title: Fuglede's theorem
text: # Fuglede's theorem
  
  In mathematics, Fuglede's theorem is a result in operator theory, named after Bent Fuglede.
  
  ## The result
  
  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:
  
  $TN^{*}=(NT)^{*}=(TN)^{*}=N^{*}T.$
  
  Tentative Proof: If the underlying Hilbert space is finite-dimensional, the spectral theorem says that N is of the form
  
  $N=\sum _{i}\lambda _{i}P_{i}$
  
  where $P_i$ are pairwise orthogonal projections. One expects that TN = NT if and only if $TP_i$ = $P_iT.$ Indeed, it can be proved to be true by elementary arguments (e.g. it can be shown that all $P_i$ are representable as polynomials of N and for this reason, if T commutes with N, it has to commute with $P_i...).$ Therefore T must also commute with
  
  $N^{*}=\sum _{i}{{\bar {\lambda }}_{i}}P_{i}.$
  
  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 $P_Ω$ to each Borel subset of σ(N). N can be expressed as
  
  $N=\int _{\sigma (N)}\lambda dP(\lambda ).$
  
  Differently from the finite dimensional case, it is by no means obvious that TN = NT implies $TP_Ω$ = $P_ΩT.$ Thus, it is not so obvious that T also commutes with any simple function of the form
  
  $\rho =\sum _{i}{\bar {\lambda }}P_{\Omega _{i}}.$
  
  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 $P_{\Omega _{i}}$, 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 generalization
  
  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) 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 $M^kT$ = $TN^k$ for all k. Thus for any λ in $\mathbb {C}$,
  
  $e^{{\bar {\lambda }}M}T=Te^{{\bar {\lambda }}N}.$
  
  Consider the function
  
  $F(\lambda )=e^{\lambda M^{*}}Te^{-\lambda N^{*}}.$
  
  This is equal to
  
  $e^{\lambda M^{*}}\left[e^{-{\bar {\lambda }}M}Te^{{\bar {\lambda }}N}\right]e^{-\lambda N^{*}}=U(\lambda )TV(\lambda )^{-1},$
  
  where $U(\lambda )=e^{\lambda M^{*}-{\bar {\lambda }}M}$ because $M$ is normal, and similarly $V(\lambda )=e^{\lambda N^{*}-{\bar {\lambda }}N}$. However we have
  
  $U(\lambda )^{*}=e^{{\bar {\lambda }}M-\lambda M^{*}}=U(\lambda )^{-1}$
  
  so U is unitary, and hence has norm 1 for all λ; the same is true for V(λ), so
  
  $\|F(\lambda )\|\leq \|T\|\ \forall \lambda .$
  
  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. 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
  
  $T'={\begin{bmatrix}0&0\\T&0\end{bmatrix}}\quad {\text{and}}\quad N'={\begin{bmatrix}N&0\\0&M\end{bmatrix}}.$
  
  The operator N' is normal and, by assumption, T' N' = N' T' . By Fuglede's theorem, one has
  
  $T'(N')^{*}=(N')^{*}T'.$
  
  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.
  
  $S^{-1}M^{*}S=N^{*}.$
  
  Take the adjoint of the above equation and we have
  
  $S^{*}M(S^{-1})^{*}=N.$
  
  So
  
  $S^{*}M(S^{-1})^{*}=S^{-1}MS\quad \Rightarrow \quad SS^{*}M(SS^{*})^{-1}=M.$
  
  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
  
  $N=S^{*}M(S^{*})^{-1}=VRMR^{-1}V^{*}=VMV^{*}.$
  
  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
  
  $(MN)(MN)^{*}=MN(NM)^{*}=MNM^{*}N^{*}.$
  
  By Fuglede, the above becomes
  
  $=MM^{*}NN^{*}=M^{*}MN^{*}N.$
  
  But M and N are normal, so
  
  $=M^{*}N^{*}MN=(MN)^{*}MN.$
  
  ## C*-algebras
  
  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*.
  
  