HAL Id: hal-02072172
https://hal.archives-ouvertes.fr/hal-02072172v2
Preprint submitted on 13 May 2020
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The atoms of the free additive convolution of two operator-valued distributions
Serban Belinschi, Hari Bercovici, Weihua Liu
To cite this version:
Serban Belinschi, Hari Bercovici, Weihua Liu. The atoms of the free additive convolution of two
operator-valued distributions. 2020. �hal-02072172v2�
SERBAN T. BELINSCHI, HARI BERCOVICI, AND WEIHUA LIU
Abstract. Suppose thatX1andX2are two selfadjoint random variables that are freely independent over an operator algebraB. We describe the possible operator atoms of the distribution of X1+X2 and, using linearization, we determine the possible eigenvalues of an arbitrary polynomial p(X1, X2) in caseB=C.
1. Introduction
Suppose that A is a von Neumann algebra, τ is a faithful normal trace state on A, and X1, X2 ∈ A are selfadjoint. Suppose, in addition, that α1, α2 ∈ R are eigenvalues of X1 and X2, respectively, and p1, p2 ∈ A are the orthogonal projections ontoker(X1−α11A)andker(X2−α21A), respectively. Ifτ(p1)+τ(p2)>
1, it follows thatp=p1∧p2 is nonzero,τ(p)≥τ(p1) +τ(p2)−1, and (X1+X2)p=X1p1p+X2p2p=α1p+α2p= (α1+α2)p.
Thus, α1+α2 is an eigenvalue of X1 +X2. It was observed in [10] that the converse statement is true ifX1 andX2 are freely independent with respect toτ.
More precisely, if α∈Ris an arbitrary eigenvalue ofX1+X2 andp denotes the orthogonal projection ontoker(X1+X2−α1), then there exist uniqueα1, α2∈R satsifying α = α1 +α2 such that (using the notation above) p = p1 ∧p2 and τ(p) =τ(p1) +τ(p2)−1. We consider the analogous question in the case in which X1 and X2 are freely independent over an algebra B ⊂ A of ‘scalars’ and the
‘eigenvalues’ themselves are selfadjoint elements of B. Denote by E : A → B the trace-preserving conditional expectation [22, Proposition V.2.36], let b ∈ B be selfadjoint, and denote by pthe orthogonal projection onto ker(X1+X2−b).
Suppose that thatp6= 0,thatX1andX2are freely independent with respect toE, andE(p)is invertible. Then there exist unique selfadjoint elementsb1, b2∈ Bsuch thatb=b1+b2andker(X1−b1)6={0} 6= ker(X2−b2). Moreover, ifpj denotes the orthogonal projection ontoker(Xj−bj), thenp=p1∧p2andτ(p) =τ(p1)+τ(p2)−1.
Similar results are true when E(ker(X −b)) is only supposed to have closed range, and this latter situation always applies if B is finite dimensional. This has consequences for variables that are freely independent with respect toτ.Suppose thatX1andX2are independent relative toτ and thatpis a selfadjoint polynomial in two noncommutative indeterminates. Then there exist n ∈ N and selfadjoint n×nscalar matrices a1, a2, bsuch thatker(p(x, y))6={0} if and only ifker(a1⊗ X1+a2⊗X2−b⊗1A)6={0}. Moreover, the variablesa1⊗X1anda2⊗X2are freely independent over Mn(C)⊗1A, thus reducing the question about a polynomial to an equivalent one concerning a sum [21, 18, 13, 16, 3].
Our results are also proved for variablesX1andX2that are possibly unbounded but affiliated withA. Some of the material below is developed forB-valued variables in the absence of a trace. The most precise results do however require a trace.
1
Earlier results in this vein were obtained in [4]. Of course, these results show that atoms rarely occur for free convolutions. Conditions under which no atoms occur at all were obtained earlier [21, 18, 13, 16, 3]. These works often deduce the lack of atoms forp(x, y)from strong regularity hypotheses onxandy and do not always require free independence.
2. Random variables and their distributions
We work in the context ofW∗ operator valued probability space. Such a space, denoted(A, E,B), consists of a von Neumann algebraA, a von Neumann subalgebra B ⊂ Athat contains the unit ofA, and a faithful conditional expectationE:A → B that we always assume to be continuous relative to the σ-weak and σ-strong topologies. When needed,Ais supposed to act on a Hilbert spaceHsuch that the σ-weak andσ-strong topologies onAare induced by the weak operator and strong operator topologies on B(H), respectively. A random variable in this probability space is a (possibly unbounded) selfadjoint operator X such that (i1A−X)−1 belongs toA. We denote byAesathe collection of all such operators, and we denote byAethe collection of formal sums of the formX+iY, whereX, Y ∈Aesa.
Given a random variableX ∈Aesa, we denote byBhXithe smallest von Neumann subalgebra ofAthat containsBand(i1A−X)−1. Two random variablesX1, X2∈ Aesaare said to have the sameB-distribution if there exists a∗-algebra isomorphism Φ :BhX1i → BhX2isuch thatΦ((i1A−X1)−1) = (i1A−X2)−1,E(Φ(Y)) =E(Y) for every Y ∈ BhX1i, and Φ(b) = b for every b ∈ B. The B-distribution of a variableX ∈Aesais simply its class relative to this equivalence relation. Naturally, it is desirable to find more concrete objects related to B that determine entirely theB-distribution of a random variable. IfX commutes withB, one may use the B-valued Cauchy transform defined by
GX(z) =E((z1A−X)−1), z∈C\R.
If, in addition, X is bounded, one can use theB-valued moments E(Xn), n∈N. The above options are inadequate in general. IfXis bounded, the noncommutative version of GX does determine the B-distribution of X. We recall the definition of this noncommutative function. We denote by H+(A) the collection of those elements a ∈ A that have a positive, invertible imaginary part; we indicate this condition by writing
ℑa= a−a∗ 2i >0.
The algebra Mn(C)⊗ A = Mn(A) of n×n matrices over A is also a von Neu- mann algebra and we write H+n(A) =H+(Mn(A)). The noncommutative version ofH+(A)is simply
H+•(A) = [
n∈N
H+n(A).
We also writeH−•(A) =−H+•(A)andH+• =H+•(C). The noncommutativeB-valued Cauchy transform of a random variableX ∈ Aesa is the function GX : H+•(B)→ H−•(B)defined by
GX(z) =En((b−1n⊗X)−1), z∈H+n(B),
whereEn:Mn(A)→Mn(B)is the conditional expectation obtained by applyingE entrywise, and1n is the unit matrix inMn(C). We also use the reciprocal Cauchy
transformFX :H+•(B)→H+•(B)defined by
FX(b) =GX(b)−1, z∈H+•(B).
It was pointed out in [29] that there are unbounded variables with different B- distributions that have identical noncommutative Cauchy transforms. However, the noncommutative functionGX does determine entirely the atoms ofX and even the B-distributions of the corresponding kernel projections. For our purposes, anatom of a random variableX ∈Aesais defined to be an elementb∈ Bsawith the property that ker(b−X) 6= 0. Here, ker(b−X) is understood as the greatest projection p ∈ A with the property that (b−X)p = 0. In order to see how these atoms are determined, we discuss briefly the concept of nontangential boundary limits for functionsf :H+→ A. Suppose thatt0∈Randa0∈ A. We write
lim
z→t0
f(z) =a0
if for every ε > 0 there exists δ > 0 such that kf(z)−a0k < ε provided that z=x+iy∈H+ satisfies|z−t0|< δ and|x−t0|/y <1/ε. Observe that for every z∈H+, the function hz:R→Cdefined by
(2.1) hz(t) = z
z−t, t∈R, satisfies
lim
z→0hz(t) =χ{0}(t) =
(1, t= 0, 0, t6= 0, and
|hz(t)|=
x+iy x−t+iy
≤ |x|+y
y = 1 +|x|
y , z=x+iy.
In other words,|hz|is uniformly bounded asz→0such that|x|/yremains bounded.
Applying these functions to an arbitrary random variableX ∈Aesa, we obtain the following result. (The second and third equalities use the σ-strong continuity of E.)
Lemma 2.1. For every X∈Aesa we have
lim
z→0z(z1A−X)−1= ker(X)andlim
z→0zGX(z1A) =E(ker(X))
in the σ-strong topology. More generally, writing u(z) = z(z1A−X)−1 and p= ker(X), we have
E((pb1)(pb2)· · ·(pbn−1)p) =lim
z→0E((u(z)b1)(u(z)b2)· · ·(u(z)bn−1)u(z)) for everyn∈Nand everyb1, . . . , bn−1∈ B.
Since the right hand side in the last equality can be written in terms of the noncommutative function GX, we see that all the moments ofp, and hence itsB- distribution, are determined byGX. The above observation, applied to the variables X−b,b∈ Bsa, shows that the distribution ofker(X−b)is entirely determined by GX.
Later, we require a slight technical variation of Lemma 2.1.
Lemma 2.2. Suppose that X∈Aesa and that f :R+→H+(A)is such that limy↓0
f(y) iy = 1A
in theσ-strong topology and yk(f(y)−X)−1k is bounded fory close to0. Then limy↓0iy(f(y)−X)−1= ker(X)
in theσ-strong topology.
Proof. By Lemma 2.1, it suffices to show that the difference iy(f(y)−X)−1−iy(iy−X)−1
convergesσ-strongly to zero asy↓0. This difference can be rewritten as iy(f(y)−X)−1
1A−f(y) iy
iy(iy−X)−1 .
The lemma follows because, asy ↓0, the first factor remains bounded, the middle factor convergesσ-strongly to zero, and the third factor converges σ-strongly to
ker(X).
Some information aboutFX(iy1A)can be obtained wheny∈R+. Observe that the functions defined by (2.1) satisfy
ℜhiy(t) = y2
y2+t2 ≥χ{0}(t), t∈R. We conclude that
(2.2) ℜ(iyGX(iy1A))≥E(ker(X)), y >0.
Lemma 2.3. Suppose that X ∈ Aesa, that b ∈ Bsa, and that E(p) is invertible, wherep= ker(X−b). Then
limy↓0
1
iyFX(b+iy1A) =E(p)−1 in theσ-strong topology.
Proof. SinceFX(b+iy1A) =FX−b(iy1A), it suffices to prove the lemma forb= 0.
In this case, the hypothesis and (2.2) imply the existence of δ > 0 such that ℜ(iyGX(iy1A))≥δ1A,and hence kFX(iy1A)k/y≤1/δ,for everyy >0. Now, if a sequence {an}n∈N⊂ A of invertible elements convergeσ-strongly to an invertible element a, and if supnka−1n k <+∞, then {a−1n }n∈N converges σ-strongly toa−1. This is easily seen from the identitya−1n −a−1=a−1n (a−an)a−1. Thus the lemma follows becauselimy↓0iyGX(iy1A) =E(p)according to Lemma 2.1.
There is a version of the preceding result that applies to the case in whichE(p) has closed range, that is, if0 is an isolated point in the spectrum ofE(p). Denote byq the support projection of E(p), that is, q= 1A−ker(E(p)). Then qAq is a von Neumann algebra, qBq is a unital von Neumann subalgebra of qAq, and the map Eq : a → qE(a)q, a ∈ qAq, is a faithful σ-strongly continuous conditional expectation fromqAqtoqBq. If one of the following conditions is satisfied:
(a) X ∈ A, or
(b) E preserves a faithful normal trace state onA, then we also haveqXq∈qAqgsa.
Corollary 2.4. Let X ∈ Aesa, and let b ∈ Bsa, be such that E(p)6= 0 has closed range, where p= ker(X−b). Set q= 1A−ker(p) soEq(p) =qE(p)q is invertible in qBq. Suppose that qXq ∈ qAqgsa and set p′ = ker(qXq−qbq). Then p′ ≥ p, Eq(p′)>0, and
limy↓0
1
iyFqXq(qbq+iyq) =Eq(p′)−1 in theσ-strong topology.
Proof. We observe first that
E((1A−q)p(1A−q)) = (1A−q)E(p)(1A−q) = 0,
and thus(1−q)p(1−q) = 0because E is faithful. This implies that(1−q)p= 0, that is, p≤q. In particular, (qXq−qbq)p=q(X−b)p= 0, and this shows that p′ ≥ p. We also have Eq(p′) ≥ Eq(p) and Eq(p) is simply E(p) regarded as an element of qBq. Thus Eq(p′)is invertible. The corollary follows now from Lemma
2.3 applied toqXqandqbq.
3. Freeness and subordination
Let(A, E,B)be a von NeumannB-valued probablility space and letX1, X2 ∈ Aesabe two random variables. We recall from [24] thatX1andX2are said to be free with respect to E, or simply E-free, if E(a1a2· · ·an) = 0 wheneveraj ∈ BhXiji are such that E(aj) = 0 for j = 1, . . . , n and ij 6= ij+1 for j = 1, . . . , n−1.
The study of E-freeness is facilitated by the fact that, in many important cases in which X =X1+X2 is defined, the noncommutative Cauchy transformGX is analytically subordinate to GX1 and toGX2. Results of this kind go back to [25].
We formulate first the relevant result from [6]. This result applies to the case in whichX1, X2 ∈ Asa, that is,X1 and X2 are bounded, and it states the existence of noncommutative analytic functionsω1, ω2:H+•(B)→H+•(B)with the following properties:
(3.1) FX(z) =FX1(ω1(z)) =FX2(ω2(z)) =ω1(z) +ω2(z)−z, z∈H+•(B).
(3.2) ℑωj(z)≥ ℑz, j= 1,2, z∈H+•(B).
In order to apply the subordination functions to the study of atoms, we also require a version [5, 8] of the Julia-Carathéodory theorem for noncommutative functions. We state below the relevant parts of this result. (An interesting point is that, while all the conclusions concern the ‘commutative’ part ω|H+(B) ofω, the proof uses the fact thatω is a noncommutative function.)
Theorem 3.1. Let ω:H+•(B)→H+•(A)be an analytic noncommutative function.
Suppose that there exist b0, c0 ∈ B such that b0 = b∗0, c0 > 0, and the quantity kℑω(b0+iyc0)k/y is bounded ify∈R+ is close to zero. Then:
(1) The limit
β= lim
y↓0
1
yℑω(b0+iyc)
exists in the σ-strong topology for every c ∈ B, c > 0, and it is strictly positive.
(2) The limit
b= lim
y↓0ω(b0+iyc)
exists in the norm topology for every c ∈ B,c > 0, it is independent of c, and it is selfadjoint.
(3) We have
limy↓0
1
y(ℜω(b0+iyc)−b) = 0 in theσ-strong topology for every c∈ B,c >0.
With these tools in hand, we can analyze the consequences ofE(ker(X−b))>0.
Theorem 3.2. Let(A, E,B)be an operator valued von Neumann probability space, letX1, X2∈ Asabe twoE-free random variables, and denoteX =X1+X2. Suppose that b ∈ Bsa is such that E(p) > 0, where p = ker(X −b). Then there exist b1, b2, β1, β2∈ Bsawith the following properties:
(i) b=b1+b2, (ii) β1, β2>0,
(iii) ker(Xj−bj)6= 0,j= 1,2,
(iv) E(ker((Xj−bj)βj−1/2)) =βj1/2E(p)βj1/2,j= 1,2,and (v) β1+β2−1A=E(p)−1.
Proof. Letω1 and ω2 be the subordination functions described earlier. Equation (3.1) shows that
ℑω1(b+iy1A)
y +ℑω2(b+iy1A)
y = 1A+ℑFX(b+iy1A)
y , y∈R+. By Lemma 2.3, the right hand side remains bounded asy ↓ 0, and thus Theorem 3.1 shows that the norm limits
bj= lim
y↓0ωj(b+iy1A), j= 1,2, and the strictly positiveσ-strong limits
βj= lim
y↓0
ℑωj(b+iy1A)
y , j= 1,2, exist and, in addition,
(3.3) βj= lim
y↓0
ωj(b+iy1A)−bj
iy , j= 1,2.
Next, we use the subordination relation to see that
(3.4) iyGX(b+iy1A) =iyGXj(ωj(b+iy1A)), j = 1,2, y∈R+. Define
fj(y) =βj−1/2(ωj(b+iy1A)−bj)β−1/2j j = 1,2, y∈R+, and observe that (3.3) implies
limy↓0
fj(y) iy = 1
in theσ-strong topology. Therefore, by Lemma 2.2,
limy↓0iy(fj(y)−βj−1/2(Xj−bj)βj−1/2)−1= ker(βj−1/2(bj−Xj)β−1/2j )
= ker((bj−Xj)βj−1/2) in theσ-strong topology. Now,
ωj(b+iy1A)−Xj=ωj(b+iy1A)−bj+bj−Xj
=β1/2j fj(y)β1/2j +bj−Xj
=β1/2j (fj(y)−βj−1/2(bj−Xj)βj−1/2)β1/2j , so
(ωj(b+iy1A)−Xj)−1=βj−1/2(fj(y)−β−1/2j (bj−Xj)βj−1/2)−1β−1/2j and
(3.5) lim
y↓0iy(ωj(b+iy1A)−Xj)−1=β−1/2j ker((bj−Xj)β−1/2j )βj−1/2 in theσ-strong topology. Similarly,
GXj(ωj(b+iy1A)) =E((ωj(b+iy1A)−Xj)−1)
=βj−1/2E((fj(y)−βj−1/2(bj−Xj)β−1/2j )−1)β−1/2j . Takingσ-strong limits in (3.4), we obtain
E(p) =βj−1/2E(ker(bj−Xj)β−1/2j )βj−1/2,
that is, (iv). In particular,ker(bj−Xj)βj−1/26= 0, and this implies (iii). We observe next that
ω1(b+iy1A) +ω2(b+iy1A) =b+iy1A+FX(iy).
The left side tends in norm tob1+b2 asy↓0, while FX(iy) =iyFX(iy)
iy →0·E(p) = 0 asy↓0. This proves (i). Finally, since
1
iy[ω1(b+iy1A)−b1] + 1
iy[ω2(b+iy1A)−b2] = 1A+ 1
iyFX(iy), we obtain by letting y↓0
(3.6) β1+β2= 1A+E(p)−1.
Remark 3.3. Since
βj1/2E(ker((Xj−bj)β−1/2j ))β−1/2j =βjE(p), j= 1,2, we obtain
β11/2E(ker[(X1−b1)β1−1/2])β1−1/2+β21/2E(ker[(X2−β2)β2−1/2]β2−1/2= 1A+E(p) upon multiplying (3.6) on the right byE(p). A similar equation is obtained when we multiply on the left byE(p)(or when we take adjoints in the above equation.)
Remark 3.4. Since ker(bj −Xj) and ker((bj−Xj)β−1/2j ) are the left and right support projections ofβ1/2j ker(bj−Xj), it follows that these two projections are Murray-von Neumann equivalent inA.
Suppose now thatAis a von Neumann algebra,B ⊂ Ais a von Neumann subal- gebra containing the unit ofA, andτ:A →Cis a normal faithful trace state. We denote byEB:A → Bthe unique trace preserving conditional expectation, that is, τ◦EB=τ, so(A, EB,B)is an operator valued probability space. In this context, the formal setAeintroduced at the beginning of Section 2 has an algebra structure;
in particular,Aesais a vector space. Thus the addition of arbitrary random variables inAesa is defined. Suppose thatX1, X2∈Aesaand setX =X1+X2. It was shown in [25, 26] that noncommutative functions ω1 and ω2 satisfying (3.1) and (3.2) do exist. In addition, the stronger subordination equation
(3.7) EBhXji((b−X)−1) = (ωj(b)−Xj)−1, j= 1,2, b∈H+(B), holds. Theorem 3.2 can be strengthened as follows.
Theorem 3.5. Let Abe a von Neumann algebra with a faithful normal trace state τ, let B ⊂ Abe a von Neumann subalgebra containing the unit of A, let X1, X2∈ Aesa be twoEB-free random variables, and setX =X1+X2. Suppose thatb∈ Bsa
is such that E(p)>0, where p= ker(X −b). Then there exist b1, b2, β1, β2 ∈ Bsa
satisfying properties (i)-(v) of Theorem 3.2 and, in addition, (vi) ker((bj−Xj)βj−1/2) =βj1/2EBhXji(p)β1/2j , and
(vii) p=p1∧p2andτ(p1) +τ(p2) = 1 +τ(p), where pj = ker(bj−Xj), j= 1,2.
Proof. Using the notation in the proof of Theorem 3.2, the equation iyEBhXji((b+iy1A−X)−1) =iy(ωj(b+iy1A)−Xj)−1 and (3.5) yield (vi) asy↓0. To prove (vii), observe that
τ(βj1/2E(ker((Xj−bj)βj−1/2))β−1/2j ) =τ(E(ker((Xj−bj)β−1/2j ))).
Remark 3.4 implies
τ(E(ker((Xj−bj)β−1/2j ))) =τ(pj).
Thus the equality τ(p1) +τ(p2) = 1 +τ(p)follows by applying τ in Remark 3.4.
Finally, we certainly havep≥p1∧p2 and
τ(p) =τ(p1) +τ(p2)−1≤τ(p1∧p2).
We conclude thatp=p1∧p2 becauseτ is faithful. This concludes the proof.
4. Matrix valued random variables
LetAbe a von Neumann algebra endowed with a faithful, normal trace state τ and letn∈N. The algebraMn(A) =Mn(C)⊗ Ais also a von Neumann algebra and the mapτn :Mn(A)→Cdefined by
τn(a) = 1 n
Xn j=1
τ(ajj), a= [aij]ni,j=1∈Mn(A),
is faithful, normal trace state. The trace-preserving conditional expectation En : Mn(C)⊗ A →Mn(C)⊗1A is given by
En(a) = [τ(aij)]ni,j=1, a= [aij]ni,j=1 ∈Mn(A),
and therefore we have
τn= trn◦En,
wheretrn denotes the normalized trace onMn(C)⊗1A. The relevance of this example comes from the following fact.
Proposition 4.1. Suppose that the variables X1, X2 ∈ Aesa are τ-free and that a1, a2∈Mn(C)are selfadjoint matrices. Then a1⊗X1 anda2⊗X2 areEn-free.
Suppose now thatX ∈Aesa is a random variable,n∈N,and a, b∈Mn(C)are selfadjoint matrices. We wish to describe the kernel of the random variable
b⊗1A−a⊗X∈M^n(A)sa.
To do this, it is convenient to view the von Neumann algebraMn(C)ha⊗Xi(that is, the algebra generated byMn(C)⊗1Aanda⊗X) as an algebra of operators on a concrete Hilbert space. Denote byµthe distribution ofX, that is,µis the Borel measure onRdefined by
µ(σ) =τ(eX(σ))
for every Borel setσ⊂R,where eX denotes the spectral measure ofX. Set H=Cn⊗L2(µ),
and let1n⊗X andMn(C)⊗1Aact onf ∈ Hvia the formulas
((1n⊗X)f)(t) =tf(t), ((b⊗1A)f)(t) =bf(t), b∈Mn(C), t∈R. Here, the elements ofH are viewed as measurable functions f : R →Cn. Using this action, the algebraMn(C)ha⊗Xiis identified with the algebra consisting of all (equivalence classes of) bounded Borel functionsh:R→Mn(C)and
(hf)(t) =h(t)f(t), h∈Mn(C)ha⊗Xi, f ∈ H.
The mapsEn andτn become simply En(h) =
Z
R
h(t)dµX(t)andτn(h) = Z
R
trn(h(t))dµX(t), h∈Mn(C)ha⊗Xi.
Returning to the operatorb⊗1A−a⊗X, we see that the equation (b⊗1A−a⊗X)f = 0
translates to
(b−ta)f(t) = 0, µX-a.e.
It is well known (see, for instance, [17, Section 2.1]) that t7→ker(b−ta)
is a Borel function fromR→Mn(C)and that the rank of the projectionker(b−ta) is equal to its minimum value outside a finite set in R. The result below follows immediately.
Lemma 4.2. Suppose that A is a von Neumann algebra with a normal, faithful trace τ, let n ∈ N, let X ∈ Aesa be a random variable, and let a, b ∈ Mn(C) be selfadjoint. Definek(t) = trn(ker(b−ta)),t∈R, and letkmin= min{k(t) :t∈R}.
Then
τn(ker(b⊗1A−a⊗X)) = Z
R
k(t)dµX(t) =kmin+X
t∈R
(k(t)−kmin)µX({t}).
The sum in the lemma above only contains finitely many nonzero terms, corre- sponding to those t∈Rsuch that k(t)> kmin and at the same timeµX({t})>0.
Theorem 3.5 yields the following result.
Corollary 4.3. Suppose thatAis a von Neumann algebra with a normal, faithful trace τ, let n∈N, letX1, X2 ∈Aesa beτ-free random variables, and leta1, a2, b∈ Mn(C)be selfadjoint. If ξ=En(ker(b⊗1A−a1⊗X1−a2⊗X2))>0, then there exist t1, . . . , tN, s1, . . . , sM ∈ R and ℓ0, ℓ1, . . . , ℓN, m0, m1, . . . , mM ∈ N such that ℓ0+ℓj≤n,m0+mi≤n, and
n(trn(ξ) + 1) =ℓ0+m0+ XN j=1
ℓjµX1({tj}) + XM i=1
miµX2({si}).
If neither µX1 nor µX2 have point masses, thenntrn(ξ)is an integer.
Proof. As noted above, the variables a1⊗X1, a2⊗X2 ∈ M^n(A) areEn-free and thus the conclusions of Theorem 3.5 apply to them. Thus, there exist selfadjoint elementsb1⊗1A, b2⊗1Asuch that
1 + trn(ξ) =τn(p1) +τn(p2), wherepj= ker(bj⊗1A−aj⊗Xj),j= 1,2. Setting
kj(t) = trn(ker(bj−taj)), t∈R, ℓ0=nmin{k1(t) :t∈R}, m0=nmin{k2(t) :t∈R}, ℓ(t) =nk1(t)−ℓ0,
m(t) =nk2(t)−m0, we conclude that
n(1 + trn(ξ)) =ℓ0+m0+X
t∈R
(ℓ(t)µX1({t}) +m(t)µX2({t})).
If neither µX1 norµX2 has any point masses, the second sum vanishes, and thus ntrn(ξ) =ℓ0+m0−nis an integer. The corollary follows.
The ordinary eigenvalues of an arbitrary polynomialP(X1, X2) in two random variables can be studied using the matrix eigenvalues of an expression of the form a1⊗X1+a2⊗X2. This is achieved by the process of linerization that we now describe briefly. Suppose that P(Z1, Z2)is a complex polynomial in two noncom- muting indeterminates and leta0, a1, a2∈Mn(C)for somen∈N.We say that the expression
L(Z1, Z2) =a0⊗1 +a1⊗Z1+a2⊗Z2
is a linearization ofP(Z1, Z2)if, given elementsz1, z2in some complex unital alge- braAandλ∈C, then the elementP(z1, z2)is invertible inAif and only ifL(z1, z2) is invertible in Mn(A). It is known that every polynomial has a linearization. As seen in [1], ifP(Z1, Z2)is selfadjoint (relative to the involution that fixesZ1andZ2), thena0, a1,anda2 can be chosen to be selfadjoint matrices. One way to construct a linearization is to findm ∈Nand polynomials B(Z1, Z2), C(Z1, Z2), D(Z1, Z2), andD′(Z1, Z2)such that
(a) B is a1×mlinear polynomial, (b) C is anm×1linear polynomial,
(c) D is anm×mlinear polynomial,
(d) D(Z1, Z2)D′(Z1, Z2) =D′(Z1, Z2)D(Z1, Z2) = 1m, and (e) B(Z1, Z2)D′(Z1, Z2)C(Z1, Z2) =P(Z1, Z2).
Once such polynomials are found,
(4.1) L(Z1, Z2) =
0 B(Z1, Z2) C(Z1, Z2) D(Z1, Z2)
is a linearization ofP with n= 1 +m. This linearization is selfadjoint if C∗=B andD∗=D.
Lemma 4.4. Let A be a von Neumann algebra with a faithful normal trace state τ, and let A, X1, X2∈Aesa be random variables. Suppose that P(Z1, Z2) is a poly- nomial in two noncommuting indeterminates and that L(Z1, Z2) is a linearization of P defined by (4.1), where B, C, D, D′ are subject to conditions (a)–(e). Denote by e1,1 ∈Mn(C) the matrix unit whose only nonzero is in the first row and first column. Then ker(A⊗e1,1+L(X1, X2)) is Murray-von Neumann equivalent to (A−kerP(X1, X2))⊕(0n−1⊗1A)inMn(A). In particular,
nτn(ker(A⊗e1,1+L(X1, X2))) =τ(ker(A−P(X1, X2))).
Proof. We first observe that
A⊗e1,1+L(X1, X2) =
A B(Z1, Z2) C(Z1, Z2) D(Z1, Z2)
, and
1A −B(X1, X2)D′(X1, X2) 0 1n−1⊗1A
(A⊗e1,1+L(X1, X2)) =
A−P(X1, X2) 0 C(X1, X2) D(X1, X2)
.
Since the first operator on the left side is injective, we deduce that ker(A⊗e1,1+L(X1, X2)) = ker
A−P(X1, X2) 0 C(X1, X2) D(X1, X2)
. Next, we note the identities
A−P(X1, X2) 0 0 1n−1⊗1A
H =
A−P(X1, X2) 0 C(X1, X2) D(X1, X2)
and
A−P(X1, X2) 0 C(X1, X2) D(X1, X2)
K=
A−P(X1, X2) 0 0 1n−1⊗1A
, where
H=
1A 0 C(X1, X2) D(X1, X2)
and
K=
1A 0
−D′(X1, X2)C(X1, X2) D′(X1, X2)
are injective operators. The first identity shows that the final space of Hker
A−P(X1, X2) 0 C(X1, X2) D(X1, X2)
=Hker(A⊗e1,1+L(X1, X2)) is less than or equal to
ker
−P(X1, X2) 0 0 1n−1⊗1A
= (ker(A−P(X1, X2)))⊕(0n−1⊗1A),
and thus
ker(A⊗e1,1+L(X1, X2))≺(ker(A−P(X1, X2)))⊕(0n−1⊗1A).
Similarly, the second equality yields
(ker(A−P(X1, X2)))⊕(0n−1⊗1A)≺ker(A⊗e1,1+L(X1, X2))
and concludes the proof.
The preceding proof yields a description of ker(A⊗e1,1+L(X1, X2)) that we note for further use.
Corollary 4.5. With the notation of Lemma 4.4,ker(A⊗e1,1+L(X1, X2))is the final support of the operator
q
−D′(X1, X2)C(X1, X2)q
, whereq= ker(A−P(X1, X2)).
Corollary 4.6. With the notation of Lemma 4.4, suppose that A∈C1A, X1 and X2 are free, 06= ker(A−P(X1, X2))6= 1A, and setp= ker(e1,1⊗A+L(X1, X2)).
If En(p)>0 then either X1 orX2 has an eigenvalue.
Proof. By Lemma 4.4, nτn(p) =τ(A−P(X1, X2))∈(0,1). In particular,nτn(p) is not an integer. The corollary follows from Corollary 4.3.
The following result allows us to treat cases in whichEn(p)is not invertible.
Lemma 4.7. Let A be a von Neumann algebra with a faithful normal trace state τ, and let X ∈M^n(A)sa. Then there exist projections q1, q2 ∈ Mn(C)⊗1A such that
(1) En(ker(q1Xq2))andEn(ker(q2Xq1))are invertible, and
(2) nτn(ker(q1Xq2))−nτn(kerX) =nτn(ker(q2Xq1))−nτn(kerX)are integers.
Proof. Setp= ker(X)and defineq1to be the support projection ofEn(p), that is, q1= 1Mn(A)−ker(En(p)).
Since
En((1Mn(A)−q1)p(1Mn(A)−q1)) = (1Mn(A)−q1)En(p)(1Mn(A)−q1) = 0, andEn is faithful, we conclude that
(1Mn(A)−q1)p(1Mn(A)−q1) = 0.
Thus,(1Mn(A)−q1)p= 0, or equivalently,p≤q1. We show next that ker(Xq1) =p+ (1Mn(A)−q1).
In fact, it is clear thatXq1p=Xp= 0andXq1(1−q1) = 0. Moreover,ker(Xq1) cannot contain any nonzero projection r orthogonal top+ (1Mn(A)−q1); such a projection satisfiesr≤q1so Xq1r=Xr6= 0becauser≤1−ker(X). We have
En(ker(Xq1)) =En(p) + (1Mn(A)−q1), soEn(ker(Xq1))is invertible and, in addition,
nτn(ker(Xq1))−nτn(p) =nτn(ker(Xq1))−nτn(En(p)) =nτn(1Mn(A)−q1) is an integer, namely the rank of1Mn(A)−q1 viewed as a projection inMn(C).
We observe next that, since q1X = (Xq1)∗, ker(q1X) is Murray-von Numann equivalent to ker(Xq1), and in particular they have the same trace. We apply the preceding observation with q1X in place of X. That is, we define q2 to be the support ofEn(ker(q1X)). Then the above arguments show thatEn(ker(q1Xq2))is invertible and nτn(ker(q1Xq2))−nτn(ker(q1X))is an integer. We conclude that (2) is true becauseker(q2Xq1)is Murray-von Neumann equivalent to ker(q1Xq2).
Finally, observe thatEn(ker(q2Xq1))≥En(ker(Xq1))must also be invertible, thus
concluding the proof of (1).
Proposition 4.8. Let A be a von Neumann algebra with a faithful normal trace state τ, let X1, X2 ∈ Aesa be two free random variables, and let P(Z1, Z2) be a polynomial in two noncommuting indeterminates such that τ(ker(P(X1, X2))) ∈/ {0,12,1}. Then either X1 orX2 has an eigenvalue.
Proof. Replacing P byP∗P does not change the kernel, so we suppose that P is selfadjoint. Let
L(Z1, Z2) =L(Z1, Z2)∗=
0 B(Z1, Z2) C(Z1, Z2) D(Z1, Z2)
be a selfadjoint linearization ofP constructed as above, and letnbe the size of its matrix coefficients. As pointed out earlier,
nτn(kerL(X1, X2)) =τ(ker(P(X1, X2)).
Set X =L(X1, X2) and let q1 and q2 be given by Lemma 4.7. Then q1Xq2 and q2Xq1are again linear polynomials in X1, X2with coefficients inMn(C), but they are not selfadjoint. However, the matrix
Y =
0 q1Xq2
q2Xq1 0
is a linear polynomial with coefficients in M2n, it is selfadjoint, and has kernel ker(q2Xq1)⊕ker(q1Xq2). Therefore, Lemma 4.7(1) implies that
2nτ2n(ker(Y)) =nτn(ker(q2Xq1)) +nτn(ker(q1Xq2))
differs from2τ(ker(P(X1, X2))by an integer, and the hypothesis implies that this is not an integer. Finally, Lemma 4.7(2) implies thatE2n(ker(Y))is invertible, and
the desired conclusion follows from Corollary 4.6.
Some conclusions about the variables X1, X2 ∈Aesa can be drawn even in case τ(ker(P(X1, X2)) = 12 and En(p) is not invertible. We use the anticommutator P(X1, X2) =X1X2+X2X1as an illustration. Suppose thatλ∈Ris an eigenvalue ofP(X1, X2)such thatq= ker(λ1A−P(X1, X2))<1A. It follows that the operator
λ1A X1 X2
X1 0 1A
X2 1A 0
has a kernelpthat is the final projection of
q X2q X1q
.
IfE3(p)is invertible, Corollary 4.6 shows that one of the operatorsX1, X2has an eigenvalue. Suppose then that neitherX1 norX2 has eigenvalues, so E3(p)is not
invertible. In this case, there exists a projectionr∈M3(C)of rank one such that (r⊗1A)E3(p) = 0. We have
E3((r⊗1A)p(r⊗1A)) = (r⊗1A)E3(p)(r⊗1A) = 0,
and we conclude that(r⊗1A)p(r⊗1A) = 0 and thus(r⊗1A)p= 0as well. If the vector(α, β, γ)∈C3 generates the range ofr, then
αq+βX1q+γX2q= 0.
SinceXj has no eigenvalues, we deduce thatβγ6= 0, soβX1+γX2has the eigen- value−α. The argument in Proposition 4.8, applied to the selfadjoint polynomial
0 βX1+γX2
βX1+γX2 0
shows now that we necessarily haveτ(ker(α+βX1+γX2)) = 12. References
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CNRS-Institute de Mathématiques de Toulouse, 118 Route de Narbonne, 31062, Toulouse, France
E-mail address: Serban.Belinschi@math.univ-toulouse.fr
Department of Mathematics, Indiana University, Bloomington, IN 47405, USA E-mail address: bercovic@indiana.edu
Department of Mathematics, The University of Arizona, 617 N. Santa Rita Ave., P.O. Box 210089, Tucson, AZ 85721-0089 USA
E-mail address: weihualiu@math.arizona.edu
