Fuglede's theorem

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In mathematics, Fuglede's theorem is a result in operator theory, named after Bent Fuglede.

[edit] 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.

Colloquially, the theorem claims that commutativity between operators is transitive under the given assumptions. The claim does not hold in general if N is not normal. A simple counterexample is provided by letting N be the unilateral shift and T = N. Also, when T is self-adjoint, the claim is trivial regardless of whether N is normal:

$TN^* = (NT)^* = (TN)^* = N^*T.\,$

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. TN = NT if and only if TP_i = P_iT. Therefore T must also commute with

$N^* = \sum_i {\bar \lambda_i} P_i.$

In general, 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

$N = \int_{\sigma(N)} \lambda d P(\lambda). \,$

As in the finite dimensinal case, TN = NT implies TP_Ω = P_ΩT. Thus T commutes with any simple function of the form

$\rho = \sum_i {\bar \lambda} P_{\Omega_i}.$

A limiting argument then shows that T commutes with

$N^* = \int_{\sigma(N)} {\bar \lambda} d P(\lambda). \,$

[edit] Putnam's generalization

The following contains Fuglede's result as a special case.

Theorem (Putnam) Let T, M, N be linear operators on a complex Hilbert space, and suppose that M and N are normal and MT = TN. Then M*T = TN*.

First proof (Rosenblum ): By induction, the hypothesis implies that M^kT = TN^k for all k. Thus for any λ in $\mathbb{C}$ ,

$e^{\bar\lambda M}T = T e^{\bar\lambda N}$ .

Consider the function

$F(\lambda) = e^{\lambda M^*} T e^{-\lambda N^*}$

This is equal to

$e^{\lambda M^*} \left[e^{-\bar\lambda M}T e^{\bar\lambda N}\right] e^{-\lambda N^*} = U(\lambda) T V(\lambda)^{-1}$ ,

where $U(\lambda) = e^{\lambda M^* - \bar\lambda M}$ and $V(\lambda) = e^{\lambda N^* - \bar\lambda N}$ . 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

$\|F(\lambda)\| \le \|T\|\ \forall \lambda$

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

$T' = \begin{bmatrix} 0 & 0\\ T & 0 \end{bmatrix} \quad \mbox{and} \quad N' = \begin{bmatrix} N & 0 \\ 0 & M \end{bmatrix}.$

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

$S^* M (S^{-1})^* = N. \,$

$S^* M (S^{-1})^* = S^{-1} M S \quad \Rightarrow \quad SS^* M (SS^*)^{-1} = M.$

Therefore, on Ran(M), SS* is the identity operator. SS* can be extended to Ran(M)^⊥ = Ker(M). Therefore, by normality of M, SS* = I, the identity operator. Similarly, S*S = I. This shows that S is unitary.