Note on Wermuth's theorem on commuting operator exponentials

TL;DR

This paper explores conditions under which commuting operator exponentials imply the commutativity of the operators themselves, extending Wermuth's theorem to Banach and Hilbert spaces with new proofs.

Contribution

It applies Wermuth's theorem to establish a criterion for operator commutativity via exponentials, providing an alternative proof for Hilbert spaces.

Findings

If $A$ has a $2 extpi i$-congruence free spectrum, then $e^A B = B e^A$ iff $AB=BA$.

The result extends to Banach spaces and offers a new proof for Hilbert spaces.

Provides insights into the relationship between operator spectra and exponential commutativity.

Abstract

We apply Wermuth's theorem on commuting operator exponentials to show that if $A, B \in B(X)$, $X$ being Banach space and $A$ of $2\pi i$-congruence free spectrum, then $e^A B = B e^A$ if and only if $AB=BA$. We employ this observation to provide alternative proof of similar result by Chaban and Mortad, applicable for $X$ being a Hilbert space.

Equations

$$\forall s_1,s_2\in S:s_1-s_2\ne kz,k\in \mathbb{Z}\setminus \{0\}.$$

$$\Delta =\underset{z_1,z_2\in U}{\sup }|z_1-z_2|$$

$$\tau =\frac{|z|}{\Delta }.$$

$$t|z_1-z_2|<\tau \Delta =|z|.$$

$$|w_1-w_2|=t|z_1-z_2|<|z|,$$

$$w_1-w_2\ne kz,k\in \mathbb{Z}\setminus \{0\}$$

$$w_1-w_2=0\ne kz,k\in \mathbb{Z}\setminus \{0\}.$$

$$e^A{e}^{tB}-e^{tB}{e}^A=0.$$

$$e^A{e}^{tB}-e^{tB}{e}^A=0\Rightarrow A\cdot tB-tB\cdot A=t(AB-BA)=0$$

$$A=\left(\begin{array}[]{cc}0&\pi\\ -\pi&0\end{array}\right),\qquad B=\left(\begin{array}[]{cc}0&a\\ a&0\end{array}\right)$$

$$\displaystyle AB-BA=\left(\begin{array}[]{cc}2\pi a&0\\ 0&-2\pi a\end{array}\right)$$

$$T=\mathrm{Re}T+i\mathrm{Im}T,$$

$$\mathrm{Re}T=\frac{1}{2}(T+T^{\ast }),\mathrm{Im}T=\frac{1}{2i}(T-T^{\ast }).$$

$$e^{A}B=B{e}^A⟺AB=BA.$$

$$\sigma (A)\subset \sigma (\mathrm{Re}A)+i\sigma (\mathrm{Im}A).$$

$$J=\inf \{[x_1,x_2]\subset \mathbb{R}:\sigma (\mathrm{Re}A)\subset [x_1,x_2]\}$$

$$\sigma (A)\subset J+i(0,\pi ).$$

$$|{\lambda }_1-{\lambda }_2|=|b_1-b_2|⩽\underset{b_1,b_2\in (0,\pi )}{\sup }|b_1-b_2|=\pi <2\pi ,$$

$${\lambda }_1-{\lambda }_2=a_1-a_2+i(b_1-b_2)\ne 2k\pi i,k\in \mathbb{Z}\setminus \{0\}.$$

Full text

1. Introduction. Wermuth’s theorem

It is well known that if two elements $a$, $b$ of noncommutative unital Banach algebra commute, then their exponentials $e^{a}$ and $e^{b}$ also commute. Although the converse statement is wrong in general, it was shown by E. Wermuth that if $A$ and $B$ are bounded operators on Banach space satisfying additional condition of being of $2\pi i$–congruence free spectrum, then the opposite implication also remains true.

Definition 1**.**

Let $S\subset\mathbb{C}$ and let $z\in\mathbb{C}$ be arbitrarily chosen. We say that $S$ is $z$-congruence free if and only if no two different elements $s_{1},s_{2}\in S$ exists such that $s_{1}=s_{2}\,(\mathrm{mod}\,z)$. Equivalently,

[TABLE]

Let $B(X)$ be a Banach algebra of all bounded linear endomorphisms over Banach space $X$. The Wermuth’s theorem is formulated for pairs of operators $A,B\in B(X)$ and involves $2\pi i$–congruence freedom of both spectra $\sigma(A)$, $\sigma(B)$ and can be stated as follows:

Theorem 1** (Wermuth).**

Let $A,B\in B(X)$ and let both $\sigma(A),\sigma(B)\subset\mathbb{C}$ to be $2\pi i$–congruence free. Then, $e^{A}e^{B}=e^{B}e^{A}$ if and only if $AB=BA$.

Original formulation [1] of Wermuth’s theorem concerned finite dimensional matrix algebra $M_{n}(\mathbb{C})$ and then was generalized [2] to the case of $B(X)$ for any Banach space $X$. Later on, several results concerning commutativity of exponentials (or their functions) in noncommutative unital algebras emerged (see e.g. [3, 4, 5, 6, 7]), with or without making explicit use of $2\pi i$–congruence freedom hypothesis; in particular, interesting result was obtained by Chaban and Mortad in [8] for a case of C*-algebra $B(H)$ for Hilbert space $H$, which roughly says that if $A\in B(H)$ is normal operator of “well-behaved” spectrum, then $e^{A}B=Be^{A}$ if and only if $AB=BA$. In this paper, we formulate a similar result in the Banach space setting in Section 2. This allows us to we present alternative proof of theorem by Chaban and Mortad in Section 3.

2. The main result

We start with a simple observation of “shrinking”, or “rescaling” property of bounded subsets of complex plane (lemma 1 below). Our main result is then formulated as Theorem 2.

Lemma 1**.**

For every bounded nonempty set $U\subset\mathbb{C}$ and every $z\in\mathbb{C}\setminus\{0\}$, there exists such $\tau>0$ small enough, that set $tU=\{tw:w\in U\}$ is $z$–congruence free for every $t\in(0,\tau)$.

Proof.

Let $U\subset\mathbb{C}$ be nonempty and bounded and let

[TABLE]

be its diameter. If $\Delta>0$, define

[TABLE]

Then, for every $t\in(0,\tau)$ and every pair of complex numbers $z_{1},z_{2}\in U$ we have $|z_{1}-z_{2}|\leqslant\Delta$ which yields

[TABLE]

Let $tU=\{tz:z\in U\}$ and take any $w_{1},w_{2}\in tU$. As $w_{1}=tz_{1}$ and $w_{2}=tz_{2}$ for some $z_{1},z_{2}\in U$, equation (2.3) implies

[TABLE]

which automatically results in

[TABLE]

for every $w_{1},w_{2}\in tU$, i.e. $w_{1}\neq w_{2}\,(\mathrm{mod}\,z)$ and $tU$ is $z$-congruence free for any $t\in(0,\tau)$. On the other hand, if $\Delta=0$, i.e. $U=\{z_{0}\}$ is a singleton, then set $tU$ is automatically $z$–congruence free for any $t\in(0,\infty)$ as $w_{1}=w_{2}=tz_{0}$ and

[TABLE]

Then, one can take any $\tau\in(0,\infty)$. ∎

Theorem 2**.**

Let $A,B\in B(X)$ and let $\sigma(A)$ be $2\pi i$–congruence free. Then, $e^{A}B=Be^{A}$ if and only if $AB=BA$.

Proof.

We only need to address the “$\Rightarrow$” direction. Assume $e^{A}B=Be^{A}$. Then, $e^{A}$ commutes also with every analytic function of $B$, so in particular, for all $t\in\mathbb{R}$,

[TABLE]

As $\sigma(B)$ is a nonempty bounded subset of $\mathbb{C}$, lemma 1 invoked for $U=\sigma(B)$ guarantees that there exists such $\tau>0$ that $t\sigma(B)$ is $2\pi i$–congruence free for any $t\in(0,\tau)$. Therefore, operator $tB$ is of $2\pi i$–congruence free spectrum. By virtue of Wermuth’s result (theorem 1),

[TABLE]

for every $t\in(0,\tau)$, so $A$ and $B$ commute. Necessity is then clear. ∎

Remark 1*.*

The assumption of $2\pi i$–congruence freedom of $\sigma(A)$ cannot be neglected, as the following (canonical) counterexample in $M_{2}(\mathbb{C})$ shows: let us choose $A$ and $B$ as, say,

[TABLE]

for some $a\in\mathbb{R}$. It is easy to verify by direct algebra that $e^{A}=-I$, so $e^{A}$ commutes with (any) $B$; however,

[TABLE]

which does not vanish unless $a=0$ and matrices $A$ and $B$ do not commute in general. This is not surprising, as one easily shows $\sigma(A)=\{\pi i,-\pi i\}$ and hence, $\sigma(A)$ is not $2\pi i$–congruence free.

3. Relation to result of Chaban and Mortad

As we mentioned in the Introduction, a similar result concerning C*-algebra $B(H)$, $H$ being a Hilbert space, was obtained some time ago by Chaban and Mortad [8]. For any $T\in B(H)$ we define its unique Cartesian decomposition of a form

[TABLE]

where $\operatorname{Re}{T}$ and $\operatorname{Im}{T}$, called the real part and the imaginary part of $T$, respectively, are self-adjoint and bounded and given by

[TABLE]

Theorem 3** (Chaban and Mortad).**

Let $A,B\in B(H)$ be such that $A$ is normal and $\sigma(\operatorname{Im}{A})\subset(0,\pi)$, we have

[TABLE]

This theorem is then proved by means of methods different to ours by employing e.g. Fuglede theorem and without making direct references to $2\pi i$–congruence freedom. However, one can easily show that the $2\pi i$–congruence freedom is in fact the case here which allows to formulate alternative proof of the above result by directly applying theorem 2 (and hence Wermuth’s theorem in consequence).

Lemma 2**.**

Let $A\in B(H)$ satisfy assumptions of theorem 3, i.e. $A$ is normal and $\operatorname{Im}{A}\subset(0,\pi)$. Then, $\sigma(A)$ is $2\pi i$–congruence free.

Proof.

If $A$ is normal, i.e. $AA^{*}=A^{*}A$, then the Cartesian decomposition of $A$ constitutes of a pair $(\operatorname{Re}{A},\operatorname{Im}{A})$ of commuting normal self-adjoint bounded operators. In such case one can show, applying the Gelfand transform, that spectrum $\sigma(A)$ satisfies

[TABLE]

For convenience, let us enclose $\sigma(A)$ by a rectangle in $\mathbb{C}$. As $\operatorname{Re}{A}=(\operatorname{Re}{A})^{*}$ and $\|\operatorname{Re}{A}\|<\infty$, its spectrum is a bounded subset of $\mathbb{R}$ and one can define

[TABLE]

i.e. $J$ is the smallest interval containing the whole spectrum of $\operatorname{Re}{A}$. Then,

[TABLE]

Take any two $\lambda_{1},\lambda_{2}\in J+i(0,\pi)$, $\lambda_{k}=a_{k}+ib_{k}$ for $a_{k}\in J$, $b_{k}\in(0,\pi)$, $k\in\{1,2\}$; there are two possible cases:

1. (1)

If it happens that $a_{1}=a_{2}$, then

[TABLE]

hence $\lambda_{1}-\lambda_{2}=i(b_{1}-b_{2})\neq 2k\pi i$ for every $b_{1},b_{2}\in(0,\pi)$ and every $k\in\mathbb{Z}\setminus\{0\}$. 2. (2)

If, on the other hand $a_{1}\neq a_{2}$, then automatically

[TABLE]

In consequence, set $J+i(0,\pi)$ is $2\pi i$–congruence free; the same can be then stated about $\sigma(\operatorname{Im}{A})$ as its subset. ∎

Finally, we present an alternative proof of theorem by Chaban and Mortad as a straightforward corollary of the above observation:

Proof of theorem 3.

If $\sigma(\operatorname{Im}{A})\subset(0,\pi)$, then $\sigma(A)$ is $2\pi i$–congruence free by lemma 2. Hence, theorem 2 applies. ∎

4. Acknowledgments

Author acknowledges support received from National Science Centre, Poland via research grant No. 2016/23/D/ST1/02043.