generates a C0-group of weighted composition operators that preserves the set of divergence free vector elds and has the form  t  (1.5) T u (x) = (?tx; t)u(?tx); x 2 X; t 2 R; where (x; t) is a continuous cocycle over t , that is for x 2 X and t;  2 R one has (x; t +  ) = (t x;  )(x; t) and (x; 0) = I . Throughout the paper we use M to denote the generator of the group fT t g, T t = etM . In particular, if (x; t) = Dt(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 results for the space L2.

2 The Spectral Mapping Theorem In this section we will prove the Spectral Mapping Theorem. Throughout the section we suppose a smooth vector eld v to be given on a compact Riemannian manifold X without boundary, that v is divergence free with respect to the Riemannian volume and that the ow t of v satis es the following standing hypotheses: the aperiodic trajectories of t are dense in X and n = dim X  3. Let M be the generator of the group fT tgt2R of weighted 4

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 divergence free vector elds. There are at least two choices for the space of continuous divergence free vector elds depending on whether the divergence is understood in the classical sense or in the sense of distributions. These spaces are de ned, respectively, as follows: (2.6) (2.7)

0 (X; T X ) = closure ff 2 C 1 (X; T X ) : divf = 0g; CND 1 (X; T X ) = ff 2 C (X; T X ) : CND Z < f; gradg > d = 0 8g 2 C 1(X; R)g: X

The closure in (2.6) is taken with respect to the sup-norm while the scalar product < ;  > and grad in (2.7) are taken with respect to a Riemannian 1 (X; T X ) is a closed metric and volume on X . We note that the space CND 0 (X; T X ). subspace of CND 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 its approximate point spectrum. Since, as noted above, 1 (X; T X )  C 0 (X; T X )  C (X; T X ) CND ND 1 (X; T X ) is an -eigenfunction for and since an -eigenfunction for T in CND 0 T in CND (X; T X ), one has 1 (X; T X ))   (T ; C 0 (X; T X ))   (T ; C (X; T X )): (2.8) ap(T ; CND ap ap ND

Throughout the remainder of the paper the space CND (X; T X ) may be 0 (X; T X ) or C 1 (X; T X ). taken to be either CND ND

Theorem 2.1 (Spectral Mapping Theorem) In the space of divergencefree vector elds CND (X; T X ) the spectrum (M ; CND (X; T X )) is invariant under vertical translations of the complex plane. Moreover, for each t = 6 0, (2.9)

(etM ) = exp t(M ): 5

Since the proof of Theorem 2.1 is quite technical, we pause to discuss our strategy. Using standard facts from the theory of C0-semigroups and by rescaling, 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 of J. 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 a vector 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 there exist 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. In fact, starting from a given -eigenfunction u for T , de ne a \bump"-function supported in D and let (j ) = (N ? jj j)=N . The \localization" y is de ned to vanish outside [Nj=?N j (D) and by y(x) = (j ) (?j x)(T j u)(x) for x 2 j (D) and jj j  N , equivalently,

y(x) =

N X j =?N

(j )(T j u)(x); x 2 X:

To prove the main assertion above, our purpose is to construct a vector eld y with zero divergence such that kMyk = O(1=N )kyk. We start with a divergence-free approximative eigenfunction u of T . Since the set of divergence free vector elds is not closed under multiplication by \bump"functions, we can not use Mather's construction directly. Instead, we will construct a divergence free vector- eld w, supported in a small neighborhood D of a given point x0 2 X in Lemma 2.3. The main part of this construction takes place in a special neighborhood D of x0 , taken to be a thin and long \ellipsoid" with the longest axis directed along u(x0). The required 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 to have value identically one on a second thin and long \ellipsoid" B contained in D. Some \ uid" leaks from the neighborhood D, but this can be recycled within a slightly larger neighborhood. The desired almost-eigenfunction y for M is given by the formula

y(x) =

Z1

?1

(t)T tw(x) dt; x 2 X; 6

where, as above, vanishes outside of [?N; N ]. Direct calculation shows that Z1

0 (t)T tw(x) dt; x 2 X: (My)(x) = ? ?1

Since suppw  D, the support of the integrand in each integral belongs to ft 2 R : jtj  N and t x 2 Dg. To obtain the desired inequality, we estimate kMyk from above and kyk from below. This requires some estimates of the sojourn time of the trajectory segment ft(x) : jtj  N g in D and B . This is 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 coordinate functions (x1 ; : : : ; xn), such that the local representation of the volume element on X is just the usual volume dx1 ^    ^ dxn on Rn . Moreover, if z 2 Tx0 X , then the coordinates can be chosen so that the local representative of z is kzk@=@x2 . Proof: Let y1 ; : : : ; yn denote local coordinates at x0 . Clearly, there is a nonvanishing density function  : Rn ! R such that volume element is given by (y1 ; : : : ; yn)dy1 ^    ^ dyn. We seek new coordinates in the form

y1 = f (x1 ; : : : ; xn); y2 = x2 ;    ; yn = xn where the volume element has the form @f (x ; : : : ; x )dx ^    ^ dx : (f (x1 ; : : : ; xn); x2 ; : : : ; xn ) @x 1 n 1 n 1 There is a smooth function f , de ned in a neighborhood of the origin in Rn , such that @f (0; : : : ; 0) 6= 0; @f (x ; : : : ; x ) = ((f (x ; : : : ; x ); x ; : : : ; x ))?1: n 1 n 2 n @x1 @x1 1 The rst condition together with the Implicit Function Theorem implies the change of coordinates is invertible; the second condition ensures the volume element in the new coordinates has the desired form. For the second statement of the lemma, note that the volume element is invariant under a rigid rotation of Euclidean space. 2 7

A coordinate chart, as in the lemma, is called adapted to the volume on X and the vector z. Of course, in the adapted coordinates, the Riemannian 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 nite symmetric matrix of smooth functions. However, we make the following observation: if u = (u1; : : : ; un) is a vector eld de ned in an adapted coordinate chart, then n @u X i divu = @x (2.10) : i=1

i

Suppose z 2 Tx0 X is a tangent vector and let (x1 ; : : : ; xn) denote local coordinates at x0 adapted to the volume on X and the vector z. If the adapted coordinate system is de ned in a coordinate ball of diameter  > 0 and if a; b 2 R are such that 0 < a < b < =8, we de ne

Da;b = f(x1; : : : ; xn) : jxj j  4b; j = 1; 2; jxj j  a; j = 3; : : : ; ng; Ba;b = f(x1 ; : : : ; xn ) : jxj j  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 bumpfunction : Rn ! [0; 1] with (x) = 1 for x 2 Ba;b and (x) = 0 for x 2= Da;b and a continuously di erentiable vector eld wn with support in Da;b such that i) The vector eld w(x) := (x)kzk@=@x2 + wn(x) is divergence free and has 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 extension.

Proof: Let z 2 Tx0 X . 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 by

x41 + x42 = 1; x41 + x42 = 2: a4 b4 a4 b4 8

Let  : R ! [0; 1] denote a smooth function such that (t) = 1 for t  1, (t) = 0 for t  2, and j0(t)j  3, t 2 R. Also, de ne the sets

R = f(x1 ; x2 ) : jx1 j  21=4 a; jx2j  21=4 bg; 1=4 4 4 S + = f(x1 ; x2 ) : (x1 ?a24 a) + xb42  2; x1  21=4 ag; 1=4 4 4 S ? = f(x1 ; x2 ) : (x1 +a24 a) + xb42  2; x1  ?21=4 ag and the functions  : R2 ! R, f : R2 ! R and  : R ! R by 4 x4 x (x1 ; x2) = ( a4 + b42 ); 3 kz k Z x1  s4 x4  4 x 0 a4 + b42 ds; f (x1 ; x2) = ? 2b4 0 Z 21 4 a  s4   4 k z k +  ( ) = ? b4 0 a4 + b4 ds; Z 0 0  s4   4 k z k 0 a4 + b4 ds; ?( ) = b4 ?21 4 a The vector eld w2 is de ned in R by f (x1 ; x2 )@=@x1 , in S + by +((x1 ? 21=4 a)4 + x42)x32 @x@ ? +((x1 ? 21=4 a)4 + x42 )(x1 ? 21=4 a)3 @x@ ; 1 2 in S ? by ?((x1 + 21=4 a)4 + x42 )x32 @x@ ? ?((x1 + 21=4a)4 + x42 )(x1 + 21=4 a)3 @x@ 1 2 and 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 eld W (x1; x2 ) := (x1 ; x2)kzk @x@ + w2(x1 ; x2) 2 is the required extension of z. A direct computation using (2.10) shows divW = 0 in the coordinate chart. Also, using the de nition of , we see the support of W is in Da;b . To =

=

9

show W is C 1 , just observe that w2 and each of its rst partial derivatives is continuous on the lines x1 = 21=4 a and on the boundary of R [ S + [ S ?. (We remark that additional smoothness can be obtained, if desired, by using the function xk1 =ak + xk2 =bk with k a suciently large positive integer in place of the choice k = 4 used here.) To obtain the required norm bound, let G(x1 ; : : : ; xn) denote the matrix of 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 by hG(x)V; V i. Thus, if kGk denotes the supremum of the matrix norms over the points in the chart, we have kV k  kGkjV j where the single bars denote the usual norm in Rn . Then, for example, using the usual estimate for the integral in the de nition of f , we estimate the norm of w2 in R by kGk sup jf (x1; x2 )j  kGk(4kzk23=4 b3 =b4)3(21=4 a)  C1a=b where the constant C1 does not depend on a or b. Similarly, we can estimate the norm of w2 in S . For example, in S + we nd the upper bound kGk sup j+((x1 ? 21=4a)4 + x42 )j ((21=4 b)6 + (21=4 a)6)1=2  C2a=b: To prove the lemma for the case n  3, we will show how to extend the vector eld W de ned above to a vector eld on Rn with the required properties. To do this, let  : R ! [0; 1] denote a smooth \bump"-function such that (t) = 1 for jtj  a2 and (t) = 0 for jtj  a, and de ne (x3 ; : : : ; xn) =

n Y

j =3

(xj ):

The required vector eld w is given by w(x1 ; : : : ; xn ) = (x3; : : : ; xn)W (x1 ; x2): The fact that the new vector eld w is continuously di erentiable, agrees with kzk@=@x2 in Ba;b, and is supported in Da;b is clear. Also, since the range of is the unit interval, the norm bound on wn is the same as the norm bound on 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. 2 To 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 v with 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  , has the following property: if x 2 D, then t x 62 D for s  jtj  N . Proof: Suppose the lemma is false and, for each positive integer k, let Dk denote the ball centered at x0 with radius 1=k. For each k, there is some xk 2 Dk and some tk in the set J := ft : 0 < s  jtj  N g with tk (xk ) in Dk . Since J  D 1 for the closure D 1 of D1 is compact, there is a convergent subsequence of the pairs (tk ; xk ). But, by the choice of Dk , the second component 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 of the ow ensures that T (x0 ) = x0 , in contradiction to the fact that x0 is not periodic. 2

Consider a vector eld v on X tangent to the ow t. For each open set U  X , each non negative integer N and each point x 2 X , de ne (2.11)

N;U (x) := ft 2 R : jtj  N; tx 2 U g; mN;U (x) := mes(N;U (x)):

Lemma 2.5 Suppose X has dimension n  3. If  > 0, then there is a

constant K > 0 such that for each non periodic point x0 and positive integer N there is a pair of numbers a; b such that  > b > a > 0 and a=b   together with a pair of open sets B; D at x0 such that B  Ba;b and Da;b  D, with the following property: for each x 2 X , mN;D (x)  K: (2.12) mN;B (x0)

c at x0 such that ty 62 D c Proof: By Lemma 2.4, there is a neighborhood D c and 1=2  jtj  2N . Suppose K; a; b; B; D, with D  D c, whenever y 2 D are given so that the inequality (2.12) holds for x 2 D. We claim that 11

(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 X and D \ft x : jtj  N g = ;, then mN;D (x) = 0 and (2.12) holds. Otherwise, x y 2 D \ ft x : jtj  N g. Since N;D (x)  2N;D (y), we have

mN;D (x)  max m (y) = max m (y)  K  mN;D (x0 ); y2D 2N;D y2D N;D as required. To complete the proof, we will construct K; a; b; B; D so that (2.12) holds for x 2 D. This will require several steps. Step 1. We will work in an adapted coordinate system at x0 with coordinate functions (x1 ; : : : ; xn). We will determine the required sets B; D for appropriate a; b in the form

( ) 4 4 x4j x x P 1 2 4 B= x: 4 + (b=2)4 + j =3 (a=2)4  1 ; ( a= 2) ) ( 4 4 4 x x x P j 2 1 + + 4j=3 1 : D= x: (4nb)4 (4nb)4 (na)4 If a < b, then, clearly, B  Ba;b and Da;b  D. We will show that there is a constant K such that for some choice of a; b and a=b all suciently 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 coordinate system, that is, at x0 , such that the Riemannian diameter diamS < , and de ne  P:= ft :  2 S ; jtj  g. Consider the local representation of v given by ni=1 vi (x)@=@xi . By a rigid rotation, if necessary, we can and will arrange the adapted coordinates so that v1 (0) = 0. Also, we de ne V by

V (x) :=

n X i=3

vi4(x); x 2 

and the number

8 !1=4 9 n < = X 4 (x) M := max :max v j v ( x ) j ; max ;: x2 1 x2 i=3 i 



12

If V (x0 ) = 0, then lim!0 M ! 0. In case V (x0) = 0 and M 6= 0 for every , we will only consider a; b such that (2.13) a = b  (M )1=2 : Of course, even under the restriction just imposed, a; b, a=b can each be chosen arbitrary small. If V (x0) 6= 0 or if V (x0 ) = 0 and M  0 for all suciently small , we ignore this restriction. Step 2. For each , Lemma 2.4 and the de nition of  together imply there is an open ball A   at the origin such that, for each x 2 A and for each time t with jtj > , the point t x 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 x0 denote the point on @D where the segment of the trajectory ftx : jtj  N g rst enters D and let x00 = tD x0 denote the point of @D where the segment of the trajectory ftx : jtj  N g last exits D. Clearly, mN;D (x)  tD . We use the Mean Value Theorem for integrals on the j th component of the vector eld v to obtain a point  j 2  such that

x00 = x0

j+

j

Zt

D

0 v

vj (tx) dt = x0j + tD vj ( j ):

For each j = 1; : : : ; n, let 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 have n x0 !4 n x0 + t v  !4 X X D i i =1 i and = 1: i i=1 i i=1

Using a standard inequality for the norm k( i)k = (Pi j ij4)1=4 , we nd 1 = Pn

i=1

 tD v 4!1=4 0P x0 i !4 11=4 P ? @ ni=1 A  ni=1 i i i !  tD v 4 1=4 P n ? 1: = i=1 i i

x0i + tD vi i

!4

This computation yields the estimate n  v  4 X i t4D (2.14)  24 : i=1 i 13

Step 3. Consider the time tB when the segment of the trajectory ftx0 : jtj  N g 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

Zt

B

x~j = vj (t(x0 )) dt = tB vj (j ): (j ),

0

De ne v~j := vj the numbers j = a=2 for j = 1; 3; : : : ; n and 2 = b=2. Since x~ 2 @B , we have n v~ !4 X j 4 tB (2.15) = 1: j =1

j

Step 4. In accordance with the previous notation, we de ne X X V~ = v~i4; V  = (vi)4 : n

n

i=3

i=3

We use (2.14)-(2.15) to obtain the estimate (tD =tB )4  216n4  d where  a 4 4 v~1 + b v~2 + V~ d :=  a 4  a 4 :   4  b v1 + b v2 + 4 V We will show that for all suciently small  > 0, there are some choices of 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 suciently small , then, since v2 (x0) 6= 0, lim!0 d = 1. Case 3. Suppose V (x0 ) = 0 and M 6= 0. However, note that we still have lim!0 M ! 0. Also, the restriction we imposed in (2.13) provides that b v~1=2  1; b V~ 1=8  1: a 1 a In this case, we have ! ! b v~ 4 + v~4 + b 4 V~ 1 2 v~2 + v~4 + V~ 1=2 a a d= !4  1 (v2)4 : 2 (v1)4 + (v2)4 + 44 b V  a 14

Passing to the limit as  ! 0, we see that the last expression converges to 1. 2 We need the following elementary fact. Lemma 2.6 Suppose A denotes an invertible bounded operator on a Banach space E and let N 2 Z. If N  2 and 1 2 ap (A), then there is a vector u 2 E with kukE = 1 such that kAk ukE  2 for each integer k with jkj  N .

0 N 1?1 X Proof: Set  = @ kAk kA . Since 1 2 ap(A), there is some u 2 E with k=?N kukE = 1 such that kAu ? uk  . Also, for 1  jkj  N , note that 0k?1 1 0 k 1 X X Ak ? I = @ Aj A (A ? I ); k > 0; Ak ? I = ? @ Aj A (A ? I ); k < 0: j =0

j =?1

Hence, for jkj  N , we have kAk u ? uk 

N X j =?N

kAj k  kAu ? uk  1 and, as

a result, kAk uk  kAk u ? uk + kuk  2. 2

The 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; T X )), then ap (M ; CND (X; T X )) contains the imaginary 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: 0 (X; T X ) or C (X; T X ) = C 1 (X; T X ): CND (X; T X ) = CND ND ND Proof: Let  2 R and let K be de ned as in Lemma 2.4. Also, for notational convenience, de ne ! := 1=(12K ). Since 1 2 ap (T ), Lemma 2.6 applied to the bounded linear operator T = T 1 ensures that, for each integer N  2, there is some vector eld u 2 C (X; T X ) such that (2.16) kukC (X;T X ) = 1; k (2.17) kT ukC (X;T X )  2 for jkj  N: 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 that

kT t u ? ukC (X;T X )  ! for jtj  s; je?it ? 1j  ! for jtj  s: Also, there is a smooth function : R ! [0; 1] such that: (2.20)

(t) = 0 for jtj  N; (2.21) j 0(t)j  N2 ; for t 2 R; (2.22)

(t) = 1 for jtj  s: (2.18) (2.19)

In view of (2.16), and the fact that the non periodic points are dense in X , there is a non periodic point x0 2 X such that ku(x0)k  21 kukC (X;T X ) = 21 : (2.23) Use Lemma 2.5 to nd small a; b with small a=b, and neighborhoods D  B 3 x0 , such that (2.12) holds. Moreover, in accordance with Lemma 2.4, we can choose a; b suciently small so that for D := Da;b , for s from (2.18){(2.19) and with c := maxjtj1 kT tk we have (2.24)

t y 62 D for any y 2 D provided s  jtj  2N

and, for some constant C , the following inequalities: ! a t (2.25) kT k  !; C b jmax tjN (2.26) max k(y; t)u(x0)k  4c: y2D;jtjN For the last inequality, we use (2.17) to show that max k(x0 ; t)u(x0 )k  jtjmax k(x; t)u(x)k N;x2X

jtjN

 jtjmax k(?t x; t)u(?tx)k = jkjNmax kT k+ uk N;x2X ?1;j j1  c jkmax kT k k  2c: jN ?1 16

Since  : (x; t) 7! (x; t) is uniformly continuous on the compact set X  [?N; N ], we have (2.26) for a suciently small neighborhood D of x0 . We use Lemma 2.3 with z = u(x0). After a rigid rotation, if necessary, we can arrange the adapted coordinates so that the component of v(x0) in the direction of the rst coordinate vanishes. Then, for this choice of adapted coordinates, there is a divergence-free vector eld of the form (2.27) w(x) = (x)u(x0) + wn(x) with and wn supported in D, kwnkC (X;T X )  C ab ; (2.28) and (x) = 1 for x 2 Ba;b . De ne the vector eld y on X by

y(x) =

Z1

?1

e?it (t)T t w(x) dt:

We see that y has zero divergence (for this remember T t preserves the divergence-free vector elds). By easy computations with My = dd T  y  =0 one has: Z1 d T twdt = iy ? Z 1 0 (t)e?itT t w dt: ? i ( t ?  ) My = ( e

( t ?  ))  =0 ?1 d ?1 To complete the proof, we must show that i 2 ap(M ; CND (X; T X )). This is an immediate consequence of the following proposition: There is a number A > 0 that does not depend on the choice of N such that

Z 1

A kyk

0(t)e?itT t w dt

 C (X;T X ) : ?1 C (X;T X ) N To prove the proposition, x x 2 X and note that

Z 1

? it 0 t

kMy(x) ? iy(x)k = ?1 e (t)(T w)(x) dt

 I1 + I2

where, by (2.20) and (2.21), RN I1 = N2 (?tx)kt (?t x; t)u(x0)k dt; ?N RN I2 = N2 k(?tx; t)k kwn(?tx)k dt: ?N 17

Since supp  D and suppwn  D, the integrations I1 and I2 can be restricted to D (x) = N;D (x), see the notation in (2.11). We use (2.26) to obtain: Z 2 0)k  2 4c  m (x): I1  N (2.29) max k ( y; t ) u ( x N;D y2D;jtjN N  (x) D

We use (2.25) and (2.28) to estimate I2: Z (2.30) max k(y; t)kkwnkC (X;T X ) dt I2  N2 y2D;jyjN D (x) t k  C a  m (x)  2 !m (x):  N2 jmax k T tjN b N;D N N;D We obtain the desired upper estimate from (2.29) and (2.30), namely, kMy ? iykC (X;T X )  AN1 max (2.31) m (x): x2X N;D To determine the lower bound, we de ne

Z 1

? it ? t 0 0

J1 = e (t) ( x )u(x ) dt

;

Z?1 h t 0 i

1 ?it ? t 0 0

J2 = e (t) ( x ) (T u)(x ) ? u(x ) dt ;



Z?1 1 ?it t

J3 = e (t)(T wn)(x0)

?1

and note that

kykC (X;T X )  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: (2.32)

J3 

Z

D (x0 )

k(T twn)(x0 )k dt  !mN;D (x0):

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 have (2.33)

J2 

Z

D (x0 )

kT tu ? ukC (X;T X ) dt  !mN;D (x0): 18

Finally, we estimate J1 from below: (2.34)

J1

Z

= ku(x0)k

where

J11 = j J12 = j

D (x0 )

Z ZD (x0 ) D (x0 )

e?it (t) (?tx0 ) dt  J11 ? J12 ;

(t) (?tx0 ) dtj  ku(x0)k; (e?it ? 1) (t) (?tx0 ) dtj  ku(x0)k:

Since, by (2.24), tx0 62 D for s  jtj  2N , equation (2.22) gives (t) = 1 for t 2 D (x0 ). As (x) = 1 for x 2 B , we use (2.23) to compute the estimate: Z J11  21 (2.35) (?tx0 ) dt  21 mN;B (x0 ):  (x0 ) B

Since ku(x0)k  1 and jtj  s for t 2 D (x0 ), the inequality (2.19) implies: Z J12  (2.36) e?it ? 1 (?tx0 ) dt  !mN;D (x0 ): D (x0 )

The estimates (2.35), (2.36), (2.33), (2.32), and (2.12) together with the our choice of ! = 1=(12K ) give the following: kykC (X;T X )  21 mN;B (x0) ? 3!mN;D (x0 )  21K max m (x) ? 3! max m (x) x2X N;D x2X N;D = 41K max m (x): x2X N;D By combining this estimate with (2.31), we have the desired result. 2 We are now in the position to prove Theorem 2.1. Proof: It is well-known (see, e.g., [19]) that the Spectral Inclusion Theorem (2.37) (etM )  exp t(M ); t 6= 0 19

holds for any C0-semigroup. Also, the spectral mapping theorem is true for the point and residual spectrum. Therefore, to prove (2.9) one needs to show that, in CND (X; T X ), ap(etM )  exp tap (M ); t 6= 0: Fix  = jjei 2 ap(etM ; CND (X; T X )). Then  = et for  = 1 ln jj + t  ? t i t . Consider the cocycle ~ (x; t) = e (x; t), and the group fT~t g, T~t = etM~ , de ned by ~ (x; t) as in (1.5). Then  2 ap (etM ; CND (X; T X )) implies that 1 2 ap (etM~ ; CND (X; T X )). By (2.8) and Lemma 2.7, we have 0 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 imaginary axis, we x  2 ap (M ) and  2 R. By the Spectral Inclusion Theorem for t = 1 we have 1 2 ap (eM~ ), also, M~ = M ? . By Lemma 2.7, one has i 2 ap (M~ ) and, as a result,  + i 2 ap (M ). 2 We will use notations ?1 tA s(A) := supfRez : z 2 (A)g and !(A) := tlim !1 t ln ke k

for the spectral bound of a generator A and the growth bound of a C0semigroup fetA g, respectively. Note [19], that for an arbitrary C0-semigroup fetAg one has s(A)  !(A), but, generally, s(A) 6= !(A). The Spectral Mapping Theorem, however, gives for the group of weighted composition operators the following fact. Corollary 2.8 In the space of continuous divergence free vector elds the spectral 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 semigroup, 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 this equality is also valid for " = 0. Remark 2.10. M. Vishik [23] has shown that lim sup"!0 !(L")  !(L0). In view 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 of the group fetL g; it is given in the terms of generators only. See Remark 3.9 below 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 perturbations for the generators of C0-semigroups. Remark 2.11. The Spectral Mapping Theorem for semigroups of weighted composition operators does not hold without the assumption that aperiodic trajectories are dense in X (see [5, 12] for examples). However, in the space C (X; T X ) (and L2 , see [5, 12]), this assumption is not required to prove the following Annular Hull Theorem: "

(2.38)

exp t(M )  (etM )  H (exp t(M )) ; t 6= 0;

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 that the equality s(M ) = !(M ) is an immediate consequence of (2.38).

3 Description of the Spectrum

In 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 Riemannian manifold with dim X  3 and the divergence-free vector eld v on X generates the ow t whose aperiodic points are dense in X . However, in fact, all the results of this section are valid provided dim X  2. By Theorem 2.1 it suces to determine (etM ) for a single value of t, say, for t = 1. As a notational convenience, we de ne the bounded operator T in C (X; T X ) by T = eM . For example, if M = L is the Lie derivative in the direction 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 the subspace CND (X; T X ). We will prove that (TND ; CND (X; T X )) is exactly one annulus, centered 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 will 21

show that both spectra are rotationally invariant and the approximate point spectra of T and TND coincide. Under the assumption that in the space of divergence-free vector elds (TND ; CND (X; T X )) has a gap, we will extend the Riesz projection for TND from CND (X; T X ) to C (X; T X ). To construct this extension, we will approximate a continuous vector eld by a linear combination of locally supported divergence-free vector elds. The Riesz projection for TND can be applied to each such divergence-free vector eld and this extension turns out to be a Riesz projection for T in C (X; T X ). By Mather's theory this Riesz projection will be an operator of multiplication by a continuous matrix-valued function. This multiplication must preserve CND (X; T X ), a contradiction. To approximate a continuous vector eld by a linear combination of locally supported divergence-free vector elds, we will need the following restricted form of Lemma 2.3: Lemma 3.1 If x0 2 X is a non periodic point of v and if z 2 Tx0 X , then, for each pair B; D of suciently small neighborhoods with D  B 3 x0 , there is a coordinate chart with coordinate functions (x1 ; : : : ; xn ) at x0 containing D and a vector eld f 2 CND (X; T X ) with suppf  D suchn that the local X @ representative of f in B is given by the constant vector eld zi whose i=1 @xi components, zi , are the components of the local representative of the vector z. By a theorem of Mather [13] (see also [5, 12]), the spectrum ap (T ) in C (X; T X ) is invariant with respect to rotations about the origin in the complex plane. As a corollary of Theorem 2.1 we have the following two assertions. 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 rotationally invariant, the assertion will be proved as soon as we show the following proposition: 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 is enough to show that 0 2 ap (MND ; CND (X; T X )) provided 1 2 ap (T ) in C (X; T X )). This is done in Lemma 2.7. 2 22

In accordance with [13] (see also [5, 12]), the set (T ; C (X; T X )) generally consists of several disjoint annuli centered at the origin. Let r? (resp., r+) denote 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. We will show the set (TND ; CND (X; T X )) is exactly this annulus. This is the content of the next theorem. Theorem 3.4 The spectrum (TND ; CND (X; T X )) of T in CND (X; T X ) is the 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 is an annulus fz : r1  jzj  r2g in the resolvent set of TND containing the unit circle T and r?  r1 < 1 < r2  r+. In this case, there is a Riesz projection P = 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, there are positive constants C1 ; C2 such that ImP = ff 2 CND (X; T X ) : (3.39) kT nf kC (X;T X )  C1r1nkf kC (X;T X ); n 2 Ng; Im(I ? P ) = ff 2 CND (X; T X ) : (3.40) kT ?nf kC (X;T X )  C2r2?nkf kC (X;T X ); n 2 Ng: We will construct a projection P in C (X; T X ) that commutes with T and 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 P f (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 zero operator, in contradiction to the fact that T is hyperbolic. To prove the 23

claim, consider local (adapted) coordinates (x1 ; : : : ; xn) so that the divergence operator is given as in (2.10). The projection P is represented by a matrix valued function with components Pij (x). For each divergence-free vector eld u, we then have X Pij j 0 = div(P u) = @@x (3.41) uj + Pij @u : @x i i i;j 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 = 0 for i 6= j . Using that fact that P is a projection, we have P (P ? I ) = 0 and it follows that each diagonal element, Pii(x), is either zero or one. Since P preserves all divergence-free vector elds, it is easy to see that all diagonal elements 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 fk gKk=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 , (3.42) (3.43)

PK  (x) = 1; k=1 k suppk n ([`=2k supp` ) 6= ;:

In particular, there is some xk 2 suppk n [`=2k supp` such that k (xk ) = 1. For each set of vectors u1; : : : ; uK , the vector eld

g(x) =

K X k=1

k (x)uk

is continuous, and kgkC (X;T X ) = supk kuk k. Indeed, if kuk0 k = supk kuk k; we can use (3.43) to choose x0 such that k0 (x0 ) = 1 for some x0 2 suppk0 n[k=2k0 suppk . Then,

X

kgk  kg(x0)k = k k (x0 )uk k = kk0 (x0 )uk0 k = kuk0 k: k

24

On the other hand, using (3.42),

X

X

k

k

kgk = max k k (x)uk k  max k (x)kuk k  sup kuk k: x2X x2X k

It is also easy to see that the set G of all such \step-functions" g = Pk k uk is dense in C (X; T X ). Step 2. We will de ne P for g 2 G. Suppose g = Pk k uk 2 G. Without loss of generality we can assume the partition of unity is so ne that Lemma 3.1 is applicable for each k. By this lemma, for each k, there is a section fk 2 CND (X; T X ) such that fk (x) = uk for x 2 suppk and kfk kC = kuk k. Then, for each x 2 X , (3.44)

g(x) =

X k

k (x)fk (x); fk 2 CND (X; T X ):

We de ne P g and Qg as follows: (3.45) P g(x) =

X k

k (x)(Pfk )(x); Qg(x) =

X k

k (x)[(I ? P )fk ](x);

where P = PND is the Riesz projection for TND in CND (X; T X ). Formally, the decomposition g = P g + Qg; depends upon the choice of fk . However, we will show that, in fact, the de nition (3.45) does not depend on this choice and that P is a bounded linear operator on G. Once this is proved, the unique bounded linear extension of P to C (X; T X ) is the desired projection. Step 3. De ne

F+ := ff 2 C (X; T X ) : limn!1 kT nf k = 0g; F? := ff 2 C (X; T X ) : limn!1 kT ?nf k = 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 have lim kT nf k = n!1 lim max x k(x; n)f (x)k = 0: n!1 In particular, there is some x0 2 X such that f (x0) 6= 0 and sup k(x0 ; n)f (x0)k < 1: n2Z

25

This implies (see [5] or [14]) that T  ap(T; C (X; T X )). Thus, by Corollary 3.3, we have T  ap (TND ; CND (X; T X )). But, this contradicts our assumption that TND is hyperbolic in CND (X; T X ). Step 4. We show that P and Q are well-de ned on G. P Suppose g = k uk 2 G and, for each k, the section fk is chosen as in (3.44). Also, de ne g+ := P g and g? := Qg. We will show g 2 F. Indeed, as Pfk 2 ImP , using the inequality (3.39), we have

X

kT ng+k = k k  nT nPfk kC (X;T X ) k

X

 sup kT nPfk kC (X;T X ) max k  n x k k  C1r1n sup kPfk kC (X;T X )  C1r1nkP k sup kfk kC (X;T X ) = k k = C1 r1nkP k sup kuk k = C1r1nkP k  kgkC (X;T X ) : k

In particular, limn!1 kT ng+k = 0 and g+ 2 F+. Similarly, using (3.40), we have

X

kT ?ng?kC (X;T X ) = k k  ?nT ?n(I ? P )fk kC (X;T X ) X

k

 max k kC (X;T X ) x k k k  C2 r2?nkI ? P k sup kfk kC (X;T X ) = C2kI ? P kr2?nkgkC (X;T X ):  ?n(x) sup kT ?n(I ? P )f k

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 decomposition g = g+ + g? with g 2 F, are uniquely de ned. If particular, the de nition of P g 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 C~1 := C1 kP k and C~2 := C2kI ? P k, we have, for n 2 N, that kT ng+kC  C~1r1nkgk; kT ?ng?kC  C~2r2?nkgk: These inequalities, for n = 0, show that P and Q are bounded on G. To complete the proof we will show these operators are linear on G. 26

Indeed, for

g=

i0 X

j0 X

~j u~j there are fi; f~j 2 CND (X; T X ) such that g = P ifi and g~ = P ~j f~j . We de ne fij = fi and f~ij = f~j for i = 1; : : : ; i0 , j = 1; : : : ; j0 , and use (3.42) to obtain: X X g = i~j fij ; g~ = i~j f~ij : i=1

iui and g~ =

j =1

i;j

i;j

Then, (3.45) gives: X P (g + g~) = i ~j (Pfij + P f~ij ) = P g + P g~; as required. 2

i;j

From 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 exactly one 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 and u 2 TxX there exist exact Lyapunov exponents  (x; u) = t!1 lim 1t ln k(x; t)uk: (3.46) For each  , there may exist n0 = n0 ( )  0n di erent Lyapunov exponents; we will denote them by 1 > 2 > : : : > n . Corollary 3.6 The boundary circles of the spectrum (TND ; CND (X; T X )) and the spectrum (MND ; CND (X; T X )) are given by ln r+ = supf1 :  2 Eg; ln r? = inf fn 0 :  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]). 2

Remark 3.7. The absence of nontrivial spectral components of (L) for the space CND (X; T X ) leads to the following observation. Consider a situation when 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 operator can 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 space CND (X; T X=[v]) and the space of sections generated by v is the union of the respective spectra. An element of the second space must be a divergence-free vector eld of the form v where is a function on the manifold. This implies grad = 0 so that is constant along the trajectories of v. But, then L v = 0 and 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, the entire imaginary axis must be in (LND ; CND (X; T X )). All the points except the 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 spectrum on the quotient. In particular, the spectrum of the quotient is the same as the spectrum on the original space. Going back to the kinematic dynamo equations (1.1), we make the following 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 following sucient condition for the non-existence of a fast kinematic dynamo: De ne the Lyapunov numbers (x; u) = lim sup 1t ln kDt(x)uk: t!1 If (3.47) supf(x; u) : x 2 X; u 2 T X g  0; 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, is used in [23]. Therefore, in view of [23], our Corollary 3.6 gives an alternate form of the sucient condition for no fast kinematic dynamo. 28

4 Acknowledgement We thank Misha Vishik for several very helpful conversations.

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