On Kadison-Schwarz Type Quantum Quadratic Operators on 𝕄₂(ℂ)

Definition 1. A linear operator Δ : 𝕄₂(ℂ) → 𝕄₂(ℂ) ⊗ 𝕄₂(ℂ) is said to be

• (a)

  a quantum quadratic operator (q.q.o.) if it satisfies the following conditions:

  • (i)

    unital, that is, Δ1 = 1 ⊗ 1;

  • (ii)

    Δ is positive, that is, Δx ≥ 0 whenever x ≥ 0;

• (b)

  a Kadison-Schwarz operator (KS) if it satisfies

  $$ ()

Remark 2. Note that if a quantum convolution Δ on 𝕄₂(ℂ) becomes a *-homomorphic map with a condition

then a pair (𝕄₂(ℂ), Δ) is called a compact quantum group [20]. It is known [20] that for any given compact quantum group there exists a unique Haar state w.r.t. Δ.

Remark 3. Let U : 𝕄₂(ℂ) ⊗ 𝕄₂(ℂ) → 𝕄₂(ℂ) ⊗ 𝕄₂(ℂ) be a linear operator such that U(x ⊗ y) = y ⊗ x for all x, y ∈ 𝕄₂(ℂ). If a q.q.o. Δ satisfies UΔ = Δ, then Δ is called a quantum quadratic stochastic operator. Such a kind of operators was studied and investigated in [17].

Each q.q.o. Δ defines a conjugate operator Δ^* : (𝕄₂(ℂ) ⊗ 𝕄₂(ℂ)) ^* → 𝕄₂(ℂ) ^* by

One can define an operator V_Δ by

which is called a quadratic operator (q.c.). Thanks to conditions (a) (i), (ii) of Definition 1 the operator V_Δ maps S(𝕄₂(ℂ)) to S(𝕄₂(ℂ)).

Lemma 4 (see [3].)The following assertions hold true:

• (a)

  x is self-adjoint if and only if w₀, w are reals;

• (b)

  Tr (x) = 1 if and only if w₀ = 0.5; here Tr  is the trace of a matrix x;

• (c)

  x > 0 if and only if ∥w∥ ≤ w₀, where $$.

Theorem 5 (see [13], Proposition 3.2.)Let Δ : 𝕄₂(ℂ) → 𝕄₂(ℂ) ⊗ 𝕄₂(ℂ) be a q.q.o. with a Haar state τ, then it has the following form:

where x = w₀ + wσ,  b_ml = (b_ml,1, b_ml,2, b_ml,3) ∈ ℝ³, m, n, k ∈ {1,2, 3}.

Proposition 6 (see [13], Proposition 3.3.)Let Δ be a q.q.o. with a Haar state τ, then |∥𝔹∥| ≤ 1.

Proposition 7 (see [13], Proposition 4.1.)Let Δ : 𝕄₂(ℂ) → 𝕄₂(ℂ) ⊗ 𝕄₂(ℂ) be a liner operator with a Haar state τ. Then Δ^*(φ ⊗ ψ) ∈ S(𝕄₂(ℂ)) for any φ, ψ ∈ S(𝕄₂(ℂ)) if and only if the following holds:

Remark 8. Note that characterizations of positive maps defined on 𝕄₂(ℂ) were considered in [24] (see also [25]). Characterization of completely positive mappings from 𝕄₂(ℂ) into itself with invariant state τ was established in [3] (see also [26]).

Theorem 9 (see [13], Theorem 3.6.)Let Δ : 𝕄₂(ℂ) → 𝕄₂(ℂ) ⊗ 𝕄₂(ℂ) be a Kadison-Schwarz operator with a Haar state τ; then it has the form (10) and the coefficients {b_ml,k} satisfy the following conditions:

for all f ∈ S,  w ∈ ℂ³. Here as before x_m = (〈b_m1, w〉, 〈b_m2, w〉, 〈b_m3, w〉); b_ml = (b_ml,1, b_ml,2, b_ml,3), and q(f, w), α_ml, and γ_ml are defined in (19), (17), and (18), respectively.

Remark 10. The provided characterization with [2, 3] allows us to construct examples of positive or Kadison-Schwarz operators which are not completely positive (see next section).

Theorem 11. Let Δ : 𝕄₂(ℂ) → 𝕄₂(ℂ) ⊗ 𝕄₂(ℂ) be a unital *-preserving linear mapping. Then Δ is a KS operator if and only if one has

for all w ∈ ℂ³ with ∥w∥ = 1.

Proof. Let x ∈ M₂(ℂ) be an arbitrary element, that is, x = w₀1 + wσ. Then $$. Therefore

Consequently, we have From (26) and (27) one gets So, the positivity of the last equality implies that Now dividing both sides by ∥w∥² we get the required inequality. Hence, this completes the proof.

Theorem 12. A linear operator Δ_ε given by (31) is a q.q.o. if and only if |ε | ≤ 1/3.

Proof. Let x = w₀1 + wσ be a positive element from 𝕄₂(ℂ). Let us show positivity of the matrix Δ_ε(x). To do it, we rewrite (31) as follows: Δ_ε(x) = w₀1 + εB; here

where positivity of x yields that w₀, ω₁, ω₂, ω₃ are real numbers. In what follows, without loss of generality, we may assume that w₀ = 1, and therefore ∥w∥ ≤ 1. It is known that positivity of Δ_ε(x) is equivalent to positivity of the eigenvalues of Δ_ε(x).

Let us first examine eigenvalues of B. Simple algebra shows us that all eigenvalues of B can be written as follows:

Now examine maximum and minimum values of the functions λ₁(w), λ₂(w), λ₃(w), λ₄(w) on the ball ∥w∥ ≤ 1.

One can see that

Note that the functions λ₃,λ₄ can reach values $$ at $$.

Now let us rewrite λ₁(w) and λ₂(w) as follows:

One can see that

where k = 1,2. Therefore, the functions λ_k(w),  k = 1,2 reach their maximum and minimum on the sphere $$ (i.e., ∥w∥ = 1). Hence, denoting t = ω₁ + ω₂ + ω₃ from (37) and (36) we introduce the following functions: where $$.

One can find that the critical values of g₁ are t = ±1, and the critical value of g₂ is t = −1. Consequently, extremal values of g₁ and g₂ on $$ are the following:

Therefore, from (37) and (38) we conclude that

It is known that for the spectrum of 1 + εB one has

Therefore, So, if then one can see 1 + ελ_k(w) ≥ 0 for all ∥w∥ ≤ 1,   $$. This implies that the matrix 1 + εB is positive for all w with ∥w∥ ≤ 1.

Now assume that Δ_ε is positive. Then Δ_ε(x) is positive whenever x is positive. This means that 1 + ελ_k(w) ≥ 0 for all ∥w∥ ≤ 1($$). From (34) and (41) we conclude that |ε | ≤ 1/3. This completes the proof.

Theorem 13. Let ε = 1/3 then the corresponding q.q.o. Δ_ε is not KS operator.

Proof. It is enough to show the dissatisfaction of (21) at some values of w (∥w∥ ≤ 1) and f = (f₁, f₁, f₂).

Assume that f = (1,0, 0); then a little algebra shows that (21) reduces to the following one:

where

Now choose w as follows:

Then calculations show that Hence, we find which means that (45) is not satisfied. Hence, Δ_ε is not a KS operator at ε = 1/3.

Theorem 14. Let Δ_ε : 𝕄₂(ℂ) → 𝕄₂(ℂ) ⊗ 𝕄₂(ℂ) be given by (31). Then Δ_ε is completely positive if and only if $$.

Lemma 15. Let V_ε be given by (58). Then V_ε maps S into itself if and only if $$ is satisfied.

Proof. “If” Part. Assume that V_ε maps S into itself. Then (16) is satisfied. Take $$, p = f. Then from (16) one finds that

which yields $$.

“Only If” Part. Assume that $$. Take any f = (f₁, f₂, f₃),  p = (p₁, p₂, p₃) ∈ S. Then one finds that

This completes the proof.

Remark 16. We stress that condition (16) is necessary for Δ to be a positive operator. Namely, from Theorem 12 and Lemma 15 we conclude that if $$ then the operator Δ_ε is not positive, while (16) is satisfied.

Proposition 17. If  $$ then V_ε has a unique fixed point (0,0, 0) in S. If  $$ then V_ε has the following fixed points: (0,0, 0) and $$ in S.

Proof. It is clear that (0,0, 0) is a fixed point of V_ε. If f_k = 0, for some k ∈ {1,2, 3} then due to $$, one can see that the only solution of (61) belonging to S is f₁ = f₂ = f₃ = 0. Therefore, we assume that f_k ≠ 0 (k = 1,2, 3). So, from (61) one finds

Denoting From (62) it follows that

According to our assumption x, y, z are nonzero, so from (64) one gets

where 2x ≠ −z and 2yz ≠ −1.

Dividing the second equality of (65) to the first one of (65) we find that

which with xz = y yields Simplifying the last equality one gets This means that either y = x or x + y = 1/2.

Assume that x = y. Then from xz = y, one finds z = 1. Moreover, from the second equality of (65) we have y + 2/y = 1 + 2y. So, y² + y − 2 = 0; therefore, the solutions of the last one are y₁ = 1,  y₂ = −2. Hence, x₁ = 1,  x₂ = −2.

Now suppose that x + y = 1/2; then x = 1/2 − y. We note that y ≠ 1/2, since x ≠ 0. So, from the second equality of (65) we find

So, 2y² − y − 1 = 0 which yields the solutions y₃ = −1/2,  y₄ = 1. Therefore, we obtain x₃ = 1, z₃ = −1/2 and x₄ = −1/2, z₄ = −2.

Consequently, solutions of (65) are the following ones:

Now owing to (63) we need to solve the following equations:

According to our assumption f_k ≠ 0, we consider cases when x_kz_k ≠ 0.

Now let us start to consider several cases.

Case  1. Let x₂ = 1, z₂ = 1. Then from (71) one gets f₁ = f₂ = f₃. So, from (61) we find $$, that is, f₁ = 1/3ε. Now taking into account $$ one gets 1/3ε² ≤ 1. From the last inequality we have $$. Due to Lemma 15 the operator V_ε is well defined if and only if $$; therefore, one gets $$. Hence, in this case a solution is $$.

Case  2. Let x₂ = 1, z₂ = −1/2. Then from (71) one finds f₁ = f₂, 2f₂ = −f₃. Substituting the last ones to (61) we get $$. Then, we have f₁ = −1/3ε, f₂ = −1/3ε, f₃ = 2/3ε. Taking into account $$ we find 1/9ε² + 4/9ε² + 1/9ε² ≤ 1. This means $$; due to Lemma 15 in this case the operator V_ε is not well defined; therefore, we conclude that there is no fixed point of V_ε belonging to S.

Using the same argument for the rest of the cases we conclude the absence of solutions. This shows that if $$ the operator V_ε has unique fixed point in S. If $$, then V_ε has three fixed points belonging to S. This completes the proof.

Theorem 18. Let V_ε be given by (58). Then the following assertions hold true:

• (i)

  if $$, then for any f ∈ S one has $$ as n → ∞.

• (ii)

  if $$, then for any f ∈ S with $$ one has $$ as n → ∞.

Proof. Let us consider the following function $$. Then we have

This means

Due to ε² ≤ 1/3 from (73) one finds that

which yields that the sequence $$ is convergent. Next we would like to find the limit of $$.

• (i)

  First we assume that $$; then from (73) we obtain

  $$ ()
  This yields that $$ as n → ∞, for all f ∈ S.

• (ii)

  Now let $$. Then consider two distinct subcases.

Case A. Let $$ and denote $$. Then one gets

Hence, we have ρ(V_ε(f)) ≤ dρ(f). This means $$. Hence, $$ as n → ∞.

Case B. Now take $$ and assume that f is not a fixed point. Therefore, we may assume that f_i ≠ f_j for some i ≠ j, otherwise from Proposition 17 one concludes that f is a fixed point. Hence, from (58) one finds

Similarly, one gets It is clear that |V_ε(f) _k | ≤|ε| (k = 1,2, 3). According to our assumption f_i ≠ f_j (i ≠ j) we conclude that one of |V_ε(f) _k| is strictly less than $$; this means $$. Therefore, from Case A, one gets that $$ as n → ∞.