Strongly continuous unitary group

A strongly continuous unitary group is a family $U(t)$ for all real $t$ that is strongly continuous, obeys the group law, and preserves the Hilbert norm. Its generator is skew-adjoint; conversely, every skew-adjoint operator generates such a group.

Every strongly continuous unitary group has a unique possibly unbounded self-adjoint operator $A$ as its generator. Its domain consists of the vectors for which the norm limit of $(U_t\xi -\xi )/(it)$ exists as $t\to 0$. Conversely, the Borel functional calculus for a normal operator constructs a strongly continuous unitary group from every self-adjoint $A$. Smooth averaging supplies a dense generator domain; Laplace integrals of $U_t$ and $U_{-t}$ prove that $A\pm i$ are surjective, which establishes self-adjointness.

An operator $A$ is skew-adjoint when $A^{\ast }=-A$. Both $A$ and $-A$ are then maximal dissipative, and the generated group is unitary.

On a finite-dimensional real inner product space, a skew-adjoint operator has even rank. In an orthonormal basis its matrix is skew-symmetric, so an invertible such matrix must have even dimension by $\det A=\det (-A)=(-1{)}^d\det A$. Moreover $\ker A=(\mathrm{im}A{)}^{\perp }$, so the image and kernel intersect trivially. The restriction of $A$ to its image is an invertible skew-adjoint map, and its dimension, the rank, is therefore even.

On a finite-dimensional positive inner-product space, a skew-adjoint nilpotent operator is zero. In an orthonormal basis, $\Vert A{\Vert }_F^2=\mathrm{tr}(A^{\ast }A)=-\mathrm{tr}(A^2)=0$, since a nilpotent matrix has zero trace and so does its square. The Frobenius norm therefore vanishes. In particular, a nonzero nilpotent adjoint map rules out a compact Lie algebra.