Proof For a positive real number $a>0$ , consider the Gateaux differentiable map $\phi _a : U\to S(H)$ defined by $\phi _a(X)=\phi (X)+aX$ . Then we have $(D\phi _a(X)) (A)> 0$ for any $X\in U$ and $A\in S(H)$ with $A> 0$ . It suffices to show the statement for $\phi _a$ instead of $\phi $ for every positive real number $a>0$ .

Let us assume $X<Y$ first. For each real number $c\in [0, 1)$ , we have $(D\phi _a((1-c)X+cY))(Y-X)>0$ . Since $(D\phi _a((1-c)X+cY))(Y-X)$ is equal to

$$ \begin{align*}\lim_{\varepsilon\to 0} \frac{\phi_a((1-c)X+cY + \varepsilon (Y-X)) - \phi_a((1-c)X+cY)}{\varepsilon} , \end{align*} $$

there exists a positive real number $ 0 <\varepsilon \le 1-c$ such that $\phi _a((1-c)X+cY + \varepsilon (Y-X))> \phi _a((1-c)X+cY)$ . In other words, if we define $X_c = (1-c)X + cY$ , $c \in [0,1]$ , then for every $c \in [0,1)$ , there exists d such that $c < d \le 1$ and $\phi _a (X_c) < \phi _a (X_d)$ . Moreover, the Gateaux differentiability of $\phi _a$ implies that the map $[0, 1]\ni c\mapsto \phi _a(X_c)$ is continuous. Thus, $\sup \{c\in (0, 1]\, : \, \phi _a((1-c)X+cY)\ge \phi _a(X)\}=1$ , so we obtain $\phi _a(X)\le \phi _a(Y)$ .

Consider the general case $X\le Y$ . Since the collection $\{(1-c)X+cY\, : \, c\in [0, 1]\}$ is compact, there exists a positive real number $\varepsilon>0$ such that $(1-c)X+c(Y +d I)\in U$ for all $c\in [0, 1], d\in (0, \varepsilon ]$ . The preceding paragraph implies $\phi _a(X)\le \phi _a(Y+dI)$ for any $d\in (0, \varepsilon ]$ . Take the limit $d\to 0$ to obtain $\phi _a(X)\le \phi _a(Y)$ .▪

Proof Let $X\in U$ and $A\in S(H)$ , $A\ge 0$ . The assumption $\phi (U)\subset S(H)$ implies

$$ \begin{align*}(D\phi(X))(A) = \lim_{t \to 0} {1 \over t} (\phi (X+ tA) - \phi (X))\in S(H). \end{align*} $$

Since $\phi $ restricts to a holomorphic map of $\Pi (H)$ into itself, we have $(D\phi (X))(iA) = i (D\phi (X))(A)\in \overline {\Pi (H)}$ , and thus $(D\phi (X))(A)\ge 0$ . It follows that $\phi $ restricted to U satisfies the assumption of the preceding lemma.▪

Proof Because $\phi : {\mathcal {U}}\to {\mathcal {V}}$ and $\phi ^{-1} : {\mathcal {V}}\to {\mathcal {U}}$ are continuous bijections and $\phi (\Pi (H)) = \Pi (H)$ and $\phi (\Pi (H)^{\ast }) = \Pi (H)^{\ast }$ , we have $\phi ({\mathcal {U}}\cap S(H)) = {\mathcal {V}}\cap S(H)$ . It follows that $\phi (U)$ is an operator domain in $S(H)$ . Let $X_0\in U$ . By the continuity of $\phi $ , we may take a neighborhood $X_0\in V\subset U$ with the following two properties:

• • For any pair $X, Y\in V$ with $X\le Y$ and any $c\in [0, 1]$ , $(1-c)X+cY\in U$ holds.

• • For any pair $X, Y\in \phi (V)$ with $X\le Y$ and any $c\in [0, 1]$ , $(1-c)X+cY\in \phi (U)$ holds.

Hence, by Theorem 2.2, $\phi $ restricts to an order isomorphism from V onto $\phi (V)$ .▪

Proof Fix $X\in U$ . Take a positive real number $c>0$ such that $X+Y\in {\cal U}$ for any $Y\in B(H)$ with $\| Y \|\le 2c$ , and $X+Y\in U$ for any $Y\in S(H)$ with $\| Y \|\le 2c$ . Fix an operator $Y\in B(H)$ with $\| Y \|\le c$ . We prove $\xi (X+Y)=0$ . Take operators $Y_1, Y_2\in S(H)$ such that $Y=Y_1+iY_2$ . Then $\| Y_1 \|, \| Y_2\|\le c$ . For each $x, y\in H$ , consider the holomorphic map $z \mapsto \langle \xi (X+Y_1+zY_2)x, y\rangle $ from some open neighborhood of the closure of ${\mathbb {D}}:=\{z\in {\mathbb {C}}\, :\, |z|<1\}$ into ${\mathbb {C}}$ . We have $X+Y_1+tY_2\in U$ for any $t\in [-1, 1]$ , and hence $\langle \xi (X+Y_1+tY_2)x, y\rangle =0$ , $t\in [-1, 1]$ . The identity theorem implies $\langle \xi (X+Y_1+zY_2)x, y\rangle = 0$ for any $z\in \overline {\mathbb {D}}$ . It follows $\xi (X+Y)=\xi (X+Y_1+iY_2)=0$ .

Consider the collection

$$ \begin{align*}{\cal U}_0 := \{X\in {\cal U}\, : \, \xi(Y)=0\text{ holds for every element }Y\text{ in some neighborhood of }X\}. \end{align*} $$

Then ${\cal U}_0\supset U$ and ${\cal U}_0$ is open. Let $X\in {\cal U}$ . Take a continuous path $\tau : [0, 1]\to {\cal U}$ such that $\tau (0)\in {\cal U}_0$ , $\tau (1)=X$ . Then the collection $V=\{t\in [0, 1] \, : \, \tau (t)\in {\cal U}_0\}$ is open in $[0, 1]$ . We prove V is closed. Let $t\in [0, 1]$ be in the closure of V. Since ${\cal U}$ is open and $\tau $ is continuous, we may take numbers $s\in [0, 1]$ and $c>0$ such that $\tau (s)\in {\cal U}_0$ , $\tau (s)+Y\in {\cal U}$ for any $Y\in B(H)$ with $\| Y\|\le c$ and $\|\tau (s)-\tau (t)\|<c$ . Consider the holomorphic function $z\mapsto \langle \xi (\tau (s)+ zY)x, y\rangle $ on ${\mathbb {D}}$ , for $Y\in B(H)$ with $\| Y\|\le c$ and $x, y\in H$ . Then the identity theorem implies that this map is the constant $0$ map, and we have $t\in V$ . We obtain $X = \tau (1) \in {\cal U}_0$ , which completes the proof.▪

Proof By the definition of local order isomorphisms, it suffices to prove the statement with the additional assumption that $\phi $ is an order isomorphism from U onto the operator domain $\phi (U)$ .

Let $X, Y\in U$ satisfy $X<Y$ . We prove $\phi (X)<\phi (Y)$ . Since $\phi $ preserves order, we have $\phi (X)\le \phi (Y)$ . Assume for a contradiction that $\phi (X)\not < \phi (Y)$ . Then we may find a vector x in H that is not in the range of $(\phi (Y)-\phi (X))^{1/2}$ . Take the rank-one projection P onto ${\mathbb {C}} x$ . By Douglas’ lemma, we have $\phi (X)+cP\nleq \phi (Y)$ for all positive real numbers $c>0$ . Fix $c>0$ small enough. We have $\phi (X)+cP\in \phi (U)$ , and hence there exists an operator $Z\in U$ such that $\phi (Z)=\phi (X)+cP$ . It follows that $\phi $ restricts to an order isomorphism from $[X, Y]\cap [X, Z]\cap U$ onto $[\phi (X), \phi (X)+cP]\cap [\phi (X), \phi (Y)]\cap \phi (U) =\{\phi (X)\}$ , which is absurd. Applying the same argument to $\phi ^{-1}$ , we see that $X<Y$ if and only if $\phi (X)<\phi (Y)$ .

Let $X\in U$ . Since ${U}$ is open, there exists a positive real number $\varepsilon $ such that $[X-\varepsilon I, X+\varepsilon I]\subset {U}$ . By the preceding paragraph, we have $\phi (X-\varepsilon I)<\phi (X)<\phi (X+\varepsilon I)$ . Since $\phi ({U})$ is open, there exists a positive real number $\varepsilon '$ such that $[\phi (X)-\varepsilon ' I, \phi (X)+\varepsilon ' I]\subset \phi ({U})\cap [\phi (X-\varepsilon I), \phi (X+\varepsilon I)]$ . It follows that $\phi ^{-1}(\phi (X)-\varepsilon ' I)<X< \phi ^{-1}(\phi (X)+\varepsilon ' I)$ , $[\phi ^{-1}(\phi (X)-\varepsilon ' I), \phi ^{-1}(\phi (X)+\varepsilon ' I)]\subset {U}$ and $\phi $ restricts to an order isomorphism from $[\phi ^{-1}(\phi (X)-\varepsilon ' I), \phi ^{-1}(\phi (X)+\varepsilon ' I)]$ onto $[\phi (X)-\varepsilon ' I, \phi (X)+\varepsilon ' I]$ .▪

Proof We define order isomorphisms $\psi : [0, I]\to [A, B]$ and $\psi ': [C, D]\to [0, I]$ by $\psi (X)=(B-A)^{1/2}X(B-A)^{1/2}+A$ and $\psi '(X)=(D-C)^{-1/2}(X-C)(D-C)^{-1/2}$ . Then $\psi '\circ \phi \circ \psi : [0, I]\to [0, I]$ is an order automorphism. It suffices to prove this lemma for $\psi '\circ \phi \circ \psi $ instead of $\phi $ . Hence, we may assume $[A, B]=[C, D]=[0, I]$ .

Let $\phi $ be an order automorphism of $[0, I]$ . By Theorem 3.3, we obtain

$$ \begin{align*}\phi(X) = f_q \left( \left( f_p (TT^{\ast} ) \right) ^{-1/2} f_p (TXT^{\ast}) \left( f_p (TT^{\ast} ) \right) ^{-1/2} \right), \ \ \ X \in [0,I]. \end{align*} $$

Note that this map can be decomposed into four order isomorphisms:

$$\begin{align*}\begin{aligned} \phi_1&: [0, I]\to [0, TT^{\ast}],\,\, X\mapsto TXT^*,\\ \phi_2&: [0, TT^{\ast}]\to [0, f_p(TT^*)],\,\, X\mapsto f_p(X),\\ \phi_3&: [0, f_p(TT^*)]\to [0, I],\,\, X\mapsto \left( f_p (TT^{\ast} ) \right) ^{-1/2} X \left( f_p (TT^{\ast} ) \right) ^{-1/2},\,\,\text{and}\\ \phi_4&: [0, I]\to [0, I],\,\, X\mapsto f_q(X). \end{aligned} \end{align*}$$

It suffices to prove the statement for $\phi _j$ , $j=1, 2, 3, 4$ . It is easy to see that $\phi _3$ continuously extends to a biholomorphic automorphism of $B(H)$ that maps $\Pi (H)$ onto itself. The same holds for $\phi _1$ if T is linear. If T is conjugate-linear, then the map $X\mapsto TX^*T^*$ on $B(H)$ extends $\phi _1$ and is a biholomorphic automorphism of $B(H)$ that maps $\Pi (H)$ onto itself.

Because for every $X+iY \in \Pi (H)$ (where $X, Y\in S(H)$ ) and every $Z \in S(H)$ we have

$$ \begin{align*}(X+iY)^{-1} = Y^{-1/2} ( Y^{-1/2} X Y^{-1/2} + iI)^{-1} Y^{-1/2} \end{align*} $$

and

$$ \begin{align*}(iI + Z)^{-1} = (I + Z^2)^{-1/2} (Z - iI) (I + Z^2)^{-1/2}, \end{align*} $$

the map $X \mapsto -X^{-1}$ is a biholomorphic automorphism of $\Pi (H)$ . Let r be a real number $<1$ , $r\not =0$ . Using

(2) $$ \begin{align} f_r (x) = {1 \over r} - {1 - r \over r^2} \left( {1-r \over r} + x \right)^{-1} \end{align} $$

and

$$ \begin{align*}f_{r}^{-1} = f_{r \over r-1}, \end{align*} $$

we see that $f_r$ extends to a biholomorphic map $X\mapsto f_r(X)$ from $\{X\in B(H)\, : \, 1- 1/r\notin \sigma (X)\}$ onto $\{X\in B(H)\, : \, 1/r\notin \sigma (X)\}$ . Since the map $X\mapsto -X^{-1}$ maps $\Pi (H)$ onto itself, we finally conclude that the map $X \mapsto f_r (X)$ is a biholomorphic automorphism of $\Pi (H)$ .▪

Proof By Lemma 3.2, for each $X\in U$ , there exist operators $A_X<X<B_X$ , $[A_X, B_X]\subset U$ , $\phi (A_X)<\phi (X)<\phi (B_X)$ , $[\phi (A_X), \phi (B_X)]\subset \phi (U)$ such that $\phi ([A_X, B_X])=[\phi (A_X), \phi (B_X)]$ and $\phi : [A_X, B_X]\to [\phi (A_X), \phi (B_X)]$ is an order isomorphism. By Lemma 3.4, $\phi $ restricted to $[A_X, B_X]$ extends to a biholomorphic map $\phi _X$ defined on some open subset of $B(H)$ containing $[A_X, B_X]\cup \Pi (H)$ which maps $[A_X, B_X]$ bijectively onto $[\phi (A_X), \phi (B_X)]$ and $\Pi (H)$ biholomorphically onto itself. Let $X, Y\in U$ . Suppose that there exist $A, B\in U$ with $A<B$ and $[A, B]\subset [A_X, B_X]\cap [A_Y, B_Y]$ . By Proposition 3.1, we have $\phi _X(Z)=\phi _Y(Z)$ , $Z\in \Pi (H)$ . Since U is an operator domain, it is path-connected. Therefore, a compactness argument implies that $\phi _X(Z)=\phi _Y(Z)$ , $Z\in \Pi (H)$ , holds for any $X, Y\in U$ . Hence, we obtain the desired conclusion.▪

Proof We refer to [Reference Harris3]. Recall that the open unit ball $B(H)_0$ of $B(H)$ is biholomorphically equivalent to ${\Pi } (H)$ through the Cayley transform $\Psi : Y\mapsto i(Y+I)(I-Y)^{-1}$ . The inverse is $X\mapsto (X-iI)(X+iI)^{-1}$ . See Theorem 12 of [Reference Harris3] with $\mathfrak {A}=B(H)$ and $V=I$ .

It is easy to see that if $A, B, C, T$ are as in the statement, then

$$ \begin{align*}X\mapsto T((X-B)^{-1}+A)^{-1}T^{\ast} +C \end{align*} $$

and

$$ \begin{align*}X\mapsto T((X^t-B)^{-1}+A)^{-1}T^{\ast} +C \end{align*} $$

are biholomorphic automorphisms of ${\Pi } (H)$ . Indeed, it is the composition of five biholomorphic automorphisms $X\mapsto X-B$ (or $X\mapsto X^t-B$ ), $X\mapsto -X^{-1}$ , $X\mapsto X-A$ , $X\mapsto -X^{-1}$ , and $X\mapsto TXT^{\ast } +C$ of $\Pi (H)$ .

Let $\phi $ be a biholomorphic automorphism of $\Pi (H)$ . We know that the map $\psi :=\Psi ^{-1}\circ \phi \circ \Psi $ is a biholomorphic automorphism of $B(H)_0$ . By [Reference Harris3, Theorems 2 and 3], $\psi $ extends (uniquely) to a homeomorphism from the closed unit ball of $B(H)$ onto itself, and moreover, such an extension maps the unitary group of H onto itself (see equation (12) of [Reference Harris3] and the formula that appears two lines above this equation). Since $\Psi ^{-1}$ (has a continuous extension that) maps $S(H)$ onto the collection

$$ \begin{align*}\{u \,:\, u\text{ is a unitary operator of }H \text{ such that }u-1\text{ is invertible}\}, \end{align*} $$

which is a dense open subset of the unitary group, we see that there exists a dense open subset $U\subset S(H)$ such that $\phi = \Psi \circ \psi \circ \Psi ^{-1}$ has a continuous extension to $\Pi (H)\cup U$ that maps U into $S(H)$ . Fix an element $X_0\in U$ . Define a biholomorphic map $\tilde {\phi }$ of $\Pi (H)$ by $\tilde {\phi }(X) = \phi (X+X_0) -\lim _{0<c\to 0}\phi (icI+X_0)$ . Then $\tilde {\phi }$ has a continuous extension to $\Pi (H)\cup \{0\}$ and $\lim _{0<c\to 0}\tilde {\phi }(icI)=0$ . Therefore, in order to complete the proof, it suffices to show the latter half of the statement of the proposition.

Let $\phi $ be a biholomorphic automorphism of $\Pi (H)$ , and suppose that there exists a sequence $(X_n)\subset \Pi (H)$ with $X_n\to 0$ and $\phi (X_n)\to 0$ . Take operators $A_1, A_2\in S(H)$ with $\phi (iI)=A_1+iA_2$ . Then $A_2>0$ . Consider the map f defined by $f(X) = T(X^{-1}+A)^{-1}T^{\ast }$ , $X\in \Pi (H)$ , where

$$ \begin{align*} A=A_2^{-1/2}A_1A_2^{-1/2},\quad T= A_2^{1/2}(A^2+I)^{1/2}. \end{align*} $$

A straightforward calculation shows that $f(iI)=\phi (iI)$ and $f^{-1}\circ \phi (X_n)\to 0$ as $n\to \infty $ . Consider biholomorphic automorphisms $\psi =\Psi ^{-1}\circ \phi \circ \Psi $ and $g=\Psi ^{-1}\circ f\circ \Psi $ of $B(H)_0$ . Since $f(iI)=\phi (iI)$ , we obtain $g^{-1}\circ \psi (0)= 0$ . By [Reference Harris3, Theorem 1] and [Reference Molnár8, Theorem A.9], there exists a pair of unitaries $u, v$ with either $g^{-1}\circ \psi (Y)= uYv$ , $Y\in B(H)_0$ , or $g^{-1}\circ \psi (Y)= uY^tv$ , $Y\in B(H)_0$ . Moreover, we see that $\Psi ^{-1}(X_n)\to -I$ and

$$ \begin{align*}g^{-1}\circ \psi (\Psi^{-1}(X_n))= \Psi^{-1}\circ f^{-1}\circ \phi (X_n) \to -I \end{align*} $$

as $n\to \infty $ . Hence, we obtain $-I = -uv$ , or equivalently, $v=u^{\ast }$ . Thus, we have either $\psi (Y) = g(uYu^{\ast })$ , $Y\in B(H)_0$ , or $\psi (Y) = g(uY^tu^{\ast })$ , $Y\in B(H)_0$ . Since $\Psi ^{-1}$ is given by functional calculus, we obtain

$$\begin{align*}\begin{aligned} \phi(X) &=\Psi\circ \psi(\Psi^{-1}(X))\\ &= \Psi\circ g(u\Psi^{-1}(X)u^*)\\ &= \Psi\circ g\circ \Psi^{-1}(uXu^*)\\ &= f(uXu^{\ast}) \\ &= T((uXu^{\ast})^{-1}+A)^{-1}T^{\ast} \\ &= Tu(X^{-1}+u^{\ast} Au)^{-1}(Tu)^{\ast},\quad X\in \Pi(H), \end{aligned} \end{align*}$$

in the former case, and similarly

$$ \begin{align*}\phi(X)= Tu((X^t)^{-1}+u^{\ast} Au)^{-1}(Tu)^{\ast},\quad X\in \Pi(H), \end{align*} $$

in the latter case. This completes the proof.▪

Proof The assumption implies $\| (X_nA+I)^{-1}\|\to \infty $ as $n\to \infty $ . Hence,

$$\begin{align*}\begin{aligned} \|\Theta_A(X_n)\|\|A\|&=\|(X_nA+I)^{-1}X_n\|\|A\|\\ &\ge \|(X_nA+I)^{-1}X_nA\| = \|I- (X_nA+I)^{-1}\|\to\infty, \end{aligned} \end{align*}$$

which completes the proof.▪

Proof Let $X\in W_A$ . Then,

$$ \begin{align*}-A\Theta_A(X) +I = -AX(AX+I)^{-1} +(AX+I)(AX+I)^{-1} = (AX+I)^{-1}. \end{align*} $$

It follows $\Theta _A(X)\in W_{-A}$ and

$$ \begin{align*}\Theta_{-A}\circ \Theta_{A} (X) = \Theta_{A}(X) (-A\Theta_A(X) +I)^{-1} = X(AX+I)^{-1}(AX+I) = X. \end{align*} $$

Thus, we have $\Theta _{-A}\circ \Theta _A = \operatorname {id}_{W_A}$ . Similarly, we can prove $\Theta _A\circ \Theta _{-A} = \operatorname {id}_{W_{-A}}$ . Hence, we obtain $(\Theta _A)^{-1}= \Theta _{-A}$ . In order to see that $\Theta _A$ is biholomorphic, it suffices to show that $\Theta _A$ is holomorphic. This is easily seen by the fact that the maps $X\mapsto X$ and $X\mapsto (XA+I)^{-1}$ are holomorphic on $W_A$ .▪

Proof Assume that the equation holds for some nonempty open set. By Proposition 3.1, the same equation holds for any $X\in \Pi (H)$ . After an easy calculation, we obtain

$$ \begin{align*}A+X^{-1}= ((T')^{-1}T)^{\ast}(A'+X^{-1})(T')^{-1}T,\quad X\in \Pi(H) \end{align*} $$

in (1) and

$$ \begin{align*}A+X^{-1}= ((T')^{-1}T)^{\ast}(A'+(X^t)^{-1})(T')^{-1}T,\quad X\in \Pi(H) \end{align*} $$

in (2). The rest of the proof is easy and is left to the reader.▪

Proof It suffices to consider the case $\phi (0)=0$ . It is clear that $\phi $ is a local order isomorphism. By Theorem 4.4, there exist $A\in S(H)$ and a bounded bijective linear operator $T: H\to H$ such that $S(H)\subset U_A$ and either

$$\begin{align*}\begin{aligned} \phi(X) &= T\Phi_A(X)T^{\ast} = T(XA+I)^{-1}XT^{\ast}, \ \ \ X\in S(H), \text{ or}\\ \phi(X) &= T\Phi_A(X^t)T^{\ast} = T(X^t A+I)^{-1}X^t T^{\ast}, \ \ \ X\in S(H). \end{aligned} \end{align*}$$

Clearly, the condition $S(H)\subset U_A$ implies $A=0$ , and hence we obtain the desired conclusion.▪

Proof Put $B=TAT^{\ast }$ . Let $X\in S(H)$ . We have $X\in \hat {U}_{B}$ if and only if $XTAT^{\ast }+I$ is invertible. On the other hand, we have $\psi (X)\in \hat {U}_A$ if and only if $T^{\ast } XTA +I$ is invertible. Hence, we obtain $X\in U_{B}$ if and only if $\psi (X)\in U_A$ . Let $X\in U_{B}$ . Then,

$$\begin{align*}\Phi_A\circ \psi (X) = T^*XT (AT^*XT +I)^{-1}, \end{align*}$$

and hence

$$\begin{align*}\begin{aligned} \psi^{-1}\circ \Phi_A\circ \psi (X) &=(T^*)^{-1}T^*XT (AT^*XT +I)^{-1}T^{-1}\\ &=X(TAT^*X+I)^{-1} = \Phi_{B}(X).\\[-24pt] \end{aligned} \end{align*}$$

▪

Proof Since $0, X\in U_A$ and $U_A$ is path-connected, there exists a continuous path $\tau : [0, 1]\to U_A$ such that $\tau (0)=0$ and $\tau (1)=X$ . We may take $0= t_0<t_1<\cdots <t_k=1$ such that $\lVert (\tau (t_{j-1})A+I)(\tau (t_j)A+I)^{-1} -I\rVert $ is sufficiently small for $j=1, \ldots , k$ . It suffices to show that there exists a bounded linear bijection $T_j: H\to H$ such that $A(\tau (t_j)A+I)^{-1} = T_jA(\tau (t_{j-1})A+I)^{-1}T_j^*$ , $j\in \{1, \ldots , k\}$ . For this, consider the biholomorphic function $f: \{z\in {\mathbb {C}} \,:\, \operatorname {Re}z>0\}\to {\mathbb {C}}\setminus (-\infty , 0]$ defined by $f(z)=z^2$ . Then the equations $f^{-1}(z) = \overline {f^{-1}(\bar {z})}$ and

$$ \begin{align*}A(\tau(t_{j-1})A+I)^{-1} \left((\tau(t_{j-1})A+I)(\tau(t_j)A+I)^{-1}\right)\end{align*} $$

$$ \begin{align*}= \left((\tau(t_{j-1})A+I)(\tau(t_j)A+I)^{-1}\right)^* A(\tau(t_{j-1})A+I)^{-1} \end{align*} $$

imply that

$$\begin{align*}\begin{aligned} &A(\tau(t_{j})A+I)^{-1}\\ &= A(\tau(t_{j-1})A+I)^{-1} \left(f^{-1}\left((\tau(t_{j-1})A+I)(\tau(t_j)A+I)^{-1}\right)\right)^{2}\\ &= S^* A(\tau(t_{j-1})A+I)^{-1} S, \end{aligned} \end{align*}$$

where $S = f^{-1}\left ( (\tau (t_{j-1})A+I)(\tau (t_j)A+I)^{-1}\right )$ .▪

Proof The condition $X+X_0\in \hat {U}_A$ means $(X+X_0)A+I$ is invertible. The condition $X\in \hat {U}_{B}$ means

$$ \begin{align*}XB+I=XA(X_0A+I)^{-1} +I = (XA+X_0A+I)(X_0A+I)^{-1} \end{align*} $$

is invertible. Hence, they are equivalent. Suppose that these equivalent conditions hold. Then we obtain

$$\begin{align*}\begin{aligned} &(X_0A+I)^{-1}\hat{\Phi}_{B}(X)(AX_0+I)^{-1} + \hat{\Phi}_A(X_0)\\ &= (X_0A+I)^{-1}\left(X(BX+I)^{-1}\right)(AX_0+I)^{-1} + (X_0A+I)^{-1}X_0\\ &= (X_0A+I)^{-1}\left(X((AX_0+I)^{-1}AX+I)^{-1}\right)(AX_0+I)^{-1} + (X_0A+I)^{-1}X_0\\ &= (X_0A+I)^{-1}X(AX+AX_0+I)^{-1} + (X_0A+I)^{-1}X_0\\ &= (X_0A+I)^{-1}(X+X_0AX+X_0AX_0+X_0) (AX+AX_0+I)^{-1}\\ &= (X+X_0)(A(X+X_0)+I)^{-1} = \hat{\Phi}_A(X+X_0).\\[-17pt] \end{aligned} \end{align*}$$

▪

Proof There exist order automorphisms $\psi _1, \psi _2: S(H)\to S(H)$ such that $\psi = \psi _2\circ \phi \circ \psi _1$ . Considering extensions to maximal local order isomorphisms of both sides, we obtain $\Phi _B = \psi _2\circ \Phi _A\circ \psi _1$ . In particular, for $X\in S(H)$ , we have $X\in U_B$ if and only if $\psi _1(X)\in U_A$ .

By Theorem 4.6, there exist bounded linear or conjugate-linear bijections $T_1, T_2: H\to H$ and $C_1, C_2\in S(H)$ such that

$$ \begin{align*}\psi_1(X)= T_1XT_1^* +C_1,\quad\psi_2(X) = T_2XT_2^* + C_2,\quad X\in S(H). \end{align*} $$

Put $D:= \hat {\Phi }_{C_1}(A) = A(C_1A+I)^{-1}$ . Let $X \in U_B$ . By Lemma 5.4, we have

$$\begin{align*}\begin{aligned} \Phi_B(X)&= \psi_2\circ\Phi_A\circ\psi_1(X) \\ &= \psi_2\circ\Phi_A(T_1XT_1^* +C_1) \\ &= \psi_2\left((C_1A+I)^{-1}\Phi_D(T_1XT_1^*)(AC_1+I)^{-1} +\Phi_A(C_1)\right) \\ &= T_2(C_1A+I)^{-1}\Phi_D(T_1XT_1^*)(AC_1+I)^{-1}T_2^* +T_2\Phi_A(C_1)T_2^* + C_2. \end{aligned} \end{align*}$$

Since $\Phi _{B}(0)=0$ , we obtain $T_2\Phi _A(C_1)T_2^* + C_2=0$ . By Lemma 5.2, we obtain

$$\begin{align*}\begin{aligned} \Phi_{B} (X) &= T_2(C_1A+I)^{-1}\Phi_D(T_1XT_1^*)(AC_1+I)^{-1}T_2^* \\ &= T_2(C_1A+I)^{-1}T_1\Phi_{T_1^*DT_1}(X)T_1^*(AC_1+I)^{-1}T_2^*. \end{aligned} \end{align*}$$

Proposition 4.5 implies $B=T_1^*DT_1=T_1^*A(C_1A+I)^{-1}T_1$ . Since $C_1=\psi _1(0)\in U_A$ , Lemma 5.3 implies that there exists a bounded linear or conjugate-linear bijection $T\in B(H)$ with $B=TAT^*$ .▪

Proof The former half is clear. In the matrix language, it can be reformulated as follows: $\Phi _A$ and $\Phi _B$ are equivalent if and only if there exists an invertible $n \times n$ complex matrix T such that $A = TBT^{\ast }$ or $A = TB^t T^{\ast }$ . Since hermitian matrices B and $B^t$ have the same eigenvalues, they are unitarily similar. Consequently, $\Phi _A$ and $\Phi _B$ are equivalent if and only if there exists an invertible $n \times n$ complex matrix T such that $A = TBT^{\ast }$ . By Sylvester’s law of inertia, this is equivalent to the condition that the number of positive eigenvalues of A and B coincide and the same is true for the number of negative eigenvalues (here, each eigenvalue is counted with its multiplicity). By Proposition 5.1, the number of equivalence classes of maximal order isomorphisms of matrix domains in $S_n$ is $(n+2)(n+1)/2$ .▪

Proof Since $X\in U_A$ , we have $-1\notin \sigma (XA)\setminus \{0\} = \sigma (X^{1/2}AX^{1/2})\setminus \{0\}$ . If $X^{1/2}AX^{1/2}\ngeq -I$ , then there exists a positive real number c such that $0<c< 1$ and $-1\in \sigma ((cX)^{1/2}A(cX)^{1/2})$ . We know that $-1\in \sigma ((cX)^{1/2}A(cX)^{1/2})$ if and only if $cXA+I$ is noninvertible, and thus $cX\notin {{U}}_A$ . If $X^{1/2}AX^{1/2}\ge -I$ , then we actually have $X^{1/2}AX^{1/2}> -I$ . It follows that $-I< TX^{1/2}AX^{1/2}T^*$ for any contractive operator T on H, i.e., for any $T\in B(H)$ with $\|T\|\le 1$ . Indeed, there exists a positive real number $0<\varepsilon \le 1$ such that

$$ \begin{align*}\langle TX^{1/2}AX^{1/2}T^* x, x\rangle = \langle X^{1/2}AX^{1/2}(T^*x), T^*x\rangle \geq -(1-\varepsilon) \lVert T^* x\rVert^2\geq -(1-\varepsilon) \end{align*} $$

for any unit vector $x\in H$ . If $Y\in [0, X]$ , then by Douglas’ lemma there exists a contractive operator $T\in B(H)$ such that $Y^{1/2}= TX^{1/2}= X^{1/2}T^*$ . We have $-I<TX^{1/2}AX^{1/2}T^* = Y^{1/2}AY^{1/2}$ , and hence $\sigma (Y^{1/2}AY^{1/2})\cap (-\infty , -1]= \emptyset $ , and $AY+I$ is invertible.

Suppose that $[0, X]\subset {{U}}_A$ . Theorem 2.2 implies $\Phi _A(X)\ge \Phi _A(0)=0$ . We have $-\Phi _A(X)A = f(XA)$ , where f is a biholomorphic function defined by

$$\begin{align*}f(z) = -\frac{z}{z+1} \end{align*}$$

on ${\mathbb {C}}\setminus \{-1\}$ . Since $\sigma (XA)\subset \sigma (X^{1/2}AX^{1/2})\cup \{0\}\subset (-1, \infty )$ , the spectral mapping theorem implies $\sigma (-\Phi _A(X)A) = f(\sigma (XA)) \subset (-1, \infty )$ . Thus, we obtain $[0, \Phi _A(X)]\subset U_{-A}$ . Since $(\Phi _A)^{-1}=\Phi _{-A}$ , Theorem 2.2 implies that $\Phi _A$ restricts to an order isomorphism from $[0, X]$ onto $[0, \Phi _A(X)]$ .▪

Proof Theorem 2.2 gives one direction. Suppose that there exist a pair $X, Y\in U$ with $X\le Y$ , $[X, Y]\not \subset U_A$ . Take a positive real number $\varepsilon>0$ such that $Y+\varepsilon I\in U$ . Then trivially we have $[X, Y+\varepsilon I]\not \subset U_A$ . By Proposition 5.4, it suffices to show: if $X\in U_A$ , $X>0$ and $[0, X]\not \subset U_A$ , then $\Phi _A(X)\ngeq 0$ . Suppose $X\in U_A$ satisfies $X>0$ and $[0, X]\not \subset U_A$ . By Proposition 5.7, the condition $[0, X]\not \subset U_A$ yields $X^{1/2}AX^{1/2}\ngeq -I$ . We obtain $X^{-1}+A \ngeq 0$ , which implies $\Phi _A(X)=(X^{-1}+A)^{-1}\ngeq 0$ .▪

Proof (1) It suffices to prove the openness of V for each m. Let $A\in V$ for a fixed m. Since the collection of all invertible operators is open, we may take a positive real number $\varepsilon $ such that any element in $[A-\varepsilon I, A+\varepsilon I]$ is invertible. Clearly, we have $A-\varepsilon I, A+\varepsilon I\in V$ . Let $B\in [A-\varepsilon I, A+\varepsilon I]$ . Since $A-\varepsilon I\le B$ , the min-max theorem implies that $m=\mathrm {rank\,} (A-\varepsilon I)_+\leq \mathrm {rank\,} B_+$ . Since $B\le A+\varepsilon I$ , we also have $\mathrm {rank\,} B_+ \leq \mathrm {rank\,} (A+\varepsilon I)_+ = m$ . Thus, we obtain $\mathrm {rank\,} B_+ = m$ .

(2) Let $A\in [A_0, A_1]$ . The min-max theorem implies $\mathrm {rank\,} A_+\geq \mathrm {rank\,} (A_0)_+=m$ . We may take a positive real number $\varepsilon>0$ such that $\mathrm {rank\,} (A_1+\varepsilon I)_+ = m$ . The min-max theorem implies $\mathrm {rank\,} (A+\varepsilon I)_+ \leq \mathrm {rank\,} (A_1+\varepsilon I)_+ = m$ . Since $m<\infty $ , we see $\mathrm {rank\,} A_+= m$ and A is invertible.▪

Proof Put $S:=s(A_+)-s(A_+)^{\perp }$ , which is a self-adjoint unitary. Let $X\in S(H)$ . We have

$$ \begin{align*}\sigma(XA)\setminus \{0\}= \sigma(X\lvert A\rvert^{1/2} S\lvert A\rvert^{1/2})\setminus \{0\} = \sigma(\lvert A\rvert^{1/2}X\lvert A\rvert^{1/2} S)\setminus \{0\}. \end{align*} $$

Hence, we have $X\in \hat {U}_A$ if and only if $\lvert A\rvert ^{1/2}X\lvert A\rvert ^{1/2} S+I$ is invertible, which is in turn equivalent to the condition $\lvert A\rvert ^{1/2}X\lvert A\rvert ^{1/2} + S$ is invertible.

Suppose that $A_+$ has rank $m<\infty $ . We prove that $\Phi _A$ is order preserving. The preceding paragraph shows that $\lvert A\rvert ^{1/2}X\lvert A\rvert ^{1/2} + S$ is invertible for any $X\in {U}_A$ . Moreover, since $0\in U_A$ is connected, Lemma 5.10 implies that the positive part $\left (\lvert A\rvert ^{1/2}X\lvert A\rvert ^{1/2} + S\right )_+$ has rank m for any $X\in {U}_A$ . Suppose that $X, Y\in {U}_A$ satisfy $X\leq Y$ . Let $Z\in [X, Y]$ . Then we have the order relation

$$ \begin{align*}\lvert A\rvert^{1/2}X\lvert A\rvert^{1/2} + S\leq \lvert A\rvert^{1/2}Z\lvert A\rvert^{1/2} + S\leq \lvert A\rvert^{1/2}Y\lvert A\rvert^{1/2} + S. \end{align*} $$

By Lemma 5.10, $\lvert A\rvert ^{1/2}Z\lvert A\rvert ^{1/2} + S$ is invertible. Hence, $ZA+I$ is invertible for any $Z\in [X, Y]$ . It follows that $[X, Y]\subset {U}_A$ , and by Proposition 5.8, $\Phi _A(X)\leq \Phi _A(Y)$ holds.

The case that $A_-$ has finite rank can be considered in a similar way. The same argument shows that $\Phi _{A}^{-1} = \Phi _{-A}$ is also order preserving.▪

Proof of Theorem 5.9

$(1)\Rightarrow (2)$ Suppose that $A_+$ is compact. We show that $\Phi _A$ is order preserving. By Lemma 5.3 and Propositions 5.4 and 5.8, all we need to do is to prove that if $X \in U_A$ and $X \geq 0$ , then $[0,X] \in U_A$ .

By $X \in U_A$ , there exists a path $\tau : [0,1] \to U_A$ such that $\tau (0) = 0$ and $\tau (1) = X$ . Since $A_+$ is compact, we can find a sequence $(A_n) \subset S(H)$ such that $\lim A_n = A$ and $(A_n)_+$ is of finite rank for every positive integer n. For every $t \in [0,1]$ , the operator $\tau (t) A + I$ is invertible. It follows that there exists $\delta _t> 0$ such that if

$$ \begin{align*}s \in (t-\delta_t , t + \delta_t) \cap [0,1] \ \ \ \mathrm{and} \ \ \ B \in S(H) \ \ \ \mathrm{and} \ \ \ \| B - A\| <\delta_t, \end{align*} $$

then $\tau (s) B +I$ is invertible. Using the compactness, we see that there exists $\delta> 0$ such that $\tau (s) B + I$ is invertible for every $s \in [0,1]$ and every $B \in S(H)$ with $\| B - A \| < \delta $ . Thus, after removing finitely many elements of the sequence, we may assume that $\tau (s) A_n +I$ is invertible for every $s \in [0,1]$ and every positive integer n. It follows that $X \in U_{A_n}$ . By Lemma 5.11 and Proposition 5.8, we conclude that $[0,X] \subset U_{A_n}$ . By Proposition 5.7, we have $X^{1/2} A_n X^{1/2} \ge -I$ for each n. Taking the limit $n \to \infty $ , we obtain $X^{1/2} A X^{1/2} \ge -I$ . By Proposition 5.7 again, we have $[0,X] \subset U_A$ , as desired.

The case that $s(A_-)$ is compact can be considered in a similar way, or we may use the equation $\Phi _{-A}(X) = (-XA+I)^{-1}X = -\Phi _A(-X)$ for $X\in U_{-A}$ as well. The same argument shows that $\Phi _{A}^{-1} = \Phi _{-A}$ is also order preserving.

$(2)\Rightarrow (1)$ Suppose that both $A_+$ and $A_-$ are noncompact. Then we may take a positive real number $\varepsilon>0$ such that both $\chi _{[\varepsilon , \infty )}(A)$ and $\chi _{(-\infty , -\varepsilon ]}(A)$ have infinite rank. We can find a projection $P\in S(H)$ with $\chi _{[\varepsilon , \infty )}(A)\leq P\leq \chi _{(-\infty , -\varepsilon ]\cup [\varepsilon , \infty )}(A)$ such that P and $\chi _{[\varepsilon , \infty )}(A)$ are unitarily equivalent in $K:=\chi _{(-\infty , -\varepsilon ]\cup [\varepsilon , \infty )}(A)H$ , $P\neq \chi _{[\varepsilon , \infty )}(A)$ , and $PA=AP$ . Take a unitary $u=e^{iB}\in B(K)$ , $B\in S(K)$ , such that $u\chi _{[\varepsilon , \infty )}(A)u^* = P$ . Put $S:= \chi _{[\varepsilon , \infty )}(A)- \chi _{(-\infty , -\varepsilon ]}(A)$ , which can be considered a self-adjoint unitary in $B(K)$ . We define a path $\tau : [0, 1]\to S(K)$ by

$$ \begin{align*}\tau(t) = \lvert A\rvert^{-1/2}(e^{itB} Se^{-itB} -S) \lvert A\rvert^{-1/2} \end{align*} $$

(which is well defined since A restricted to K is invertible). We identify each element $X\in S(K)$ with $\chi _{(-\infty , -\varepsilon ]\cup [\varepsilon , \infty )}(A)X\chi _{(-\infty , -\varepsilon ]\cup [\varepsilon , \infty )}(A)\in S(H)$ . Then,

$$ \begin{align*}\lvert A \rvert^{1/2}\tau(t)\lvert A\rvert^{1/2} + (s(A_+)-s(A_+)^{\perp})= e^{itB} Se^{-itB} +(\chi_{(0, \varepsilon)}(A)- \chi_{(-\varepsilon, 0]}(A)) \end{align*} $$

is invertible in $B(H)$ for any $t\in [0, 1]$ . In particular, we have $\tau (1)\in U_A$ (see the first paragraph of the proof of Lemma 5.11). However, we have

$$\begin{align*}\tau(1) = \lvert A\rvert^{-1/2} 2(P-\chi_{[\varepsilon, \infty)}(A)) \lvert A\rvert^{-1/2} = 2\lvert A\rvert^{-1}(P-\chi_{[\varepsilon, \infty)}(A))\geq 0 \end{align*}$$

and

$$\begin{align*}\begin{aligned} \Phi_A(\tau(1)) &= \Phi_A(2\lvert A\rvert^{-1}(P-\chi_{[\varepsilon, \infty)}(A)))\\ &= 2\lvert A\rvert^{-1}(P-\chi_{[\varepsilon, \infty)}(A))\cdot (A\cdot 2\lvert A\rvert^{-1}(P-\chi_{[\varepsilon, \infty)}(A))+I)^{-1}\\ &= -2\lvert A\rvert^{-1}(P-\chi_{[\varepsilon, \infty)}(A))\ngeq 0=\Phi_A(0).\\[-17pt] \end{aligned} \end{align*}$$

▪

Proof After applying a translation $X \mapsto X-A$ , we may assume with no loss of generality that $A=0$ and $B \in S_n$ satisfies $B \ge 0$ and $\mathrm {rank }\, B =r$ . Furthermore, there is no loss of generality in assuming that the matrix B is diagonal with the first r diagonal entries positive and all others equal to zero. Applying a congruence transformation $X \mapsto TXT^{\ast }$ , where T is a suitable diagonal invertible element in $M_n$ , we can finally assume that B is the diagonal projection with the first r diagonal entries equal to one and all others equal to zero. It is now clear that $[A,B]$ is the set of all $n \times n$ matrices whose upper left $r \times r$ corner is an element of $E_r$ and all entries outside the upper left $r \times r$ corner are zero.

Proof Consider the order embedding $f : [0,1] \to \mathbb {R}$ defined by $f(t) := \mathrm {tr}\, (\phi (tI))$ . Here, $\mathrm {tr}\, (\cdot )$ stands for the trace function. We know that the set N of all points $c \in (0,1)$ at which f is not continuous is at most countable. Let $c \in (0,1) \setminus N$ and $\varepsilon>0$ . Then we can find $\delta $ , $0 < \delta < \min \{ c, 1-c \}$ , such that for every real $t \in [ c- \delta , c + \delta ]$ we have $| f(c) - f(t) | \le \varepsilon $ . In particular, $| \mathrm {tr}\, (\phi (cI) - \phi ((c \pm \delta ) I)) | \le \varepsilon $ . Since $\phi ((c - \delta )I) \le \phi (cI)$ and $\phi ((c + \delta )I) \ge \phi (cI)$ , we conclude that $\phi (cI) - \varepsilon I \le \phi ((c-\delta )I) \le \phi ((c+\delta )I) \le \phi (cI) + \varepsilon I$ (note that we are using the same symbol I to denote the $n \times n$ identity matrix and the $m \times m$ identity matrix). This further implies that for every $X \in [ (c-\delta )I , ( c+ \delta )I ]$ we have $\phi (cI) - \varepsilon I \le \phi (X) \le \phi (cI) + \varepsilon I$ , and thus $\phi $ is continuous at $cI$ .▪

Proof By induction on n. The case $n=1$ is trivial. Assume that the statement holds for $n= k-1$ ( $ \ge 1$ ). We prove the statement for $n=k$ . By Lemma 6.2, we can find a real number c, $0 < c < 1$ , such that $\phi $ is continuous at $cI$ . We take any projection $P\in E_n$ of rank $k-1$ . Observe that the rank of $(P+cP^{\perp }) - cI$ is $k-1$ . By Lemma 6.1 and the induction hypothesis, the rank of $\phi (P + cP^{\perp }) - \phi (cI)$ is no smaller than $k-1$ .

We claim that $(\phi (P + cP^{\perp }) - \phi (cI)) + (\phi (cP + dP^{\perp }) - \phi (cI))$ has rank $\ge k$ for every d, $c < d \le 1$ . If this does not hold, then there exists some $d> c$ such that the range of $\phi (cP + dP^{\perp }) - \phi (cI)$ is contained in the range of $\phi (P + cP^{\perp }) - \phi (cI)$ . Since $\phi $ is continuous at $cI$ , there exists $d'$ with $c < d' < d$ such that $\phi (cP + d' P^{\perp }) - \phi (cI) \le \phi (P + cP^{\perp }) - \phi (cI)$ , and therefore $\phi (cP + d' P^{\perp }) \le \phi (P + cP^{\perp })$ . Since $\phi $ is an order embedding, we necessarily have $cP + d' P^{\perp } \le P + cP^{\perp }$ , a contradiction. Thus, we have shown that $m \ge k$ .▪

Proof By Lemma 6.1, the interval $[A,B]$ is order isomorphic to $E_n$ , whereas the interval $[\phi (A), \phi (B)]$ is order isomorphic to $E_m$ , where m is the rank of $\phi (B) - \phi (A)$ . Since $\phi $ maps $[A,B]$ into $[\phi (A), \phi (B)]$ , Lemma 6.3 yields that $m=n$ .▪

Proof Assume first that $R \le A$ . Then the range of R is a subspace of the range of A, and we have

$$ \begin{align*}(A^{\dagger})^{1/2} R (A^{\dagger})^{1/2} \le (A^{\dagger})^{1/2} A (A^{\dagger})^{1/2} = P, \end{align*} $$

where P denotes the projection onto the range of A. After replacing $A, A^{\dagger } , R,P$ with their restrictions to the range of A, we can rewrite the above inequality as $A^{-1/2} R A^{-1/2} \le I$ , and since $A^{-1/2} R A^{-1/2}$ is a positive rank-one operator, this is equivalent to $\mathrm {tr}\, (A^{-1} R ) = \mathrm { tr}\, (A^{-1/2} R A^{-1/2} ) \le 1$ .

The proof of the other direction is now easy and is left to the reader.▪

Proof The statement is trivial when $c=0$ or $d=1$ . Thus, we assume that $0< c < d <1$ . Taking an appropriate orthonormal basis of $\mathbb {C}^2$ , we may assume $P= \begin {bmatrix}1 & 0 \\ 0 & 0 \end {bmatrix}$ , $Q= \begin {bmatrix}0 & 0 \\ 0 & 1 \end {bmatrix}$ , and $E= \begin {bmatrix}s^2 & s\sqrt {1-s^2} \\ s\sqrt {1-s^2} & 1-s^2 \end {bmatrix}$ for some real number $0\leq s\leq 1$ . Note that $P-E$ is a trace zero $2 \times 2$ hermitian matrix and therefore $\| P-E\| = \sqrt { - \det (P-E)} = \sqrt { 1 - s^2}$ , and

$$ \begin{align*}(P+cQ)^{-1} = \begin{bmatrix}1 & 0 \\ 0 & c^{-1} \end{bmatrix}. \end{align*} $$

The proof can be completed by a straightforward application of Lemma 6.7.▪

Proof It is easy to see that $\psi $ preserves antipodes. Let $x\in {\cal S}$ and $0<c<1$ . Then $C_{x, c}:=\{y\in {\cal S}\, : \, \lvert y-x\rvert = c\}$ is mapped onto $C_{\psi (x), g_x(c)}=\{y\in {\cal S}\, : \, \lvert y-\psi (x)\rvert = g_x (c)\}$ . (These collections are circles in ${\cal S}$ .) Therefore, for any circle C in ${\cal S}$ , the image $\psi (C)$ is a circle in ${\cal S}$ .

Let $x_1, x_2$ be distinct elements in ${\cal S}$ . Suppose that $x_1$ and $x_2$ are not antipodal in ${\cal S}$ . Take the great circle $C_0$ in ${\cal S}$ that contains the two points $x_1, x_2$ . Then the image $\psi (C_0)$ is a circle in ${\cal S}$ . Let $x\in C_0\setminus \{x_1, x_2, -x_1, -x_2\}$ . An elementary calculation shows that we can take $c_1, c_2\in (0, 1)$ such that $C_{x_1, c_1}\cap C_{x_2, c_2} = \{x\}$ . It follows that $C_{\psi (x_1), g_{x_1}(c_1)}\cap C_{\psi (x_2), g_{x_2}(c_2)} = \{\psi (x)\}$ . This equation implies that $\psi (x)$ lies in the great circle that contains the two points $\psi (x_1), \psi (x_2)$ . Hence, the image $\psi (C_0)$ is the great circle that contains the two points $\psi (x_1), \psi (x_2)$ .

We consider the geodesic distance $d_g$ for a while. We already know that $\psi $ preserves the geodesic distance $\pi /2$ . That is, we have $d_g(\psi (x_1), \psi (x_2))= \pi /2$ if $d_g(x_1, x_2)= \pi /2$ . Suppose that the geodesic distance d with $0<d\leq \pi /2$ is preserved by $\psi $ . We prove that the geodesic distance $d/2$ is also preserved by $\psi $ . Let $x_1, x_2\in {\cal S}$ satisfy $d_g(x_1, x_2)=d/2$ . Take the great circle $C_0$ in ${\cal S}$ that contains the two points $x_1, x_2$ . Take the unique point $x_0\in C_0\setminus \{x_2\}$ such that $d_g(x_1, x_0)=d/2$ . Then we have $d_g(x_0, x_2) = d$ , and hence $d_g(\psi (x_0), \psi (x_2))=d$ . It follows that

$$\begin{align*}\{\psi(x_0), \psi(x_2)\} \subset \psi(C_{x_1, \sin(d/2)}\cap C_0) = \psi(C_{x_1, \sin(d/2)})\cap\psi(C_0). \end{align*}$$

Note that $\psi (C_{x_1, \sin (d/2)})$ is a circle with center $\psi (x_1)\in \psi (C_0)$ . Note also that $\psi $ restricted to $C_0$ is a homeomorphism onto $\psi (C_0)$ and preserves antipodes. Hence, we obtain $d_g(\psi (x_1), \psi (x_2))= d/2$ .

We next prove the following in a similar way: if $\psi $ preserves the geodesic distances $d_1, d_2>0$ with $d_1+d_2<\pi /2$ , then $\psi $ also preserves the geodesic distance $d_1+d_2$ . Let $x_1, x_2\in {\cal S}$ satisfy $d_g(x_1, x_2)=d_1+d_2$ . Take the great circle $C_0$ in ${\cal S}$ that contains the two points $x_1, x_2$ . Take the unique point $x_0\in C_0$ such that $d_g(x_1, x_0)=d_1$ and $d_g(x_0, x_2)=d_2$ . Then we have $d_g(\psi (x_1), \psi (x_0))=d_1$ and $d_g(\psi (x_0), \psi (x_2))=d_2$ . Since $\psi $ restricted to $C_0$ is a homeomorphism onto $\psi (C_0)$ and preserves antipodes, we obtain $d_g(\psi (x_1), \psi (x_2)) = d_1+d_2$ .

It follows that $\psi $ preserves the geodesic distance $k\pi /2^l$ whenever $k, l$ are positive integers with $k/2^l\leq 1/2$ . Since $\psi $ is a homeomorphism, we see that $\psi $ is an isometry with respect to the geodesic distance, thus also an isometry with respect to the euclidean distance in $\mathbb {R}^3$ .▪

Proof By Lemma 6.2, there exists a real number $0<t<1$ such that $\phi $ is continuous at $tI$ . By Corollary 6.4, we have $\phi (0)<\phi (tI)<\phi (I)$ . By Corollary 6.6, there exist continuous order isomorphisms $\psi _1: E_n\to E_n$ and $\psi _2: [\phi (0), \phi (I)]\to E_n$ such that $\psi _1(I/2) = tI$ and $\psi _2(\phi (tI)) = I/2$ . Considering $\psi _2\circ \phi \circ \psi _1: E_n\to E_n$ instead of $\phi $ , we may assume that $\phi $ is an order embedding from $E_n$ into itself, and also fixes the three points $0, I/2, I$ , and is continuous at these points.

We next prove that $\phi $ maps every projection of rank m to a projection of rank m, $0\leq m\leq n$ . By Corollary 6.4, we know that $\phi ((0, I))\subset (0, I)$ . If P is a rank-one projection, then $P\nleq X$ holds for any $X\in (0, I)$ . By the assumption that $\phi $ is continuous at I, there exists a sequence $(X_k)_{k\ge 1}$ in $\phi ((0, I))\,\, (\subset (0, I))$ such that $X_k\to I$ . It follows that $\phi (P)\nleq X_k$ for any $k\ge 1$ , and hence $\phi (P)$ has $1$ as its eigenvalue. Since the subset $[P, I]$ , which is order isomorphic to $E_{n-1}$ , is mapped into $[\phi (P), \phi (I)] = [\phi (P), I]$ , by Lemmas 6.1 and 6.3, the multiplicity of $1$ as an eigenvalue of $\phi (P)$ is one. In other words, $[\phi (P), I]$ is order isomorphic to $E_{n-1}$ . Consider the restriction $\phi |_{[P, I]}: [P, I] \to [\phi (P), I]$ . Then the same discussion shows the following: for any rank-one projection Q with $PQ=0$ , $\phi (P+Q)$ has $1$ as an eigenvalue with multiplicity $2$ . Iterate the same, then we have: for any $0\leq m\leq n$ and any rank-m projection P, $\phi (P)$ has $1$ as an eigenvalue with multiplicity m. Let $0\leq m\leq n$ . Using the continuity of $\phi $ at $0$ , a similar argument ensures that if P is a projection with rank m, then $\phi (P)$ has $0$ as an eigenvalue with multiplicity $n-m$ . It follows that if P is a projection with rank m, then so is $\phi (P)$ . By the assumption that $\phi $ fixes $I/2$ and is continuous at $I/2$ , we also see that for any projection P, there exists a projection Q such that $\mathrm {rank\,} P= \mathrm {rank\,} Q$ and $\phi (P/2)=Q/2$ . Since $\phi $ is an order embedding, we obtain $\phi (P/2)=\phi (P)/2$ .

Let us show that the restriction $\phi |_{P_{n}^1}$ is an isometry with respect to the metric induced by the operator norm. For any subspace $V\subset \mathbb {C}^n$ , we denote by $P^1(V) \subset P_{n}^1$ the subset of all rank-one projections whose ranges are contained in V. It suffices to show that $\phi |_{P^1(V)}$ is an isometry for any two-dimensional subspace $V\subset \mathbb {C}^n$ . Let $P\in P^1(V)$ . Take the unique projection $Q\in P^1(V)$ such that $PQ=0$ . By the preceding paragraph, $\phi (P+Q)$ is a projection of rank two. Denote by W the range of $\phi (P+Q)$ , and take the unique projection $Q'\in P^1(W)$ such that $\phi (P)Q'=0$ . For a real number $0\leq c\leq 1$ , we have $P\leq P+cQ\leq P+Q$ , and therefore $\phi (P) \leq \phi (P+cQ)\leq \phi (P)+ Q'$ . Thus, there exists a monotone increasing function $f_P: [0, 1]\to [0, 1]$ such that $\phi (P+cQ) = \phi (P) + f_P(c) Q'$ . We prove that $f_P(c)\to 0$ as $c\to 0$ . By the continuity of $\phi $ at $0$ , we know that $\phi (2c(P+Q))\to 0$ as $c\to 0$ . However, by $P+cQ\ngeq 2c(P+Q)$ , $c>0$ , we obtain $\phi (P)+ f_P(c) Q'=\phi (P+cQ)\ngeq \phi (2c(P+Q))$ . Hence, we conclude that $f_P(c)\to 0$ as $c\to 0$ .

Since $\phi (E/2)= \phi (E)/2$ for any $E\in P^1(V)$ , Lemma 6.8 with $d=1/2$ assures the continuity of $\phi |_{P^1(V)}$ at P. Hence, $\phi |_{P^1(V)}$ is an injective continuous map from $P^1(V)$ into $P^1(W)$ . Since both $P^1(V)$ and $P^1(W)$ can be identified with the two-dimensional compact connected manifold ${\cal S}$ without boundary, by the invariance of domain theorem, we actually have $\phi (P^1(V)) = P^1(W)$ . Fix $P\in P^1(V)$ , Q as above again. It follows that for any $0< c< 1/2$ , $\phi $ maps the collection $\{E\in P^1(V)\, : \, E/2 \leq P+dQ \,\,\text { for any }\,\, d>c,\,\, E/2 \nleq P+dQ\,\, \text { for any }\,\,d<c\}$ into the collection $\{E\in P^1(W)\, : \, E/2\leq \phi (P) + f_P(d)Q'\,\, \text { for any }\,\,d>c,\,\, E/2\nleq \phi (P) + f_P(d)Q' \,\,\text { for any }\,\,d<c\}$ . By Lemma 6.8, the former collection is equal to $\{E\in P^1(V)\, : \, \lVert P-E\rVert = \sqrt {c/(1-c)}\}$ and hence homeomorphic to the circle. The latter collection is

$$ \begin{align*}\{E\in P^1(W) : \lVert P-E\rVert \in [\lim_{d\uparrow c} \sqrt{f_P(d)/(1-f_P(d))}, \lim_{d\downarrow c} \sqrt{f_P(d)/(1-f_P(d))}]\}. \end{align*} $$

Since $\phi $ restricts to a homeomorphism from $P^1(V)$ onto $P^1(W)$ , we see that $f_P$ is a continuous monotone increasing bijection on $[0, 1/2]$ . Hence, we may apply Lemma 6.9 for

$$ \begin{align*}\psi=\phi|_{P^1(V)}: P^1(V)\to P^1(W), \end{align*} $$

and

$$ \begin{align*}g_P(\sqrt{c/(1-c)})=\sqrt{f_P(c)/(1-f_P(c))}, \ \ \ 0\leq c\leq 1/2, \end{align*} $$

to obtain that $\phi |_{P^1(V)}$ is an isometry.

Since $P_{n}^1$ is a compact connected manifold without boundary, we see that $\phi $ restricts to a surjective isometry on $P_{n}^1$ by the invariance of domain theorem. Applying Wigner’s unitary-antiunitary theorem, we see that there exists a unitary or antiunitary $u: \mathbb {C}^n \to \mathbb {C}^n$ such that $\phi (P) = uPu^*$ , $P\in P_{n}^1$ . We will prove that $\phi (A) = uAu^*$ for any $A\in E_n$ . It suffices to show that the map $\phi _0: E_n\to E_n$ given by $\phi _0(A) := u^*\phi (A) u$ is the identity map on $E_n$ . It is easy to see that $\phi _0$ fixes every (not necessarily rank-one) projection.

Take any two-dimensional subspace $V\subset \mathbb {C}^n$ and mutually orthogonal projections $P, Q\in P^1(V)$ . Fix a real number $0\leq c<1/2$ . By Lemma 6.8 with $d=1/2$ , for $E\in P^1(V)$ , the condition $\lVert P-E\rVert \leq \sqrt {c/(1-c)}$ is equivalent to $E/2\leq P+cQ$ , which is in turn equivalent to $\phi _0(E/2)\leq \phi _0(P+cQ)$ . We know that $\phi _0(E/2)=E/2$ . Moreover, since $P=\phi _0(P)\leq \phi _0(P+cQ)\leq \phi _0(P+Q)=P+Q$ , there exists a real number $c'\in [0, 1]$ such that $\phi _0(P+cQ)= P+c'Q$ . Hence, by Lemma 6.8 again, the condition $\phi _0(E/2)\leq \phi _0(P+cQ)$ is equivalent to $\lVert P-E\rVert \leq \sqrt {c'/(1-c')}$ . Therefore, for $E\in P^1(V)$ , two conditions $\lVert P-E\rVert \leq \sqrt {c/(1-c)}$ and $\lVert P-E\rVert \leq \sqrt {c'/(1-c')}$ are equivalent. We obtain $c=c'$ , that is, $\phi _0(P+cQ)=P+cQ$ for any $0\leq c<1/2$ .

We fix a two-dimensional subspace $V\subset \mathbb {C}^n$ again, but fix $E\in P^1(V)$ instead of P. Let $0<d< 1$ be a real number. By Lemma 6.8 with $c=d/2$ , for $P\in P^1(V)$ , the condition $\lVert P-E\rVert \leq \sqrt {(1-d)/(2-d)}$ is equivalent to $dE\leq P+dQ/2$ , where Q is the unique projection in $P^1(V)$ with $PQ=0$ . Then this condition is in turn equivalent to $\phi _0(dE)\leq \phi _0(P+dQ/2)$ . The preceding paragraph implies $\phi _0(P+dQ/2)=P+dQ/2$ . Since $0 =\phi _0(0)\leq \phi _0(dE)\leq \phi _0(E)=E$ , there exists $d'\in (0, 1)$ such that $\phi _0(dE)=d'E$ . Hence, by Lemma 6.8 again, the condition $\phi _0(dE)\leq \phi _0(P+dQ/2)$ is equivalent to $\lVert P-E\rVert \leq \sqrt {d(1-d')/(d' (2-d))}$ . It follows that $(1-d)/(2-d) = d(1-d')/(d' (2-d))$ , and consequently, $d=d'$ .

We have shown that $\phi _0(dE)=dE$ for any $E\in P_{n}^1$ and $d\in [0, 1]$ . Let $A\in E_n$ . Lemma 6.7 implies that the ranges of A and $\phi _0 (A)$ coincide and that $\mathrm {tr}\, (A^{\dagger } P) = \mathrm {tr}\, (\phi _0 (A)^{\dagger } P)$ for every $P \in P_{n}^1$ whose range is contained in the range of A. It follows that $A^{\dagger } = \phi _0 (A)^{\dagger }$ , and therefore $A = \phi _0 (A)$ .▪

Proof By the invariance of domain theorem, it suffices to show continuity. Let $A\in U$ . Lemma 6.2 implies that we can take real numbers $c_0, c_1>0$ such that $[A-c_0 I, A+c_1 I]\subset U$ and $\phi $ is continuous at the two points $A-c_0 I, A+c_1 I$ . By Corollary 6.4, $\phi (A-c_0 I) < \phi (A+c_1 I)$ . Using the preceding theorem, we see that $\phi $ restricts to an order isomorphism from $[A-c_0 I, A+c_1 I]$ onto $[\phi (A-c_0 I), \phi (A+c_1 I)]$ . We know that every order isomorphism between two operator intervals is continuous, and hence $\phi $ is continuous at A.▪