Invariant maximal positive subspaces and polar decompositions (Q2509110)

Let \({\mathcal H}\) be a complex Hilbert space with the inner product \(\langle \cdot ,\cdot \rangle\) and let \(J\) be a bounded invertible selfadjoint operator on \({\mathcal H}\). This operator induces a Krein space structure on \({\mathcal H}\) in a standard way: the (generally indefinite) inner product on \({\mathcal H}\) is defined by \([x,y]=\langle Jx,y \rangle\). A subspace \({\mathcal M}\subset{\mathcal H}\) is called uniformly \(J\)-positive if there exists \(\varepsilon>0\) such that \([x,x]\geq\varepsilon\langle x,x \rangle\) for all \(x\in {\mathcal M}\). The authors introduce the notion of maximal uniformly \(J\)-positive subspace in an obvious way. A~\(J\)-polar decomposition of an operator \(X\) is the representation \(X=UA\) where \(U\) is \(J\)-unitary (i.e., \(U^{-1}=U^{[\ast]}\), \(U^{[\ast]}\) is the conjugate operator with respect to \([\cdot,\cdot]\)) and \(A\) is \(J\)-selfadjoint. The following assertion is the main result of the paper. Let \(X\) be an invertible operator on \({\mathcal H}\), and suppose that \(X\) has an invariant maximal uniformly \(J\)-positive subspace \({\mathcal M}\) such that \(X(M^{[\perp]})\) is \(J\)-nonpositive. Then \(X\) allows a \(J\)-polar decomposition \(X=UA\) such that \[ \sigma(A)\subseteq\{ z\in C:\text{Re}z\geq|\text{Im}z|\}\setminus \{ 0\}. \] Moreover, the \(J\)-polar decomposition \(X=UA\) with the previous property of \(\sigma(A)\) is unique. If in addition, the restriction of \(X\) to \({\mathcal M}\) is invertible and the subspace \(X({\mathcal M}^{[\perp]})\) is uniformly \(J\)-negative, then for the unique polar decomposition of the type mentioned above, we actually have \[ \sigma(A)\subseteq\{ z\in C:\text{Re}z>| \text{Im}z|\}. \] As a corollary, several classes of operators are shown to allow polar decomposition.