Fuglede's theorem
In mathematics, Fuglede's theorem is a result in operator theory, named after Bent Fuglede.
The resultEdit
Theorem (Fuglede) Let T and N be bounded operators on a complex Hilbert space with N being normal. If TN = NT, then TN* = N*T, where N* denotes the adjoint of N.
Normality of N is necessary, as is seen by taking T=N. When T is self-adjoint, the claim is trivial regardless of whether N is normal:
<math display="block">TN^* = (NT)^* = (TN)^* = N^*T.</math>
Tentative Proof: If the underlying Hilbert space is finite-dimensional, the spectral theorem says that N is of the form <math display="block">N = \sum_i \lambda_i P_i </math> where Pi are pairwise orthogonal projections. One expects that TN = NT if and only if TPi = PiT. Indeed, it can be proved to be true by elementary arguments (e.g. it can be shown that all Pi are representable as polynomials of N and for this reason, if T commutes with N, it has to commute with Pi...). Therefore T must also commute with <math display="block">N^* = \sum_i {\bar \lambda_i} P_i.</math>
In general, when the Hilbert space is not finite-dimensional, the normal operator N gives rise to a projection-valued measure P on its spectrum, σ(N), which assigns a projection PΩ to each Borel subset of σ(N). N can be expressed as <math display="block">N = \int_{\sigma(N)} \lambda d P(\lambda). </math>
Differently from the finite dimensional case, it is by no means obvious that TN = NT implies TPΩ = PΩT. Thus, it is not so obvious that T also commutes with any simple function of the form <math display="block">\rho = \sum_i {\bar \lambda} P_{\Omega_i}.</math>
Indeed, following the construction of the spectral decomposition for a bounded, normal, not self-adjoint, operator T, one sees that to verify that T commutes with <math>P_{\Omega_i}</math>, the most straightforward way is to assume that T commutes with both N and N*, giving rise to a vicious circle!
That is the relevance of Fuglede's theorem: The latter hypothesis is not really necessary.
Putnam's generalizationEdit
The following contains Fuglede's result as a special case. The proof by Rosenblum pictured below is just that presented by Fuglede for his theorem when assuming N=M.
Theorem (Calvin Richard Putnam)<ref name="Putnam1951">Template:Cite journal</ref> Let T, M, N be linear operators on a complex Hilbert space, and suppose that M and N are normal, T is bounded and MT = TN. Then M*T = TN*.
First proof (Marvin Rosenblum): By induction, the hypothesis implies that MkT = TNk for all k. Thus for any λ in <math>\Complex</math>, <math display="block">e^{\bar\lambda M}T = T e^{\bar\lambda N}.</math>
Consider the function <math display="block">F(\lambda) = e^{\lambda M^*} T e^{-\lambda N^*}.</math> This is equal to <math display="block">e^{\lambda M^*} \left[e^{-\bar\lambda M}T e^{\bar\lambda N}\right] e^{-\lambda N^*} = U(\lambda) T V(\lambda)^{-1},</math> where <math>U(\lambda) = e^{\lambda M^* - \bar\lambda M}</math> because <math>M</math> is normal, and similarly <math>V(\lambda) = e^{\lambda N^* - \bar\lambda N}</math>. However we have <math display="block">U(\lambda)^* = e^{\bar\lambda M - \lambda M^*} = U(\lambda)^{-1}</math> so U is unitary, and hence has norm 1 for all λ; the same is true for V(λ), so <math display="block">\|F(\lambda)\| \le \|T\|\ \forall \lambda.</math>
So F is a bounded analytic vector-valued function, and is thus constant, and equal to F(0) = T. Considering the first-order terms in the expansion for small λ, we must have M*T = TN*.
The original paper of Fuglede appeared in 1950; it was extended to the form given above by Putnam in 1951.<ref name="Putnam1951"/> The short proof given above was first published by Rosenblum in 1958; it is very elegant, but is less general than the original proof which also considered the case of unbounded operators. Another simple proof of Putnam's theorem is as follows:
Second proof: Consider the matrices
<math display="block">T' = \begin{bmatrix} 0 & 0 \\ T & 0 \end{bmatrix} \quad \text{and} \quad N' = \begin{bmatrix} N & 0 \\ 0 & M \end{bmatrix}.</math>
The operator N' is normal and, by assumption, T' N' = N' T' . By Fuglede's theorem, one has <math display="block">T' (N')^* = (N')^*T'. </math>
Comparing entries then gives the desired result.
From Putnam's generalization, one can deduce the following:
Corollary If two normal operators M and N are similar, then they are unitarily equivalent.
Proof: Suppose MS = SN where S is a bounded invertible operator. Putnam's result implies M*S = SN*, i.e. <math display="block">S^{-1} M^* S = N^*. </math>
Take the adjoint of the above equation and we have <math display="block">S^* M (S^{-1})^* = N. </math>
So <math display="block">S^* M (S^{-1})^* = S^{-1} M S \quad \Rightarrow \quad SS^* M (SS^*)^{-1} = M.</math>
Let S*=VR, with V a unitary (since S is invertible) and R the positive square root of SS*. As R is a limit of polynomials on SS*, the above implies that R commutes with M. It is also invertible. Then <math display="block">N = S^*M (S^*)^{-1}=VRMR^{-1}V^*=VMV^*. </math>
Corollary If M and N are normal operators, and MN = NM, then MN is also normal.
Proof: The argument invokes only Fuglede's theorem. One can directly compute <math display="block">(MN) (MN)^* = MN (NM)^* = MN M^* N^*. </math>
By Fuglede, the above becomes <math display="block">= M M^* N N^* = M^* M N^*N. </math>
But M and N are normal, so <math display="block">= M^* N^* MN = (MN)^* MN. </math>
C*-algebrasEdit
The theorem can be rephrased as a statement about elements of C*-algebras.
Theorem (Fuglede-Putnam-Rosenblum) Let x, y be two normal elements of a C*-algebra A and z such that xz = zy. Then it follows that x* z = z y*.