#  Multiplicatively spectrum-preserving maps on $C^{*}$-algebras

Michiya Mori  Graduate School of Mathematical Sciences, The University of
Tokyo, 3-8-1 Komaba, Meguro-ku, Tokyo, 153-8914, Japan; Interdisciplinary
Theoretical and Mathematical Sciences Program (iTHEMS), RIKEN, 2-1 Hirosawa,
Wako, Saitama, 351-0198, Japan<EMAIL_ADDRESS>and  Shiho Oi
Department of Mathematics, Faculty of Science, Niigata University, Niigata
950-2181, Japan<EMAIL_ADDRESS>
###### Abstract.

We study surjective maps between the sets of all self-adjoint elements of
unital $C^{*}$-algebras which satisfy the multiplicatively spectrum-preserving
property. We show that such maps are characterized by Jordan isomorphisms and
central symmetries. This is an answer to a problem posed by Molnár.

###### Key words and phrases:

$C^{*}$-algebras, spectra, Jordan $*$-isomorphisms, preserver problems

###### 2020 Mathematics Subject Classification:

46L05, 47A10, 47B49

##  1\. Introduction

Let $A$ and $B$ be unital $C^{*}$-algebras and let $\sigma(\cdot)$ denote the
spectrum of elements. For any $a,b\in A$, we define the Jordan product $\circ$
by $a\circ b=\frac{1}{2}(ab+ba)$. We denote the set of all self-adjoint
elements of $A$ (resp. $B$) by $A_{sa}$ (resp. $B_{sa}$). As we have $a\circ
b\in A_{sa}$ for any $a,b\in A_{sa}$, the self-adjoint part $A_{sa}$ of any
$C^{*}$-algebra $A$ is a Jordan algebra, which is a real commutative (non-
associative) algebra whose product $\circ$ satisfies $a\circ(b\circ
a^{2})=(a\circ b)\circ a^{2}$. We call a real-linear map $J\colon A_{sa}\to
B_{sa}$ a Jordan homomorphism if $J(a\circ b)=J(a)\circ J(b)$ holds for any
$a,b\in A_{sa}$. Moreover, if $J\colon A_{sa}\to B_{sa}$ is bijective, then it
is called a Jordan isomorphism. For any Jordan homomorphism $J\colon A_{sa}\to
B_{sa}$, one may uniquely obtain the complexification of $J$, which is a
complex-linear map $J_{\mathbb{C}}$ from $A$ into $B$ that satisfies
$J=J_{\mathbb{C}}$ on $A_{sa}$. Then $J_{\mathbb{C}}$ is a Jordan
$*$-isomorphism, that is, it satisfies $J_{\mathbb{C}}(a\circ
b)=J_{\mathbb{C}}(a)\circ J_{\mathbb{C}}(b)$, and
$J_{\mathbb{C}}(a^{*})=J_{\mathbb{C}}(a)^{*}$ for any $a,b\in A$. Jordan
$*$-isomorphisms appear ubiquitously when studying mappings between
$C^{*}$-algebras preserving a certain structure. See for example [7]. On the
other hand, the quantum-mechanical observables are expected to be self-adjoint
in quantum theory. From this perspective, studying Jordan isomorphisms between
the self-adjoint parts of $C^{*}$-algebras is equally interesting.

Størmer [8, Theorem 3.3] showed that any Jordan $*$-homomorphism between two
unital $C^{*}$-algebras is the sum of a $*$-homomorphism and a $*$-anti-
homomorphism. This implies that any Jordan isomorphism $J\colon A_{sa}\to
B_{sa}$ satisfies

$\sigma(J(a)J(b))=\sigma(ab),\quad a,b\in A_{sa}.$ (1)

For a detailed proof, see the first part of the proof of Theorem 2.2 in [6].
We call (1) the multiplicatively spectrum-preserving property of a Jordan
isomorphism.

Molnár initiated the study of the multiplicatively spectrum-preserving
property in [5]. In [6], he considered whether a surjection (not assumed to be
linear) between self-adjoint parts of unital $C^{*}$-algebras which has the
multiplicatively spectrum-preserving property induces a Jordan isomorphism or
not. He obtained a positive solution in the contexts of von Neumann algebras
([6, Theorem 2.2]) and standard $C^{*}$-algebras ([6, Theorem 2.5]). Indeed,
in order to prove the theorem [6, Theorem 2.2], he applied the lemmas [6,
Lemma 2.3] ([7, Lemma 22]) and [6, Lemma 2.4] ([7, Lemma 23]), which provide
spectral characterizations of elements. However, he noted that the proofs of
the lemmas were valid for only von Neumann algebras. Consequently, he posed a
problem of whether the theorem [6, Theorem 2.2] can survive for general
$C^{*}$-algebras (see [6, p.194], [7, Problem 4]). It can be seen from [7,
p.35] that the answer is yes if one may obtain extensions of the lemmas [6,
Lemma 2.3] ([7, Lemma 22]) and [6, Lemma 2.4] ([7, Lemma 23]) to general
$C^{*}$-algebras. The purpose of this paper is to actually carry it out in
Lemma 3 and Lemma 6 below. This gives a positive answer to the problem, which
reads as follows. Recall that a self-adjoint unitary is called a symmetry.

######  Theorem 1.

Let $A$ and $B$ be unital $C^{*}$-algebras. Suppose that a surjective map
$\phi\colon A_{sa}\to B_{sa}$ satisfies

$\sigma(\phi(a)\phi(b))=\sigma(ab),\ \ a,b\in A_{sa}.$

Then there is a central symmetry $s\in B_{sa}$ and a Jordan isomorphism
$J\colon A_{sa}\to B_{sa}$ such that

$\phi(a)=sJ(a),\quad a\in A_{sa}.$

Note that the problem we study is one of two problems posed by Molnár in [6,
p.194]. Another problem has an affirmative solution, too (see [7, p.36] and
[4, Theorem 5.5]). Also note that multiplicatively spectrum-preserving
mappings defined on the whole algebra (instead of the self-adjoint part) are
studied in [1] in the setting of unital semisimple Banach algebras.

##  2\. Results

We present notations to be used throughout the paper. For a unital
$C^{*}$-algebra $A$, $1$ stands for its unit. The symbols $A_{+}$ and
$A_{+}^{-1}$ denote the set of all positive elements and the set of all
positive invertible elements of $A$, respectively. For $x\in A_{sa}$, let
$x=x_{+}-x_{-}$ denote the decomposition into its positive and negative parts.
Note that $x_{+}x_{-}=x_{-}x_{+}=0$ holds. If $x\in A_{+}$, then $x^{1/2}\in
A_{+}$ stands for its positive square root.

######  Lemma 2.

Let $x\in A_{+}$ and $y\in A$. If $xy=0$, then $x^{1/2}y=0$. If $yx=0$, then
$yx^{1/2}=0$.

###### Proof.

This follows from the basic fact that there is a sequence of polynomials
$P_{n}$ satisfying $P_{n}(0)=0$ and $P_{n}(x)\to x^{1/2}$ as $n\to\infty$. ∎

Throughout the paper, we use the following well-known fact several times.

$\sigma(ab)\setminus\\{0\\}=\sigma(ba)\setminus\\{0\\}\text{ whenever }a,b\in
A.$ (2)

This is called Jacobson’s lemma.

The next lemma is an extension of [6, Lemma 2.3] ([7, Lemma 22]). This might
be of independent interest.

######  Lemma 3.

Let $x\in A_{sa}$. Then the following two conditions are equivalent.

  * •

$\sigma(xy)\subset\mathbb{R}$ for every $y\in A_{sa}$.

  * •

$x_{+}yx_{-}=0$ for every $y\in A$.

###### Proof.

Firstly we assume that $x_{+}yx_{-}=0$ for every $y\in A$. Let $y\in A_{sa}$.
By Lemma 2, we have
$x_{+}-x_{-}=(x_{+}^{1/2}+x_{-}^{1/2})(x_{+}^{1/2}-x_{-}^{1/2})$ and
$x_{+}^{1/2}yx_{-}^{1/2}=0$. We also have
$x_{-}^{1/2}yx_{+}^{1/2}=(x_{+}^{1/2}yx_{-}^{1/2})^{*}=0$. Thus we get

$xy=(x_{+}-x_{-})y=(x_{+}^{1/2}+x_{-}^{1/2})(x_{+}^{1/2}-x_{-}^{1/2})y$

and

$(x_{+}^{1/2}-x_{-}^{1/2})y(x_{+}^{1/2}+x_{-}^{1/2})=x_{+}^{1/2}yx_{+}^{1/2}-x_{-}^{1/2}yx_{-}^{1/2}\in
A_{sa}.$

Combining these with (2), we get $\sigma(xy)\subset\mathbb{R}$.

Conversely, assume that $\sigma(xy)\subset\mathbb{R}$ for every $y\in A_{sa}$.
Let $y\in A$. We prove $x_{+}yx_{-}=0$. It suffices to consider the case $y\in
A_{sa}$. In this case, we have
$z:=x_{+}^{1/2}yx_{-}^{1/2}+x_{-}^{1/2}yx_{+}^{1/2}\in A_{sa}$. From
$\sigma(xz)\subset\mathbb{R}$,

$xz=(x_{+}-x_{-})z=(x_{+}^{1/2}+x_{-}^{1/2})(x_{+}^{1/2}-x_{-}^{1/2})z,$
$(x_{+}^{1/2}-x_{-}^{1/2})z(x_{+}^{1/2}+x_{-}^{1/2})=x_{+}yx_{-}-x_{-}yx_{+}\in
iA_{sa},$

and (2), we get $\sigma(x_{+}yx_{-}-x_{-}yx_{+})=\\{0\\}$ and
$x_{+}yx_{-}=x_{-}yx_{+}$. Hence $x_{+}yx_{-}$ is self-adjoint and

$(x_{+}yx_{-})^{2}=(x_{+}yx_{-})(x_{+}yx_{-})=x_{+}y\cdot 0\cdot yx_{-}=0,$

so we obtain $x_{+}yx_{-}=0$ as desired. ∎

Our proofs of Lemma 4 and Lemma 6 are similar to the proof of Theorem 2.2 in
[2], which shows that for any semisimple Banach algebra $X$ if
$\sigma(ax)=\sigma(bx)$ for all elements $x\in X$ then $a=b$ holds.

######  Lemma 4.

Let $a,b\in A_{sa}$. Suppose that $\sigma(ay)=\sigma(by)$ for any $y\in
A_{+}^{-1}$. Let $x\in A_{sa}$ and let $\lambda$ be a real number with
$|\lambda|>\lVert x\rVert$. Then $\lambda\in\sigma(x+a)$ if and only if
$\lambda\in\sigma(x+b)$.

###### Proof.

Suppose that $\lambda>\lVert x\rVert$. Then $\lambda-x\in A_{+}^{-1}$. We have
$\lambda-(x+a)=(\lambda-x)[1-(\lambda-x)^{-1}a]$. This implies that
$\lambda\in\sigma(x+a)\iff 1\in\sigma((\lambda-x)^{-1}a)\iff
1\in\sigma((\lambda-x)^{-1}b)\iff\lambda\in\sigma(x+b)$. Suppose that
$-\lambda>\lVert x\rVert$. Then $-\lambda+x\in A_{+}^{-1}$. Thus the
conclusion follows from a similar procedure as described above. ∎

######  Corollary 5.

Let $a,b\in A_{sa}$. Suppose that $\sigma(ay)=\sigma(by)$ for any $y\in
A_{+}^{-1}$. Let $x\in A_{sa}$. Then either $\max\\{\lVert x+a\rVert,\lVert
x+b\rVert\\}\leq\lVert x\rVert$ or $\lVert x+a\rVert=\lVert x+b\rVert$ holds.

The next lemma extends [6, Lemma 2.4] ([7, Lemma 23]).

######  Lemma 6.

Let $a,b\in A_{sa}$. If $\sigma(ay)=\sigma(by)$ for any $y\in A_{+}^{-1}$,
then we have $a=b$.

###### Proof.

We prove by induction that

$\lVert nb-(n-1)a\rVert\leq\lVert b\rVert\text{ and }\lVert
n(b-a)\rVert\leq\lVert b\rVert$ (3)

for $n\geq 1$. Applying Corollary 5 with $x=-b\in A_{sa}$, we get either
$\lVert-b+a\rVert\leq\lVert-b\rVert$ or $\lVert-b+a\rVert=\lVert-b+b\rVert=0$.
In both cases, we get $\lVert b-a\rVert\leq\lVert b\rVert$. Thus we get (3)
when $n=1$.

Let $k\geq 1$ and suppose that (3) holds with $n=k$. Applying Corollary 5 with
$x=k(b-a)\in A_{sa}$, we get either $\lVert k(b-a)+b\rVert\leq\lVert
k(b-a)\rVert$ or $\lVert k(b-a)+a\rVert=\lVert k(b-a)+b\rVert$. Therefore, in
both cases, the assumption implies

$\lVert(k+1)b-ka\rVert\leq\lVert b\rVert.$ (4)

Applying Corollary 5 with $x=ka-(k+1)b\in A_{sa}$, we have either $\lVert
ka-(k+1)b+a\rVert\leq\lVert ka-(k+1)b\rVert$ or $\lVert
ka-(k+1)b+a\rVert=\lVert ka-(k+1)b+b\rVert$. In both cases, the assumption
together with (4) shows

$\lVert(k+1)(b-a)\rVert\leq\lVert b\rVert.$

Thus we have obtained the equation $\lVert n(b-a)\rVert\leq\lVert b\rVert$ for
any $n\geq 1$. This implies $a=b$. ∎

Having obtained Lemma 3 and Lemma 6, we are ready to prove the main theorem of
this paper according to the observations of [7, p.35]. Although the proof is
very similar to that of [6, Theorem 2.2], we present it for the sake of
readability.

###### Proof of Theorem 1.

Let $s=\phi(1)$. Then $s\in B_{sa}$. We have
$\sigma(s^{2})=\sigma(1)=\\{1\\}$, thus we obtain that
$\sigma(s)\subset\\{1,-1\\}$, so $s$ is a symmetry. As $\phi$ is a surjection,
for any $y\in B_{sa}$ there is $x\in A_{sa}$ such that $\phi(x)=y$. We have
$\sigma(sy)=\sigma(\phi(1)\phi(x))=\sigma(x)\subset\mathbb{R}$. By Lemma 3, we
obtain $s_{+}ys_{-}=0$ for any $y\in B$. For any $y\in B_{sa}$, we get
$s_{+}ys_{-}=0$ and $s_{-}ys_{+}=(s_{+}ys_{-})^{*}=0$, thus

$sy=(s_{+}-s_{-})y(s_{+}+s_{-})=s_{+}ys_{+}-s_{-}ys_{-}\in B_{sa}.$

It follows that $sy=(sy)^{*}=ys$. Hence $s$ is central.

We define a map $T\colon A_{sa}\to B_{sa}$ by $T(x)=s\phi(x)$ for any $x\in
A_{sa}$. For any $y\in B_{sa}$, as $sy\in B_{sa}$ and $\phi$ is surjective,
there is $x\in A_{sa}$ such that $\phi(x)=sy$. Thus $T(x)=ssy=y$. Therefore,
$T\colon A_{sa}\to B_{sa}$ is a surjection. We have

$\sigma(T(x)T(y))=\sigma(s\phi(x)s\phi(y))=\sigma(\phi(x)\phi(y))=\sigma(xy)$
(5)

for any $x,y\in A_{sa}$. By taking $y=1$, we get $\sigma(T(x))=\sigma(x)$ for
any $x\in A_{sa}$. It follows that $x\in A_{+}^{-1}\iff T(x)\in B_{+}^{-1}$.
Therefore, $T|_{A_{+}^{-1}}\colon A_{+}^{-1}\to B_{+}^{-1}$ is a surjective
map which satisfies (5).

For any $x\in A_{+}^{-1}$ we have

$\\{1\\}=\sigma(xx^{-1})\stackrel{{\scriptstyle\eqref{multiplicative}}}{{=}}\sigma(T(x)T(x^{-1}))\stackrel{{\scriptstyle\eqref{abba}}}{{=}}\sigma(T(x)^{1/2}T(x^{-1})T(x)^{1/2}).$

Thus we obtain $T(x)^{1/2}T(x^{-1})T(x)^{1/2}=1$ and

$T(x)^{-1}=T(x^{-1}).$ (6)

It follows that

$\begin{split}\sigma(x^{-{1/2}}yx^{-{1/2}})\stackrel{{\scriptstyle\eqref{abba}}}{{=}}\sigma(yx^{-1})&\stackrel{{\scriptstyle\eqref{multiplicative}}}{{=}}\sigma(T(y)T(x^{-1}))\\\
&\stackrel{{\scriptstyle\eqref{-1}}}{{=}}\sigma(T(y)T(x)^{-1})\stackrel{{\scriptstyle\eqref{abba}}}{{=}}\sigma(T(x)^{-{1/2}}T(y)T(x)^{-{1/2}})\end{split}$

for any pair $x,y\in A_{+}^{-1}$. Combining this with a fact that the Thompson
metric $d_{T}$ on $A_{+}^{-1}$ satisfies
$d_{T}(x,y)=\|\log(x^{-{1/2}}yx^{-{1/2}})\|$ for any $x,y\in A_{+}^{-1}$, we
have $T|_{A_{+}^{-1}}\colon A_{+}^{-1}\to B_{+}^{-1}$ is a surjective Thompson
isometry with $T(1)=1$. For any $t>0$ and $x,y\in A_{+}^{-1}$, we see that
$\sigma(T(x)T(y))=\sigma(xy)$ and $\sigma(txy)=\sigma(T(tx)T(y))$ by (5),
hence

$\sigma(tT(x)T(y))=\sigma(txy)=\sigma(T(tx)T(y)).$

Thus, Lemma 6 shows that $tT(x)=T(tx)$ for any $x\in A_{+}^{-1}$. In other
words, $T|_{A_{+}^{-1}}$ is positive homogeneous. By applying [3, Theorem 9]
([7, Theorem 8]) with $T|_{A_{+}^{-1}}$, we see that there is a Jordan
$*$-isomorphism $J\colon A\to B$ such that $T=J$ on $A_{+}^{-1}$. Fix $a\in
A_{sa}$. For any $y\in B_{+}^{-1}$, there is $x\in A_{+}^{-1}$ such that
$T(x)=J(x)=y$. We have

$\sigma(J(a)y)=\sigma(J(a)J(x))\stackrel{{\scriptstyle\eqref{jajb}}}{{=}}\sigma(ax)\stackrel{{\scriptstyle\eqref{multiplicative}}}{{=}}\sigma(T(a)T(x))=\sigma(T(a)y).$

Since $J(a),T(a)\in B_{sa}$, Lemma 6 implies that $J(a)=T(a)$. Thus
$\phi(a)=sJ(a)$ for any $a\in A_{sa}$. ∎

### Acknowledgments

The first and the second authors were supported by JSPS KAKENHI Grant Numbers
22K13934 and 21K13804, respectively. The authors appreciate Professors Masaki
Izumi and Narutaka Ozawa for giving them an opportunity to visit Kyoto, where
this joint work began.

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