Commutators on l. D. Dosev and W. B. Johnson

Transcription

1 Submitted exclusively to the London Mathematical Society doi: /0000/ Commutators on l D. Dosev and W. B. Johnson Abstract The oerators on l which are commutators are those not of the form λi + S with λ 0 and S strictly singular. 1. Introduction The commutator of two elements A and B in a Banach algebra is given by [A, B] = AB BA. A natural roblem that arises in the study of derivations on a Banach algebra A is to classify the commutators in the algebra. Using a result of Wintner([18], who roved that the identity in a unital Banach algebra is not a commutator, with no effort one can also show that no oerator of the form λi + K, where K belongs to a norm closed ideal I(X of L(X and λ 0, is a commutator in the Banach algebra L(X of all bounded linear oerators on the Banach sace X. The latter fact can be easily seen just by observing that the quotient algebra L(X /I(X also satisfies the conditions of Wintner s theorem. In 1965 Brown and Pearcy ([5] made a breakthrough by roving that the only oerators on l 2 that are not commutators are the ones of the form λi + K, where K is comact and λ 0. Their result suggests what the classification on the other classical sequence saces might be, and, in 1972, Aostol ([3] roved that every non-commutator on the sace l for 1 < < is of the form λi + K, where K is comact and λ 0. One year later he roved that the same classification holds in the case of X = c 0 ([4]. Aostol roved some artial results on l 1, but only 30 year later was the same classification roved for X = l 1 by the first author ([6]. Note that if X = l (1 < or X = c 0, the ideal of comact oerators K(X is the largest roer ideal in L(X ([8], see also [17, Theorem 6.2]. The classification of the commutators on l, 1 <, and artial results on other saces suggest the following Conjecture 1. Let X be a Banach sace such that X ( X, 1 or = 0 (we say that such a sace admits a Pe lczyński decomosition. Assume that L(X has a largest ideal M. Then every non-commutator on X has the form λi + K, where K M and λ 0. In [3] Aostol obtained a artial result regarding the commutators on l. He roved that if T L(l and there exists a sequence of rojections (P n n=1 on l such that P n (l l for n = 1, 2,... and P n T 0 as n, then T is a commutator. This condition is clearly 2000 Mathematics Subject Classification 47B47 (rimary, 47L20 (secondary. Research for both authors is suorted in art by NSF grant DMS This article is art of the first author s doctoral dissertation, which is being reared at Texas A&M University under the direction of the second author.

2 Page 2 of 15 D. DOSEV AND W. B. JOHNSON satisfied if T is a comact oerator, but, as the first author showed in [6], it is also satisfied if T is strictly singular, which is an essential ste for roving the conjecture for l. In order to give a ositive answer to the conjecture one has to rove Every oerator T M is a commutator If T L(X is not of the form λi + K, where K M and λ 0, then T is a commutator. In this aer we will give ositive answer to this conjecture for the sace l. The authors would like to thank the referee and Dr. Niels Jakob Lausten for their comments and suggestions that hel imrove the manuscrit. 2. Notation and basic results For a Banach sace X denote by the L(X, K(X, C(X and S X the sace of all bounded linear oerators, the ideal of comact oerators, the set of all finite co-dimensional subsaces of X and the unit shere of X. By ideal we always mean closed, non-zero, roer ideal. A ma from a Banach sace X to a Banach sace Y is said to be strictly singular if whenever the restriction of T to a subsace M of X has a continuous inverse, M is finite dimensional. In the case where X Y, the set of strictly singular oerators forms an ideal which we will denote by S(X. Recall that for X = l, 1 <, S(X = K(X ([8] and on l the ideals of strictly singular and weakly comact oerators coincide ([1, Theorem 5.5.1]. A Banach sace X is called rime if each infinite-dimensional comlemented subsace of X is isomorhic to X. The saces l, 1, are all rime (cf. [13, Theorem 2.a.3 and Theorem 2.a.7]. We say that a linear oerator between two Banach saces T : X Y is an isomorhism if T is injective bounded linear ma. If in addition T is surjective we will say that T is an onto isomorhism. For any two subsaces (ossibly not closed X and Y of a Banach sace Z let d(x, Y = inf{ x y : x S X, y Y }. A well known consequence of the oen maing theorem is that for any two closed subsaces X and Y of Z, d(x, Y > 0 if and only if X Y = {0} and X + Y is a closed subsace of Z. Note also that d(x, Y = 0 if and only if d(y, X = 0. First we rove a roosition that will later allow us to consider translations of an oerator T by a multile of the identity instead of the oerator T itself. Proosition 2.1. Let X be a Banach sace and T L(X be such that there exists a subsace Y X for which T is an isomorhism on Y and d(y, T Y > 0. Then for every λ C, (T λi Y is an isomorhism and d(y, (T λiy > 0. Proof. First, note that the two hyotheses on Y (that T is an isomorhism on Y and d(y, T Y > 0 are together equivalent to the existence of a constant c > 0 s.t. for all y S Y, d(t y, Y > c. To see this, let us first assume that the hyotheses of the theorem are satisfied. Then there exists a constant C such that T y C for every y S Y. For an arbitrary y S Y, let z y = and then we clearly have T y T y d(t y, Y = T y d(z y, Y Cd(T Y, Y =: c > 0. To show the other direction note that for y S Y, 0 < c < d(t y, Y = T y d(z y, Y T d(z y, Y. Taking the infimum over all z y S T Y in the last inequality, we obtain that d(t Y, Y > 0 and hence d(y, T Y > 0. On the other hand, for all y S Y we have 0 < c < d(t y, Y T y c 2 y T y + c 2, hence T y c 2, which in turn imlies that T is an isomorhism on Y.

3 COMMUTATORS ON l Page 3 of 15 Now it is easy to finish the roof. The condition d(t y, Y > c for all y S Y is clearly satisfied if we substitute T with T λi since for a fixed y S Y, d((t λiy, Y = inf (T λiy z = inf T y z = d(t y, Y, z Y z Y hence (T λi Y is an isomorhism and d(y, (T λiy > 0. Note the following two simle facts: If T : X X is a commutator on X and S : X Y is an onto isomorhism, then ST S 1 is a commutator on Y. Let T : X X be such that there exists X 1 X for which T X1 is an isomorhism and d(x 1, T X 1 > 0. If S : X Y is an onto isomorhism, then there exists Y 1 Y, Y 1 X 1, such that ST S 1 Y 1 is an isomorhism and d(y 1, ST S 1 Y 1 > 0 (in fact Y 1 = SX 1. Note also that if X 1 is comlemented in X, then Y 1 is comlemented in Y. Using the two facts above, sometimes we will relace an oerator T by an oerator T 1 which is similar to T and ossibly acts on another Banach sace. If {Y i } i=0 is a sequence of arbitrary Banach saces, by ( i=0 Y i we denote the sace of all sequences {y i } i=0 where y i Y i, i = 0, 1,..., such that ( y i Yi l with the norm (y i = y i Yi (if Y i Y for every i = 0, 1,... we will use the notation ( Y. We will only consider the case where all the saces Y i, i = 0, 1..., are uniformly isomorhic to a Banach sace Y, that is, there exists a constant λ > 0 and sequence of onto isomorhisms {T i : Y i Y } i=0 such that T 1 = 1 and T λ. In this case we define an onto isomorhism U : ( i=0 Y i ( Y via (T i by U(y 0, y 1,... = (T 0 (y 0, T 1 (y 1,..., (2.1 ( and it is easy to see that U λ and U 1 = 1. Sometimes we will identify the sace i=0 Y i with ( Y via the isomorhism U when there is no ambiguity how the roerties of an oerator T on ( i=0 Y i translate to the roerties of the oerator UT U 1 on ( Y. For y = (y i ( Y, y i Y, define the following two oerators : R(y = (0, y 0, y 1,..., L(y = (y 1, y 2,.... The oerators L and R are, resectively, the left and the right shift on the sace ( Y. Denote by P i, i = 0, 1,..., the natural, norm one, rojection from ( Y onto the i-th comonent of ( Y, which we denote by Y i. We should note that if Y ( Y, then some of the results in this aer are similar to results in [6], but initially we do not require this condition, and, in articular, some of the results we rove here have alications to saces like ( l q for arbitrary 1, q. Our first roosition shows some basic roerties of the left and the right shift as well as the fact that all the owers of L and R are uniformly bounded, which will lay an imortant role in the sequel. Since the roof follows immediately from the definitions we will omit it. Proosition 2.2. Consider the Banach sace ( Y. We have the following identities L n = 1, R n = 1 for every n = 1, 2,... (2.2 LP 0 = P 0 R = 0, LR = I, RL = I P 0, RP i = P i+1 R, P i L = LP i+1 for i 0. (2.3 Note that we can define a left and right shift on ( i=0 Y i by L = U 1 LU and R = U 1 RU, and, using the above roosition, we immediately have R n λ and L n λ. If there is no ambiguity, we will denote the left and the right shift on ( i=0 Y i simly by L and R.

4 Page 4 of 15 D. DOSEV AND W. B. JOHNSON Following the ideas in [3], for 1 < and = 0 define the set and for T A define ( ( A = {T L Y : T A = R n T L n is strongly convergent}, (2.4 R n T L n. Now using the fact that an oerator T is a commutator if and only if T is in the range of D S for some S, where D S is the inner derivation determined by S, defined by D S (T = ST T S, it is easy to see ([6, Lemma 3] that if T A then hence T is a commutator. T = D L (RT A = D R (T A L, ( Commutators on ( Y The ideas in this section are similar to the ideas in [6], but here we resent them from a different oint of view, in a more general setting and we also include the case =. The following lemma is a generalization of [3, Lemma 2.8] in the case = and [6, Corollary 7] in the case 1 < and = 0. The roof resented here follows the ideas of the roof in [6]. Of course, some of the ideas can be traced back to the classic aer of Brown and Pearcy ([5] and to Aostol s aers [3], [4], and the references therein. Lemma 3.1. Let T L (( Y. Then the oerators P0 T and T P 0 are commutators. Proof. The roof shows that P 0 T is in the range of D L and T P 0 is in the range of D R. We will consider two cases deending on. Case I : = In this case we first observe that the series S 0 = R n P 0 T L n is ointwise convergent coordinatewise. Indeed, let x ( Y and define y n = R n P 0 T L n x for n = 0, 1,.... Note that from the definition we immediately have y n Y n so the sum y n converges in the roduct toology on ( Y to a oint in ( Y since y n R n P 0 T L n x T x. Secondly, we observe that S 0 and L commute. Because L and R are continuous oerators on ( Y with the roduct toology and LR = I, we have ( ( S 0 Lx = R n P 0 T L n+1 x = L R n P 0 T L n x = L R n P 0 T L n x LP 0 T x (3.1 = LS 0 x 0 n=1 since LP 0 = 0. That is, D L S 0 = 0, as desired.

5 COMMUTATORS ON l Page 5 of 15 On the other hand, again using LP 0 = 0, (I RLS 0 x = (I RLR n P 0 T L n x = (I RLP 0 T x + (I RLR n P 0 T L n x Therefore = (I RLP 0 T x = P 0 T x. n=1 } {{ } 0 (3.2 D L (RS 0 = (D L RS 0 + R(D L S 0 = (I RLS = P 0 T. (3.3 The roof of the statement that T P 0 is a commutator involves a similar modification of the roof of [3, Lemma 2.8]. Again, consider the series S = R n P 0 T P 0 L n. This is ointwise convergent coordinatewise and SL = LS (from the above reasoning alied to the oerator T P 0, and Now it is easy to see that D R ( SL = D R (LS = RLS + LSR = (I P 0 S + LSR = S + P 0 S + SLR = S + P 0 S + S = P 0 T P 0. D R (LT P 0 SL = RLT P 0 LT P 0 R +P }{{} 0 T P 0 = (I P 0 T P 0 + P 0 T P 0 = T P 0. 0 Case II : 1 < or = 0 In this case the roof is similar to the roof of [6, Lemma 6 and Corollary 7] and we include it for comleteness. Let us consider the case 1 first. For any y ( Y we have Since m+r m+r R n P i T P j L n y = R n P i T P j L n P j+n y = m+r m+r P i T P j P j+n y P i T P j P j+n y 0 as m we have that P i T P j A. For = 0 a similar calculation shows m+r m+r R n P i T P j L n y = P i T P j R n P i T P j L n P j+n y P j+n y. R n P i T P j L n is strongly convergent and R n P i T P j L n P j+n y = max m n m+r P j+ny max m n m+r Rn P i T P j L n P j+n y and since max P j+ny 0 as m we aly the same argument as in the case 1 m n m+r to obtain P i T P j A. Using P i T P j A for i = j = 0 and (2.5 we have P 0 T P 0 = D L (R(P 0 T P 0 A = D R ((P 0 T P 0 A L. Again, as in [6, Corollary 7], via direct comutation we obtain T P 0 = D R (LT P 0 (P 0 T P 0 A L (3.4 P 0 T = D L ( P 0 T R + R(P 0 T P 0 A. (3.5

6 Page 6 of 15 D. DOSEV AND W. B. JOHNSON Now we switch our attention to Banach saces which in addition satisfy X ( X for some 1 or = 0. Note that the Banach sace ( Y satisfies this condition regardless of the sace Y, hence we will be able to use the results we roved so far in this section. We begin with a definition. Definition 1. Let X be a Banach sace such that X ( X, 1 or = 0. We say that D = {X i } i=0 is a decomosition of X if it forms an l or c 0 decomosition of X into subsaces which are uniformly isomorhic to X ; that is, if the following three conditions are satisfied: There are uniformly bounded rojections P i on X with P i X = X i and P i P j = 0 for i, j = 0, 1,... and i j There exists a collection of isomorhisms ψ i : X i X, i N 0 (we denote N 0 = N {0}, such that ψ 1 i = 1 and λ = su ψ i < i N 0 The formula Sx = (ψ i P i x defines a surjective isomorhism from X onto ( X If D = {X i } i=0 is a decomosition of X we have X ( X ( i=0 X i, where the second isomorhic relation is via the isomorhism U defined in (2.1. Using this simle observation we will often identify X with ( i=0 X i. Our next theorem is similar to [6, Theorem 16] and [3, Theorem 4.6], but we state it and rove it in a more general setting and also include the case =. Theorem 3.2. Let X be a Banach sace such that X ( X, 1 or = 0. Let T L(X be such that there exists a subsace X X such that X X, T X is an isomorhism, X + T (X is comlemented in X and d(x, T (X > 0. Then there exists a decomosition D of X such that T is similar to a matrix oerator of the form ( L on X X, where L is the left shift associated with D. Proof. Clearly X = X T (X Z where Z is comlemented in X. Note that without loss of generality we can assume that Z is isomorhic to X. Indeed, if this is not the case, let X = X 1 X 2, X X 1 X 2 and X 1, X 2 comlemented in X (hence also comlemented in X. Then d(x 1, T (X 1 > 0 and X = X 1 T (X 1 Z 1 where Z 1 is a comlemented subsace of X, which contains the subsace X 2 X, such that X 2 is isomorhic to X and comlemented in Z. Alying the Pe lczýnski decomosition technique ([14, Proosition 4], we conclude that Z 1 is isomorhic to X. This observation lays an imortant role and will allow us to construct the decomositions we need during the rest of the roof. Denote by I P the rojection onto T (X with kernel X + Z. Consider two decomositions D 1 = {X i } i=0, D 2 = {Y i } i=0 of X such that T (X = Y 0 = X 1 X 2, X 0 = Y 1 Y 2, Y 1 = X, and Z = Y 2 Y 3. Define a ma S Sϕ = L D1 ϕ L D2 ϕ, ϕ X from X to X X. The ma S is invertible (S 1 (a, b = R D1 a + R D2 b. Just using the definition of S and the formula for S 1 we see that

7 COMMUTATORS ON l Page 7 of 15 hence Let ST S 1 (a, b = ST (R D1 a + R D2 b = S(T R D1 a + T R D2 b = (L D1 T R D1 a + L D1 T R D2 b (L D2 T R D1 a + L D2 T R D2 b, ( ST S 1 LD1 T R = D2. A = P Y0 T R D2 = (I P T R D2 (3.6 and note that A PY0 X A (I P X : (I P X (I P X is onto and invertible since R D2 is an isomorhism on P Y0 X and R D2 (P Y0 X = Y 1 = X. Here we used the fact that P Y0 T is an isomorhism on X (P X = X. Denote by T 0 : (I P X (I P X the inverse of A PY0 X (note that T 0 is an automorhism on (I P X and consider G : X X defined by G = I + T 0 (I P T 0 A. We will show that G 1 = A + P. In fact, from the definitions of A and T 0 it is clear that AT 0 (I P = T 0 A(I P = I P, P T 0 (I P = P A = 0, (I P A = A (3.7 and since A mas onto (I P X and AT 0 = I (I P X we also have Now using (3.7 and (3.8 we comute A AT 0 A = 0. (3.8 (A + P G = (A + P (I + T 0 (I P T 0 A = A + AT 0 (I P AT 0 A + P = I P + P = I G(A + P = (I + T 0 (I P T 0 A(A + P = A + P + T 0 (I P A + T 0 (I P P T 0 AA T 0 AP = A + P + T 0 A T 0 AA T 0 AP = P + (I T 0 AA + T 0 A(I P = P + (I T 0 A(I P A + (I P = I + ((I P T 0 A(I P A = I + (I P (I P A = I. Using a similarity we obtain ( ( I 0 LD1 T R D2 0 G 1 ( I 0 0 G ( LD1 T R = D2 G It is clear that we will be done if we show that L D1 = L D1 T R D2 G. In order to do this consider the equation (A + P G = I AG + P G = I. Multilying both sides of the last equation on the left by L D1 gives us L D1 AG + L D1 P G = L D1. Using L D1 P L D1 P X0 = 0 we obtain L D1 AG = L D1. Finally, substituting A from (3.6 in the last equation yields L D1 = L D1 AG = L D1 P Y0 T R D2 G = L D1 (I P X0 T R D2 G = L D1 T R D2 G, which finishes the roof. The following theorem was roved in [3] for X = l, 1 < <, but inessential modifications give the result in these general settings..

8 Page 8 of 15 D. DOSEV AND W. B. JOHNSON Theorem 3.3. Let X be a Banach sace such that X ( X. Let D be a decomosition of X and let L be the left shift associated with it. Then the matrix oerator ( T1 L acting on X X is a commutator. T 2 T 3 Proof. Let D = {X i } be the given decomosition. Consider a decomosition D 1 = {Y i } such that Y 0 = X i and X 0 = Y i. Now there exists an oerator G such that i=1 i=1 D LD G = R D1 L D1 (T 1 + T 3. This can be done using Lemma 3.1, since R D1 L D1 = I P Y0 = P X0. Note that we have T 1 + T 3 LG + GL = T 1 + T 3 D L G = T 1 + T 3 R D1 L D1 (T 1 + T 3 = P Y0 (T 1 + T 3, and using Lemma 3.1 again, we deduce that T 1 + T 3 LG + GL is a commutator. Thus by making the similarity ( I 0 T := G I ( T1 L ( T 2 T 3 G I I 0 ( T1 LG L = T 3 + GL and relacing T by T we can assume that T 1 + T 3 is a commutator, say T 1 + T 3 = AB BA and A < 1/(2 R (this can be done by scaling. Denote by M S left multilication by an oerator S. Then M R D A < 1 where R is the right shift associated with D. The oerator T 0 = (M I M R D A 1 M R (T 3 B T 2 is well defined and it is easy to see that ( A 0 A L T 3 This finishes the roof. ( B I T 0 0 ( B I T 0 0 ( A 0 = A L T 3 ( T1 L. T 2 T 3 4. Oerators on l Definition 2. The left essential sectrum of T L(X is the set ([2] Def 1.1 σ l.e. (T = {λ C : inf x S Y (λ T x = 0 for all Y X s.t. codim(y < }. Aostol [2, Theorem 1.4] roved that for any T L(X, σ l.e. (T is a closed non-void set. The following lemma is a characterization of the oerators not of the form λi + K on the classical Banach sequence saces. The roof resented here follows Aostol s ideas [3, Lemma 4.1], but it is resented in a more general way. Lemma 4.1. Let X be a Banach sace isomorhic to l for 1 < or c 0 and let T L(X. Then the following are equivalent (1 T λi is not a comact oerator for any λ C. (2 There exists an infinite dimensional comlemented subsace Y X such that Y X, T Y is an isomorhism and d(y, T (Y > 0. Proof. (2 = (1 Assume that T = λi + K for some λ C and some K K(X. Clearly λ 0 since T Y is an isomorhism. Now there exists a sequence {x i } i=1 S Y such that Kx n 0 as n. Let

9 COMMUTATORS ON l Page 9 of 15 ( xn y n = T and note that λ ( xn ( xn xn Kx n x n y n = x n (λi + K = x n K = 0 λ λ λ as n which contradicts the assumtion d(y, T (Y > 0. Thus T λi is not a comact oerator for any λ C. (1 = (2 The roof in this direction follows the ideas of the roof of Lemma 4.1 from [3]. Let λ σ l.e. (T. Then T 1 = T λi is not a comact oerator and 0 σ l.e (T 1. Using just the definition of the left essential sectrum, we find a normalized block basis sequence {x i } i=1 of the standard unit vector basis of X such that T 1 x n < 1 2 n for n = 1, 2,.... Thus if we denote Z = san{x i : i = 1, 2,...} we have Z X and T 1 Z is a comact oerator. Let I P be a bounded rojection from X onto Z ([14, Lemma 1] so that T 1 (I P is comact. Now consider the oerator T 2 = (I P T 1 P. We have two ossibilities: Case I. Assume that T 2 = (I P T 1 P is not a comact oerator. Then there exists an infinite dimensional subsace Y 1 P X on which T 2 is an isomorhism and hence using [14, Lemma 2] if necessary, we find a comlemented subsace Y P X, such that T 2 is an isomorhism on Y. By the construction of the oerator T 2 we immediately have d(y, (I P T 1 P (Y > 0 and hence d(y, T 1 (Y > 0. Note that since X is rime and Y is comlemented in X, Y X is automatic. Now we are in osition to use Proosition 2.1 to conclude that d(y, T (Y > 0. Case II. Now we can assume that the oerator (I P T 1 P is comact. Since T 1 (I P is comact and using T 1 = T 1 (I P + (I P T 1 P + P T 1 P we conclude that the oerator P T 1 P is not comact. Using X P X (I P X, we identify P X (I P X with X X via an onto isomorhism U, such that U mas P X onto the first coy of X in the sum X X. Without loss of generality we assume that T 1 = ( T 11 T 12 T 21 T 22 is acting on X X. Denote by P = ( I the rojection from X X onto the first coy of X. In the new settings, we have that T 11 is not comact and T 21, T 22 and T 12 are comact oerators. Define the oerator S on X X in the following way: ( I I 2S =. I I Clearly S 2 = I hence S = S 1. Now consider the oerator 2(I P S 1 T 1 SP. A simle calculation shows that ( 2(I P S T 1 SP = T 11 + T 12 T 21 T 22 0 hence (I P S 1 T 1 SP is not comact. Now we can continue as in the revious case to conclude that there exists a comlemented subsace Y X in the first coy of X X for which d(y, S 1 T 1 S(Y > 0 and hence d(sy, T 1 (SY > 0. Again using Proosition 2.1, we conclude that d(sy, T (SY > 0. Remark 1. We should note that the two conditions in the receding lemma are equivalent to a third one, which is the same as (2 lus the additional condition that Y T (Y is comlemented in X. This is essentially what was used for roving the comlete classification of the commutators on l 1 in [6], and l, 1 < <, and c 0 in [3] and [4]. The last mentioned condition will also lay an imortant role in the roof of the comlete classification of the commutators on l, but we should oint out that once we have an infinite dimensional subsace Y l such that Y l, T Y is an isomorhism and d(y, T (Y > 0, then Y and Y T (Y will be automatically comlemented in l.

10 Page 10 of 15 D. DOSEV AND W. B. JOHNSON Lemma 4.2. Let T L(l and denote by I the identity oerator on l. Then the following are equivalent (a For each subsace X l, X c 0, there exists a constant λ X and a comact oerator K X : X l deending on X such that T X = λ X I X + K X. (b There exists a constant λ such that T = λi + S, where S S(l. Proof. Clearly (b imlies (a, since every strictly singular oerator from c 0 to any Banach sace is comact ([1, Theorem ]. For roving the other direction we will first show that for every two subsaces X, Y such that X Y c 0 we have λ X = λ Y. We have several cases. Case I. X Y = {0}, d(x, Y > 0. Let {x i } i=1 and {y i } i=1 be bases for X and Y, resectively, which are equivalent to the usual unit vector basis of c 0. Consider the sequence {z i } i=1 such that z 2i = x i, z 2i 1 = y i for i = 1, 2,.... If we denote Z = san{z i : i = 1, 2,...}, then clearly Z c 0, and, using the assumtion of the lemma, we have that T Z = λ Z I Z + K Z. Now using X Z we have that λ X I X + K X = (λ Z I Z + K Z X, hence (λ X λ Z I X = (K Z X K X. The last equation is only ossible if λ X = λ Z since the identity is never a comact oerator on a infinite dimensional subsace. Similarly λ Y = λ Z and hence λ X = λ Y. Case II. X Y = {0}, d(x, Y = 0. Again let {x i } i=1 and {y i } i=1 be bases of X and Y, resectively, which are equivalent to the usual unit vector basis of c 0 and assume also that λ X λ Y. There exists a normalized block basis {u i } i=1 of {x i } i=1 and a normalized block basis {v i } i=1 of {y i } i=1 such that u i v i < 1 i. Then u i v i 0 T u i T v i 0 λ X u i + K X u i λ Y v i K Y v i 0. Since u i 0 weakly (as a bounded block basis of the standard unit vector basis of c 0 we have K X u i 0 and using u i v i 0 we conclude that (λ X λ Y v i K Y v i 0. Then there exists N N such that K Y v i > λ X λ Y 2 v i for i > N, which is imossible because K Y is a comact oerator. Thus, in this case we also have λ X = λ Y. Case III. X Y = Z {0}, dim(z =. In this case we have (λ X I X + K X Z = (λ Y I Y + K Y Z and, as in the first case, we rewrite the receding equation in the form (λ X I X λ Y I Y Z = (K Y K X Z. Again, as in Case I, the last equation is only ossible if λ X = λ Y since the identity is never a comact oerator on a infinite dimensional subsace. Case IV. X Y = Z {0}, dim(z <. Let X = Z X 1 and Y = Z Y 1. Then X 1 Y 1 = {0}, X 1 Y 1 c 0 and we can reduce to one of the revious cases. Let us denote S = T λi where λ = λ X for arbitrary X l, X c 0. If S is not a strictly singular oerator, then there is a subsace Z l, Z l such that S Z is an isomorhism ([16, Corollary 1.4], hence we can find Z 1 Z l, Z 1 c 0, such that S Z1 is an isomorhism. This contradicts the assumtion that S Z1 is a comact oerator. The following corollary is an immediate consequence of Lemma 4.2.

11 COMMUTATORS ON l Page 11 of 15 Corollary 4.3. Suose T L(l is such that T λi / S(l for any λ C. Then there exists a subsace X l, X c 0 such that (T λi X is not a comact oerator for any λ C. Theorem 4.4. Let T L(l be such that T λi / S(l for any λ. Then there exists a subsace X l such that X c 0, T X is an isomorhism and d(x, T (X > 0. Proof. By Corollary 4.3 we have a subsace X l, X c 0 such that (T λi X is not a comact oerator for any λ. Let Z = X T (X and let P be a rojection from Z onto X (such exists since Z is searable and X c 0. We have two cases: Case I. The oerator T 1 = (I P T P is not comact. Since T 1 is a non-comact oerator from X c 0 into a Banach sace we have that T 1 is an isomorhism on some subsace Y X, Y c 0 ([1, Theorem ]. Clearly, from the form of the oerator T 1 we have d(y, T 1 (Y = d(y, (I P T P (Y > 0 and hence d(y, T (Y > 0. Case II. If (I P T P is comact and λ C, then (I P T P + P T P λi Z = T P λi Z is not comact and hence P T P λi Z is not comact. Now for T 2 := P T P : X X we aly Lemma 4.1 to conclude that there exists a subsace Y X, Y c 0 such that d(y, P T (Y = d(y, P T P (Y > 0 and hence d(y, T (Y > 0. The following theorem is an analog of Lemma 4.1 for the sace l. Theorem 4.5. Let T L(l be such that T λi / S(l for any λ C. Then there exists a subsace X l such that X l, T X is an isomorhism and d(x, T (X > 0. Proof. From Theorem 4.4 we have a subsace Y l, Y c 0 such that T Y is an isomorhism and d(y, T (Y > 0. Let N k = {3i + k : i = 0, 1,...} for k = 1, 2, 3. There exists an isomorhism S : Y T Y c 0 (N 1 c 0 (N 2 such that S(Y = c 0 (N 1 and S(T Y = c 0 (N 2. Note that the sace Y T Y is indeed a closed subsace of l due to the fact that d(y, T (Y > 0. Now we use [12, Theorem 3] to extend S to an automorhism S on l. Let T 1 = ST S 1 and consider the oerator (P N2 T 1 l (N 1 : l (N 1 l (N 2, where P N2 is the natural rojection onto l (N 2. Since T 1 (c 0 (N 1 = c 0 (N 2, by [16, Proosition 1.2] there exists an infinite set M N 1 such that (P N2 T 1 l (M is an isomorhism. This immediately yields d(l (M, P N2 T 1 (l (M > 0. If x l (M, x = 1 and y l (M is arbitrary, then x T y = max( x P M T 1 y, P M ct 1 y max( x P M T 1 y, P N2 T 1 y If y < 1 2 T then x P 1 M T 1 y 1 2. Otherwise P 1 N 2 T 1 y 2 T (P N2 T 1 1 where the norm of the inverse of P N2 T 1 in the receding inequality is taken considering P N2 T 1 as an oerator from l (M to P N2 T 1 (l (M. Now it is clear that ( 1 x T y max 2, 1 2 T (P N2 T 1 1 for all x l (M, x = 1 and y l (M hence Finally, recall that T 1 = ST S 1, thus d(l (M, T 1 (l (M > 0. (4.1 d(l (M, ST S 1 (l (M > 0

12 Page 12 of 15 D. DOSEV AND W. B. JOHNSON and hence d(s 1 (l (M, T S 1 (l (M > 0. Finally, we can rove our main result. Theorem 4.6. any λ 0. An oerator T L(l is a commutator if and only if T λi / S(l for Proof. Note first that if T is a commutator, from the remarks we made in the introduction it follows that T λi cannot be strictly singular for any λ 0. For roving the other direction we have to consider two cases: Case I. If T S(l (λ = 0, the statement of the theorem follows from [6, Theorem 23]. Case II. If T λi / S(l for any λ C, then we aly Theorem 4.5 to get X l such that X l, T X an isomorhism and d(x, T X > 0. The subsace X + T X is isomorhic to l and thus( is comlemented in l. Theorem 3.2 now yields that T is similar to an oerator L of the form. Finally, we aly Theorem 3.3 to comlete the roof. 5. Remarks and roblems We end this note with some comments and questions that arise from our work. First consider the set M X = {T L(X : I X does not factor through T }. This set comes naturally from our investigation of the commutators on l for 1. We know ([6, Theorem 18], [3, Theorem 4.8], [4, Theorem 2.6] that the non-commutators on l, 1 < and c 0 have the form λi + K where K M X and λ 0, where M X = K(l is actually the largest ideal in L(l ([8], and, in this aer we showed (Theorem 4.6 that the non-commutators on l have the form λi + S where S M X and λ 0, where M X = S(l. Thus, it is natural to ask the question for which Banach saces X is the set M X the largest ideal in L(X? Let us also mention that in addition to the already mentioned saces, if X = L (0, 1, 1 <, then M X is again the largest ideal in L(X (cf. [7] for the case = 1 and [9, Proosition 9.11] for > 1. First note that the set M X is closed under left and right multilication with oerators from L(X, so the question whether M X is an ideal is equivalent to the question whether M X is closed under addition. Note also that if M X is an ideal then it is automatically the largest ideal in L(X and hence closed, so the question we will consider is under what conditions we have M X + M X M X. (5.1 The following roosition gives a sufficient condition for (5.1 to hold. Proosition 5.1. Let X be a Banach sace such that for every T L(X we have T / M X or I T / M X. Then M X is the largest (hence closed ideal in L(X.

13 COMMUTATORS ON l Page 13 of 15 Proof. Let S, T M X and assume that S + T / M X. By our assumtion, there exist two oerators U : X X and V : X X which make the following diagram commute: X U X S + T I X V X Denote W = (S + T U(X and let P : X W be a rojection onto W (we can take P = (S + T UV. Clearly V P (S + T U = I. Now S, T M X imlies V P SU, V P ST M X which is a contradiction since V P SU + V P T U = I. Let us just mention that the conditions of the roosition above are satisfied for X = C([0, 1] ([11, Proosition 2.1] hence M X is the largest ideal in L(C([0, 1] as well. We should oint out that there are Banach saces for which M X is not an ideal in L(X. In the sace l l q, 1 < q <, there are exactly two maximal ideals ([15], namely, the closure of the ideal of the oerators that factor through l, which we will denote by α, and the closure of the ideal of the oerators that factor through l q, which we will denote by α q. In this articular sace, the first author roved a necessary and sufficient condition for an oerator to be a commutator: Theorem 5.2. ([6, Theorem 20] Let P l and P lq be the natural rojections from l l q onto l and l q, resectively. Then T is a commutator if and only of P l T P l and P lq T P lq are commutators as oerators acting on l and l q resectively. If we denote T = ( T 11 T 12 T 21 T 22, the last theorem imlies that T is not a commutator if and only if T 11 or T 22 is not a commutator as an oerator acting on l or l q resectively. Now using the classification of the commutators on l for 1 < and the results in [15], it is easy to deduce that an oerator on l l q is not a commutator if and only if it has the form λi + K where λ 0 and K α α q. We can generalize this fact, but first we need a definition and a lemma that follows easily from [6, Corollary 21]. Proerty P. We say that a Banach sace X has roerty P if T L(X is not a commutator if and only if T = λi + S, where λ 0 and S belongs to some roer ideal of L(X. All the Banach saces we have considered so far have roerty P and our goal now is to show that roerty P is closed under taking finite sums under certain conditions imosed on the elements of the sum. Lemma 5.3. Let {X i } n i=1 be a finite sequence of Banach saces that have roerty P. Assume also that all oerators A: X i X i that factor through X j are in the intersection of all maximal ideals in L(X i for each i, j = 1, 2,..., n, i j. Let X = X 1 X 2 X n and let P i be the natural rojections from X onto X i for i = 1, 2,..., n. Then T L(X is a commutator if and only if for each 1 i n, P i T P i is a commutator as an oerator acting on X i. Proof. The roof is by induction ( and it mimics the roof of [6, Corollary 21]. First A B consider the case n = 2. Let T = where A : X C D 1 X 1, D : X 2 X 2, B : X 2 X 1, C : X 1 X 2. If T is a commutator, then T = [T 1, T 2 ] for some T 1, T 2 L(X. Write

14 Page 14 of 15 D. DOSEV AND W. B. JOHNSON T i = ( Ai B i for i = 1, 2. A simle comutation shows that C i D i ( [A T = 1, A 2 ] + B 1 C 2 B 2 C 1 A 1 B 2 + B 1 D 2 A 2 B 1 B 2 D 1. C 1 A 2 + D 1 C 2 C 2 A 1 D 2 C 1 [D 1, D 2 ] + C 1 B 2 C 2 B 1 From the fact that X 1 and X 2 have roerty P, and the fact that the B 1 C 2, B 2 C 1 lie in the intersection of all maximal ideals in L(X 1 and C 1 B 2, C 2 B 1 lie in the intersection of all maximal ideals in L(X 2 we immediately deduce that the diagonal entries in the last reresentation of T are commutators. In the receding argument we used the fact that a erturbation of a commutator on a Banach sace Y having roerty P by an oerator that lies in the intersection of all maximal ideals in L(Y is still a commutator. To show this fact assume that A L(Y is a commutator, B L(Y lies in the intersection of all maximal ideals in L(Y and A + B = λi + S where S is an element of some ideal M in L(Y. Now using the simle observation that every ideal is contained in some maximal ideal, we conclude that S B is contained in a maximal ideal, say M containing M hence A λi M, which is a contradiction with the assumtion that Y has roerty P. For the other direction we aly [6, Lemma 19] which concludes the roof in the case n = 2. The general case follows from the same considerations as in the case n = 2 in a obvious way. Our last corollary shows that roerty P is reserved under taking finite sums of Banach saces having roerty P and some additional assumtions as in Lemma 5.3. Corollary 5.4. Let {X i } n i=1 be a finite sequence of Banach saces that have roerty P. Assume also that all oerators A: X i X i that factor through X j are in the intersection of all maximal ideals in L(X i for each i, j = 1, 2,..., n, i j. Then X = X 1 X 2 X n has roerty P. Proof. Assume that T L(X is not a commutator. Using Lemma 5.3, this can haen if and only if P i T P i is not commutator on X i for some i {1, 2,..., n} and without loss of generality assume that i = 1. Since P 1 T P 1 is not a commutator and X 1 has roerty P then P 1 T P 1 = λi X1 + S where S belongs to some maximal ideal J of L(X 1. Consider M = {B L(X : P 1 BP 1 J}. (5.2 Clearly, if B M and A L(X, then AB, BA M because of the assumtion on the oerators from X 1 to X 1 that factor through X j. It is also obvious that M is closed under addition, hence M is an ideal. Now it is easy to see that T λi M which shows that all non-commutators have the form λi + S, where λ 0 and S belongs to some roer ideal of L(X. The other direction follows from our comment in the beginning of the introduction that no oerator of the form λi + S can be a commutator for any λ 0 and any oerator S which lies in a roer ideal of L(X. References 1. F. Albiac and N. Kalton, Toics in Banach sace theory, Graduate Texts in Mathematics, 233. Sringer, New York, C. Aostol, On the left essential sectrum and non-cyclic oerators in Banach saces, Rev. Roum. Math. Pures Al. 17 ( C. Aostol, Commutators on l saces, Rev. Roum. Math. Pures Al. 17 (

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