Products of n-roots of the identity

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TeoriaOperatorów Kraków,22-28 Seplcmber 2003

pp. 35-42

PRODUCTS OF n-ROOTS OF THE IDENTITY

PIOTR BUDZYŃSKI

Abstract. We present recent results concerning decomposition of operators into products of finite number of n-roots of the identity. We show that every invertible bounded linear operator on a complex infinite-dimensional Hilbert space is a product of five n-roots of the identity for every n > 2. In the case when n = 2 seven factors suffice.

1. Prelminaries

Assume that all operators appearing in this paper act on an infinite-dimensional separable Hilbert space. All results are given only in this settings. Generalizations to the nonseparablecase will be obvious for the reader.

Definition 1.1. Let n €N. Operator A € S(W) is called an n-root of the identity (n-root) if:

By root we mean 2-root.

Definition 1.2. Let T be an operator acting on a Hilbert space H. We call this operator:

(i) normal iff TT* = T*T, (ii) unitary iff TT* = T*T = I, (iii) self-adjoint iff T* = T

(iv) compact iff the set T({f€ H : ||/|| < 1}) iscompact in H.

Corollary 1.3. Let A € B(Tt) bea product offinite number of n-roots of the identity.

If operator B g B(7C) is similar to A (i.e. B = SAS-1 for some invertible S) then itselfis the product ofthe same numberofn-roots.

Proof. Assume that there exist operator S e B(H,IC) and n-roots of the identity TUT2,...,7fc € B(H) such that B = SAS-1 i A = Tj.. .Tk. Then we have:

and every operator STiS \ i = l,...,fc, is n-root. If S is unitary operator and B = {ST1S-l)...{STkS~1)

□

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3G PIOTR BUDZYŃSKI

Definition 1.4. Banach algebra is (complex) unital algebra A equipped with a norm satisfying followingconditions:

(0 ll^yll < lklll|y||,for every x,y eA (ii) ||e|| = 1, (e is neutral element of>1), (iii) (.4,|| • ||) is Banach space.

Spectrum of element a of algebra Ais defined by :

<7.4 (a) = {A G C : (Ae— a) is not invertible in .4}.

From the spectral theorem weget:

Corollary 1.5. Let N G be a normal operator . For every natural n > 2 there exist closed subspacesTCi,...,ICn C H, with the same dimension and operators Nj G B(JCj) (j = 1,... ,n) such that N = Ni ffi... ® Nn.

2. Roots

Theorem 2.1. Everyunitary operator on H is a product of fourself-adjoint roots of the identity.

Proof. Decompose space 7Y intoorthogonal sum Tt = ©■¿Hn of infinite-dimensional subspaces, reducingoperator U. Find unitary operator S acting on H satisfying:

SHn = Hn+1. (2.1)

Ifwe define T = S*U, then obviously:

U = ST and

THn = S*UHn = S*Hn=Hn-1. (2.2) Since the adjoint of every operator T satisfying (2.2) satisfies condition (2.1), it is enough to show that operatorSis aproduct of two self-adjoint roots ofthe identity.

Define unitary operators P, Q on Tl by:

P\h„ =51-2n|«„ (2.3)

and

Q\nn =S~2n\-Hn. (2.4)

Simple calculations show that P, Q are roots ofthe identity andS = PQ. □ Example 2.2. On every Hilbert space H there exists a unitaryoperatorU that is not a product of three roots. Consider a G C : a3 = 1, 1 anddefine U = al.

Such operator commutes with any other operator, hence ifX, Y, Z are roots of the identity suchthat U= XYZthen:

[74 = U3U = UXUYUZ = U(XU)Y(UZ) = U(YZ)Y(XY) =

= Y(UZY){XY) = Y(X)(XY) = Y2 = I.

what is contradiction.

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Lemma 2.3. If B G B(H) is an invertible operator, then B © B 1 is a product of two roots.

Proof. Observe that:

(B 0 \ = / 0 BWO A

b-1) v#-1 oJv oj’

□ Lemma 2.4. If B G B(TT) is an invertible operator then B © 1-n is aproduct of four roots.

Proof. Since 7Y is infinite-dimensionalthen:

B®I-h=B®I®I®I®---

= (B © B_1 ©B®B_1 ©•••)(/© B © B_1 © B® B~! © ■ ••).

From the previous lemma we get B © B_1 = TS, where T2 = S2 = I and S* = S therefore equalities:

B © In = (TS © TS ©•••)(/© TS © TS © • ••)

= (T ® T ® •••)(S ® S ffi •••)(/©T © T©••• )(Z © S ® S©•••),

finish the proof. □

Theorem 2.5. IfN G Bffi) is an invertible, normal operator then N is a product of six roots.

Proof. We can decompose N into orthogonal sum:

N = M ffi N2 “B © C e B(K. © K).

We have

BffiC = (C“1 ©C) (CB®I).

Apply both lemmas. □

Using polar decompositionweget:

Theorem 2.6. Every invertible operator A G B(W) is a product ofsixroots. Moreover four of them are self-adjoint operator.

3. n-ROOTS

Definition 3.1. Aninvertible operator T acting ona Hilbertspace H is commuta

tor if there exist invertibleoperators A and B on H suchthat:

T = ABA~1B~1.

We arenowto present and provelemma which we will frequentlyuseinthissection.

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38 PIOTR BUDZYŃSKI

Theorem 3.2. LetA3,... ,An G B(TL), n > 2, be such that aproduct Ai ••• ATl is a commutator. Then there exist Ti,T2,T3 G B(TC) such that:

Ai ffi... ffi An = TiT2T3 and

fJ-tTl _ rjvn _ _ Moreover operator T3 can be chosen unitary.

Proof. Let Ai... An = UVU 1V '. Define unitary operator W by nx n operator matrix:

0 0

^I

0 0 I

0\

0

w =

0 0 0 I

V

⁰ ⁰

Now put

Zi = ([/-'AMj® A^Af'UVAi ffi 7®...ffi/ffiAf’V“1),

Z2 = (VAiA2 ® A71 Af^Ai ffiAf1 AiA2A3 ® A4 ffi ... ® An), and

T^ZrW, T2 = Z2W, T3 = W~2. Elementary calculationsshow that T" = I, j = 1,2,3. Moreover

Ai ffi A3ffi A4 ffi ... ffiAnffi A2 = T\T2T3.

Since operators Ai ffi... ffi An and Ai ffiA3ffi A4 ffi... ffiAn ffiA2 are unitary equivalent the proof isfinished (n-root73 is unitary so is Tb = UTsl/-1, if U is unitary). □ Before wegettothe specific application of above lemma we briefly discuss technique of decomposinginvertiblenormal operator into a product of TV-roots. By Corollary 1.5 any normaloperator N on an infinite-dimensional Hilbert space can be split into the orthogonal sum of arbitrarily many normal operators acting on infinite-dimensional subspaces:

N = Mffi ... ffi Nn.

Onecan immediately notice connection withthe lemma— knowingthat the product Ni ••• Nn is a commutator we follow theproofand construct proper n-roots. Wesee that the main difficulty is containedin the question: which properties ofsummands Niy..., Nn imply that the product N3 ■ ■■Nn is a commutator? It turns out that the answer is based on exact analysis ofessential spectrum <7ess(N) of operator N and behavior of N onreducing subspaces.

Theorem 3.3. If T is an invertible operator on B(TL) and K, is closed, infinite dimensional subspaceofH which reduces T tounitary operatorthen Tis commutator.

A simple demonstration of mentioned techniqueoffers:

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Theorem 3.4. Let N = M ffi U & B(Lt®Lt) be an invertible operator such that jVj

is normal and U isunitary. Then operator N is a product ofthree n-rootsof identity, for every n >2.

Proof. We can decompose N into thesum:

N = N] ffi U2 ffiU3 ffi... ffi U2n and find unitary operator V so:

VNV~r =M ffi U2 ffi U3 ffi ...ffi U2n € B(W2n), where Ui € B(Tt), i= 2,..., 2n, are unitary. Operator:

T = (M ffi U2)(U3ffi U4)... (U2n_iffi U2n)

= (MC/3 . . .U2n-1) ffi(U2U4 ... U2n)

has infinite-dimensional subspace reducing to unitaryoperator,hence is commutator

by previous theorem. □

Corollary 3.5. Every unitary operator is aproduct of three unitary n-roots, for every natural n > 2.

This resultis best possible as shows following:

Example 3.6. Let a € C : |a| = 1 and an / 1.

Unitary operator al cannot be a product oftwo n-roots of identity. Suppose that X, Y are such roots. From the equalities a = XY and Xn = Yn = I we get that X and Y are commuting (X = aYn~!'j so:

on = (XY)n = XnYn = I, what is obvious contradiction.

We said that one of the possible ways to answer the question if the product TVi• • ■ Nn is to analyze essential spectrum of N. This notion is strictly connected with Calkin algebra.

Definition 3.7. Calkin algebra (over Hilbertspace Lt) is quotient Banach algebra defined:

Ac(Li) = B(Lt)/C(Lt), where C(Li) is two-sidedideal ofcompact operatorson Lt.

Definition 3.8. Let A be an operatoron Li. Essential spectrum of A is the set:

<Tess(j4) = &Ac ([-^]~)-

Theorem 3.9. Let N € B(Li) be an invertible normal operator. If there exist a, fl € Oess{N) such that |a| < |/3| then for every natural n > 2 operator N is aproduct of

three n-roots ofidentity.

Wedon’tgivethe proof, mainly because it is to long,howeverwementionfollowing theorem of Brown and Pearcy which plays crucial role in it.

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40 PIOTRBUDZYŃSKI

Theorem 3.10. If invertible operators F,G, acting on a Hilbert space H, are not of the form A + К, where К is compact operator, then orthogonal sum F ф G is commutator.

It is also used in the next one.

Theorem 3.11. Let N = Nф (a + /<) G фH) be an invertible and normal operator such that: cress(N) C [z G C : |z| = r} for some r > 0 and К is compact operator. Ifr=l or there exists /3 G cress(N) such that/3 a then N is a product of threen-roots, for every naturaln > 2.

Proof. Considerfirst the case when r= 1. If a isthe only point inessential spectrum cre.S5(N), then N = a+ К with some compact operator К and |ct| = 1. This comes from spectral mapping theorem and:

аЛс([^ - a[/]~) = {0}.

Use spectral theorem to get:

N ~ (a + K4) Ф ... ф (a +Kn) € B(fCu).

a product (o + Ki)... (a+ Kn) = an +K' is commutatorby Theorem3.10,sinceK' is compact. Now it suffices to use Theorem 3.2 to prove this case.

Assume now that there exists (3 a contained in essential spectrum Decompose:

where N-¡,N3 are normal, K2,K4,... ,K4n are compact, each of them acts on an infinite-dimensional Hilbert space /С and:

/3 £ &ess(Ni) О СГе.чз(А^з).

Put:

Ti =N\(a + Кз)(а+Jfg)... (a + К4п_з)

ф(а + K2)(a + K§)(a + /Сю)... (a+ К4п_2)

Т2 =N3(a + К7)(а + Кп)... (a + K4n-i) ф (а + /С4)(а + Kg,) (а+ К42)... (а + К4п)

={ап~1 N3 + К'2) ®(ап + K'f).

Observethat neither T\ nor T2cannot beof the form X+K for some compact operator K. If we suppose that then computing essentialspectrum:

{A}=aess(T1)D{an,an-1^},

similarly for T2. It is obvious contradiction to the fact that (3 / a. Thus from Theorem 3.10 follows that operator T\ ф T2 is commutator. Theorem 3.2 gives the

way offinding adequate n-roots. □

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The last twotheorems refer to the case when essential spectrumcontains at least two points. Now it’s time to consider the simplest situation: essential spectrum is just asingleton.

Corollary 3.12. Let N = a + K G B(?Z), where K is compact and |o| / 1 be normal, invertible operator. Then for every naturaln > 2 operator N is aproduct of four n-roots and at least two ofthem are unitary operator.

Proof. Notice first that a 0 since any compact operator is not invertible. Let at G C : wn = 1. Standardprocedure gives:

Now define:

One can easily check that N1 is n-root of the identity. As for N2, it comes from Lemma ?? that one is a product of three n-roots (one of which is unitary). To see this compute essential spectrum:

Ccs.^M’)= {o,cua,iv2a,...,wn_1o} C {zG C : |z| = |o|}

and observe that card(cre.,s(X2))> 2. MoreoveryViX2 = (a + Z<o)®- • .©(a+ZG-i) = N which finally impliesthatN is aproductof four n-roots of theidentityand at least

twoofthem are unitary. □

Example 3.13. Every invertible normal operator N = a+ K G B(H), where |a| / 1 and K is compact cannot be a product of three 3-roots ofthe identity. Supposethat there exist three 3-rootsX, Y, Z such that a + K = XYZ. It implies that:

aZ~r+ KZ-1 =XY, and then:

a3+ K' = (XY)3,

where K' is compactoperator. Observe that for any pair ofoperators X, Ywe have formula:

(

xy

) 3 = [[x,

^y

],

yx

2 ]

^yx

3

y

2 ,

where [X, Y] denotes commutator XYX~1Y~1. Immediately we get that if X3 = Y3 = I then:

(XY)3 = [[X,Y],YX2],

thus a3 + K' = (XY)3 is a commutatorwhat is impossible (x) .

Theorem 3.14. Let N =rU ©FG B(H®J-) be aninvertible normal operator where U is unitary, 1 / r > 0 and Hilbert space F isfinite dimensional. If there are no isolated points in spectrumcr(U) then N is a product ofthreen-roots for every natural n > 2.

From Theorems?? follows general theorem for normaloperators:

(') If T is commutator and T = X + K, where K is compact operator, then |A| = 1.

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42 PIOTR BUDZYŃSKI

Theorem 3.15. Let N be an invertible, normal operator and n > 2 be natural. Iffor some a € C, such that |a| 1, and some compact operator K we have N — a+ K thenN is aproduct offourn-roots of the identity and atleast two of them are unitary.

In opposite case we can decompose N into a product of threen-roots, one ofthemis unitary.

Use the polar decomposition toobtain:

Theorem 3.16. If T is an invertible operator and n > 2 is natural number then T is a product of five n-roots of the identity. At least three of them are unitary.