Editing 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 operator|normal]]. If ''TN'' = ''NT'', then ''TN*'' = ''N*T'', where ''N*'' denotes the [[Hermitian adjoint|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 ''P<sub>i</sub>'' are pairwise orthogonal projections. One expects that ''TN'' = ''NT'' if and only if ''TP<sub>i</sub>'' = ''P<sub>i</sub>T''.
Indeed, it can be proved to be true by elementary arguments (e.g. it can be shown that all ''P<sub>i</sub>'' are representable as polynomials of ''N'' and for this reason, if ''T'' commutes with ''N'', it has to commute with ''P<sub>i</sub>''...).
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 ''P''<sub>Ω</sub> 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 ''TP''<sub>Ω</sub> = ''P''<sub>Ω</sub>''T''. 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 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)'''<ref name="Putnam1951">{{cite journal
 | last1   = Putnam
 | first1  = C. R.
 | date    = April 1951
 | title   = On Normal Operators in Hilbert Space
 | url     = http://www.jstor.org/stable/2372180
 | journal = American Journal of Mathematics
 | volume  = 73
 | issue   = 2
 | pages   = 357–362
 | doi     = 10.2307/2372180 | jstor = 2372180
 | url-access= subscription
 }}</ref> Let ''T'', ''M'', ''N'' be [[linear operator]]s on a complex [[Hilbert space]], and suppose that ''M'' and ''N'' are [[normal operator|normal]], ''T'' is bounded and ''MT'' = ''TN''.
Then ''M''*''T'' = ''TN''*.

'''First proof (Marvin Rosenblum)''':
By induction, the hypothesis implies that ''M''<sup>''k''</sup>''T'' = ''TN''<sup>''k''</sup> 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*''-algebras ==
The theorem can be rephrased as a statement about elements of [[C*-algebra]]s.

'''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 ==
{{Reflist}}

* Fuglede, Bent. [http://www.pnas.org/content/36/1/35.full.pdf+html?ck=nck A Commutativity Theorem for Normal Operators — PNAS]
* {{citation
 | last = Berberian | first = Sterling K.
 | isbn = 0-387-90080-2
 | location = New York-Heidelberg-Berlin
 | mr = 0417727
 | page = 274
 | publisher = Springer-Verlag
 | series = Graduate Texts in Mathematics
 | title = Lectures in Functional Analysis and Operator Theory
 | volume = 15
 | year = 1974}}.
* {{Rudin Walter Functional Analysis|edition=1}} <!-- {{sfn | Rudin | 1973 | p=}} -->

{{Spectral theory}}
{{Functional analysis}}

[[Category:Operator theory]]
[[Category:Theorems in functional analysis]]