Intersection homology

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:

$T{N}^{*}=(NT{)}^{*}=(TN{)}^{*}=N^{*}T.$

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 aspects 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

$N^{*}=\sum _i{\overline{\lambda }}_i{P}_i.$

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

$N=\underset{\sigma (N)}{\int }\lambda dP(\lambda ).$

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

$\rho =\sum _i\overline{\lambda }{P}_{{\mathrm{\Omega }}_i}.$

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 $P_{{\mathrm{\Omega }}_i}$, 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, M 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 $\mathbb{ℂ}$,

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

Consider the function

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

This is equal to

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

where $U(\lambda )=e^{\lambda M^{*}-\overline{\lambda }M}$ and $V(\lambda )=e^{\lambda N^{*}-\overline{\lambda }N}$. However we have

$U(\lambda {)}^{*}=e^{\overline{\lambda }M-\lambda M^{*}}=U(\lambda {)}^{-1}$

so U is unitary, and hence has norm 1 for all λ; the same is true for V(λ), so

$\Vert F(\lambda )\Vert \le \Vert T\Vert \mathrm{\forall }\lambda .$

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

$T^{\prime }=\left[\begin{array}{cc}0& 0\\ T& 0\end{array}\right]\text{and}{N}^{\prime }=\left[\begin{array}{cc}N& 0\\ 0& M\end{array}\right].$

The operator N' is normal and, by assumption, T' N' = N' T' . By Fuglede's theorem, one has

$T^{\prime }(N^{\prime }{)}^{*}=(N^{\prime }{)}^{*}{T}^{\prime }.$

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

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

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.

Corollary If M and N are normal operators, and MN = NM, then MN is also normal.

Proof: The argument invokes only Fuglede's theoerm. 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
• Rudin, Walter. Functional Analysis. Mc Graw-Hill (1973).