Tingley's problem for spaces of trace class operators

TL;DR

This paper solves Tingley's problem for trace class operator spaces by showing that every surjective isometry on their unit spheres extends uniquely to a linear or conjugate linear isometry of the entire space.

Contribution

It provides the first positive solution to Tingley's problem for a new class of operator algebras, specifically trace class spaces.

Findings

Every surjective isometry between unit spheres extends uniquely to a linear or conjugate linear isometry.

The result applies to trace class spaces, expanding the classes where Tingley's problem is solved.

This advances understanding of the geometric structure of operator algebras.

Abstract

We prove that every surjective isometry between the unit spheres of two trace class spaces admits a unique extension to a surjective complex linear or conjugate linear isometry between the spaces. This provides a positive solution to Tingley's problem in a new class of operator algebras.

Equations

$$\Vert a{\Vert }_1:=\left(\sum _{n=1}^{\infty }|{\mu }_n(a)|\right)<\infty .$$

$$a=\sum _{n=1}^{\infty }{\lambda }_n{\eta }_n\otimes {\xi }_n,$$

$$s(a)=\sum _{n=1}^{\infty }{\eta }_n\otimes {\xi }_n,$$

$$\left(\Vert a{\Vert }_1+\Vert b{\Vert }_1\right)\le \Vert a+b{\Vert }_1+\Vert a-b{\Vert }_1\le 2\left(\Vert a{\Vert }_1+\Vert b{\Vert }_1\right)$$

$$\Vert a+b{\Vert }_1+\Vert a-b{\Vert }_1=2\left(\Vert a{\Vert }_1+\Vert b{\Vert }_1\right)$$

$$\Vert a\pm b{\Vert }_1=2\iff a\perp b(in{C}_1(H))\iff s(a)\perp s(b)(inB(H)).$$

$$\Vert x{\Vert }_1^2\ge \Vert pxp{\Vert }_1^2+\Vert px(1-p){\Vert }_1^2+\Vert (1-p)xp{\Vert }_1^2+\Vert (1-p)x(1-p){\Vert }_1^2$$

$$\Vert f(a)-f(b){\Vert }_1=2.$$

$$F^{\prime }=\{a\in {\mathcal{B}}_{X^{\ast }}:a(x)=1\forall x\in F\},G_{\prime }=\{x\in {\mathcal{B}}_X:a(x)=1\forall a\in G\}.$$

$$\Lambda_{C}^{+}:=\Big{\{}\mathcal{M}:\mathcal{M}\hbox{ is a maximal proper face of $\mathcal{B}_{X}$ containing $C$}\Big{\}}.$$

$$C=\bigcap _{\mathcal{M}\in {\Lambda }_C^{+}}\mathcal{M}.$$

$$C\subseteq f^{-1}\left(\bigcap _{\mathcal{M}\in {\Lambda }_C^{+}}f\left(\mathcal{M}\right)\right)\subseteq \bigcap _{\mathcal{M}\in {\Lambda }_C^{+}}{f}^{-1}\left(f\left(\mathcal{M}\right)\right)=\bigcap _{\mathcal{M}\in {\Lambda }_C^{+}}\mathcal{M}=C.$$

$$f\left(\mathbb{T}{e}_0\right)=f\left(\{e_0{\}}^{\perp \perp }\cap S(C_1(H))\right)=\{f(e_0){\}}^{\perp \perp }\cap S(C_1(H^{\prime }))=\mathbb{T}f(e_0).$$

$$|1-\mu |=\Vert f(e_0)-f(i{e}_0){\Vert }_1=\Vert e_0-i{e}_0{\Vert }_1=|1-i|=\sqrt{2},$$

$$|1-\mu |=\Vert f(e_0)-f(\lambda e_0){\Vert }_1=\Vert e_0-\lambda e_0{\Vert }_1=|1-\lambda |,$$

$$|i-\mu |=\Vert if(e_0)-f(\lambda e_0){\Vert }_1=\Vert f(i{e}_0)-f(\lambda e_0){\Vert }_1=\Vert i{e}_0-\lambda e_0{\Vert }_1=|i-\lambda |,$$

$$\mathcal{F}=\{w{\}}_{{}_{\prime }}=\{x\in C_1(H)=B(H{)}_{\ast }:\Vert x{\Vert }_1=1=x(w)\},$$

$$\mathcal{M}=\{u{\}}_{{}_{\prime }}=\{x\in C_1(H)=B(H{)}_{\ast }:\Vert x{\Vert }_1=1=x(u)\},$$

$$f({\lambda }_1{e}_1+{\lambda }_2{e}_2)={\lambda }_{1}f(e_1)+{\lambda }_{2}f(e_2).$$

$$g({\varphi }_j)={\varphi }_j,where{\varphi }_j={\xi }_j\otimes {\xi }_j,\forall 1\le j\le 2.$$

$$\mathcal{M}=\{u{\}}_{{}_{\prime }}=\{x\in C_1(H)=B(H{)}_{\ast }:\Vert x{\Vert }_1=1=x(u)=1\},$$

$$\left\|\left(\begin{array}[]{cc}t&c\\ \overline{c}&1-t\\ \end{array}\right)-\left(\begin{array}[]{cc}1&0\\ 0&0\\ \end{array}\right)\right\|_{1}=\left\|g(\lambda_{1}\phi_{1}+\lambda_{2}\phi_{2})-g(\phi_{1})\right\|_{1}$$

$$=\Vert {\lambda }_1{\varphi }_1+{\lambda }_2{\varphi }_2-{\varphi }_1{\Vert }_1=1-{\lambda }_1+{\lambda }_2=2(1-{\lambda }_1).$$

$$1-{\lambda }_1=\sqrt{(1-t{)}^2+|c{|}^2}.$$

$$2\sqrt{t^{2}+|c|^{2}}=\left\|\left(\begin{array}[]{cc}t&c\\ \overline{c}&1-t\\ \end{array}\right)-\left(\begin{array}[]{cc}0&0\\ 0&1\\ \end{array}\right)\right\|_{1}=\left\|g(\lambda_{1}\phi_{1}+\lambda_{2}\phi_{2})-g(\phi_{2})\right\|_{1}$$

$$=\Vert {\lambda }_1{\varphi }_1+{\lambda }_2{\varphi }_2-{\varphi }_2{\Vert }_1={\lambda }_1+1-{\lambda }_2=2(1-{\lambda }_2)=2{\lambda }_1,$$

$${\lambda }_1=1-{\lambda }_2=\sqrt{t^2+|c{|}^2}.$$

$$g(\lambda_{1}\phi_{1}+\lambda_{2}\phi_{2})=\left(\begin{array}[]{cc}\lambda_{1}&0\\ 0&\lambda_{2}\\ \end{array}\right)=\lambda_{1}\phi_{1}+\lambda_{2}\phi_{2}=\lambda_{1}g(\phi_{1})+\lambda_{2}g(\phi_{2}),$$

$$f({\lambda }_1{e}_1+{\lambda }_2{e}_2)={\lambda }_{1}f(e_1)+{\lambda }_{2}f(e_2),$$

$$f({\lambda }_1{e}_1+{\lambda }_2{e}_2)=f(|{\lambda }_1|{\sigma }_1{e}_1+|{\lambda }_2|{\sigma }_2{e}_2)=|{\lambda }_1|f({\sigma }_1{e}_1)+|{\lambda }_2|f({\sigma }_2{e}_2)$$

Full text

1. Introduction

In 1987, D. Tingley published a study on surjective isometries between the unit spheres of two finite dimensional Banach spaces, showing that any such mappings preserves antipodes points (see [38]). A deep and difficult geometric problem has been named after Tingley’s contribution. Namely, let $f:S(X)\to S(Y)$ be a surjective isometry between the unit spheres of two Banach spaces $X$ and $Y$. Is $f$ the restriction of a (unique) isometric real linear surjection $T:X\to Y$? This problem remains unsolved even if we asume that $X$ and $Y$ are 2-dimensional spaces. Despite Tingley’s problem still being open for general Banach spaces, positive solutions have been found for specific cases (see, for example, [3, 6, 7, 8, 9, 10, 5, 11, 12, 13, 17, 19, 20, 21, 24, 27, 31, 32, 33, 34, 35, 36, 37, 39] and [40]), and each particular case has required strategies and proofs which are more or less independent.

For the purposes of this note, we recall that positive solutions to Tingley’s problem include the following cases: $f:S(c_{0})\to S(c_{0})$ [7], $f:S(\ell_{1})\to S(\ell_{1})$ [9], $f:S(\ell_{\infty})\to S(\ell_{\infty})$ [8], $f:S(K(H))\to S(K(H^{\prime}))$, where $H$ and $H^{\prime}$ are complex Hilbert spaces [27], and $f:S(B(H))\to S(B(H^{\prime}))$ [20]. It is well known that the natural dualities $c_{0}^{*}=\ell_{1}$, and $\ell_{1}^{*}=\ell_{\infty}$ admit a non-commutative counterparts in the dualities $K(H)^{*}=C_{1}(H)$ and $C_{1}(H)^{*}=B(H)$, where $C_{1}(H)$ is the space of all trace class operators on $H$. So, there is a natural open question concerning Tingley’s problem in the case of surjective isometries between the unit spheres of two trace class spaces. In this paper we explore this problem and we prove that every surjective isometry between the unit spheres of two trace class spaces admits a unique extension to a surjective complex linear or conjugate linear isometry between the spaces (see Theorem 4.1).

The results are distributed in three main sections. In section 2 we establish new geometric properties of a surjective isometry $f:S(X)\to S(Y)$ in the case in which norm closed faces of the closed unit balls of $X$ and $Y$ are all norm-semi-exposed, and weak∗ closed faces of the closed unit balls of $X^{*}$ and $Y^{*}$ are all weak∗-semi-exposed (see Corollary 2.5). Applying techniques of geometry and linear algebra, in section 3 we present a positive answer to Tingley’s problem for surjective isometries $f:S(C_{1}(H))\to S(C_{1}(H^{\prime}))$ when $H$ and $H^{\prime}$ are finite dimensional (see Theorem 3.7). The result in the finite dimensional case play a fundamental role in the proof of our main result.

2. Facial stability for surjective isometries between the unit spheres of trace class spaces

Let $H$ be a complex Hilbert space. We are interested in different subclasses of the space $K(H)$ of all compact operators on $H$. We briefly recall the basic terminology. For each compact operator $a$, the operator $a^{*}a$ lies in $K(H)$ and admits a unique square root $|a|=(a^{*}a)^{\frac{1}{2}}$. The characteristic numbers of the operator $a$ are precisely the eigenvalues of $|a|$ arranged in decreasing order and repeated according to multiplicity. Since $|a|$ belongs to $K(H)$, only an at most countably number of its eigenvalues are greater than zero. According to the standard terminology, we usually write $\mu_{n}(a)$ for the $n$-th characteristic number of $a$. It is well known that $(\mu_{n}(a))_{n}\to 0$.

The symbol $C_{1}=C_{1}(H)$ will stand for the space of trace class operators on $H$, that is, the set of all $a\in K(H)$ such that

[TABLE]

We set $\|a\|_{\infty}=\|a\|$, where the latter stands for the operator norm of $a$. The set $C_{1}$ is a two-sided ideal in the space $B(H)$ of all bounded linear operators on $H$, and $(C_{1},\|.\|_{1})$ is a Banach algebra. If tr$(.)$ denotes the usual trace on $B(H)$ and $a\in K(H)$, we know that $a\in C_{1}$ if, and only if, tr$(|a|)<\infty$ and $\|a\|_{1}=\hbox{tr}(|a|)$. It is further known that the predual of $B(H)$ and the dual of $K(H)$ both can be identified with $C_{1}(H)$ under the isometric linear mapping $a\mapsto\varphi_{a}$, where $\varphi_{a}(x):=\hbox{tr}(ax)$ ($a\in C_{1}(H),x\in B(H)$). The dualities $K(H)^{*}=C_{1}(H)$ and $C_{1}(H)^{*}=B(H)$ can be regarded as a non-commutative version of the natural dualities between $c_{0}$, $\ell_{1}$ and $\ell_{\infty}$.

It is known that every element $a$ in $C_{1}(H)$ can be written uniquely as a (possibly finite) sum

[TABLE]

where $(\lambda_{n})\subset\mathbb{R}_{0}^{+}$, $(\xi_{n})$, $(\eta_{n})$ are orthonormal systems in $H$, and $\displaystyle\|a\|_{1}=\sum_{n=1}^{\infty}\lambda_{n}$.

Along the paper, we shall try to distinguish between $C_{1}(H)$ and $C_{1}(H)^{*}\equiv B(H)$, however the reader must be warned that we shall regard $K(H)$ and $C_{1}(H)$ inside $B(H)$. For example, when an element $a$ in $C_{1}(H)$ is regarded in the form given in (1), the element

[TABLE]

is a partial isometry in $B(H)$ (called the support partial isometry of $a$ in $B(H)$), and it is precisely the smallest partial isometry $e$ in $B(H)$ satisfying $e(a)=\|a\|_{1}$.

We recall that two elements $a,b\in B(H)$ are orthogonal ($a\perp b$ in short) if and only if $ab^{*}=b^{*}a=0$. The relation “being orthogonal” produces a partial order $\leq$ in the set $\mathcal{U}(B(H))$ of all partial isometries given by $w\leq s$ if and only if $s-w$ is a partial isometry with $s-w\perp w$ (this is the standard order employed, for example, in [1, 16]).

We refer to [23, Chapter III], [14, §9], [30, Chapter II] and [29, §1.15] for the basic results and references on the spaces $K(H)$, $C_{1}(H)$ and $B(H)$.

It is worth recalling that, by a result due to B. Russo [28] every surjective complex linear isometry $T:C_{1}(H)\to C_{1}(H)$ is of the form $T(x)=vxu$ or of the form $T(x)=vx^{t}u$ ($x\in C_{1}(H)$), where $u$ and $v$ are unitary elements in $B(H)$ and $x^{t}$ denotes the transpose of $x$ (compare [22, Theorem 11.2.2]).

The non-commutative Clarkson-McCarthy inequalities (see [25, Theorem 2.7]) can be written as follows:

[TABLE]

holds for every $a$ and $b$ in $C_{1}(H)$. It is further known that equality

[TABLE]

holds in (2) if and only if $(a^{*}a)(b^{*}b)=0$, which is equivalent to say that $a$ and $b$ are orthogonal as elements in $B(H)$ ($a\perp b$ in short), or in other words $s(a)$ and $s(b)$ are orthogonal partial isometries in $B(H)$ (i.e. $(s(a)^{*}s(a),s(b)^{*}s(b))$ and $(s(a)s(a)^{*},s(b)s(b)^{*})$ are two pairs of orthogonal projections in $B(H)$). Consequently, if we fix $a,b\in S(C_{1}(H))$ we can conclude that

[TABLE]

Let us recall a technical result due to X.N. Fang, J.H. Wang and G.G. Ding, who established it in [17] and [10], respectively.

Lemma 2.1**.**

([17, Corollary 2.2], [10, Corollary 1])* Let $X$ and $Y$ be normed spaces and let $f:S(X)\to S(Y)$ be a surjective isometry. Then for any $x,y$ in $S(X)$, we have $\|x+y\|=2$ if and only if $\|f(x)+f(y)\|=2$.$\hfill\Box$*

Throughout the paper, the extreme points of a convex set $C$ will be denoted by $\partial_{e}(C)$, and the symbol $\mathcal{B}_{X}$ will stand for the closed unit ball of a Banach space $X$. We shall write $\mathbb{T}$ for the unit sphere of $\mathbb{C}$. Following standard notation, the elements in $\partial_{e}(\mathcal{B}_{C_{1}(H)})$ are called pure atoms. It is known that every pure atom in $C_{1}(H)$ is an operator of the form $\eta\otimes\xi$, where $\xi$ and $\eta$ are elements in $S(H)$.

Given $\xi,\eta$ in a Hilbert space $H$, the symbol $\eta\otimes\xi$ will denote the rank one operator on $H$ defined by $\eta\otimes\xi(\zeta)=\langle\zeta|\xi\rangle\eta$ $(\zeta\in H$). Clearly $\eta\otimes\xi\in C_{1}(H)$. When $\eta\otimes\xi$ is regarded as an element in $C_{1}(H)$, we shall identify it with the normal functional on $B(H)$ given by $\eta\otimes\xi(x)=\langle x(\eta)|\xi\rangle$ ($x\in B(H)$). As it is commonly assumed, given $\phi\in B(H)_{*}$ and $z\in B(H)$ we define $\phi z,z\phi\in B(H)_{*}$ by $(\phi z)(x)=\phi(zx)$ and $(z\phi)(x)=\phi(xz)$ ($x\in B(H)$). Accordingly with this notation, for $\eta\otimes\xi$ in $C_{1}(H)=B(H)_{*}$, we have $(\eta\otimes\xi)z=\eta\otimes z^{*}(\xi)$ and $z(\eta\otimes\xi)=z(\eta)\otimes\xi,$ for every $z\in B(H)$.

We also recall an inequality established by J. Arazy in [2, Proposition in page 48]: For each projection $p$ in $B(H)$ and every $x\in C_{1}(H)$, we have

[TABLE]

Suppose $\{\xi_{i}\}_{i\in I}$ is an orthonormal basis of $H$. The elements in the set $\{\xi_{i}\otimes\xi_{i}:i\in I\}$ are mutually orthogonal in $C_{1}(H)$. Actually, the dimension of $H$ is precisely the cardinal of the biggest set of mutually orthogonal pure atoms in $C_{1}(H)$.

We can state now a non-commutative version of [9, Lemma 3].

Lemma 2.2**.**

Let $H$ and $H^{\prime}$ be complex Hilbert spaces, and let $f:S(C_{1}(H))\to S(C_{1}(H^{\prime}))$ be a surjective isometry. Then $f$ preserves orthogonal elements in both directions, that is, $a\perp b$ in $S(C_{1}(H))$ if and only if $f(a)\perp f(b)$ in $S(C_{1}(H^{\prime}))$.

Proof.

Take $a,b$ in $S(C_{1}(H))$. We have already commented that $a\perp b$ if and only if $\|a\pm b\|_{1}=2$ (compare (3)). Since $f$ is an isometry we deduce that

[TABLE]

Lemma 2.1 implies that $\|f(a)+f(b)\|_{1}=2$, and hence $\|f(a)\pm f(b)\|_{1}=2$, which assures that $f(a)\perp f(b)$. ∎

Among the ingredients and prerequisites needed in our arguments we highlight the following useful geometric result which is essentially due to L. Cheng and Y. Dong [3, Lemma 5.1] and R. Tanaka [35] (see also [34, Lemma 3.5], [36, Lemmas 2.1 and 2.2]).

Proposition 2.3**.**

([3, Lemma 5.1]**, [35, Lemma 3.3], [34, Lemma 3.5])* Let $f:S(X)\to S(Y)$ be a surjective isometry between the unit spheres of two Banach spaces, and let $\mathcal{M}$ be a convex subset of $S(X)$. Then $\mathcal{M}$ is a maximal proper face of $\mathcal{B}_{X}$ if and only if $f(\mathcal{M})$ is a maximal proper (closed) face of $\mathcal{B}_{Y}$.$\hfill\Box$*

The previous result emphasizes the importance of a “good description” of the facial structure of a Banach space. A basic tool to understand the facial structure of the closed unit ball of a complex Banach space $X$ and that of the unit ball of its dual space is given by the “facear” and “pre-facear” operations. Following [16], for each $F\subseteq\mathcal{B}_{X}$ and $G\subseteq\mathcal{B}_{X^{*}}$, we define

[TABLE]

Then, $F^{\prime}$ is a weak∗ closed face of $\mathcal{B}_{X^{*}}$ and $G_{\prime}$ is a norm closed face of $\mathcal{B}_{X}$. The subset $F$ is said to be a norm-semi-exposed face of $\mathcal{B}_{X}$ if $F=(F^{\prime})_{\prime}$, while the subset $G$ is called a weak∗-semi-exposed face of $\mathcal{B}_{X^{*}}$ if $G=(G_{\prime})^{\prime}$. The mappings $F\mapsto F^{\prime}$ and $G\mapsto G_{\prime}$ are anti-order isomorphisms between the complete lattices $\mathcal{S}_{n}(\mathcal{B}_{X})$ of norm-semi-exposed faces of $\mathcal{B}_{X}$ and $\mathcal{S}_{w^{*}}(\mathcal{B}_{X^{*}})$ of weak∗-semi-exposed faces of $\mathcal{B}_{X^{*}}$ and are inverses of each other.

Our next result is a generalization of the above Proposition 2.3.

Proposition 2.4**.**

Let $f:S(X)\to S(Y)$ be a surjective isometry between the unit spheres of two Banach spaces, and let $C$ be a convex subset of $S(X)$. Suppose that for every extreme point $\phi_{0}\in\partial_{e}(\mathcal{B}_{X^{*}})$, the set $\{\phi_{0}\}$ is a weak∗-semi-exposed face of $\mathcal{B}_{X^{*}}$. Then $C$ is a norm-semi-exposed face of $\mathcal{B}_{X}$ if and only if $f(C)$ is a norm-semi-exposed face of $\mathcal{B}_{Y}$.

Proof.

We begin with an observation. By Eidelheit’s separation theorem [26, Theorem 2.2.26], every maximal proper face of $\mathcal{B}_{X}$ is a norm-semi-exposed face (compare [35, Lemma 3.3]).

Suppose $C\in\mathcal{S}_{n}(\mathcal{B}_{X})$. We set

[TABLE]

Let us observe that $\bigcap_{\mathcal{M}\in\Lambda_{C}^{+}}\mathcal{M}$ is a proper face of $\mathcal{B}_{X}$ containing $C$. Since a non-empty intersection of proper norm-semi-exposed faces of $\mathcal{B}_{X}$ (respectively, weak∗-semi-exposed face of $\mathcal{B}_{X^{*}}$) is a proper norm-semi-exposed face of $\mathcal{B}_{X}$ (respectively, a proper weak∗-semi-exposed face of $\mathcal{B}_{X^{*}}$), the set $\bigcap_{\mathcal{M}\in\Lambda_{C}^{+}}\mathcal{M}$ is a proper norm-semi-exposed face of $\mathcal{B}_{X}$.

We shall next show that

[TABLE]

The inclusion $\subseteq$ is clear. To see the other inclusion we argue by contradiction, and thus we assume that $C\subsetneqq\bigcap_{\mathcal{M}\in\Lambda_{C}^{+}}\mathcal{M}.$ Therefore, $C^{\prime}\supsetneqq\left(\bigcap_{\mathcal{M}\in\Lambda_{C}^{+}}\mathcal{M}\right)^{\prime}$. The sets $C^{\prime}$ and $\left(\bigcap_{\mathcal{M}\in\Lambda_{C}^{+}}\mathcal{M}\right)^{\prime}$ are weak∗-closed convex faces of $\mathcal{B}_{X^{*}}$, and hence, by the Krein-Milman theorem $C^{\prime}=\overline{\hbox{co}}^{w^{*}}(\partial_{e}(C^{\prime}))$. Therefore, having in mind that $C^{\prime}$ is a face, we can find an extreme point $\phi_{0}\in\partial_{e}(C^{\prime})\subset\partial_{e}(\mathcal{B}_{X^{*}})$ such that $\phi_{0}\notin\left(\bigcap_{\mathcal{M}\in\Lambda_{C}^{+}}\mathcal{M}\right)^{\prime}$. Since, by hypothesis, $\{\phi_{0}\}$ is a weak∗-semi-exposed face of $\mathcal{B}_{X^{*}}$, we can easily check that $\mathcal{M}_{0}=\{\phi_{0}\}_{\prime}$ is a maximal proper face of $\mathcal{B}_{X}$. Furthermore, $\phi_{0}\in C^{\prime}$ implies that $C\subseteq\{\phi_{0}\}_{\prime}$ and hence $\mathcal{M}_{0}\in\Lambda_{C}^{+}$. Clearly $\phi_{0}\in\mathcal{M}_{0}^{\prime}\subseteq\left(\bigcap_{\mathcal{M}\in\Lambda_{C}^{+}}\mathcal{M}\right)^{\prime},$ which is impossible. We have thus proved (5).

Clearly, $f(C)\subseteq f(\mathcal{M})$ for every $\mathcal{M}\in\Lambda_{C}^{+}$, and thus $f(C)\subseteq\bigcap_{\mathcal{M}\in\Lambda_{C}^{+}}f\left(\mathcal{M}\right).$ Applying $f^{-1}$ and (5) we get

[TABLE]

Therefore, the identity $f(C)=\bigcap_{\mathcal{M}\in\Lambda_{C}^{+}}f\left(\mathcal{M}\right)$ follows from the bijectivity of $f$.

Since, by Proposition 2.3, for each $\mathcal{M}\in\Lambda_{C}^{+}$, $f(\mathcal{M})$ is a maximal proper face of $\mathcal{B}_{X}$ and hence norm-semi-exposed, the set $f(C)$ coincides with a non-empty intersection of norm-semi-exposed faces, and hence $f(C)$ is a norm-semi-exposed face too. ∎

In certain classes of Banach spaces where norm closed faces are all norm-semi-exposed and weak∗ closed faces in the dual space are all weak∗-semi-exposed, the previous proposition becomes meaningful and guarantees the stability of the facial structure under surjective isometries of the unit spheres. For example, when $X$ is a C∗-algebra or a JB∗-triple, every proper norm closed face of $\mathcal{B}_{X}$ is norm-semi-exposed, and every weak∗ closed proper face of $\mathcal{B}_{X^{*}}$ is weak∗-semi-exposed (see [1], [15], and [18]). The same property holds when $X$ is the predual of a von Neumann algebra or the predual of a JBW∗-triple (see [16]). Suppose $X$ and $Y$ are Banach spaces satisfying the just commented property, and $f:S(X)\to S(Y)$ is a surjective isometry. Clearly $f$ maps proper norm closed faces of $\mathcal{B}_{X}$ to proper norm closed faces of $\mathcal{B}_{Y}$ and preserves the order given by the natural inclusion. In this particular setting, for each extreme point $e\in\mathcal{B}_{X}$, the set $\{e\}$ is a minimal norm-semi-exposed face of $\mathcal{B}_{X}$ and hence $\{f(e)\}=f(\{e\})$ must be a minimal norm closed face of $\mathcal{B}_{Y}$, and thus $f(e)\in\partial_{e}(\mathcal{B}_{Y})$. All these facts are stated in the next corollary.

Corollary 2.5**.**

Let $X$ and $Y$ be Banach spaces satisfying the following two properties

1. $(1)$

*Every norm closed face of $\mathcal{B}_{X}$ *(respectively, of $\mathcal{B}_{Y}$) is norm-semi-exposed; 2. $(2)$

Every weak∗* closed proper face of $\mathcal{B}_{X^{*}}$ (respectively, of $\mathcal{B}_{Y^{*}}$) is weak∗-semi-exposed.*

Let $f:S(X)\to S(Y)$ be a surjective isometry. The following statements hold:

1. $(a)$

Let $\mathcal{F}$ be a convex set in $S(X)$. Then $\mathcal{F}$ is a norm closed face of $\mathcal{B}_{X}$ if and only if $f(\mathcal{F})$ is a norm closed face of $\mathcal{B}_{Y}$; 2. $(b)$

Given $e\in S(X)$, we have that $e\in\partial_{e}(\mathcal{B}_{X})$ if and only if $f(e)\in\partial_{e}(\mathcal{B}_{Y})$.$\hfil\Box$**

We have already commented that Corollary 2.5 holds when $X$ and $Y$ are von Neumann algebras, or predual spaces of von Neumann algebras, or more generally, JBW∗-triples or predual spaces of JBW∗-triples. It is well known that $C_{1}(H)$ is the predual of $B(H)$.

Proposition 2.6**.**

Let $f:S(C_{1}(H))\to S(C_{1}(H^{\prime}))$ be a surjective isometry, where $H$ and $H^{\prime}$ are complex Hilbert spaces. Then the following statements hold:

1. $(a)$

A subset $\mathcal{F}\subset S(C_{1}(H))$ is a proper norm-closed face of $\mathcal{B}_{C_{1}(H)}$ if and only if $f(\mathcal{F})$ is. 2. $(b)$

$f$* maps $\partial_{e}(\mathcal{B}_{C_{1}(H)})$ into $\partial_{e}(\mathcal{B}_{C_{1}(H^{\prime})})$;* 3. $(c)$

dim$(H)=$dim$(H^{\prime})$. 4. $(d)$

For each $e_{0}\in\partial_{e}(\mathcal{B}_{C_{1}(H)})$ we have $f(ie_{0})=if(e_{0})$ or $f(ie_{0})=-if(e_{0})$; 5. $(e)$

*For each $e_{0}\in\partial_{e}(\mathcal{B}_{C_{1}(H)})$ if $f(ie_{0})=if(e_{0})$ *(*respectively, $f(ie_{0})=-if(e_{0})$) then $f(\lambda e_{0})=\lambda f(e_{0})$ *(respectively, $f(\lambda e_{0})=\overline{\lambda}f(e_{0})$) for every $\lambda\in\mathbb{C}$ with $|\lambda|=1$.

Proof.

$(a)$ and $(b)$ are consequences of Corollary 2.5. Having in mind that the dimension of $H$ is precisely the cardinal of the biggest set of mutually orthogonal pure atoms in $C_{1}(H)$, statement $(c)$ follows from $(b)$ and Lemma 2.2.

$(d)$ Let $e_{0}\in\partial_{e}(\mathcal{B}_{C_{1}(H)})$. Let us set $\{e_{0}\}^{\perp}:=\{x\in C_{1}(H):x\perp e_{0}\}$. By Lemma 2.2 we have $f\left(\{e_{0}\}^{\perp}\cap S(C_{1}(H))\right)=\{f(e_{0})\}^{\perp}\cap S(C_{1}(H^{\prime}))$ and

[TABLE]

Therefore $f(ie_{0})=\mu f(e_{0})$ for a suitable $\mu\in\mathbb{T}$. Since

[TABLE]

we deduce that $\mu\in\{\pm i\}$, as desired.

$(e)$ Suppose $e_{0}\in\partial_{e}(\mathcal{B}_{C_{1}(H)})$ and $f(ie_{0})=if(e_{0})$. Let $\lambda\in\mathbb{T}$. Arguing as above, we prove that $f(\lambda e_{0})=\mu f(e_{0})$ for a suitable $\mu\in\mathbb{T}$. The identities

[TABLE]

and

[TABLE]

prove that $\mu=\lambda$. ∎

Whenever we have a surjective isometry $f:S(C_{1}(H))\to S(C_{1}(H^{\prime}))$, we deduce from the above proposition that $H$ and $H^{\prime}$ are isometrically isomorphic, we can therefore restrict our study to the case in which $H=H^{\prime}$.

We complete this section by recalling a result established by C.M. Edwards and G.T. Ruttimann in [16] (later rediscovered in [1]). More concretely, as a consequence of the result proved by C.M. Edwards and G.T. Rüttimann in [16, Theorem 5.3], we know that every proper norm-closed face $\mathcal{F}$ of $\mathcal{B}_{C_{1}(H)}$ is of the form

[TABLE]

for a unique partial isometry $w\in B(H)$. Furthermore the mapping $w\mapsto\{w\}_{{}_{\prime}}$ is an order preserving bijection between the lattices of all partial isometries in $B(H)$ and all norm closed faces of $\mathcal{B}_{C_{1}(H)}.$

If $H$ is a finite dimensional complex Hilbert space, a maximal (or complete) partial isometry $w\in B(H)$ is precisely a unitary element. Therefore by the just commented result ([16, Theorem 5.3], see also [1, Theorem 4.6]) every maximal proper (norm-closed) face $\mathcal{M}$ of $\mathcal{B}_{C_{1}(H)}$ is of the form

[TABLE]

for a unique unitary element $u\in B(H)$.

3. Surjective isometries between the unit spheres of two finite dimensional trace class spaces

In this section we present a positive solution to Tingley’s problem for surjective isometries $f:S(C_{1}(H))\to S(C_{1}(H))$, in the case in which $H$ is a finite dimensional complex Hilbert space.

Our next result is a first step towards a solution to Tingley’s conjecture in $M_{2}(\mathbb{C})$ when the latter is equipped with the trace norm.

Proposition 3.1**.**

Let $f:S(C_{1}(H))\to S(C_{1}(H))$ be a surjective isometry, where $H$ is a two dimensional complex Hilbert space. Suppose $e_{1},e_{2}$ is a (maximal) set of mutually orthogonal pure atoms in $S(C_{1}(H))$ and $\lambda_{1},\lambda_{2}$ are real numbers satisfying $|\lambda_{1}|+|\lambda_{2}|=1$. Then

[TABLE]

Proof.

Under these assumptions $B(H)$ is $M_{2}(\mathbb{C})$ with the spectral or operator norm, and $C_{1}(H)$ is $M_{2}(\mathbb{C})$ with the trace norm. We can assume the existence of orthonormal basis of $H$ $\{\eta_{1},\eta_{2}\},$ $\{\xi_{1},\xi_{2}\}$, $\{\widetilde{\eta}_{1},\widetilde{\eta}_{2}\}$ and $\{\widetilde{\xi}_{1},\widetilde{\xi}_{2}\}$ such that $e_{j}=\eta_{j}\otimes\xi_{j}$ and $f(\eta_{j}\otimes\xi_{j})=\widetilde{\eta}_{j}\otimes\widetilde{\xi}_{j}$ for every $j=1,2$ (compare Proposition 2.6$(b)$ and Lemma 2.2).

Proposition 2.6($(d)$ and $(e)$) shows that the desired statement is true when $\lambda_{1}\lambda_{2}=0$.

To simplify the notation, let $u_{1}$, $u_{2}$ and $v_{2}^{*}$ be the unitaries in $B(H)$ mapping the basis $\{\eta_{1},\eta_{2}\},$ $\{\widetilde{\eta}_{1},\widetilde{\eta}_{2}\}$ and $\{\widetilde{\xi}_{1},\widetilde{\xi}_{2}\}$ to the basis $\{{\xi}_{1},\xi_{2}\},$ respectively. Let $T_{1},T_{2}:C_{1}(H)\to C_{1}(H)$ be the surjective complex linear isometries defined by $T_{1}(x)=u_{1}x$ and $T_{2}(x)=u_{2}xv_{2}$ ($x\in C_{1}(H)$). We set $g=T_{2}|_{S(C_{1}(H))}\circ f\circ T_{1}^{-1}|_{S(C_{1}(H))}:S(C_{1}(H))\to S(C_{1}(H))$. Then $g$ is a surjective isometry satisfying

[TABLE]

We can chose a matricial representation (i.e. the representation on the basis $\{\xi_{1},\xi_{2}\}$) such that \phi_{1}=\left(\begin{array}[]{cc}1&0\\ 0&0\\ \end{array}\right) and \phi_{2}=\left(\begin{array}[]{cc}0&0\\ 0&1\\ \end{array}\right).

We recall that every maximal proper (norm-closed) face $\mathcal{M}$ of $\mathcal{B}_{C_{1}(H)}$ is of the form

[TABLE]

for a unique unitary element $u\in B(H)$ (see (7) or [16, Theorem 5.3]).

We note that $\phi_{1},\phi_{2}\in\{1\}_{{}_{\prime}}$ and $\{1\}_{{}_{\prime}}$ is precisely the set of all normal states on $B(H)$.

By Proposition 2.3 we know that $g(\{1\}_{{}_{\prime}})=\{u\}_{{}_{\prime}},$ for a unique unitary element $u\in B(H)$. Since $\phi_{j}=g(\phi_{j})\in\{u\}_{{}_{\prime}}$ we can easily check that $u=1$.

We shall first assume that $0\leq\lambda_{j}$ for every $j$. Since $\lambda_{1}+\lambda_{2}=1$ we have $g(\lambda_{1}\phi_{1}+\lambda_{2}\phi_{2})\in g(\{1\}_{{}_{\prime}})=\{1\}_{{}_{\prime}}$. Therefore, the element $g(\lambda_{1}\phi_{1}+\lambda_{2}\phi_{2})$ must be a positive matrix \left(\begin{array}[]{cc}t&c\\ \overline{c}&1-t\\ \end{array}\right) with trace and trace norm equal to one (i.e., $0\leq t\leq 1$ and \Big{\|}\left(\begin{array}[]{cc}t&c\\ \overline{c}&1-t\\ \end{array}\right)\Big{\|}_{1}=1.

Since, by hypothesis, we have

[TABLE]

[TABLE]

It is easy to check that the eigenvalues of the matrix \left(\begin{array}[]{cc}t-1&c\\ \overline{c}&1-t\\ \end{array}\right)^{2} are exactly $\{(1-t)^{2}+|c|^{2},(t-1)^{2}+|c|^{2}\}\},$ and hence \left\|\left(\begin{array}[]{cc}t-1&c\\ \overline{c}&1-t\\ \end{array}\right)\right\|_{1}=2\sqrt{(1-t)^{2}+|c|^{2}}. Therefore,

[TABLE]

The equality

[TABLE]

[TABLE]

implies

[TABLE]

Combining (8) and (9) we get $\lambda_{1}=t$ and $c=0$, which shows that

[TABLE]

equivalently, since $g=T_{2}|_{S(C_{1}(H))}\circ f\circ T_{1}^{-1}|_{S(C_{1}(H))}$ we obtain

[TABLE]

for every $\lambda_{1},\lambda_{2}\geq 0$, with $\lambda_{1}+\lambda_{2}=1$ and every set $\{e_{1},e_{2}\}$ of mutually orthogonal rank one elements in $S(C_{1}(H))$.

Finally, suppose $\lambda_{1},\lambda_{2}\in\mathbb{R}$, with $|\lambda_{1}|+|\lambda_{2}|=1$. Set $\sigma_{i}\in\{\pm 1\}$ such that $\lambda_{i}=\sigma_{i}|\lambda_{i}|$. Given and arbitrary set $\{e_{1},e_{2}\}$ of mutually orthogonal rank one elements in $S(C_{1}(H))$, the set $\{\sigma_{1}e_{1},\sigma_{2}e_{2}\}$ satisfies the same properties. It follows from (10) that

[TABLE]

[TABLE]

∎

Corollary 3.2**.**

Let $f:S(C_{1}(H))\to S(C_{1}(H))$ be a surjective isometry, where $H$ is a two dimensional complex Hilbert space. Suppose $e_{1},e_{2}$ is a (maximal) set of mutually orthogonal rank one elements in $S(C_{1}(H))$. The following statements hold:

1. $(a)$

If $f(ie_{1})=if(e_{1})$ then $f(ie_{2})=if(e_{2})$; 2. $(b)$

$f(ie_{1})=-if(e_{1})$* then $f(ie_{2})=-if(e_{2})$;*

Proof.

$(a)$ Suppose $f(ie_{1})=if(e_{1})$. Let us find two orthonormal basis $\{\eta_{1},\eta_{2}\},$ $\{\xi_{1},\xi_{2}\}$ such that $e_{1}=\eta_{1}\otimes\xi_{1}$ and $e_{2}=\eta_{2}\otimes\xi_{2}$. We set $u_{1}=\eta_{1}\otimes\xi_{2}$ and $u_{2}=\eta_{2}\otimes\xi_{1}$. The elements $x=\frac{1}{2}(e_{1}+e_{2}+u_{1}+u_{2})$ and $y=\frac{1}{2}(e_{1}+e_{2}-u_{1}-u_{2})$ are rank one elements in $S(C_{1}(H))$.

By Proposition 2.6$(d)$, $f(ix)=\pm if(x)$ and $f(iy)=\pm if(y)$. If $f(ix)=-if(x)$ we have

[TABLE]

and

[TABLE]

[TABLE]

which gives a contradiction. Therefore, $f(ix)=if(x)$. We similarly prove $f(iy)=if(y)$.

Since $ix\perp iy$ in $S(C_{1}(H))$, we can apply Proposition 3.1 to deduce that

[TABLE]

[TABLE]

Since a new application of Proposition 3.1 gives $f\left(\frac{1}{2}e_{1}+\frac{1}{2}e_{2}\right)=\frac{1}{2}f\left(e_{1}\right)+\frac{1}{2}f\left(e_{2}\right)$ and $f\left(\frac{i}{2}e_{1}+\frac{i}{2}e_{2}\right)=\frac{1}{2}f\left(ie_{1}\right)+\frac{1}{2}f\left(ie_{2}\right)$, the equality $f(ie_{1})=if(e_{1})$ proves that $f(ie_{2})=if(e_{2})$.

We can actually prove that $f(iu_{1})=if(u_{1})$ and $f(iu_{2})=if(u_{2})$.

Statement $(b)$ follows by similar arguments. ∎

Let $f:S(C_{1}(H))\to S(C_{1}(H))$ be a surjective isometry, where $H$ is a two dimensional complex Hilbert space. Suppose $e_{1},e_{2}$ is a (maximal) set of mutually orthogonal rank one elements in $S(C_{1}(H))$ and $\lambda_{1},\lambda_{2}$ are real numbers satisfying $|\lambda_{1}|+|\lambda_{2}|=1$. Then we have proved in Proposition 3.1 that

[TABLE]

Arguing as in the proof of Proposition 3.1 we can find two surjective complex linear isometries $T_{1},T_{2}:S(C_{1}(H))\to S(C_{1}(H))$ such that $g=T_{2}|_{S(C_{1}(H))}\circ f\circ T_{1}^{-1}|_{S(C_{1}(H))}:S(C_{1}(H))\to S(C_{1}(H))$ is a surjective isometry satisfying $g(\phi_{j})=\phi_{j},$ where \phi_{1}=\left(\begin{array}[]{cc}1&0\\ 0&0\\ \end{array}\right) and \phi_{2}=\left(\begin{array}[]{cc}0&0\\ 0&1\\ \end{array}\right). Now, we claim that

[TABLE]

or

[TABLE]

for suitable $\mu\in\mathbb{T}$. Indeed, we know from previous arguments that g\left(\left(\begin{array}[]{cc}0&1\\ 0&0\\ \end{array}\right)\right) and g\left(\left(\begin{array}[]{cc}0&0\\ 1&0\\ \end{array}\right)\right) are rank one orthogonal elements in $S(C_{1}(H))$. Let us write g\left(\left(\begin{array}[]{cc}0&1\\ 0&0\\ \end{array}\right)\right)=\eta\otimes\xi=\left(\begin{array}[]{cc}\alpha&\beta\\ \gamma&\delta\\ \end{array}\right), with $\alpha\delta=\gamma\beta$ and $|\alpha|^{2}+|\beta|^{2}+|\gamma|^{2}+|\delta|^{2}=1$.

By hypothesis,

[TABLE]

[TABLE]

Since \left\|\left(\begin{array}[]{cc}\alpha\pm 1&\beta\\ \gamma&\delta\\ \end{array}\right)\right\|_{1}^{2}=2, we deduce from the inequality in (4) that

[TABLE]

which assures that $0\geq\pm 2\Re\hbox{e}(\alpha)$ and hence $\Re\hbox{e}(\alpha)=0$. Replacing g\left(\left(\begin{array}[]{cc}1&0\\ 0&0\\ \end{array}\right)\right) with g\left(\left(\begin{array}[]{cc}i&0\\ 0&0\\ \end{array}\right)\right) and having in mind that g\left(i\left(\begin{array}[]{cc}1&0\\ 0&0\\ \end{array}\right)\right)\in\left\{\pm ig\left(\left(\begin{array}[]{cc}1&0\\ 0&0\\ \end{array}\right)\right)\right\}, we obtain $\Im\hbox{m}(\alpha)=0$, and thus $\alpha=0$.

Similar arguments applied to g\left(\left(\begin{array}[]{cc}0&0\\ 0&1\\ \end{array}\right)\right) instead of g\left(\left(\begin{array}[]{cc}1&0\\ 0&0\\ \end{array}\right)\right) prove $\delta=0$. Since $\gamma\beta=0$ and $|\beta|^{2}+|\gamma|^{2}=1$ we deduce that

[TABLE]

or

[TABLE]

with $\delta,\mu\in\mathbb{T}$.

We shall show next that $\delta=\overline{\mu}$. Indeed, let us assume that $g$ satisfies (13). Then, applying Proposition 3.1 we have

[TABLE]

[TABLE]

[TABLE]

Since

[TABLE]

we have

[TABLE]

which implies $2=|\overline{\delta}+\mu|,$ and hence $\overline{\delta}=\mu$.

Similarly, when we are in case (14) we get $\overline{\delta}=\mu$ and hence (12) holds.

Let us assume we are in the case derived from (11), that is,

[TABLE]

Consider the unitary v=\left(\begin{array}[]{cc}\sqrt{\mu}&0\\ 0&\overline{\sqrt{\mu}}\\ \end{array}\right) and the surjective linear isometry $T:S_{1}(H)\to S_{1}(H)$, $T(x)=v^{*}xv$. It is easy to see that T\left(\left(\begin{array}[]{cc}1&0\\ 0&0\\ \end{array}\right)\right)=\left(\begin{array}[]{cc}1&0\\ 0&0\\ \end{array}\right), T\left(\left(\begin{array}[]{cc}0&0\\ 0&1\\ \end{array}\right)\right)=\left(\begin{array}[]{cc}0&0\\ 0&1\\ \end{array}\right), T\left(\left(\begin{array}[]{cc}0&\mu\\ 0&0\\ \end{array}\right)\right)=\left(\begin{array}[]{cc}0&1\\ 0&0\\ \end{array}\right) and T\left(\left(\begin{array}[]{cc}0&0\\ \overline{\mu}&0\\ \end{array}\right)\right)=\left(\begin{array}[]{cc}0&0\\ 1&0\\ \end{array}\right). Therefore, replacing $g$ with $h=T|_{S(S_{1}(H))}g$, we obtain a surjective isometry $h:S(S_{1}(H))\to S(S_{1}(H))$ satisfying h\left(\left(\begin{array}[]{cc}1&0\\ 0&0\\ \end{array}\right)\right)=\left(\begin{array}[]{cc}1&0\\ 0&0\\ \end{array}\right), h\left(\left(\begin{array}[]{cc}0&0\\ 0&1\\ \end{array}\right)\right)=\left(\begin{array}[]{cc}0&0\\ 0&1\\ \end{array}\right), h\left(\left(\begin{array}[]{cc}0&1\\ 0&0\\ \end{array}\right)\right)=\left(\begin{array}[]{cc}0&1\\ 0&0\\ \end{array}\right) and h\left(\left(\begin{array}[]{cc}0&0\\ 1&0\\ \end{array}\right)\right)=\left(\begin{array}[]{cc}0&0\\ 1&0\\ \end{array}\right).

When (12) holds we can find surjective linear isometry $T:S_{1}(H)\to S_{1}(H)$ such that $h=T|_{S(S_{1}(H))}g$ is a surjective isometry satisfying h\left(\left(\begin{array}[]{cc}1&0\\ 0&0\\ \end{array}\right)\right)=\left(\begin{array}[]{cc}1&0\\ 0&0\\ \end{array}\right), h\left(\left(\begin{array}[]{cc}0&0\\ 0&1\\ \end{array}\right)\right)=\left(\begin{array}[]{cc}0&0\\ 0&1\\ \end{array}\right), h\left(\left(\begin{array}[]{cc}0&1\\ 0&0\\ \end{array}\right)\right)=\left(\begin{array}[]{cc}0&0\\ 1&0\\ \end{array}\right) and h\left(\left(\begin{array}[]{cc}0&0\\ 1&0\\ \end{array}\right)\right)=\left(\begin{array}[]{cc}0&1\\ 0&0\\ \end{array}\right).

We shall establish now a technical proposition to measure the distance between two pure atoms in $C_{1}(H)$. For each pure atom $e=\eta\otimes\xi$ in $S(C_{1}(H))$, as before, let $s(e)=\eta\otimes\xi$ be the unique minimal partial isometry in $B(H)$ satisfying $e(s(e))=1$. For any other $x\in S(C_{1}(H))$ the evaluation $x(s(e))\in\mathbb{C}$. For each partial isometry $s$ in $B(H)$, the Bergmann operator $P_{0}(s):B(H)\to B(H),$ $x\mapsto(1-ss^{*})x(1-s^{*}s)$, is weak∗ continuous and hence $P_{0}(s)^{*}(x)\in C_{1}(H)$ for every $x\in C_{1}(H)$.

Lemma 3.3**.**

Let $H$ be a complex Hilbert space. Suppose $e_{1},e_{2}$ are two rank one pure atoms in $S(C_{1}(H))$. Then the following formula holds:

[TABLE]

where $\alpha=\alpha(e_{1},e_{2})=e_{2}(s(e_{1}))$ and $\hat{\delta}=\hat{\delta}(e_{1},e_{2})=\|P_{0}(s(e_{1}))^{*}(e_{2})\|_{1}$.

Proof.

By choosing an appropriate matrix representation we can find two orthonormal systems $\{\eta_{1},\eta_{2}\}$ and $\{\xi_{1},\xi_{2}\}$ to represent $e_{1}$ and $e_{2}$ in the form $e_{1}=\eta_{1}\otimes\xi_{1}$, $e_{2}=\widetilde{\eta}_{1}\otimes\widetilde{\xi}_{1}$ and

[TABLE]

where $e_{1}=v_{11}$, $v_{12}={\eta}_{2}\otimes{\xi}_{1}$, $v_{21}={\eta}_{1}\otimes{\xi}_{2}$, $v_{22}={\eta}_{2}\otimes{\xi}_{2}$, $\alpha=\langle\xi_{1}/\widetilde{\xi}_{1}\rangle\langle\widetilde{\eta}_{1}/\eta_{1}\rangle,$ $\beta=\langle\xi_{1}/\widetilde{\xi}_{1}\rangle\langle\widetilde{\eta}_{1}/\eta_{2}\rangle,$ $\gamma=\langle\xi_{2}/\widetilde{\xi}_{1}\rangle\langle\widetilde{\eta}_{1}/\eta_{1}\rangle,$ $\delta=\langle\xi_{2}/\widetilde{\xi}_{1}\rangle\langle\widetilde{\eta}_{1}/\eta_{2}\rangle\in\mathbb{C},$ with $|\alpha|^{2}+|\beta|^{2}+|\gamma|^{2}+|\delta|^{2}$ $=|\langle\xi_{1}/\widetilde{\xi}_{1}\rangle|^{2}\|\widetilde{\eta}_{1}\|^{2}+|\langle\xi_{2}/\widetilde{\xi}_{1}\rangle|^{2}\|\widetilde{\eta}_{1}\|^{2}=\|\widetilde{\xi}_{1}\|^{2}=1$, and $\alpha\delta=\beta\gamma$. In an appropriate matrix representation we can identify $e_{1}$ and $e_{2}$ with \left(\begin{array}[]{cc}1&0\\ 0&0\\ \end{array}\right), and \left(\begin{array}[]{cc}\alpha&\beta\\ \gamma&\delta\\ \end{array}\right), respectively.

Following the arguments in the proof of [19, Proposition 3.3] we deduce that the eigenvalues of the element (e_{2}-e_{1})(e_{2}-e_{1})^{*}=\left(\begin{array}[]{cc}\alpha-1&\beta\\ \gamma&\delta\\ \end{array}\right)\left(\begin{array}[]{cc}\alpha-1&\beta\\ \gamma&\delta\\ \end{array}\right)^{*} are precisely $(1-\Re\hbox{e}(\alpha))\pm\sqrt{(1-\Re\hbox{e}(\alpha))^{2}-|\delta|^{2}}$ and hence

[TABLE]

which proves the desired formula. ∎

We are now in position to solve Tingley’s problem for the case of trace class operators on a two dimensional Hilbert space.

Theorem 3.4**.**

Let $f:S(C_{1}(H))\to S(C_{1}(H))$ be a surjective isometry, where $H$ is a two dimensional complex Hilbert space. Then there exists a surjective complex linear or conjugate linear isometry $T:C_{1}(H)\to C_{1}(H)$ satisfying $f(x)=T(x)$ for every $x\in S(C_{1}(H))$. More concretely, there exist unitaries $u,v\in M_{2}(\mathbb{C})$ such that one of the following statements holds:

1. $(a)$

$f(x)=uxv$, for every $x\in S(C_{1}(H))$; 2. $(b)$

$f(x)=ux^{t}v$, for every $x\in S(C_{1}(H))$; 3. $(c)$

$f(x)=u\overline{x}v$, for every $x\in S(C_{1}(H))$; 4. $(d)$

$f(x)=ux^{*}v$, for every $x\in S(C_{1}(H))$,

where $\overline{(x_{ij})}=(\overline{x_{ij}})$.

Proof.

By the comments preceding this theorem, we can find two surjective linear isometries $U,V:C_{1}(H)\to C_{1}(H)$ such that the mapping $h=U|_{S(C_{1}(H))}\circ f\circ V|_{S(C_{1}(H))}:S(C_{1}(H))\to S(C_{1}(H))$ is a surjective isometry satisfying precisely one of the next statements

[TABLE]

[TABLE]

or

[TABLE]

[TABLE]

Proposition 2.6$(d)$ and $(e)$ implies that $f(-z)=-f(z)$ and $h(-z)=-h(z)$ for every pure atom $z$ in $C_{1}(H)$.

We assume that (15) holds. Let us denote v_{11}=\left(\begin{array}[]{cc}1&0\\ 0&0\\ \end{array}\right), v_{12}=\left(\begin{array}[]{cc}0&1\\ 0&0\\ \end{array}\right), v_{21}=\left(\begin{array}[]{cc}0&0\\ 1&0\\ \end{array}\right), and v_{22}=\left(\begin{array}[]{cc}0&0\\ 0&1\\ \end{array}\right).

By Proposition 2.6$(d)$ and Corollary 3.2, we know that one of the next statements holds:

1. $(a)$

$h(iv_{11})=ih(v_{11})$ and $h(iv_{22})=ih(v_{22})$; 2. $(b)$

$h(iv_{11})=-ih(v_{11})$ and $h(iv_{22})=-ih(v_{22})$.

The proof will be splitted into two cases corresponding to the above statements.

Case $(a)$. Let us assume that $(a)$ holds. We consider the pure atom e_{1}=\left(\begin{array}[]{cc}\frac{1}{2}&\frac{1}{2}\\ \frac{1}{2}&\frac{1}{2}\\ \end{array}\right). Proposition 2.6 assures that $h(e_{1})$ is a pure atom, and hence it must be of the form h(v)=\left(\begin{array}[]{cc}\alpha^{\prime}&\beta^{\prime}\\ \gamma^{\prime}&\delta^{\prime}\\ \end{array}\right) with $|\alpha^{\prime}|^{2}+|\beta^{\prime}|^{2}+|\gamma^{\prime}|^{2}+|\delta^{\prime}|^{2}=1$, $\alpha^{\prime}\delta^{\prime}=\beta^{\prime}\gamma^{\prime}$. By the hypothesis on $h$ and Lemma 3.3 we get

[TABLE]

[TABLE]

and

[TABLE]

[TABLE]

Taking squares in both sides we get

[TABLE]

and

[TABLE]

which gives $\Re\hbox{e}(\alpha^{\prime})=\frac{1}{2}$ and $|\delta^{\prime}|=\frac{1}{2}$.

When in the above arguments we replace $v_{11}$ with $v_{12}$, $v_{21}$ and $v_{22}$ we obtain $\Re\hbox{e}(\beta^{\prime})=\frac{1}{2}$, $\Re\hbox{e}(\gamma^{\prime})=\frac{1}{2}$ and $\Re\hbox{e}(\delta^{\prime})=\frac{1}{2}$. Since $|\alpha^{\prime}|^{2}+|\beta^{\prime}|^{2}+|\gamma^{\prime}|^{2}+|\delta^{\prime}|^{2}=1$, we deduce that $\alpha^{\prime}=\beta^{\prime}=\gamma^{\prime}=\delta^{\prime}=\frac{1}{2}$, and hence h(e_{1})=e_{1}=\left(\begin{array}[]{cc}\frac{1}{2}&\frac{1}{2}\\ \frac{1}{2}&\frac{1}{2}\\ \end{array}\right).

We can similarly show that taking e_{2}=\left(\begin{array}[]{cc}\frac{1}{2}&-\frac{1}{2}\\ -\frac{1}{2}&\frac{1}{2}\\ \end{array}\right), we have $h(e_{2})=e_{2}$.

Now since $e_{1}$ and $e_{2}$ are orthogonal pure atoms, Proposition 2.6$(d)$ and $(e)$ and Corollary 3.2 imply that exactly one of the next statements holds:

1. $(a.1)$

$h(ie_{1})=ih(e_{1})$ and $h(ie_{2})=ih(e_{2})$; 2. $(a.2)$

$h(ie_{1})=-ih(e_{1})$ and $h(ie_{2})=-ih(e_{2})$.

Let us show that the conclusion in $(a.2)$ is impossible. Indeed, if $(a.2)$ holds, Proposition 3.1 implies

[TABLE]

[TABLE]

which is impossible.

Since $(a.1)$ holds, we deduce, via Proposition 3.1, that

[TABLE]

[TABLE]

We know from Corollary 3.2 that $h\left(iv_{jk}\right)\in\{\pm ih(v_{jk})\}=\{\pm iv_{jk}\}$, for every $k,j=1,2$. Thus,

[TABLE]

We shall prove that

[TABLE]

Let $v$ be a pure atom (i.e. a rank one partial isometry) in $S(C_{1}(H))$. By Proposition 2.6$(b)$, $h(v)$ is a pure atom in $S(C_{1}(H))$. Arguing as in the proof of Lemma 3.3, we may assume that v=\left(\begin{array}[]{cc}\alpha&\beta\\ \gamma&\delta\\ \end{array}\right), and h(v)=\left(\begin{array}[]{cc}\alpha^{\prime}&\beta^{\prime}\\ \gamma^{\prime}&\delta^{\prime}\\ \end{array}\right), with $|\alpha|^{2}+|\beta|^{2}+|\gamma|^{2}+|\delta|^{2}=1$, $|\alpha^{\prime}|^{2}+|\beta^{\prime}|^{2}+|\gamma^{\prime}|^{2}+|\delta^{\prime}|^{2}=1$, $\alpha^{\prime}\delta^{\prime}=\beta^{\prime}\gamma^{\prime}$, and $\alpha\delta=\beta\gamma$.

Applying the hypothesis on $h$ and Lemma 3.3 we get the following equations

[TABLE]

[TABLE]

and

[TABLE]

[TABLE]

Taking squares in both sides we get

[TABLE]

and

[TABLE]

which gives $\Re\hbox{e}(\alpha^{\prime})=\Re\hbox{e}(\alpha)$ and $|\delta|=|\delta^{\prime}|$.

Then applying the above arguments to $v$, $iv_{11}$ and $-iv_{11}$ we obtain $\Im\hbox{m}(\alpha^{\prime})=\Im\hbox{m}(\alpha)$, and hence $\alpha=\alpha^{\prime}$.

Having in mind that in case $(a)$ we have $h(iv_{jk})=iv_{jk}$ for every $j,k$ (see (17)), a similar reasoning applied to $v$, $v_{22}$ and $iv_{22}$ (respectively, $v$, $v_{12}$ and $iv_{12}$ or $v$, $v_{21}$ and $iv_{21}$) gives $\delta=\delta^{\prime}$ (respectively, $\beta=\beta^{\prime}$ or $\gamma=\gamma^{\prime}$). We have therefore shown that $h(v)=v$, for every pure atom $v$.

Proposition 3.1 assures that $h(x)=x$ for every $x\in S(C_{1}(H))$, and hence $f=U^{-1}V^{-1}|_{S(C_{1}(H))},$ where $U^{-1}V^{-1}=C_{1}(H)\to C_{1}(H)$ is a surjective complex linear isometry.

In Case $(b)$, we can mimic the above arguments to show that

[TABLE]

where \overline{\left(\begin{array}[]{cc}x_{11}&x_{12}\\ x_{21}&x_{21}\\ \end{array}\right)}=\left(\begin{array}[]{cc}\overline{x_{11}}&\overline{x_{12}}\\ \overline{x_{21}}&\overline{x_{21}}\\ \end{array}\right), and consequently, $f(x)=U^{-1}(\overline{V^{-1}(x)})=T(x)$, for every $x\in S(C_{1}(H))$, where $T:C_{1}(H)\to C_{1}(H)$, $T(x)=U^{-1}(\overline{V^{-1}(x)})$ ($x\in C_{1}(H)$) is a surjective conjugate linear isometry.

Finally, if we assume (16), then there exist two surjective linear isometries $U,V:C_{1}(H)\to C_{1}(H)$ such that

[TABLE]

or

[TABLE]

for every $x\in S(C_{1}(H)).$ ∎

Before dealing with surjective isometries between the unit spheres of trace class spaces over a finite dimensional complex Hilbert space, we shall present a technical result.

Proposition 3.5**.**

Let $f:S(C_{1}(H))\to S(C_{1}(H))$ be a surjective isometry, where $H$ is a complex Hilbert space with dim$(H)=n$. Suppose $f$ satisfies the following property: given a set $\{e_{1},e_{2},\ldots,e_{k}\}$ of mutually orthogonal pure atoms in $S(C_{1}(H))$ with $k<n$, and real numbers $\lambda_{1},\lambda_{2},\ldots,\lambda_{k}$ satisfying $\displaystyle\sum_{j=1}^{k}|\lambda_{j}|=1$ we have $\displaystyle f\left(\sum_{j=1}^{k}\lambda_{j}e_{j}\right)=\sum_{j=1}^{k}\lambda_{j}e_{j}$. Then $f(x)=x,$ for every $x\in S(C_{1}(H))$.

Proof.

Let $\{\xi_{1},\ldots,\xi_{n}\}$ be an orthonormal basis of $H$. We set $v_{j}=\xi_{j}\otimes\xi_{j}$ ($j\in\{1,\ldots,n\}$). We claim that the identity

[TABLE]

holds for every $\mu_{1},\ldots,\mu_{n}$ in $\mathbb{R}^{+}_{0}$ with $\displaystyle\sum_{j=1}^{n}\mu_{j}=1$. We observe that we can assume that $\mu_{j}>0$ for every $j$, otherwise the statement is clear from the hypothesis on $f$.

By the hypothesis on $f$ we know that $f(v_{j})=v_{j}$ for every $j\in\{1,\ldots,n\}$. Let $1$ denote the identity in $B(H)$. Clearly $\{v_{1},v_{2},\ldots,v_{n}\}\subseteq\{1\}_{\prime}.$ By Proposition 2.6 (see also (7)) there exists a unique unitary $w\in B(H)$ such that $f(\{1\}_{\prime})=\{w\}_{\prime}$. Since, for each $j=1,\ldots,n$, $v_{j}=f(v_{j})\in\{w\}_{\prime},$ we can easily deduce that $w=1$ and hence $f(\{1\}_{\prime})=\{1\}_{\prime}$ is the face of all states (i.e., positive norm-one functional on $B(H)$).

The element $\displaystyle f\left(\sum_{j=1}^{n}\mu_{j}v_{j}\right)\in f(\{1\}_{\prime})=\{1\}_{\prime}$, and hence there exists a positive matrix $a=(a_{ij})\in M_{n}(\mathbb{C})$ with $\|a\|_{1}=1=\hbox{tr}(a)$ such that $\displaystyle f\left(\sum_{j=1}^{n}\mu_{j}v_{j}\right)=a.$ It should be remarked that we can also identify each $v_{j}$ with the matrix in $M_{n}(\mathbb{C})$ with entry $1$ in the $(j,j)$ position and zero otherwise.

Let us fix a projection $p\in B(H)$. The mapping $M_{p}:C_{1}(H)\to C_{1}(H)$, $M_{p}(x)=pxp+(1-p)x(1-p)$ is linear, contractive and positive. Let $p_{j}$ denote the projection $\xi_{j}\otimes\xi_{j}=s(v_{j})\in B(H)$. By hypothesis

[TABLE]

[TABLE]

Having in mind that $(1-{p_{j}})a(1-{p_{j}})$, ${p_{j}}a{p_{j}},$ and $a$ are positive functionals with $a(1)=\|a\|_{1}=1$, we get

[TABLE]

[TABLE]

[TABLE]

This shows that $a_{jj}\geq\mu_{j}$ for every $j=1,\ldots,n$. Since

[TABLE]

we deduce that $a_{jj}=\mu_{j}$ for every $j=1,\ldots,n$.

We shall now show that $a_{ij}=0$ for every $i\neq j$. For this purpose, fix $i\neq j$ and set $q=p_{i}+p_{j}\in B(H)$ and $x=\mu_{i}v_{i}+(1-\mu_{i})v_{j}\in S(C_{1}(H)).$ By hypothesis $f(x)=x$, and

[TABLE]

Let us observe that

[TABLE]

whose eigenvalues are precisely

[TABLE]

and thus

[TABLE]

[TABLE]

Therefore, we have

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

which implies that $1-\mu_{i}-\mu_{j}\geq\sqrt{(1-\mu_{i}-\mu_{j})^{2}+4|a_{ij}|^{2}},$ and hence $a_{ij}=0$ as desired. We have thus proved that

[TABLE]

which concludes the proof of (20).

Finally, let us take $x\in S(C_{1}(H))$. We can find a set $\{e_{1},e_{2},\ldots,e_{n}\}$ of mutually orthogonal pure atoms in $S(C_{1}(H))$ and real numbers $\lambda_{1},\lambda_{2},\ldots,\lambda_{n}$ such that $\displaystyle x=\sum_{j=1}^{n}\lambda_{j}e_{j}.$ We observe that, replacing each $e_{j}$ with $\pm e_{j}$ we can always assume that $\lambda_{j}\geq 0$ for every $j$. Let us pick two unitary matrices $u_{1},w_{1}\in B(H)$ satisfying $u_{1}v_{j}w_{1}=e_{j}$ for every $j=1,\ldots,n$. The operator $T_{u_{1},w_{1}}:C_{1}(H)\to C_{1}(H)$, $T(y)=u_{1}yw_{1}$ is a surjective isometry mapping elements of rank $k$ to elements of the same rank. Consequently, the mapping $f_{2}:S(C_{1}(H))\to S(C_{1}(H)),$ $f_{2}(y)=u_{1}^{*}f(u_{1}yw_{1})w_{1}^{*}$ is a surjective isometry. For each $y\in S(C_{1}(H))$ with rank$(y)<n$, we have $u_{1}yw_{1}\in S(C_{1}(H))$ with rank$(u_{1}yw_{1})<n$ and thus, by the hypothesis on $f$, $f(u_{1}yw_{1})=u_{1}yw_{1}$, which implies $f_{2}(y)=y$. Therefore $f_{2}$ satisfies the same hypothesis of $f$. Applying (20) we get

[TABLE]

[TABLE]

which proves that $\displaystyle f(x)=f\left(\sum_{j=1}^{n}\lambda_{j}e_{j}\right)=\sum_{j=1}^{n}\lambda_{j}e_{j}=x,$ as desired. ∎

Remark 3.6**.**

Let $T:C_{1}(H)\to C_{1}(H)$ be a surjective real linear isometry, where $H$ is a complex Hilbert space. Since $T^{*}:B(H)\to B(H)$ is a surjective real linear isometry, $T^{*}$ and $T$ must be complex linear or conjugate linear (see [4, Proposition 2.6]). We therefore deduce from [28] (see also [22, §11.2]) that there exist unitary matrices $u,v\in B(H)$ such that one of the next statements holds:

1. $(a.1)$

$T(x)=uxv$, for every $x\in S(C_{1}(H))$; 2. $(a.2)$

$T(x)=ux^{t}v$, for every $x\in S(C_{1}(H))$; 3. $(a.3)$

$T(x)=u\overline{x}v$, for every $x\in S(C_{1}(H))$; 4. $(a.4)$

$T(x)=ux^{*}v$, for every $x\in S(C_{1}(H))$,

where $\overline{(x_{ij})}=(\overline{x_{ij}})$.

We can find a more concrete description under additional hypothesis. Suppose dim$(H)=n$. The symbol $e_{ij}\in C_{1}(H)$ will denote the elementary matrix with entry $1$ at position $(i,j)$ and zero otherwise.

Case A Suppose that $T(\zeta e_{kk})=\zeta e_{kk}$ for every $k=1,\ldots,n-1$, and all $\zeta\in\mathbb{T}$, $T(e_{n1})=\alpha e_{n1},$ and $T(e_{1n})=\mu e_{1n}$ with $\alpha,\mu\in\mathbb{T}$. Then $T$ has the form described in case $(a.1)$ above with

[TABLE]

If we also assume $T(e_{nn})=e_{nn},$ then $\alpha=\overline{\mu}$.

For the proof we simply observe that since $T(ie_{11})=ie_{11}$, we discard cases $(a.3)$ and $(a.4)$. The assumption $T(e_{1n})=\mu e_{1n}$ shows that case $(a.2)$ is impossible too. Since $T$ has the form described in $(a.1)$ for suitable $u,v$. Now, $T(e_{kk})=e_{kk}$ implies that $u_{kk}v_{kk}=1$, for all $k=1,\ldots,n-1$. Finally, it is straightforward to check that $T(e_{1n})=\mu e_{1n}$ and $T(e_{n1})=\alpha e_{n1}$ give the desired statement.

We can present now the main result of this section.

Theorem 3.7**.**

Let $f:S(C_{1}(H))\to S(C_{1}(H))$ be a surjective isometry, where $H$ is a finite dimensional complex Hilbert space. Then there exists a surjective complex linear or conjugate linear isometry $T:C_{1}(H)\to C_{1}(H)$ satisfying $f(x)=T(x)$ for every $x\in S(C_{1}(H))$. More concretely, there exist unitary elements $u,v\in M_{n}(\mathbb{C})=B(H)$ such that one of the following statements holds:

1. $(a)$

$f(x)=uxv$, for every $x\in S(C_{1}(H))$; 2. $(b)$

$f(x)=ux^{t}v$, for every $x\in S(C_{1}(H))$; 3. $(c)$

$f(x)=u\overline{x}v$, for every $x\in S(C_{1}(H))$; 4. $(d)$

$f(x)=ux^{*}v$, for every $x\in S(C_{1}(H))$,

where $\overline{(x_{ij})}=(\overline{x_{ij}})$.

Proof.

We shall argue by induction on $n=$dim$(H)$. The case $n=2$ has been proved in Theorem 3.4. Let us assume that the desired conclusion is true for every surjective isometry $f:S(C_{1}(K))\to S(C_{1}(K))$, where $K$ is a finite dimensional complex Hilbert space of dimension $\leq n$. Let $f:S(C_{1}(H))\to S(C_{1}(H))$ be a surjective isometry, where $H$ is an $(n+1)$-dimensional complex Hilbert space.

Let $\{\xi_{1},\ldots,\xi_{n+1}\}$ be an orthonormal basis of $H$, and let $e_{ij}=\xi_{j}\otimes\xi_{i}$. Clearly $e_{ij}$ is a pure atom for every $i,j$, and $\{e_{11},\ldots,e_{(n+1)(n+1)}\}$ is a maximal set of mutually orthogonal pure states in $C_{1}(H)$. By Proposition 2.6 and Lemma 2.2 $\{f(e_{11}),\ldots,f(e_{(n+1)(n+1)})\}$ is a maximal set of mutually orthogonal pure atoms in $C_{1}(H)$ too. We can find unitary matrices $u_{1},w_{1}\in M_{n+1}(\mathbb{C})$ such that $u_{1}f(e_{ii})w_{1}=e_{ii}$ for every $i=1,\ldots,n+1$. We set $f_{1}=u_{1}fw_{1}$. We observe that $f$ admits an extension to a surjective real linear isometry if and only if $f_{1}$ does.

Let us note that $\{e_{(n+1)(n+1)}\}^{\perp}:=\{x\in C_{1}(H):x\perp e_{(n+1)(n+1)}\}\cong C_{1}(K)$ for a suitable $n$-dimensional complex Hilbert subspace of $H$. We regard $C_{1}(K)$ as a complemented subspace of $C_{1}(H)$ under the appropriate identification. Lemma 2.2 implies that

[TABLE]

[TABLE]

and hence $f_{1}|_{S(C_{1}(K))}:S(C_{1}(K))\to S(C_{1}(K))$ is a surjective isometry. By the induction hypothesis, there exist unitaries $u_{n},v_{n}\in M_{n}(\mathbb{C})=B(K)$ such that one of the following statements holds:

1. $(1)$

$f_{1}(x)=u_{n}xv_{n}$, for every $x\in S(C_{1}(K))$; 2. $(2)$

$f_{1}(x)=u_{n}x^{t}v_{n}$, for every $x\in S(C_{1}(K))$; 3. $(3)$

$f_{1}(x)=u_{n}\overline{x}v_{n}$, for every $x\in S(C_{1}(K))$; 4. $(4)$

$f_{1}(x)=u_{n}x^{*}v_{n}$, for every $x\in S(C_{1}(K))$.

Let u_{n+1}=\left(\begin{array}[]{cc}u_{n}&0\\ 0&1\\ \end{array}\right) and v_{n+1}=\left(\begin{array}[]{cc}v_{n}&0\\ 0&1\\ \end{array}\right). In each case from $(1)$ to $(4)$, we can define via the unitaries $u_{n+1},v_{n+1}$ in $B(H)=M_{n+1}(\mathbb{C})$, the involution ∗, the transposition and the conjugation $\overline{\,\cdot\,}$, a surjective complex linear or conjugate linear isometry $T_{1}:C_{1}(H)\to C_{1}(H)$ such that $T_{1}f_{1}(x)=x,$ for every $x\in S(C_{1}(K))$ and $T_{1}f_{1}(e_{(n+1)(n+1)})=e_{(n+1)(n+1)}$.

We deal now with the mapping $f_{2}=T_{1}f_{1},$ which is a surjective isometry from $S(C_{1}(H))$ onto itself and satisfies

[TABLE]

and $f_{2}(e_{(n+1)(n+1)})=e_{(n+1)(n+1)}.$

We claim that

[TABLE]

for a suitable $\mu$ in $\mathbb{T}$. Indeed, since $e_{1(n+1)}\perp e_{22},\ldots,e_{nn},e_{21}$ and $e_{(n+1)1}\perp e_{22},\ldots,e_{nn},e_{12}$, Lemma 2.2 implies that

[TABLE]

and

[TABLE]

which implies that

[TABLE]

and

[TABLE]

with $|\alpha|^{2}+|\beta|^{2}=1$ and $|\lambda|^{2}+|\mu|^{2}=1$ (compare Proposition 2.6$(b)$). Since $e_{(n+1)1}\perp e_{1(n+1)}$, a new application of Lemma 2.2 proves that $f_{2}(e_{1(n+1)})\perp f_{2}(e_{(n+1)1})$, and consequently $\lambda=\beta=0$. We have thus proved that

[TABLE]

with $\mu,\alpha$ in $\mathbb{T}$. We shall show next that $\alpha=\overline{\mu}$. To this end, we observe that the subspace

[TABLE]

is isometrically isomorphic to $C_{1}(K_{2})$ for a suitable $2$-dimensional complex Hilbert subspace $K_{2}$ of $H$, contains $e_{11},$ $e_{1(n+1)},$ $e_{(n+1)1},$ and $e_{(n+1)(n+1)}$, and by Lemma 2.2, $f_{2}|_{S(C_{1}(K_{2}))}:S(C_{1}(K_{2}))\to S(C_{1}(K_{2}))$ is a surjective isometry. So, by the induction hypothesis, there exist unitaries $u_{3},v_{3}\in B(K_{2})$ such that $f_{2}|_{S(C_{1}(K_{2}))}$ satisfies one of the statements from $(a)$ to $(d)$ in our theorem. Having in mind that $f_{2}(\zeta e_{11})=\zeta e_{11}$ for every $\zeta\in\mathbb{T}$, $f_{2}(e_{(n+1)(n+1)})=e_{(n+1)(n+1)}$, $f_{2}(e_{1(n+1)})=\mu e_{1(n+1)}$ and $f_{2}(e_{(n+1)1})=\alpha e_{(n+1)1}$, it can be easily seen that we can identify $u_{3}$ and $v_{3}$ with \left(\begin{array}[]{cc}1&0\\ 0&u_{33}\\ \end{array}\right) and \left(\begin{array}[]{cc}1&0\\ 0&\overline{u_{33}}\\ \end{array}\right), where $u_{33}\in\mathbb{T}$, respectively (see Remark 3.6 Case A). This shows that $\mu=\overline{u_{33}}$ and $\alpha={u_{33}},$ which finishes the proof of (22).

Now, the subspace $\{e_{(n+1)1}\}^{\perp}\subset C_{1}(H)$ is isometrically isomorphic to $C_{1}(K_{3})$ for a suitable $n$-dimensional complex Hilbert space $K_{3}$, and since $f_{2}(e_{(n+1)1})=\overline{\mu}e_{(n+1)1}$ (see (22)), Lemma 2.2 implies that

[TABLE]

is a surjective isometry. By the induction hypothesis there exists a surjective real linear isometry $T_{3}:C_{1}(K_{3})\to C_{1}(K_{3})$ such that $f_{2}|_{S(C_{1}(K_{3}))}\equiv T_{3}|_{S(C_{1}(K_{3}))}$. Since $T_{3}(\zeta e_{ij})=f_{2}(\zeta e_{ij})=\zeta e_{ij}$ for every $i,j\in\{2,\ldots,n\}$ and all $\zeta\in\mathbb{T}$ (see (21)) and $T_{3}(e_{1(n+1)})=f_{2}(e_{1(n+1)})={\mu}e_{1(n+1)}$, Remark 3.6 Case A shows that $T_{3}(x)=xv_{3}$, where $v_{3}$ identifies with the matrix \left(\begin{array}[]{cccc}1&\ldots&0&0\\ \vdots&&\vdots&\vdots\\ 0&\ldots&1&0\\ 0&\ldots&0&{\mu}\\ \end{array}\right). This implies that

[TABLE]

for every $z\in S(C_{1}(K_{3}))$ with $z=zp_{n+1}$, where $p_{n+1}$ is the rank one projection $\xi_{n+1}\otimes\xi_{n+1}\in B(H)$, that is, for every $z\in S(C_{1}(H))$ with $z\perp e_{(n+1)1}$ and $z=zp_{n+1}$.

Similar arguments prove that

[TABLE]

for every $z\in S(C_{1}(H))$ with $z\perp e_{1(n+1)}$ and $z=p_{n+1}z$.

Let us consider the unitaries u_{4}=\left(\begin{array}[]{cccc}1&\ldots&0&0\\ \vdots&&\vdots&\vdots\\ 0&\ldots&1&0\\ 0&\ldots&0&{\mu}\\ \end{array}\right)\in B(H), and v_{4}=\left(\begin{array}[]{cccc}1&\ldots&0&0\\ \vdots&&\vdots&\vdots\\ 0&\ldots&1&0\\ 0&\ldots&0&\overline{\mu}\\ \end{array}\right)\in B(H), and the surjective complex linear isometry defined by $T_{4}(x)=u_{4}xv_{4}$. Let $f_{3}:S(C_{1}(H))\to S(C_{1}(H))$ be the surjective isometry defined by $f_{3}(x)=T_{4}(f_{2}(x))$ ($x\in S(C_{1}(H))$). Since $T_{4}(y)=y$ for every $y\in\{e_{(n+1)(n+1)}\}^{\perp}$, $T_{4}(e_{(n+1)(n+1)})=e_{(n+1)(n+1)}$, $T_{4}(z)=\overline{\mu}z,$ for all $z\in S(C_{1}(H))$ with $z\perp e_{(n+1)1}$ and $z=zp_{n+1}$, and $T_{4}(z)={\mu}z,$ for all $z\in S(C_{1}(H))$ with $z\perp e_{1(n+1)}$ and $z=p_{n+1}z$, we deduce from (21), (23), and (24) that

[TABLE]

for all $x$ in the intersection of $S(C_{1}(H))$ with the set

[TABLE]

We shall show next that

[TABLE]

for every pure atom $v\in S(C_{1}(H)).$ Let $v=\eta\otimes\zeta$ be a pure atom in $C_{1}(H)$. We can always write $\zeta=\lambda_{1}\zeta_{1}+\lambda_{2}\xi_{n+1}$ and $\eta=\mu_{1}\eta_{1}+\mu_{2}\xi_{n+1}$, with $\lambda_{j},\mu_{j}\in\mathbb{C},$ $|\mu_{1}|^{2}+|\mu_{2}|^{2}=1,$ $|\lambda_{1}|^{2}+|\lambda_{2}|^{2}=1,$ and $\eta_{1},\zeta_{1}$ are norm one elements in $\{\xi_{n+1}\}^{\perp}$. As before, we can write

[TABLE]

where $v_{11}=\eta_{1}\otimes\zeta_{1}$, $v_{12}={\eta}_{1}\otimes{\xi}_{n+1}$, $v_{21}={\xi}_{n+1}\otimes{\zeta}_{1}$, $v_{22}={\xi}_{n+1}\otimes{\xi}_{n+1}$, $|\alpha|^{2}+|\beta|^{2}+|\gamma|^{2}+|\delta|^{2}=1$, and $\alpha\delta=\beta\gamma$.

Let us note that $\mathbb{T}v_{11}\subseteq\{e_{(n+1)(n+1)}\}^{\perp}$, $\mathbb{T}v_{12}\subseteq\{z\perp e_{(n+1)1}\&z=zp_{n+1}\}\cap S(C_{1}(H))$, and $\mathbb{T}v_{12}\subseteq\{z\perp e_{1(n+1)}\&z=p_{n+1}z\}\cap S(C_{1}(H))$, it follows from (25) that

[TABLE]

Let us consider the space $v\in\{v_{11},v_{12},v_{21},v_{22}\}^{\perp\perp}=\hbox{Span}\{v_{11},v_{12},v_{21},v_{22}\}\cong M_{2}(\mathbb{C})$ whose unit sphere is fixed by $f_{3}$ (compare Lemma 2.2). It follows from the induction hypothesis (i.e. $f_{3}$ satisfies one of the statements from $(a)$ to $(d)$ in the statement of the theorem for every element in the unit sphere of $\{v_{11},v_{12},v_{21},v_{22}\}^{\perp\perp}$) and (27) that

[TABLE]

which finishes the proof of (26).

We shall next prove that $f_{3}$ satisfies the hypothesis of the above Proposition 3.5. Let $\{v_{1},v_{2},\ldots,v_{k}\}$ be a set of mutually orthogonal pure atoms in $S(C_{1}(H))$ with $k<n+1$, and real numbers $\lambda_{1},\lambda_{2},\ldots,\lambda_{k}$ satisfying $\displaystyle\sum_{j=1}^{k}|\lambda_{j}|=1$. Since dim$(H)=n+1$, we can find a non-empty finite set of pure atoms $\{v_{k+1},\ldots,v_{n+1}\}$ such that $\{v_{k+1},\ldots,v_{n+1}\}^{\perp}=\{v_{1},\ldots,v_{k}\}$. By (26) $f_{3}(v_{j})=v_{j}$ for every $j$, and then Lemma 2.2 assures that $f_{3}\left(\{v_{k+1},\ldots,v_{n+1}\}^{\perp}\cap S(C_{1}(H))\right)=\{v_{k+1},\ldots,v_{n+1}\}^{\perp}\cap S(C_{1}(H)).$ Having in mind that $\{v_{k+1},\ldots,v_{n+1}\}^{\perp}\cong C_{1}(H^{\prime})$, where $H^{\prime}$ is a complex Hilbert space with dim$(H^{\prime})=k<n+1$, and $f_{3}|_{S(C_{1}(H^{\prime}))}:S(C_{1}(H^{\prime}))\to S(C_{1}(H^{\prime}))$, it follows from the induction hypothesis the existence of a surjective real linear isometry $R:C_{1}(H^{\prime})\to C_{1}(H^{\prime})$ such that $f_{3}(x)=R(x)$ for all $x\in S(C_{1}(H^{\prime}))$. Applying (26) we get

[TABLE]

Finally, since $f_{3}$ satisfies the hypothesis of the above Proposition 3.5, we deduce from this result that $f_{3}(x)=x$, for every $x\in S(C_{1}(H))$. ∎

4. Surjective isometries between the unit spheres of two arbitrary trace class spaces

In this section we consider the trace class operators on a complex Hilbert space $H$ of arbitrary dimension. The answers obtained in the finite dimensional case can be now applied to simplify the study.

Theorem 4.1**.**

Let $f:S(C_{1}(H))\to S(C_{1}(H))$ be a surjective isometry, where $H$ is an arbitrary complex Hilbert space. Then there exists a surjective complex linear or conjugate linear isometry $T:C_{1}(H)\to C_{1}(H)$ satisfying $f(x)=T(x)$ for every $x\in S(C_{1}(H))$.

Proof.

Let $\{\xi_{k}:k\in\mathcal{I}\}$ be an orthonormal basis of $H$. As before, we set $e_{k}:=\xi_{k}\otimes\xi_{k}$. Then the set $\{e_{k}:k\in\mathcal{I}\}$ is a maximal set of mutually orthogonal pure atoms in $S(C_{1}(H)).$ By Lemma 2.2 and Proposition 2.6$(b)$ the elements in the set $\{f(e_{k}):k\in\mathcal{I}\}$ are mutually orthogonal pure atoms in $C_{1}(H)$. We can therefore find orthonormal systems $\{\eta_{k}:k\in\mathcal{I}\}$ and $\{\zeta_{k}:k\in\mathcal{I}\}$ in $H$ such that $f(e_{k})=\eta_{k}\otimes\zeta_{k}$ for every $k\in\mathcal{I}$.

We claim that at least one of $\{\eta_{k}:k\in\mathcal{I}\}$ and $\{\zeta_{k}:k\in\mathcal{I}\}$ must be an orthonormal basis of $H$. Otherwise, we can find norm one elements $\eta_{0}$ and $\zeta_{0}$ in $H$ such that $\eta_{0}\perp\eta_{k}$ and $\zeta_{0}\perp\zeta_{k}$ (in $H$) for every $k$. Then the element $v_{0}=\eta_{0}\otimes\zeta_{0}$ is a pure atom in $S(C_{1}(H))$ which is orthogonal to every $f(e_{k})$. Applying Lemma 2.2 to $f^{-1}$ we deduce that $f^{-1}(v_{0})\perp e_{k}$ for every $k\in\mathcal{I}$, which is impossible because $\{\xi_{j}:j\in\mathcal{I}\}$ is a basis of $H$. We can therefore assume that $\{\eta_{k}:k\in\mathcal{I}\}$ is an orthonormal basis of $H$.

In a second step we shall show that $\{\zeta_{k}:k\in\mathcal{I}\}$ also is an orthonormal basis of $H$. If that is not the case, there exists $\zeta_{0}$ in $H$ such that $\zeta_{0}\perp\zeta_{k}$ (in $H$) for every $k$. Fix an index $k_{0}$ in $\mathcal{I}$ and set $v_{0}:=\eta_{k_{0}}\otimes\zeta_{0}\in\partial_{e}(\mathcal{B}_{C_{1}(H)})$. Clearly, $v_{0}\perp f(e_{k})$ for every $k\neq k_{0}$. Lemma 2.2 and Proposition 2.6$(b)$ imply that $f^{-1}(v_{0})$ is a pure atom in $C_{1}(H)$ which is orthogonal to $e_{k}$ for every $k\neq k_{0}$. Since $\{\xi_{j}:j\in\mathcal{I}\}$ is a basis of $H$, we can easily see that $f^{-1}(v_{0})=\lambda e_{k_{0}}$ for a unique $\lambda\in\mathbb{T}$. We deduce from Proposition 2.6$(d)$ and $(e)$ that $\eta_{k_{0}}\otimes\zeta_{0}=v_{0}=ff^{-1}(v_{0})=\mu f(e_{k_{0}})=\eta_{k_{0}}\otimes\zeta_{k_{0}}$ with $\mu\in\mathbb{T}$, which contradicts the fact $\zeta_{0}\perp\zeta_{k}$ (in $H$) for every $k$.

We have therefore shown that $\{\eta_{k}:k\in\mathcal{I}\}$ and $\{\zeta_{k}:k\in\mathcal{I}\}$ both are orthonormal basis of $H$.

Let us pick two unitary elements $u_{1},w_{1}\in B(H)$ such that $u_{1}f(e_{k})w_{1}=e_{k}$ for every $k\in\mathcal{I}$. The mapping $f_{2}:S(C_{1}(H))\to S(C_{1}(H)),$ $f_{2}(x)=u_{1}f(x)w_{1}$ is a surjective isometry and $f_{2}(e_{k})=e_{k}$ for every $k\in\mathcal{I}$. Let $T_{1}$ denote the surjective complex linear isometry on $C_{1}(H)$ given by $T_{1}(x)=u_{1}xw_{1}$ ($x\in C_{1}(H)$).

Now, let $F$ be a finite subset of $\mathcal{I}$, and let $q_{F}$ denote the orthogonal projection of $H$ onto $H_{F}=\hbox{span}\{\xi_{k}:k\in F\}$. The set $\{e_{k}:k\notin F\}$ is invariant under $f_{2}$. Lemma 2.2 assures that $f_{2}\left(\{e_{k}:k\notin F\}^{\perp}\cap S(C_{1}(H))\right)=\{e_{k}:k\notin F\}^{\perp}\cap S(C_{1}(H)),$ where $\{e_{k}:k\notin F\}^{\perp}\cap S(C_{1}(H))=S(C_{1}(H_{F}))$, and

[TABLE]

is a surjective isometry. By Theorem 3.7 there exists a surjective real linear isometry $T_{{}_{F}}:C_{1}(H_{F})\to C_{1}(H_{F})$ such that $f_{2}(x)=T_{{}_{F}}(x)$ for all $x\in S(C_{1}(H_{F}))$.

Let $T_{2}:C_{1}(H)\to C_{1}(H)$ denote the homogeneous extension of $f_{2}$ defined by $T_{2}(x):=\|x\|_{1}f_{2}\left(\frac{x}{\|x\|_{1}}\right)$ if $x\neq 0$ and $T_{2}(0)=0$. Clearly $T_{2}$ is surjective. We shall show that $T_{2}$ is an isometry. To this end, let us fix $x,y\in C_{1}(H)\backslash\{0\}$ and $\varepsilon>0$. Since every element in $C_{1}(H)(\subseteq K(H))$ can be approximated in norm by elements $x\in S(C_{1}(H))$ with $x=q_{F}xq_{F}$, where $F$ is a finite subset of $\mathcal{I}$, we can find a finite set $F\subset\mathcal{I}$, $x_{\varepsilon}$ and $y_{\varepsilon}$ in $S(C_{1}(H))$ such that $x_{\varepsilon}=q_{F}x_{\varepsilon}q_{F}$, $y_{\varepsilon}=q_{F}y_{\varepsilon}q_{F}$, $\|\frac{x}{\|x\|_{1}}-x_{\varepsilon}\|_{1}<\frac{\varepsilon}{2(\|x\|_{1}+\|y\|_{1})}$ and $\|\frac{y}{\|y\|_{1}}-y_{\varepsilon}\|_{1}<\frac{\varepsilon}{2(\|x\|_{1}+\|y\|_{1})}$. By the triangular inequality we have

[TABLE]

[TABLE]

and since $f_{2}$ is an isometry we get

[TABLE]

[TABLE]

[TABLE]

[TABLE]

On the other hand, since $y_{\varepsilon},x_{\varepsilon}\in S(C_{1}(H_{F}))$, we can consider the surjective real linear isometry $T_{{}_{F}}:C_{1}(H_{F})\to C_{1}(H_{F})$ satisfying $f_{2}(x)=T_{{}_{F}}(x)$ for all $x\in S(C_{1}(H_{F}))$ to deduce that

[TABLE]

[TABLE]

Combining this identity with the previous two inequalities we obtain

[TABLE]

The arbitrariness of $\varepsilon$ shows that $\left\|T_{2}(x)-T_{2}(y)\right\|_{1}=\left\|x-y\right\|_{1}$, and hence $T_{2}$ is an isometry.

Finally, since $T_{2}$ is a surjective isometry with $T_{2}(0)=0$, the Mazur-Ulam theorem guarantees that $T_{2}$ is a surjective real linear isometry, and hence $f(x)=T_{1}^{-1}T_{2}(x)$ for all $x\in S(C_{1}(H))$, witnessing the desired conclusion. ∎

Acknowledgements First, second and third author were partially supported by the Spanish Ministry of Economy and Competitiveness (MINECO) and European Regional Development Fund project no. MTM2014-58984-P and Junta de Andalucía grant FQM375. Fourth author partially supported by grants MTM2014-54240-P, funded by MINECO and QUITEMAD+-CM, Reference: S2013/ICE-2801, funded by Comunidad de Madrid.

The authors are indebted to the anonymous reviewer for a thorough report, insightful comments, and suggestions.