A Trace Inequality for Commuting d-Tuples of Operators

Abstract

For a commuting d-tuple of operators $\varvec{T}$ defined on a complex separable Hilbert space $\mathcal H$, let $\big [ \!\!\big [ \varvec{T}^*, \varvec{T} \big ]\!\!\big ]$ be the $d\times d$ block operator $\big (\!\!\big (\big [ T_j^* , T_i\big ]\big )\!\!\big )$ of the commutators $[T^*_j , T_i] := T^*_j T_i - T_iT_j^*$. We define the determinant of $\big [ \!\!\big [ \varvec{T}^*, \varvec{T} \big ]\!\!\big ]$ by symmetrizing the products in the Laplace formula for the determinant of a scalar matrix. We prove that the determinant of $\big [ \!\!\big [ \varvec{T}^*, \varvec{T} \big ]\!\!\big ]$ equals the generalized commutator of the 2d - tuple of operators, $(T_1,T_1^*, \ldots , T_d,T_d^*)$ introduced earlier by Helton and Howe. We then apply the Amitsur–Levitzki theorem to conclude that for any commuting d-tuple of d-normal operators, the determinant of $\big [ \!\!\big [ \varvec{T}^*, \varvec{T} \big ]\!\!\big ]$ must be 0. We show that if the d-tuple $\varvec{T}$ is cyclic, the determinant of $\big [ \!\!\big [ \varvec{T}^*, \varvec{T} \big ]\!\!\big ]$ is non-negative and the compression of a fixed set of words in $T_j^*$ and $T_i$—to a nested sequence of finite dimensional subspaces increasing to $\mathcal H$—does not grow very rapidly, then the trace of the determinant of the operator $\big [\!\! \big [ \varvec{T}^* , \varvec{T}\big ] \!\!\big ]$ is finite. Moreover, an upper bound for this trace is given. This upper bound is shown to be sharp for a class of commuting d-tuples. We make a conjecture of what might be a sharp bound in much greater generality and verify it in many examples.