The Hilbert Schidt version of the coutator theore for zero trace atrices Oer Angel Gideon Schechtan March 205 Abstract Let A be a coplex atrix with zero trace. Then there are atrices B and C such that A = [B, C] and B C 2 (log + O()) /2 A 2 where D is the nor of D as an operator on l 2 and D 2 is the Hilbert Schidt nor of D. Moreover, the atrix B can be taken to be noral. Conversely there is a zero trace atrix A such that whenever A = [B, C], B C 2 log O() /2 A 2 for soe absolute constant c > 0. Introduction As is well known (or see e.g. [Fi]) a coplex atrix A is a coutator (i.e., there are atrices B and C of the sae diensions as A such that A = [B, C] = BC CB) if and only if A has zero trace. Let E denote the operator nor of an atrix (as a ap E : l 2 l 2 ) and let be any other nor on the space of atrices satisfying EF E F and F E E F for all atrices. In such a situation clearly if A = [B, C] then A 2 B C. We are interested in the reverse inequality: If A has zero trace are there atrices B and C such that A = [B, C] and B C K A for AMS subject classification: 47B47, 5A60. Key words: coutators, zero trace, Hilbert Schidt nor of atrices Supprted in part by NSERC, the Isaac Newton Institute and Sions Foundation. Supported in part by the Israel Science Foundation.
soe absolute constant K? If not what is the behavior of the best K as a function on? In [JOS] this question was dealt with for being the operator nor. An upper bound on K which is saller than any power of was given. Here we deal with being the Hilbert Schidt nor which we denote 2. We give atching upper and lower bounds (up to a constant factor). Theore. Let A be an atrix with zero trace, then there are atrices B and C such that A = [B, C] and B C 2 (c + log ) /2 A 2. Moreover, the atrix B can be taken to be noral. Conversely, for each there is a zero trace atrix A such that for any atrices B, C with A = [B, C], B C 2 2 (c + log ) /2 A 2, where c, c are soe universal constants. The proof of the upper bound which is done by quite a siple rando choice is given in Section 2. The lower bound is a bit ore involved and is based on an idea fro [DFWW] and in particular on a variation on a lea of Brown [Br]. The proof is given in Section 3. 2 The upper bound Since both nors and 2 are unitarily invariant and since any zero trace atrix is unitarily equivalent to a atrix with zero diagonal, we ay and shall assue that A has zero diagonal. In that case we shall find a diagonal atrix B = (b, b 2,..., b ) with the desired property. Note that translating back and assuing A has erely zero trace, the resulting B is noral, being unitarily equivalent to a diagonal atrix. Let >. If A = [B, C] with A with zero diagonal and B = (b, b 2,..., b ) with all its diagonal entries distinct, then necessarily c i,j = a i,j b i b j for i j. Let G Z + Zı be the points with sallest absolute values, so that ax z G { z } + /π. Let {b i } be a uniforly rando perutation of these points, so that necessarily B + /π. We now evaluate the expectation of the resulting C 2 2. E C 2 2 = E i j a i,j 2 b i b j 2 = i j a i,j 2 E b i b j = 2 A 2 2E b b 2. () 2 2
To evaluate E b b 2 2 fix b G. The expectation conditioned on b is b 2 G b 2 b b b 2 2 [ a 0 + a + π log z b 2 /π ] dz 2 z b 2 for soe absolute constants a 0, a. Plugging this into () we get that E C 2 2 a +π log A 2 2 and thus there is a realization of the b i -s which gives B C 2 c + log A 2, for soe absolute constant c, as desired. Reark. One can clearly replace the points of G by another set of points in the sae disc about zero. Sets iniizing such an energy function are a well studied subject. However, no significant iproveent can be gained by replacing G with another set, and in particular our choice of G achieves the optial leading ter π log. See for exaple [HS] in which tight bounds are given for a related quantity on the two diensional sphere. 3 The lower bound We begin with a Lea which is a variation on a lea of Brown [Br] Lea. Assue S, T are atrices,, and M is a finite diensional subspace of l 2 (where l 2 = l 2 ) such that for soe λ C ([S, T ] + λi)(l 2 ) M. Then there are orthogonal subspaces H n l 2, n = 0,,..., with H 0 = M, dih n (n + )dim, n =, 2,..., and P i SP j = P i T P j = 0 for all i > j +, j = 0,,.... Here P l is the orthogonal projection onto H l. Moreover, n=0 H n is invariant under S and T. Proof. Let V 0 = H 0 = M and for n let V n be the linear span of {S k T l M; k + l n}. For n put H n = V n V n. Clearly, dih n (n + )dim and n=0 H n is invariant under S and T. To show that P i SP j = P i T P j = 0 for all i > j + it is enough to show that T V n V n+ and SV n V n+ for all n. 3
The second containent is obvious. To prove the first it is enough to show that for all k and k + l n, T S k T l M V n+. Now, T S k T l = S k T l+ + i<k S i [T, S]S k i T l = S k T l+ kλs k T l + i<k S i ([T, S] + λi)s k i T l. Now, the first ter here has range in V n+ and the second in V n V n+. Since [T, S] + λi has range in M the ith ter in the last su has range in S i M V i V n+, and the proof is coplete. Let P be the rank one orthogonal projection onto the first coordinate in l 2, <, given by the atrix 0... 0 0 0... 0 P =...... 0 0... 0 and let A = P I. Obviously A has zero trace and Hilbert Schidt nor. We now show that this A gives the lower bound of Theore. Moreover, our arguent gives bounds on the leading singular values of C, based on the proof of Theore 7.3 in [DFWW], which also gives a lower bound on C 2. Specifically, we get the following: Theore 2. Assue A = [B, C] with A as above, and the operator nor of B equals. Denote the singular values of C as s, s 2,..., s, arranged in non-increasing order. Then for all l, l s i l/6. In particular the Hilbert Schidt nor of C is at least c log for soe absolute constant c > 0. Proof of the lower bound in Theore. Let M be the one diensional subspace of l 2 spanned by the first coordinate. Applying Lea to this 4
subspace with S = B, T = C and λ = / we get orthogonal subspaces M = H 0, H,... (which of course are eventually the zero subspace) with dih n n +, so that n=0 H n is invariant under B and C and P i BP j = P i CP j = 0 for all i > j + > 0, where P l is the orthogonal projection onto H l. Note that P 0 = P. Note also that n=0 H n is R. Indeed, a proper subspace of R containing H 0 which is invariant under B and C is also invariant under A, and the restriction of A to such a subspace has zero trace which clearly can t hold. Now, on H 0, A is just ( ), so ( ) P 0 = P 0 [B, C]P 0 Siilarly, for k > 0, since A Hk other i, j, = P 0 BP 0 P 0 CP 0 P 0 CP 0 P 0 BP 0 + P 0 BP P CP 0 P 0 CP P BP 0. = I H k, and using P i BP j = P i CP j = 0 for P k = P k [B, C]P k = P k BP k P k CP k P k CP k P k BP k + P k BP k P k CP k P k CP k P k BP k + P k BP k+ P k+ CP k P k CP k+ P k+ BP k. Using the trace property (e.g., Tr(P k BP k P k CP k ) = Tr(P k CP k P k BP k )), we get that for all n, n n rankp k = Tr(P k [B, C]P k ) So, since B =, = Tr(P n BP n+ CP n ) Tr(P n CP n+ BP n ). n rankp k P n+ CP n + P n CP n+. (2) Since rankp k k +, this gives a lower bound on the nors of P n CP n+ and P n+ CP n : P n+ CP n + P n CP n+ ( ) n + 2. (3) 2 5
The atrices P n+ CP n and P n CP n+ have rank at ost n +, so changing to other nors is not too costly, which allows us to bound fro below the Hilbert Schidt nor of C. To coplete the proof of the lower bound of Theore, note that for a atrix M of rank r we have M 2 2 r M 2, so C 2 2 n n n n P n CP n+ 2 2 + P n+ CP n 2 2 ( Pn CP n+ 2 + P n+ CP n 2 n + ) 2(n + ) ( P ncp n+ + P n+ CP n ) 2 ( ( )) 2 n + 2. 2(n + ) 2 We take the su over n with ( ) n+2 2 <. It is straightforward to see that the last su is log + O(), giving the claied lower bound. 4 Proof of Theore 2. Lea 7.9 in [DFWW] (whose proof is siple, based on polar decoposition) says that there are partial isoetries V, W on l 2 such that Consequently, and by (2), P n V CP n = P n+ CP n and P n W C P n = P n+ C P n. P n (V C + W C )P n = P n+ CP n + P n+ C P n Tr(P n (V C + W C )P n ) n rankp k. Fix a positive integer k and let E k = k i=0 P i and r k = ranke k (k + )(k + 2)/2. Denoting by s i (R) the singular values of the operator R, we get that as long as (k + )(k + 2), (k+)(k+2)/2 s i (V C + W C ) r k k n=0 6 s i (E k (V C + W C )E k ) Tr(P n (V C + W C )P n ) k + 2.
Where we have used Weyl s inequality to deduce the second inequality. It follows that for all k as above (k+)(k+2)/2 s i (C) k+. The ain assertion 4 of the theore follows easily fro that. As for the last assertion, it is well known that it follows fro the first. Indeed, the non-increasing sequence s, s 2,..., s ajorizes a sequence equivalent (with universal constants) to, 2/ 2, / 3,..., /. Consequently, ( s 2 i ) /2 c( /i) /2 c (log ) /2. Acknowledgeent. We benefitted a lot fro discussions with Bill Johnson concerning the aterial of this note. In particular, he is the one who pointed [DFWW] to us. References [Br] L. G. Brown, Traces of coutators of Schatten-von Neuann class operators. J. Reine Angew. Math. 45, 7 74 (994). [DFWW] K. Dykea, T. Figiel, G. Weiss, M. Wodzicki, Coutator structure of operator ideals. Adv. Math. 85, no., 79 (2004). [Fi] P. A. Fillore, On siilarity and the diagonal of a atrix. Aer. Math. Monthly 76: 67 69 (969). [HS] D. P. Hardin, E. B. Saff, Discretizing anifolds via iniu energy points. Notices Aer. Math. Soc. 5 (2004), no. 0, 86 94. [JOS] W. B. Johnson, N. Ozawa, G. Schechtan, A quantitative version of the coutator theore for zero trace atrices. Proc. Natl. Acad. Sci. USA 0 (203), no. 48, 925 9255. O. Angel G. Schechtan Departent of Matheatics Departent of Matheatics University of British Colubia Weizann Institute of Science Vancouver, BC, V6T Z2, Canada Rehovot 7600, Israel angel@ath.ubc.ca gideon@weizann.ac.il 7