Fuglede's theorem
# 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*.

