]]> 2001 American Mathematical Society Proceedings of the American Mathematical Society Proc. Amer. Math. Soc. proc 129 10 20010315 February 16, 2000 March 15, 2001 2965 2965-2972 8 S0002-9939-01-06035-X 10.1090/S0002-9939-01-06035-X 0002-9939 1088-6826 English 42A99 2000 http://www.ams.org/jourcgi/jour-getitem?pii=S0002-9939-01-06035-X The results intro A Borel set n of positive measure is said to tile n by translations if there is a discrete set T n such that, up to sets of measure 0, the sets t, tT, are disjoint and tT ( t) n . We may rescale so that 1 . We say that k: k n is a spectrum for if equation e 2i kx k is an orthonormal basis for L 2(). a.e00 equation A spectral set is a domain n such that ( a.e00 ) holds for some . Fuglede Fug conjectured that a domain n is a spectral set if and only if it tiles n by translations, and proved this conjecture under the assumption that either or T is a lattice. The conjecture is related to the question of the existence of commuting self-adjoint extensions of the operators -i x j , j 1,,n Fug , , P1 ; other relations between the tiling and spectral properties of subsets of n have been conjectured and, in some cases, proved; see IP , JP1 , JP2 , K2 , LRW , LW2 . Recently there has been significant progress on the special case of the conjecture when is assumed to be convex K1 , IKP , IKT1 , and in particular the 2-dimensional convex case appears to be nearly resolved IKT2 . The non-convex case is considerably more complicated, and is not understood even in dimension 1. The strongest results yet in that direction seem to be those of Lagarias and Wang LW1 , LW2 , who proved that all tilings of by a bounded region must be periodic, and that the corresponding translation sets are rational up to affine transformations. This in turn leads to a structure theorem for bounded tiles. It was also observed in LW2 that the tiling implies spectrum" part of Fuglede's conjecture for compact sets in would follow from a conjecture of Tijdeman Tij concerning factorization of finite cyclic groups; however, Tijdeman's conjecture is now known to fail without additional assumptions (see LS for a discussion). See also New , CM for partial results on the related problem of characterizing all tilings of by a finite set, and LW2 , P2 for a classification of domains in n which have L n as a spectrum for some finite set L . Another recent result PW is that sets which tile 0,) by translations are spectral sets. The purpose of the present article is to address the following special case of Fuglede's conjecture in one dimension. Let I 1I 2 , where I 1,I 2 are disjoint intervals of non-zero length. By scaling, translation, and symmetric reflection, we may assume that equation (0,r)(a,a 1-r), 0 r, ar. e.omega1 equation Our first theorem characterizes all 's of the form ( e.omega1 ) which are spectral sets. theoreme Suppose that is a spectrum for , 0 . Then at least one of the following holds: enumerate (i) a-r and ; (ii) r , a 2 for some n , and 2( 2) for some odd integer p . enumerate Conversely, if and satisfy e.omega1 and if either (i) or (ii) holds, then is a spectrum for . s.thm1 theoreme As a corollary, we prove that Fuglede's conjecture holds for a union of two intervals. theoreme Let be a union of two disjoint intervals, 1 . Then has a spectrum if and only if it tiles by translations. s.thm2 theoreme Theorem s.thm2 follows easily from Theorem s.thm1 . We may assume that is as in ( e.omega1 ). Suppose that is a spectrum for ; without loss of generality, we may assume that 0 . Then by Theorem s.thm1 one of the conclusions (i), (ii) must hold, and in each of these cases tiles by translations. Conversely, if tiles by translations, by Proposition tis.prop1 must satisfy Theorem s.thm1 (i) or (ii) , hence by Theorem s.thm1 again it is a spectral set. Theorem s.thm1 will be proved as follows. Suppose that k: k is a spectrum for ; we may assume that 0 0 . Let kk' k - k' , - kk' : k,k' , and equation Z 0 : () 0 . e.Z equation Then the functions e 2i k x are mutually orthogonal in L 2() , hence -Z . This will lead to a number of restrictions on the possible values of k . Next, let equation (x) (0,r) e 2ix , s.e11 equation where (0,r) denotes the characteristic function of (0,r) . By Parseval's formula, the Fourier coefficients c k 0 r e 2i(- k)x dx of satisfy equation k c k 2 (0,r) e 2ix L 2() 2 r. s.e10 equation Given that the k 's are subject to the orthogonality restrictions mentioned above, we will find that there are not enough k 's for ( s.e10 ) to hold, unless the conditions of Theorem s.thm1 are satisfied. The author is grateful to Alex Iosevich for helpful conversations about spectral sets and Fuglede's conjecture, and to Jeffrey Lagarias and Yang Wang for bringing the references LS and PW to her attention. Tiling implies spectrum tis proposition If as in e.omega1 tiles by translations, it must satisfy (i) or (ii) of Theorem s.thm1 . tis.prop1 proposition proof Suppose that may be tiled by translates of . Assume first that r . Then any copy of used in the tiling has a gap" of length a-r a- , which must be covered by non-overlapping intervals of length ; hence a as in Theorem s.thm1 (ii) . Assume now that 0 r . Let I 1 (0,r) , I 2 (a,a 1-r) . We will prove that translates of I 1 and I 2 must alternate in any tiling of by translates of ; this implies immediately that a-r as in Theorem s.thm1 (i) . itemize If contained two consecutive translates (, r) and ( r, 2r) of I 1 , it would also contain the matching translates ( a, a 1-r) and ( a r, a 1) of I 2 , which is impossible since the latter two intervals overlap. Suppose now that contains two consecutive translates ( a, a 1-r) and ( a 1-r, a 2-2r) of I 2 . Then must also contain the matching translates I' 1 (, r) and I'' 1 ( 1-r, 2-2r) of I 1 . The gap between I' 1 and I'' 1 has length 1-2r , which is strictly less than 1-r I 2 , so that I' 1 must be followed by another translate of I 1 . But this has just been shown to be impossible. itemize proof Next, we prove the second part of Theorem s.thm1 . This easy result appears to have been known to several authors; see, e.g., the examples in Fug , JP1 , LW2 . Since we will rely on it later on in the proof of the hard" part of the theorem, we include the short proof. proposition If and are as in Theorem s.thm1 (i) or (ii) , then is a spectrum for . tis.prop2 proposition proof If (i) holds, then is a fundamental domain for , and, consequently, is a spectrum Fug . Suppose now that (ii) holds. For any function f on , we define functions f ,f - : f (x) (f(x) f(x')), f -(x) (f(x)-f(x')), x, where x' x a if x(0,) , and x' x-a if x(a,a ) . Then f(x) f (x) f -(x), f (x) f (x'), f -(x) -f -(x'). It therefore suffices to prove that equation g(x) k c ke 4kix for any g(x)L 2() such that g(x) g(x'), s.e20 equation equation h(x) k c' ke (4k 2p n)ix for any h(x)L 2() such that h(x) -h(x'). s.e21 equation Since e 4kix , k , is a spectrum for (0,) , we have g(x) k c ke 4kix , h(x) e 2p nix k c' ke 4kix , x(0,). ( s.e20 ) follows immediately by periodicity. From the second equation above we find that ( s.e21 ) holds for all x(0,) , and that for such x e 2p ni(x a) k c' ke 4ki(x a) -e 2p nix k c' ke 4kix -h(x) h(x a), where we used that 2p na p is odd. Hence, ( s.e21 ) also holds for x(a,a ) . This ends the proof of Proposition tis.prop2 . proof Orthogonality 2 We now begin the proof of the first part of Theorem s.thm1 . Throughout the rest of the paper, is assumed to satisfy ( e.omega1 ), k: k is a spectrum for , 0 0 , kk' k- k' , - kk' : k,k' , and Z is defined by ( e.Z ). lemme Z Z 1Z 2Z 3 , where array l Z 1 : a , (2r-1) , 2mm Z 2 : r , 2mm Z 3 : (a-r) . array s.lemma1 lemme proof Suppose that 0 , Z . Then e 2ix dx e 2ir -1 e 2i(a 1-r) -e 2ia 0. All solutions to z 1 z 2 z 3 1 0 , z i 1 , must be of the form z 1,z 2,z 3 -1,z ,-z . Hence, Z if and only if one of the following holds: itemize e 2ia -1 and e 2ir e 2i(a 1-r) 0 , hence Z 1 ; e 2ir 1 and e 2i(1-r) 1 , hence Z 2 ; e 2i(a 1-r) 1 and e 2ia e 2ir , hence Z 3 . itemize proof Observe that Z 2 , Z 3 are additive subgroups of . lemme At least one of the following holds: gather Z 1Z 2, s.e1 Z 1Z 3. s.e2 gather s.lemma2 lemme -.75pc proof By Lemma s.lemma1 , -Z Z 1Z 2Z 3 . If Z 2Z 3 , ( s.e2 ) holds. Suppose therefore that there is a iZ 2Z 3 . It suffices to prove that for any jZ 3 we must have jZ 1 or jZ 2 . Let jZ 3 , then ij i- jZ by orthogonality. By Lemma s.lemma1 , ij Z 1Z 2Z 3 . If ij Z 2 , then jZ 2 and we are done, and if ij Z 3 , then iZ 3 , which contradicts the above assumption on i . Therefore assume that ij Z 1 . Then ij , ij a , ij (2r-1), hence 2 j r 2 i r- ij (2r-1)- ij . If jr , then jZ 2 ; if j r , then j a by the definition of Z 3 and j(2r-1) , so that jZ 1 . proof lemme (i) Z 2 is not possible; (ii) Z 3 is possible only if a-r and Z 3 . s.lemma3 lemme proof Suppose that Z i for i 2 or 3 . Since Z i is an additive subgroup of , we must have Z i p for some integer p 0 . Furthermore, if there was a p , we would have k-p and e 2ix would be orthogonal to e 2i k x for all k , which would contradict ( a.e00 ). Hence, Z i p . We also observe that, if p were 2 , any function of the form f(x) k c ke 2i k x would be periodic with period 1p , which again would contradict ( a.e00 ). Thus Z i . If i 2 , this is not possible, since nr cannot be an integer for all n if r . If i 3 , we obtain that n(a-r) for all n ; letting n 1 , we find that a-r . proof If , are as in Lemma s.lemma3 (ii) , then Theorem s.thm1 (i) is satisfied and we are done. Thus we may assume throughout the sequel that equation Z 2, Z 3. s.e5 equation lemme If s.e5 holds, then Z 1(Z 2Z 3) . s.lemma4 lemme proof By Lemma s.lemma2 , it suffices to prove that equation if (Z 1Z 2), then Z 2Z 3; s.e6 equation equation if (Z 1Z 3), then Z 3Z 2; s.e7 equation We will only prove ( s.e6 ), since the proof of ( s.e7 ) is almost identical. Suppose that iZ 1Z 2 , and let jZ 2 . By Lemma s.lemma1 , ij belongs to at least one of Z 1 , Z 2 , Z 3 . We may not have ij Z 2 , since then i would also be in Z 2 . Thus we only need consider the following two cases. itemize Let ij Z 1 . Then i a, ij a , hence j a and jZ 2Z 3 . Assume now that ij Z 3 . Then i , hence 2 i r . Moreover, i r would imply iZ 2 , hence i r . It follows that i(a-r) ; since also ij (a-r) , we obtain that j(a-r) and jZ 2Z 3 . itemize proof lemme Assume s.e5 . Then: (i) -Z 1(Z 2Z 3) ; (ii) Z 1 r -1 for some . s.lemma5 lemme proof For k , let k - k jk : j . Then k is also a spectrum for , 0 k , hence all of the results obtained so far apply with replaced by k . Thus (i) follows from Lemmas s.lemma3 and s.lemma4 . To prove (ii) , it suffices to verify that ij r whenever i, jZ 1 . Indeed, if i, jZ 1 , then ij a , hence ij Z 1 and, by (i) , ij Z 2Z 3 . But this implies that ij r . proof Completeness Fix j,n , and consider the function defined by ( s.e11 ) with j-nr -1 . The Fourier coefficients of are c k 0 r e 2i(- k)x dx 0 r e 2i( jk -nr -1 )x dx, hence c k r if jk nr -1 , and equation c k 1 2i( jk -nr -1 ) (e 2i( jk r-n) -1), jk nr -1 . c.e1 equation Define jk jk r . Plugging ( c.e1 ) into ( s.e10 ), we obtain that for all j , equation 1r 1 k: jk 1 4 2 jk 2 e 2i jk -1 2, c.e2 equation and for all n,j , equation 1r n,j k: jk 1 4 2 ( jk -n) 2 e 2i( jk -n) -1 2, c.e3 equation where n,j 1 if there is a k such that jk n , and n,j 0 otherwise. We define the equivalence relation between the indices k,k' : kk' kk' , and denote by A 1,A 2,,A m, the (non-empty and disjoint) equivalence classes with respect to this relation. Hence k,k' belong to the same A m if and only if kk' ; in particular, A m: k: kA m m for some m 0,1). lemme Let M denote the number of distinct and non-empty A m 's. Then equation Mr -1 . c.e4 equation Moreover, if one of the A m 's skips a number (i.e. A m m ), then Mr -1 1 . c.lemma1 lemme proof For each m,m' , let mm' m- m' ; note that mm' 0 if mm' . Fix m' and jA m' , then ( c.e2 ) may be rewritten as equation 1r 1 mm' S mm' , c.e5 equation where S mm' kA m 1 4 2 jk 2 e 2i mm' -1 2. Clearly, equation S mm' S( mm' ), c.e7 equation where equation S() k 1 4 2( k) 2 e 2i -1 2. c.e6 equation Hence ( c.e4 ) follows from ( c.e5 ) and Lemma c.lemma2 below. Suppose now that A m' skips a number. Then we may find jA m' and n such that n,j 0 , and ( c.e4 ) may be improved to M1 r -1 by using ( c.e3 ) instead of ( c.e2 ). proof lemme Let S() be as in c.e6 , then S() 1 for all 0 1 . c.lemma2 lemme proof By Proposition tis.prop2 , 2( 2) , where n and p is an odd integer, is a spectrum for (0,)(2, n 1 2) . Plugging this back into ( c.e2 ) we obtain that: 1 k 1 4 2( k) 2 e 2i -1 2 for 2n . However, the set of of this form is dense in , hence by continuity the lemma holds for all (0,1) . proof Conclusion proof Proof of Theorem s.thm1 If is as in Lemma s.lemma3 (ii) , then Theorem s.thm1 (i) is satisfied. We may therefore assume that ( s.e5 ) holds. From Lemma s.lemma5 we have: equation -Z 1(Z 2Z 3), Z 2Z 3r -1 , Z 1( r -1 ), s.e15 equation for some , hence M2 (using the notation in Section ). However, by Lemma c.lemma1 Mr -1 2 , and this may be improved to M3 if one of the A m 's skips a number. Therefore we must have r , and equation - 2( 2), Z 2Z 3 2, Z 1 2. s.e16 equation Pick ij , kl Z 1 such that ij - kl 2 . From the definition of Z 1 we have ij a, kl a , hence 2a ij a- kl a, so that a 2 for some n . Finally, we have a n , hence n p for some odd integer p . Thus and satisfy (ii) of Theorem s.thm1 . proof