Symmetric and Antisymmetric Tensor Products for the Function-Theoretic Operator Theorist

We study symmetric and antisymmetric tensor products of Hilbert-space operators, focusing on norms and spectra for some well-known classes favored by function-theoretic operator theorists. We pose many open questions that should interest the field.


INTRODUCTION
Tensor products and their symmetrization have appeared in the literature since the midnineteenth century, such as in Riemann's foundational work on differential geometry [26,27].Tensors describe many-body quantum systems [24] and symmetric tensors underpin the foundations of general relativity [3].In a separate yet overlapping vein, multilinear algebra [16] and representation theory [11] utilize symmetric tensor product spaces.
Decomposing a symmetric tensor into a minimal linear combination of tensor powers of the same vector arises in mobile communications, machine learning, factor analysis of kway arrays, biomedical engineering, psychometrics, and chemometrics; see [4,6,9,30,33] and the references therein.We refer the reader to [5] for a study of this decomposition problem.Symmetric tensors also arise in statistics [23].
In quantum mechanics, many-body systems are represented in terms of tensor products of wave functions.In the simplest case, where the systems do not interact, the Hamiltonian of the many-body system corresponds to a symmetric tensor product of operators [20,Ch. 4,Sec. 9].Recently there has been an endeavour within the physics community to study selfadjoint extensions of symmetric tensor products of operators [18,19,22].Furthermore, the symmetric part of a quantum geometric tensor can be exploited as a tool to detect quantum phase transitions in PT -symmetric quantum mechanics [34].
Unfortunately, there is little literature about symmetric tensor products of non-normal operators.The purpose of this paper is to introduce the basic ideas to the function-theoretic operator theory community.We study some fundamental operator-theoretic questions in this area, such as finding the norm and spectrum of symmetric tensor products of operators.We work through some examples with familiar operators, such as the unilateral shift, its adjoint, and diagonal operators.Given the ramifications of symmetric tensor products in a broad spectrum of fields, we hope that initiating this study will shed new light on classical problems and lead to new directions of study for function-theoretic operator theorists.
The layout of this paper is as follows.Section 2 introduces symmetric and antisymmetric tensor power spaces, the domains for the operators in Section 3. In Section 4 we collect results on operator-theoretic properties of symmetric tensor products of bounded operators.Section 5 is devoted to the norm of such operators, while Section 6 focuses on the spectrum.We study symmetric tensor products of diagonal operators in Section 7, the forward and backward shift operators in Section 8, and the symmetric tensor product of shifts and diagonal operators in Section 9. We conclude in Section 10 with a host of open questions that should appeal to researchers in function-theoretic operator theory.

SYMMETRIC AND ANTISYMMETRIC TENSOR POWER SPACES
Symmetric and antisymmetric tensor powers are familiar in mathematical physics, but less so in function-theoretic operator theory.We summarize the basics, with abbreviated explanations or without proof; see [1,Sec. I.5] or [32,Section 3.8] for the details.
Let H be a complex Hilbert space, in which the inner product •, • is linear in the first argument and conjugate linear in the second.We assume that H has a countable orthonormal basis.A superscript − denotes the closure with respect to the norm of H.
Let B(H) denote the space of bounded linear operators on H.For u 1 , u 2 . . ., u n ∈ H, the simple tensor u 1 ⊗ u 2 ⊗ • • • ⊗ u n : H n → C acts as follows: A simple tensor is a conjugate-multilinear function of its arguments.The map taking an n-tuple in H n to the corresponding simple tensor is linear in each argument.
Let H ⊗n denote the C-vector space spanned by the simple tensors.There is a unique inner product on H ⊗n such that Definition 2.2 (Tensor powers of Hilbert spaces).Let H ⊗0 := C. For n = 1, 2, . . ., let H ⊗n denote the completion of H ⊗n with respect to the inner product (2.1).
For n = 2, we may write H ⊗ H instead of H ⊗2 .If e 1 , e 2 , . . . is an orthonormal basis for H, then e i1 ⊗ e i2 ⊗ • • • ⊗ e in for (i 1 , i 2 , . . ., i n ) ∈ N n is an orthonormal basis for H ⊗n .
Let Σ n be the group of permutations of [n] := {1, 2, . . ., n}.For all π ∈ Σ n and u 1 , u 2 , . . ., u n ∈ H, define The density of the span of the simple tensors ensures that π extends to a bounded linear map on H ⊗n .
Proof.(a) Since the span of the simple tensors is dense in H ⊗n , it suffices to observe that Therefore, π * = π −1 and hence π −1 = π * by (a).
We now define certain subspaces of H ⊗n that respect the action of the operators π.
We may write H ⊙ H and H ∧ H instead of H ⊙2 and H ∧2 , respectively.In this case, there is only one nonidentity π ∈ Σ 2 .
Example 2.5.Let H 2 (D) denote the Hardy space on the unit disk D. The monomials 1, z, z 2 , . . .are an orthonormal basis for H 2 (D), so the simple tensors z i ⊗ z j for i, j = 0, 1, . . .are an orthonormal basis for ), the Hardy space on the bidisk D 2 [10].Thus, we identify and respectively.We freely use these identifications in what follows.
We say "orthogonal" instead of "orthonormal" because the vectors described in the previous proposition need not be unit vectors.Let m ℓ denote the number of occurrences of ℓ in (i give rise to the same simple tensor.Thus, . If dim H = d is finite, then (using the notation for binomial coefficients) The case n = 2 is special since dim , which suggests Proposition 2.11.The simple symmetric and antisymmetric tensors are Proof.Let π be the nonidentity permutation in Σ 2 .If x ∈ H ⊗2 , then and hence H ⊗2 = H ⊙2 + H ∧2 .Since π is unitary (Proposition 2.3) and involutive (since where the direct summands are defined by (2.6) and (2.7), respectively.In this context, z i w i and (z i w j + z j w i )/ √ 2 for 0 i < j form an orthonormal basis for H 2 sym (D 2 ) and (z i w j − z j w i )/ √ 2 for i < j form an orthonormal basis for H 2 asym (D 2 ).
Proof.Proposition (2.10) ensures that i j Lemma 2.15.For u, v ∈ H, we have 1 √ 2 u v u⊙v u v ; both inequalities are sharp.In particular, the symmetric tensor product of two nonzero vectors is nonzero.
Proof.The Cauchy-Schwarz inequality provides the upper inequality since In (2.16), | u, v | 2 is nonnegative, so we obtain the lower inequality.The upper inequality is sharp if u = v and the lower inequality is sharp if u ⊥ v.

SYMMETRIC AND ANTISYMMETRIC TENSOR PRODUCTS OF OPERATORS
This extends by linearity to the linear span H ⊗n of the simple tensors.The density of H ⊗n in H ⊗n ensures that [32, (3.8.17)]: We may write Then H ⊙n and H ∧n are invariant under a sum of elements in H ⊙n .The density of the simple symmetric tensors in H ⊙n ensures that T H ⊙n ⊆ H ⊙n .The proof that T H ∧n ⊆ H ∧n is similar.
The proposition above suggests the following definition.

Definition 3.3 (Symmetric tensor products of operators). Let
to H ⊙n and H ∧n , respectively.We may write A ⊙n and A ∧n instead of Symmetric tensor products are permutation invariant: (b) First we have A ⊙n A n from (a).Then ), the linear extension of the map z i w j → T g (z i )T g (w j ), has norm g 2 ∞ .Proposition 3.4 says that T g ⊙ T g , the restriction of T g ⊗ T g to H 2 sym (T 2 ), also has norm g 2 ∞ .

BASIC PROPERTIES
In this section we collect some results on the operator-theoretic properties of symmetric tensor products of bounded Hilbert-space operators.
Proof.This follows from Proposition 3.2.
For example, for the 2 × 2 matrix A in (4.3), the formula (4.4) gives . ., A n are normal and commute, then Proof.(a) This follows from Proposition 4.6.
(b) The Fuglede-Putnam theorem [25] ensures that A i A * j = A * j A i for 1 i, j n.Proposition 4.6 and a computation establish the normality of is not normal.Thus, the commutativity hypothesis is necessary in (b).
is not.
Proof.Since P and Q are selfadjoint, 2S 2 (P, Q) is selfadjoint.Since P Q = QP = 0, we have (2S 2 (P, Q)) 2 = 2S 2 (P, Q).Thus, 2S 2 (P, Q) is an orthogonal projection, so One can define tensor powers of bounded conjugate-linear operators.An analogue of Proposition 3.2 shows that H ⊙n is invariant under S n (C 1 , C 2 , . . ., C n ) for any bounded conjugate-linear operators C 1 , C 2 , . . ., C n .We say that C is a conjugation on H if C is conjugate linear, isometric, and involutive.We say that it follows that C ⊙n is a well-defined conjugation on H ⊙n .The desired result follows from part (a) and Proposition 4.6.

NORMS AND SPECTRAL RADIUS
In this section we provide various bounds for the norm of symmetric tensor products of operators, as well as a spectral-radius formula for symmetric tensor powers.
Equality is attained for A = 1 0 0 0 , B = 0 0 1 0 , and x = 1 0 .Indeed, (4.4) ensures that If there is a unit vector x such that Ax = 0 and Bx = 0, then (a) ensures that 0 < 1 √ 2 Ax Bx A ⊙ B .So suppose that Ax = 0 or Bx = 0 for all x ∈ H. Pick u such that Au = 0 and v such that Bv = 0. Then Bu = 0 and Av = 0; moreover, u = −v.Let x = u+v u+v , then (a) ensures that In both cases, A ⊙ B = 0 since it has positive norm.
In contrast to symmetric tensor products, the antisymmetric products of nonzero operators may be 0. If P is a rank-one orthogonal projection, then P ∧n = 0 for n 2.
Theorem 5.2.(a) If A 1 , A 2 , . . ., A n ∈ B(H) and the A i have orthogonal ranges, then The inequalities are sharp.
Proof.(a) Recall that the set of finite sums of simple tensors are dense in H ⊗n .Take the supremum over such vectors and observe that The prepenultimate equality above follows because the A i have orthogonal ranges; the final inequality is due to (3.1).For n = 2, the matrices from the proof of Theorem 5.1.ahave orthogonal ranges and demonstrate that the inequality is sharp.
(b) Part (a) ensures that the desired upper inequality holds and is sharp.It suffices to examine the lower inequality.For u ∈ (ker A) ⊥ and v ∈ ker A, and (ker B) ⊥ ⊆ ker A. To see that the lower inequality is sharp, let A = 1 0 0 0 and B = I − A, so (ker B) ⊥ ⊆ ker A and ran B ⊆ (ran A) ⊥ .Then Proposition 4.12 yields A ⊙ B = 1 2 .

SPECTRUM
Here we present results on the spectrum of symmetric products of Hilbert-space operators (the finite-dimensional case is simpler; see [1, p. 18]).We find a complete description in some special cases.In what follows, σ(A), σ p (A), and σ ap (A) denote the spectrum, point spectrum and approximate point spectrum of A, respectively [15, Def.2.4.5].For X, Y ⊆ C, let X + Y := {x + y : x ∈ X, y ∈ Y } and XY := {xy : x ∈ X, y ∈ Y }.
Proof.This follows from the direct-sum decomposition (3.6).

DIAGONAL OPERATORS
Since diagonal operators are among the most elementary operators one encounters in the infinite-dimensional setting [15,Ch. 2], it makes sense to consider their symmetric tensor products.Let e 1 , e 2 , . . .be an orthonormal basis for H and suppose that L, M ∈ B(H) satisfy Le i = λ i e i and M e i = µ i e i for i 1.For i, j 1, 4 (λ i e i ⊗ µ j e j + µ i e i ⊗ λ j e j + λ j e j ⊗ µ i e i + µ j e j ⊗ λ i e i ) = 1  2 (λ i µ j + λ j µ i )(e i ⊙ e j ).(7.1) Thus, L ⊙ M is a diagonal operator with For symmetric products of diagonal operators, we can improve upon Theorem 5.1.a.
and one of the following holds: (a) The lower bound is attained by Then σ(D) has planar measure zero, but an argument similar to that in (a) ensures that σ(D ⊙ D) is the annulus centered at 0 with radii 1 and e 2 , which has positive measure.
Below the brackets { { and } } indicate a multiset; that is, a set that permits multiplicity.

THE SHIFT OPERATOR AND ITS ADJOINT
In this section we find the spectrum of the symmetric tensor product of the unilateral shift and its adjoint.Let (Sf )(z) = zf (z) denote the unilateral shift on H 2 (D) [15,Ch. 5].Its adjoint is the backward shift

and consider
and note that dim ) is an orthogonal direct sum.We have With respect to the orthonormal basis {z k−i w i } k i=0 of V k , we identify the restriction T | V k with the (k + 1) × (k + 1) matrix (by convention A 0 = [0]) From [21, Prop.2.1], we have For k odd, with respect to the orthonormal basis For k even, with respect to the orthonormal basis , with the convention B 0 = [0].We identify the spectrum of the B k later.

SHIFTS AND DIAGONAL OPERATORS
We consider here the symmetric tensor product of shift operators and diagonal operators.This setting suggests working on the sequence space ℓ 2 instead of H 2 (D) [15,Sec. 1.2].Let e 0 , e 1 , . . .be the standard basis of ℓ 2 and consider the unilateral shift Se i = e i+1 [15,Ch. 3].Its adjoint is given by S * e i = e i−1 for i 1 and S * e 0 = 0. (b) Suppose that the set of nonzero µ i is bounded away from zero.Note that for all i, j 0, If some µ i = 0, then (9.2) ensures that 0 ∈ σ p (S ⊙ M ) since (S ⊙ M )(e i ⊙ e i ) = 0. Thus, we may assume that |µ i | δ > 0 for all i 0. Define C = M /δ.Let i j |a ij | 2 < ∞ and let v = 2 0 i j<∞ a ij e i ⊙ e j , which is well defined by Lemma 2.14.Then (9.2) ensures that Then (S ⊙ M )v = 0 if and only if the a k,ℓ are square summable and (9.4) -(9.7) vanish for all ℓ k 0. We define such a k,ℓ , not all zero, in four steps; see Figure 1.
(3) Let a 1,1 = 0 and, for each k 2, let a k,k and a k−1,k+1 be such that (4) For k 2, let Then (9.7) vanishes for all k 1 with odd ℓ k + 3.This completes the definition of the a k,ℓ .We must prove that they are square summable.
For k 1, (9.8) yields (9.10) Then (9.9), and then (9.10) with k + 1 in place of k ensure that for k 1.Next (9.9) and then (9.11) with k + 1 in place of k imply that (9.12) Then Step 1, Step 2, and (9.12) ensure that which is finite by a standard argument in the study of elliptic functions [31,Prop. 10.4.2]. 1  Thus, v is a well-defined vector in the kernel of S ⊙ M .
(c) Suppose that λ = 0 and (S ⊙ M )v = λv, in which v = 2 0 i j<∞ a ij e i ⊙ e j and The double sum can be explicitly evaluated.Write the summands in an array with r indexing the columns and k the rows.Sum each column and simplify to reduce the double sum to the well-known ∞ k=1 1 Lemma 2.15 prompts us to consider symmetric tensor products of more than two vectors Thus, only lower bounds on Here is a partial answer.Problem 2. For A 1 , A 2 , . . ., A n ∈ B(H) is Proposition 7.2 provides the sharp inequalities L M ( √ 2 − 1) L ⊙ M L M for diagonal operators L, M (with respect to the same orthonormal basis).Since the upper bound easily generalizes, the lower bound is of greater interest.The general problem suggested by the previous questions is the following.Problem 9.For A 1 , A 2 , . . ., A n ∈ B(H), describe the norm and spectrum (and its parts) of There are countless other questions that can be raised.For example, what can be said about symmetric tensor products of composition operators?

Problem 3 .
Let A 1 , A 2 , . . ., A n ∈ B(H) be diagonal operators (with respect to the same orthonormal basis).Find a sharp lower bound, in the spirit of Proposition 7.2, onA 1 ⊙ A 2 ⊙ • • • ⊙ A n in terms of A 1 , A 2 , . . ., A n .The Weyl-von Neumann-Berg theorem asserts that every normal operator on a separable Hilbert space is the sum of a diagonal operator and a compact operator of arbitrarily small norm[8, Cor.II.4.2].This suggests possible extensions to normal operators.

Problem 4 .Problem 6 .
Let A 1 , A 2 , . . ., A n ∈ B(H) be commuting normal operators.Find a sharp lower bound, in the spirit of Proposition 7.2, onA 1 ⊙ A 2 ⊙ • • • ⊙ A n in terms of A 1 , A 2 , . . ., A n .Proposition 7.5 suggests the following.Problem 5. Let A 1 , A 2 , . . ., A n ∈ B(H) be commuting normal operators.Describe σ(A 1 ⊙ A 2 ⊙ • • • ⊙ A n ) (and its parts).Let us now consider the unilateral shift S and its adjoint.Theorem 8.1 identified the norm and spectrum of S ⊙ S * and S ∧ S * .What can be said about other combinations?Identify the norm and spectrum of arbitrary symmetric or antisymmetric tensor products of S and S * (for example, consider S 2 ⊙ S ⊙ S * 3 and S 2 ∧ S ∧ S * 3 ).Problem 7. Describe the norm and spectrum of S α ⊙ S * α and S α ∧ S * α , in which S α is a weighted shift operator.What can be said if more factors are included?Theorems 9.1 and 9.19 answer some questions about S ⊙ M and S * ⊙ M , in which M = diag(µ 0 , µ 1 , . ..) is a diagonal operator.However, a complete picture eludes us.Problem 8. Identify the norm and spectrum (and its parts) for S ⊙ M and S * ⊙ M .
S n is the orthogonal projection from H ⊗n onto H ⊙n .(b) A n is the orthogonal projection from H ⊗n onto H ∧n .In particular, H ⊙n and H ∧n are closed subspaces of H ⊗n .Proof.(a) Use Proposition 2.3 and the fact that πS n = S n for all π ∈ Σ n to show that S 2 n = S n = S * n and ran S n = H ⊙n .The proof of (b) is similar.Let v 1 , v 2 , . . ., v n ∈ H and define the simple symmetric and antisymmetric tensors Definition 2.8 (Symmetrization and antisymmetrization operators).Let sgn π denote the sign of a permutation π ∈ Σ n .Define A n : H ⊗n → H ⊗n and S n : H ⊗n → H ⊗n by e 2 , e 3 , . . . is an orthonormal basis for H, then ⊙ e j ) for i < j and e i ⊙ e i for i 1 form an orthonormal basis for H ⊙2 , and ∧ e j ) for i < j form an orthonormal basis for H ∧2 .Proposition 2.11.H ⊗2 = H ⊙2 ⊕ H ∧2 is an orthogonal decomposition.