Exponential growth

In mathematics, a trace class operator is a compact operator for which a trace may be defined, such that the trace is finite and independent of the choice of basis. Trace class operators are essentially the same as nuclear operators, though many authors reserve the term "trace class operator" for the special case of nuclear operators on Hilbert spaces, and reserve nuclear (=trace class) operators for more general Banach spaces.

Definition

Mimicking the definition for matrices, a bounded linear operator A over a separable Hilbert space H is said to be in the trace class if for some (and hence all) orthonormal bases {e_k}_k of H the sum of positive terms

$\Vert A{\Vert }_1=\mathrm{T}\mathrm{r}\left|A\right|:=\sum _k⟨(A^{*}A{)}^{1/2}{e}_k,e_k⟩$

is finite. In this case, the sum

$\mathrm{T}\mathrm{r}A:=\sum _k⟨A{e}_k,e_k⟩$

is absolutely convergent and is independent of the choice of the orthonormal basis. This value is called the trace of A. When H is finite-dimensional, every operator is trace class and this definition of trace of A coincides with the definition of the trace of a matrix.

By extension, if A is a non-negative self-adjoint operator, we can also define the trace of A as an extended real number by the possibly divergent sum

$\sum _k⟨A{e}_k,e_k⟩.$

Properties

1.	If A is a non-negative self-adjoint, A is trace class if and only if Tr(A) < ∞. Therefore a self adjoint operator A is trace class if and only if its positive part A+ and negative part A− are both trace class. (The positive and negative parts of a self adjoint operator are obtained via the continuous functional calculus.)
2.	The trace is a linear functional over the space of trace class operators, i.e. $\mathrm{Tr}(aA+bB)=a\mathrm{Tr}(A)+b\mathrm{Tr}(B).$ The bilinear map $⟨A,B⟩=\mathrm{Tr}(A^{*}B)$ is an inner product on the trace class; the corresponding norm is called the Hilbert-Schmidt norm. The completion of the trace class operators in the Hilbert-Schmidt norm are called the Hilbert-Schmidt operators.
3.	If $A$ is bounded and $B$ is trace class, $AB$ and $BA$ are also trace class and $\Vert AB{\Vert }_1=\mathrm{Tr}(|AB|)\le \Vert A\Vert \Vert B{\Vert }_1,\Vert BA{\Vert }_1=\mathrm{Tr}(|BA|)\le \Vert A\Vert \Vert B{\Vert }_1$ besides, under the same hypothesis, $\mathrm{Tr}(AB)=\mathrm{Tr}(BA)$ The last assertion also holds under the weaker hypothesis that $A$ and $B$ are Hilbert Schmidt.
4.	If $A$ is trace class, then one can define the Fredholm determinant of $1+A$ $\mathrm{d}\mathrm{e}\mathrm{t}(I+A):=\prod _{n\ge 1}[1+{\lambda }_n(A)]$ for $\{{\lambda }_n(A){\}}_n$ the elements of the spectrum of $A$; the trace class condition on $A$ guarantees that the infinite product is finite: indeed $\mathrm{d}\mathrm{e}\mathrm{t}(I+A)\le e^{\Vert A{\Vert }_1};$ it also guarantees that $\mathrm{d}\mathrm{e}\mathrm{t}(I+A)\ne 0$ if and only if $(I+A)$ is invertible.

Lidskii's theorem

Let $A$ be a trace class operator in a separable Hilbert space $H$, and let $\{{\lambda }_n(A){\}}_{n=1}^N,$ $N\le \mathrm{\infty }$ be the eigenvalues of $A$. Let us assume that ${\lambda }_n(A)$ are enumerated with algebraic multiplicities taken into account (i.e. if the algebraic multiplicity of $\lambda$ is $k$ then $\lambda$ is repeated $k$ times in the list ${\lambda }_1(A),{\lambda }_2(A),\dots$). Lidskii's theorem (named after Victor Borisovich Lidskii) states that

$\sum _{n=1}^N{\lambda }_n(A)=\mathrm{Tr}(A).$

Note that the series in the left hand side converges absolutely due to Weyl's inequality

$\sum _{n=1}^N|{\lambda }_n(A)|\le \sum _{m=1}^M{s}_m(A)$

between the eigenvalues $\{{\lambda }_n(A){\}}_{n=1}^N$ and the singular values $\{s_m(A){\}}_{m=1}^M$ of a compact operator $A$. See e.g.

Relationship between some classes of operators

One can view certain classes of bounded operators as noncommutative analogue of classical sequence spaces, with trace-class operators as the noncommutative analogue of the sequence space l¹(N). Indeed, applying the spectral theorem, every normal trace-class operator on a separable Hilbert space can be realized as a l¹ sequence. In the same vein, the bounded operators are noncommutative versions of l^∞(N), the compact operators that of c₀ (the sequences convergent to 0), Hilbert-Schmidt operators correspond to l²(N), and finite-rank operators the sequences that have only finitely many non-zero terms. To some extent, the relationships between these classes of operators are similar to the relationships between their commutative counterparts.

Recall that every compact operator T on a Hilbert space takes the following canonical form

$\mathrm{\forall }h\in H,Th=\sum _{i=1}{\alpha }_i⟨h,v_i⟩u_i\text{where}{\alpha }_i\ge 0\text{and}{\alpha }_i\to 0$

for some orthonormal bases {u_i} and {v_i}. Making the above heuristic comments more precise, we have that T is trace class if the series ∑_i α_i is convergent, T is Hilbert-Schmidt if ∑_i α_i² is convergent, and T is finite rank if the sequence {α_i} has only finitely many nonzero terms.

The above description allows one to obtain easily some facts that relate these classes of operators. For example, the following inclusions hold and they are all proper when H is infinite dimensional: {finite rank} ⊂ {trace class} ⊂ {Hilbert-Schmidt} ⊂ {compact}.

The trace-class operators are given the trace norm ||T||₁ = Tr [ (T*T)^½ ] = ∑_i α_i. The norm corresponding to the Hilbert-Schmidt inner product is ||T||₂ = (Tr T*T)^½ = (∑_iα_i²)^½. Also, the usual operator norm is ||T|| = sup_i(α_i). By classical inequalities regarding sequences,

$\Vert T\Vert \le \Vert T{\Vert }_2\le \Vert T{\Vert }_1,$

for appropriate T.

It is also clear that finite-rank operators are dense in both trace-class and Hilbert-Schmidt in their respective norms.

Trace class as the dual of compact operators

The dual space of c₀ is l¹(N). Similarly, we have that the dual of compact operators, denoted by K(H)*, is the trace-class operators, denoted by C₁. The argument, which we now sketch, is reminiscent of that for the corresponding sequence spaces. Let f ∈ K(H)*, we identify f with the operator T_f defined by

$⟨T_{f}x,y⟩=f(S_{x,y}),$

where S_x,y is the rank-one operator given by

$S_{x,y}(h)=⟨h,y⟩x.$

This identification works because the finite-rank operators are norm-dense in K(H). In the event that T_f is a positive operator, for any orthonormal basis u_i, one has

$\sum _i⟨T_f{u}_i,u_i⟩=f(I)\le \Vert f\Vert ,$

where I is the identity operator

$I=\sum _i⟨\cdot ,u_i⟩u_i.$

But this means T_f is trace-class. An appeal to polar decomposition extend this to the general case where T_f need not be positive.

A limiting argument via finite-rank operators shows that ||T_f ||₁ = || f ||. Thus K(H)* is isometrically isomorphic to C₁.

As the predual of bounded operators

Recall that the dual of l¹(N) is l^∞(N). In the present context, the dual of trace-class operators C₁ is the bounded operators B(H). More precisely, the set C₁ is a two-sided ideal in B(H). So given any operator T in B(H), we may define a continuous linear functional φ_T on $C_1$ by φ_T(A)=Tr(AT). This correspondence between elements φ_T of the dual space of $C_1$ and bounded linear operators is an isometric isomorphism. It follows that B(H) is the dual space of $C_1$. This can be used to define the weak-* topology on B(H).

References

1. Dixmier, J. (1969). Les Algebres d'Operateurs dans l'Espace Hilbertien. Gauthier-Villars.