Extracting square roots of Bergmann operators

Ottmar Loos

Institut für Mathematik Universität Innsbruck, Technikerstr. 25/7 A-6072 Innsbruck, Austria

In [2], W. Kaup and J. Sauter use the Gelfand-Naimark-Friedmann-Russo Theorem for JB^{∗}-triples to show that the positive square root of the Bergmann operator B(x, x̄), where x runs in a bounded symmetric domain 𝒟 determined by a JB^{∗}-triple, can be extended continuously to the boundary of 𝒟. The purpose of this note is to give an elementary proof of this fact, based only on the binomial series. Our method does not require holomorphic or continuous functional calculus, it yields explicit formulas for B(x, x̄)^{1/2}, and it is applicable to real bounded symmetric domains as well.

Regarding Jordan theory, we follow the notations and conventions of [3, 4] and not of [2]; in particular, the Jordan triple product and the quadratic operators are related by D(x, y)z = {xyz} = Q_{x+z}y − Q_{x}y − Q_{z}y, and the Bergmann operator is B(x, y) = Id − D(x, y) + Q_{x}Q_{y}.

Recall the binomial series expansion

g(t) = 1 − \sqrt{1 − t} = \frac{t}{1 + \sqrt{1 − t}} = ∑_{n=1}^{∞} β_{n}t^{n}  (1)

with positive coefficients β_{n} = (−1)^{n−1}\binom{1/2}{n}. It is well known that the series (1) converges absolutely and uniformly on I = [−1, 1]; in particular,

1 = g(1) = ∑_{n=1}^{∞} β_{n} = \frac{1}{2} + \frac{1}{8} + \frac{1}{16} + \frac{5}{128} + \frac{7}{256} + · · · .  (2)

From the relation (1 − g(t))^{2} = 1 − t we obtain the recursion formula

2β_{1} = 1,  2β_{n} = ∑_{i=1}^{n−1} β_{i}β_{n−i} (n ≥ 2)  (3)

which also shows that β_{n} ∈ Z[\frac{1}{2}]. — Next, we consider the continuous function

f(t) = \frac{t}{\sqrt{1 + \sqrt{1 − t}}}  (4)

on I and prove the following

Lemma. (a) On the interval I, the function f has the normally convergent power series expansion

f(t) = ∑_{n=1}^{∞} α_{n}t^{n}  (5)

with positive coefficients α_{n} ∈ Z[\frac{1}{\sqrt{2}}]; in particular,

1 = f(1) = ∑_{n=1}^{∞} α_{n} = \frac{1}{\sqrt{2}} (1 + \frac{1}{8} + \frac{7}{128} + \frac{33}{1024} + \frac{715}{32768} + ⋯) .  (6)

(b) The α_{n} are related to the binomial coefficients β_{n} by the recursion formula

∑_{i=1}^{n} α_{i}α_{n+1−i} = β_{n}.  (7)

Proof. From (1) and (4), we see that f(t) = t(β_{1} +tβ_{2} +t^{2}β_{3} +⋯)^{1/2}. Since β_{1} = 1/2, equation (2) shows ∑_{n=1}^{∞} 2β_{n+1} = 1. Hence the power series g_{1}(t) = ∑_{n=1}^{∞} 2β_{n+1}t^{n} converges absolutely on I and takes values in I. By substituting this expansion into the binomial series, it follows that f(t) = 2^{−1/2}t(1 + g_{1}(t))^{1/2} has a power series expansion converging absolutely on I.

Now we show that the coefficients α_{n} of this expansion are all positive (this proof is due to G. Kirchner). For real α, β ≥ 0, consider the function h_{αβ}(t) = (1 + \sqrt{1 − t})^{−α} · (1 − s)^{−β/2}. A computation shows 2h′_{αβ} = αh_{α+1,β+1} +βh_{α,β+2}. Clearly, h_{αβ}(0) = 2^{−α} > 0. It follows inductively that all derivatives of h_{αβ} at the origin are positive. Since f(t) = t · h_{\frac{1}{2} ,0}(t), we see that the α_{n} are positive. Hence, ∑_{n=1}^{∞} α_{n} majorizes (5) on I, proving normal convergence.

From (4) and (1) it is clear that f(t)^{2} = tg(t) so the recursion formula (7) follows from the expansions (1) and (5) by comparing coefficients. Since (7) may also be written as 2α_{1}α_{n} = β_{n} − ∑_{i=2}^{n−1} α_{i}α_{n+i} and α_{1}^{2} = β_{1} = 1/2, we see from β_{n} ∈ Z [\frac{1}{2}] that α_{n} ∈ Z[\frac{1}{\sqrt{2}}].

Consider now a real or complex Banach Jordan pair V = (V^{+}, V^{−}); i.e., a Jordan pair where the spaces V^{±} are Banach spaces and the triple product satisfies the inequality ∥{xyz}∥ ≤ 2∥x∥·∥y∥·∥z∥. It is easy to see that the n-th power of an element x ∈ V^{+} in the y-homotope V_{y}^{+} satisfies the estimate

∥x^{(n,y)}∥ ≤ ∥x∥^{n} · ∥y∥^{n−1}.  (8)

We denote the open and closed unit balls in V^{±} by 𝒟^{±} and 𝒦^{±}, respectively.

Theorem. The formulas

ψ(x, y) = ∑_{n=1}^{∞} β_{n} x^{(n,y)}, φ_{+}(x, y) = ∑_{n=1}^{∞} α_{n} x^{(n,y)}, φ_{−}(x, y) = ∑_{n=1}^{∞} α_{n} y^{(n,x)},  (9)

define continuous maps ψ: 𝒦^{+}×𝒦^{−} → 𝒦^{+} and φ_{±}: 𝒦^{+}×𝒦^{−} → 𝒦^{±} whose restrictions to 𝒟^{+}×𝒟^{−} are of class C^{ω} and take values in 𝒟^{±}. For all (x, y) ∈ 𝒦^{+} × 𝒦^{−} we have

B(φ_{+}(x, y), φ_{−}(x, y)) = B(ψ(x, y), y),  (10)
B(ψ(x, y), y)^{2} = B(x, y),  (11)

Proof. Because of (2), (6) and (8), the series (9) converge normally on 𝒦^{+} × 𝒦^{−} and hence define continuous maps which are of class C^{ω} on 𝒟^{+} × 𝒟^{−}. For (10), we expand both sides and use the formulas [5, (3.11)] as well as (7):

\begin{aligned}D(φ_{+}(x, y), φ_{−}(x, y)) & = ∑_{i,j=1}^{∞} α_{i}α_{j}D(x^{(i,y)}, y^{(j,x)}) = ∑_{i,j=1}^{∞} α_{i}α_{j}D(x^{(i+j−1,y)}, y) \\ & = ∑_{n=1}^{∞} (∑_{i=1}^{n} α_{i}α_{n+1−i}) D(x^{(n,y)}, y) = ∑_{n=1}^{∞} β_{n}D(x^{(n,y)}, y) = D(ψ(x, y), y). \end{aligned}

A similar computation using [5, (3.10)] shows Q(φ_{+}(x, y))Q(φ_{−}(x, y)) = Q(ψ(x, y))Q(y). This proves (10). In the same way, expansion and the recursion formula (3) show that 2ψ(x, y) − ψ(x, y)^{(2,y)} = x for all (x, y) ∈ 𝒦^{+} × 𝒦^{−}. Now (11) follows from the identity JP25 of [3].

Finally, let E be a JB^{∗}-triple with triple product {xy^{∗}z}, complex linear in x and z and antilinear in y, and let 𝒟 be the open unit ball of E. Also let E^{−} be the complex conjugate vector space of E; i.e., E = E^{−} as a real Banach space but scalar multiplication in E^{−} is twisted with complex conjugation. The maps E → E^{−} and E^{−} → E which are the identity on the underlying real vector spaces are antilinear and will be denoted by x ↦ x̄. Then the pair V = (E, E^{−}) is a complex Banach Jordan pair with quadratic operators 2Q_{x}y = \{xȳ^{∗}x\} and 2Q_{y}x = \overline{\{ȳx^{∗}ȳ\}} (x ∈ E, y ∈ E^{−}). For x ∈ 𝒟, the operator B(x, x̄) has strictly positive spectrum [1, Cor. 3.4.ii], so by holomorphic functional calculus (in the Banach algebra of all bounded linear operators on E), it has a welldefined square root B(x, x̄)^{1/2} whose spectrum is again strictly positive.

Corollary. With E and 𝒟 as above, we have

B(ψ(x, x̄), x̄) = B(φ_{+}(x, x̄), φ_{−}(x, x̄)) = B(x, x̄)^{1/2}  (12)

for all x ∈ 𝒟. Hence the function x ↦ B(x, x̄)^{1/2} on 𝒟 extends continuously the boundary of 𝒟.

Proof. Let x ∈ 𝒟. By the theorem, φ_{+}(x, x̄) = \overline{φ_{−}(x, x̄)} = ∑_{n=1}^{∞} α_{n}x^{2n−1} ∈ 𝒟. Here the x^{2n−1} = x^{(n,x̄)} are the odd powers of x in the Jordan triple system E. By (11), S := B(ψ(x, x̄), x̄) is a square root of T := B(x, x̄), and (10) and [1, Cor. 3.4.ii] show that it has strictly positive spectrum. Hence, it agrees with the square root T^{1/2} obtained from holomorphic functional calculus. Indeed, let Sp A denote the spectrum of a continuous linear operator A on E. Then the squaring function ( )^{2}: Sp S → Sp S^{2} = Sp T is a homeomorphism with inverse ( )^{1/2}: Sp T → Sp S, and both these functions extend to holomorphic functions in suitable open neighborhoods of the spectra. Now it follows from standard properties of holomorphic functional calculus that S = (S^{2})^{1/2} = T^{1/2}.

Example. In case of the open unit disc in the complex plane it is easy to see that

ψ(z, z̄) = \frac{z}{1 + \sqrt{1 − zz̄}}, φ_{+}(z, z̄) = \frac{z}{\sqrt{1 + \sqrt{1 − zz̄}}}.

References

183 (1983), 503–529.

1977.

[1] W. Kaup, A Riemann mapping theorem for bounded symmetric domains in complex Banach spaces., Math. Z.

[2] W. Kaup and J. Sauter, Boundary structure of bounded symmetric domains, preprint, 1999.

[3] O. Loos, Jordan pairs, Lecture Notes in Math., vol. 460, Springer-Verlag, 1975.

[4] , Bounded symmetric domains and Jordan pairs, Lecture notes, The University of California at Irvine, [5] H. P. Petersson, Zur Arithmetik der Jordan-Paare, Math. Z. 147 (1976), 139–161.