Fuglede's theorem explained

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$$N\; =\; \backslash sum\_i\; \backslash lambda\_i\; P\_i$$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$$N^*\; =\; \backslash sum\_i\; P\_i.$$

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$$N\; =\; \backslash int\_\; \backslash lambda\; d\; P(\backslash lambda).$$

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$$\backslash rho\; =\; \backslash sum\_i\; P\_.$$

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)^[1] 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

\Complex ,$$e^T\; =\; T\; e^.$$

Consider the function$$F(\backslash lambda)\; =\; e^\; T\; e^.$$This is equal to$$e^\; \backslash left[e^\{-\backslash bar\backslash lambda\; M\}T\; e^\{\backslash bar\backslash lambda\; N\}\backslash right]\; e^\; =\; U(\backslash lambda)\; T\; V(\backslash lambda)^,$$where

U( λ ) = e ^{λ M ^{*} - \bar λ M} because M is normal, and similarly V( λ ) = e ^{λ N ^{*} - \bar λ N} . However we have$$U(\backslash lambda)^*\; =\; e^\; =\; U(\backslash lambda)^$$so U is unitary, and hence has norm 1 for all λ; the same is true for V(λ), so$$\backslash |F(\backslash lambda)\backslash |\; \backslash le\; \backslash |T\backslash |\backslash \; \backslash forall\; \backslash lambda.$$

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.^[1] 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

The operator N' is normal and, by assumption, T' N' = N' T' . By Fuglede's theorem, one has$$T\text{'}\; (N\text{'})^*\; =\; (N\text{'})^*T\text{'}.$$

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.$$S^\; M^*\; S\; =\; N^*.$$

Take the adjoint of the above equation and we have$$S^*\; M\; (S^)^*\; =\; N.$$

So$$S^*\; M\; (S^)^*\; =\; S^\; M\; S\; \backslash quad\; \backslash Rightarrow\; \backslash quad\; SS^*\; M\; (SS^*)^\; =\; M.$$

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$$N\; =\; S^*M\; (S^*)^=VRMR^V^*=VMV^*.$$

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$$(MN)\; (MN)^*\; =\; MN\; (NM)^*\; =\; MN\; M^*\; N^*.$$

By Fuglede, the above becomes$$=\; M\; M^*\; N\; N^*\; =\; M^*\; M\; N^*N.$$

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*.

References

• Fuglede, Bent. A Commutativity Theorem for Normal Operators — PNAS
• .

Notes and References

1. Putnam . C. R. . April 1951 . On Normal Operators in Hilbert Space . American Journal of Mathematics . 73 . 2 . 357–362 . 10.2307/2372180 .