A trace inequality for commuting tuple of operators

Abstract

For a commuting $d$- tuple of operators $\boldsymbol T$ defined on a complex separable Hilbert space $\mathcal H$, let $\big [ \!\!\big [ \boldsymbol T^*, \boldsymbol T \big ]\!\!\big ]$ 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 $\big [ \!\!\big [ \boldsymbol T^*, \boldsymbol T \big ]\!\!\big ]$ by symmetrizing the products in the Laplace formula for the determinant of a scalar matrix. We prove that the determinant of $\big [ \!\!\big [ \boldsymbol T^*, \boldsymbol T \big ]\!\!\big ]$ 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 $\big [ \!\!\big [ \boldsymbol T^*, \boldsymbol T \big ]\!\!\big ]$ must be $0$. We show that if the $d$- tuple $\boldsymbol T$ is cyclic, the determinant of $\big [ \!\!\big [ \boldsymbol T^*, \boldsymbol T \big ]\!\!\big ]$ 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 $\big [\!\! \big [ \boldsymbol T^* , \boldsymbol T\big ] \!\!\big ]$ 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.