# Joint spectrum shrinking maps on projections

Wenhua Qian1 1\. School of Mathematical Sciences, Chongqing Normal University,
Chongqing, China  ,  Dandan Xiao1 Email<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS> ,  Tanghong Tao1<EMAIL_ADDRESS>,
Wenming Wu1,† and  Xin Yi1

###### Abstract.

Let $\mathcal{H}$ be a finite dimensional complex Hilbert space with dimension
$n\geq 3$ and $\mathcal{P}(\mathcal{H})$ the set of projections on
$\mathcal{H}$. Let
$\varphi:\mathcal{P}(\mathcal{H})\to\mathcal{P}(\mathcal{H})$ be a surjective
map. We show that $\varphi$ shrinks the joint spectrum of any two projections
if and only if it is joint spectrum preserving for any two projections and
thus is induced by a ring automorphism on $\mathbb{C}$ in a particular way. In
addition, for an arbitrary $k\geq 3$, $\varphi$ shrinks the joint spectrum of
any $k$ projections if and only if it is induced by a unitary or an anti-
unitary. Assume that $\phi$ is a surjective map on the Grassmann space of rank
one projections. We show that $\phi$ is joint spectrum preserving for any $n$
rank one projections if and only if it can be extended to a surjective map on
$\mathcal{P}(\mathcal{H})$ which is spectrum preserving for any two
projections. Moreover, for any $k>n$, $\phi$ is joint spectrum shrinking for
any $k$ rank one projections if and only if it is induced by a unitary or an
anti-unitary.

###### Key words and phrases:

joint spectrum preserving, joint spectrum shrinking, Kaplansky’s problem,
projections

###### 2010 Mathematics Subject Classification:

Primary 47B49; Secondary 47A25

Qian is supported in part by the Natural Science Foundation of Chongqing
Science and Technology Commission(cstc2020jcyj-msxmX0723) and the Research
Foundation of Chongqing Educational Committee (No. KJQN2021000529). Wu(†
corresponding author) is supported in part by NSF of China (No. 11871127, No.
11971463).

##  1\. Introduction

The well-known Gleason-Kahane-Żelazko theorem ([8, 12]) states that a nonzero
linear functional $\rho:\mathcal{A}\to\mathbb{C}$ on a unital complex Banach
algebra $\mathcal{A}$ is an algebra homomorphism if and only if $\rho$ maps
every element inside its spectrum. It is easy to verify that a nonzero linear
functional $\rho$ on $\mathcal{A}$ is an algebra homomorphism if and only if
$\rho$ is a Jordan homomorphism, that is, $\rho(I)=1$ where $I$ is the unit of
$\mathcal{A}$ and $\rho$ preserves the squares. Motivated by this classical
result, in [13] Kaplansky asked whether a unital linear map
$\varphi:\mathcal{A}\to\mathcal{B}$ between unital complex Banach algebras
which shrinks spectrum (i.e., $\sigma(\varphi(A))\subseteq\sigma(A),\ \forall\
A\in\mathcal{A}$) is a Jordan homomorphism. Notice that a unital linear map
$\varphi:\mathcal{A}\to\mathcal{B}$ is spectrum shrinking if and only if it is
invertibility preserving.

It is well-known that in general Kaplansky problem has a negative answer. A
counterexample can be found in [2]. A lot of work has been done on Kaplansky
Problem by additional assumptions (see [3, 10] for some survey). Aupetit
conjectured that Kaplansky Problem has a positive answer when both Banach
algebras are semi-simple and the map $\varphi$ is surjective and he confirmed
this conjecture for von Neumann algebras [4]. This problem is still open, even
for C∗-algebras [5, 9]. It was proved in [6] that the conjecture is true for
C∗-algebras if in addition $\varphi$ is positive. In particular, some related
maps on matrix algebras are also considered [7, 15].

Recall that [17] the joint spectrum of a tuple of operators
$A_{1},A_{2},\ldots,A_{l}$ acting on a Hilbert space $\mathcal{H}$ is the set

$\sigma([A_{1},\dots,A_{l}])=\\{(c_{1},\dots,c_{l})\in\mathbb{C}^{l}:c_{1}A_{1}+\dots+c_{l}A_{l}~{}\text{is
not invertible in}~{}\mathcal{B}(\mathcal{H})\\}.$

It is an interesting issue to discuss the mapping which shrinks or preserves
the joint spectrum of operators. It is easy to verify that a unital map
$\varphi:\mathcal{A}\to\mathcal{B}$ is spectrum shrinking if and only if it
shrinks the joint spectrum of the $2$-tuple $[I,A]$ for any element
$A\in\mathcal{A}$. Therefore according to Aupetit’s results [4], we can obtain
the form of the mapping preserving the joint spectrum of any two operators in
$\mathcal{B}(\mathcal{H})$.

In this paper we will characterize the mappings which shrink or preserve the
joint spectrum of a tuple of projections.

Assume that $\mathcal{H}$ is a finite dimensional Hilbert space. We first
consider a surjective map $\varphi$ on the set $\mathcal{P}(\mathcal{H})$ of
projections on $\mathcal{H}$ which shrinks the joint spectrum of any two
projections. We first show that $\varphi$ leaves every Grassmann space
invariant. By showing that the restriction of $\varphi$ on each Grassmann
space is bijective, we get that $\varphi$ is bijective. A mathematical
induction gives that $\varphi$ is determined by its action on rank $n-1$
projections and as a consequence we obtain that $\varphi$ is a lattice
isomorphism which preserves the joint spectrum of any two projections. If
$n=2$, it is easy to verify that a surjective map
$\varphi:\mathcal{P}(\mathcal{H})\to\mathcal{P}(\mathcal{H})$ is joint
spectrum preserving for any two projections if and only if $\varphi$ is
bijective with $\varphi(I)=I,\varphi(0)=0$. If $n\geq 3$, some further
calculations in Section 2 give the following result.

######  Theorem 1.1.

Assume that $3\leq n(=\dim(\mathcal{H}))<+\infty$ and
$\varphi:\mathcal{P}(\mathcal{H})\to\mathcal{P}(\mathcal{H})$ is a surjective
map. Then the followings are equivalent.‘

  1. (1)

$\varphi$ shrinks the joint spectrum of any two projections;

  2. (2)

$\varphi$ preserves the joint spectrum of any two projections;

  3. (3)

there exist an orthonormal basis $\alpha_{1},\alpha_{2},\dots,\alpha_{n}$, a
basis $\beta_{1},\beta_{2},\dots,\beta_{n}$ for $\mathcal{H}$ and a ring
automorphism $f$ of $\mathbb{C}$ such that

$\varphi(P)(\mathcal{H})=\\{\hat{f}(P\xi):\xi\in\mathcal{H}\\},\forall\
P\in\mathcal{P}(\mathcal{H}),$

where $\hat{f}:\mathcal{H}\to\mathcal{H}$ is induced by $f$ with

$\hat{f}(z_{1}\alpha_{1}+z_{2}\alpha_{2}+\dots+z_{n}\alpha_{n})=f(z_{1})\beta_{1}+f(z_{2})\beta_{2}+\dots+f(z_{n})\beta_{n}.$

Moreover, we consider a surjective map $\varphi$ on the set
$\mathcal{P}(\mathcal{H})$ which shrinks the joint spectrum of any $k(\geq 3)$
projections. We will further prove that $\varphi$ preserves the orthogonality
of projections (i.e., $PQ=0$ if and only if $\varphi(P)\varphi(Q)=0$) and
obtain the following equivalent characterizations.

######  Theorem 1.2.

Assume that $3\leq n(=\dim(\mathcal{H}))<+\infty$ and
$\varphi:\mathcal{P}(\mathcal{H})\to\mathcal{P}(\mathcal{H})$ is a surjective
map. Then the followings are equivalent.

  1. (1)

there exists $k_{0}\geq 3$ such that $\varphi$ shrinks the joint spectrum of
any $k_{0}$ projections;

  2. (2)

there exists $k_{0}\geq 3$ such that $\varphi$ preserves the joint spectrum of
any $k_{0}$ projections;

  3. (3)

for any $k\geq 3$, $\varphi$ shrinks the joint spectrum of any $k$
projections;

  4. (4)

for any $k\geq 3$, $\varphi$ preserves the joint spectrum of any $k$
projections;

  5. (5)

there exists a unitary or anti-unitary $U$ such that $\varphi(P)=U^{*}PU,\
\forall\ P\in\mathcal{P}(\mathcal{H})$.

We also investigate a surjective map $\phi$ on the set
$\mathcal{P}_{1}(\mathcal{H})$ of rank one projections which preserves the
joint spectrum of any $n$ rank one projections. It is shown that $\phi$ is
order-preserving in the sense that
$\phi(P)\leq\phi(P_{1})\vee\phi(P_{2})\vee\dots\vee\phi(P_{k})$ if and only if
$P\leq P_{1}\vee P_{2}\vee\dots\vee P_{k}$ for any positive integer $k$ and
any $P,P_{1},\dots,P_{k}\in\mathcal{P}_{1}(\mathcal{H})$. It follows that
$\phi$ can be extended to a surjective map on the set
$\mathcal{P}(\mathcal{H})$ of projections on $\mathcal{H}$ which preserves the
joint spectrum of any two projections and we obtain the following result.

######  Theorem 1.3.

Assume that $3\leq n(=\dim(\mathcal{H}))<+\infty$ and
$\phi:\mathcal{P}_{1}(\mathcal{H})\to\mathcal{P}_{1}(\mathcal{H})$ is a
surjective map. Then the followings are equivalent.

  1. (1)

$\phi$ preserves the joint spectrum of any $n$ rank one projections;

  2. (2)

there exist an orthonormal basis $\alpha_{1},\alpha_{2},\dots,\alpha_{n}$, a
basis $\beta_{1},\beta_{2},\dots,\beta_{n}$ for $\mathcal{H}$ and a ring
automorphism $f$ of $\mathbb{C}$ such that

$\phi(P)(\mathcal{H})=\\{\hat{f}(P\xi):\xi\in\mathcal{H}\\},\forall\
P\in\mathcal{P}_{1}(\mathcal{H}),$

where $\hat{f}:\mathcal{H}\to\mathcal{H}$ is induced by $f$ as in Theorem 1.1.

Moreover, if
$\phi:\mathcal{P}_{1}(\mathcal{H})\to\mathcal{P}_{1}(\mathcal{H})$ is
surjective and shrinks the joint spectrum of any $n+1$ projections, we can
show that $\phi$ preserves the orthogonality of projections and obtain the
following theorem.

######  Theorem 1.4.

Assume that $3\leq n(=\dim(\mathcal{H}))<+\infty$ and
$\phi:\mathcal{P}_{1}(\mathcal{H})\to\mathcal{P}_{1}(\mathcal{H})$ is a
surjective map. Then the followings are equivalent.

  1. (1)

there exists $k_{0}\geq n+1$ such that $\phi$ shrinks the joint spectrum of
any $k_{0}$ projections;

  2. (2)

there exists $k_{0}\geq n+1$ such that $\phi$ preserves the joint spectrum of
any $k_{0}$ projections;

  3. (3)

for any $k\geq n+1$, $\phi$ shrinks the joint spectrum of any $k$ projections;

  4. (4)

for any $k\geq n+1$, $\phi$ preserves the joint spectrum of any $k$
projections;

  5. (5)

there exist a unitary or anti-unitary $U$ such that $\phi(P)=U^{*}PU,\
\forall\ P\in\mathcal{P}_{1}(\mathcal{H})$.

##  2\. Maps shrinking the joint spectrum of any two projections

Let $\mathcal{H}$ be a Hilbert space with dimension $n<+\infty$. Denote by
$\mathcal{P}(\mathcal{H})$ and $\mathcal{P}_{r}(\mathcal{H})$ (i.e., the order
$r$ Grassmann space) the set of projections and the set of rank $r$
projections on $\mathcal{H}$. In this section we assume that
$\varphi:\mathcal{P}(\mathcal{H})\to\mathcal{P}(\mathcal{H})$ is a surjective
map which shrinks the joint spectrum of any two projections, i.e.,
$\sigma([\varphi(P),\varphi(Q)])\subseteq\sigma([P,Q]),\ \forall\
P,Q\in\mathcal{P}(\mathcal{H})$.

######  Lemma 2.1.

$\varphi(I)=I,\varphi(0)=0$.

###### Proof.

For any $Q\in\mathcal{P}(\mathcal{H})$, $(1,0)\notin\sigma([I,Q])$. Hence
$(1,0)\notin\sigma([\varphi(I),\varphi(Q)])$ and by the surjection of
$\varphi$ we have $\varphi(I)=I$. Since $(1,-1)\notin\sigma([I,0])$, we have
$(1,-1)\notin\sigma([\varphi(I),\varphi(0)])=\sigma([I,\varphi(0)])$. Hence
$\varphi(0)=0$. ∎

For any $P,Q\in\mathcal{P}(\mathcal{H})$, it is easy to verify that $P\vee
Q=I$ if and only if $(1,1)\notin\sigma([P,Q])$. Thus the following lemma is
obvious.

######  Lemma 2.2.

Let $P,Q\in\mathcal{P}(\mathcal{H})$. If $P\vee Q=I$, then
$\varphi(P)\vee\varphi(Q)=I$.

######  Lemma 2.3.

Let $P,Q\in\mathcal{P}(\mathcal{H})$. If $P\vee Q=I,P\wedge Q=0$, then
$\varphi(P)\vee\varphi(Q)=I,\varphi(P)\wedge\varphi(Q)=0$.

###### Proof.

If $P=I,Q=0$ or $P=0,Q=I$, then Lemma 2.1 gives the result. Assume that
$P,Q\in\mathcal{P}(\mathcal{H})\setminus\\{0,I\\}$. Then it follows from [16]
that $(1,1)\notin\sigma([P,Q]),(1,-1)\notin\sigma([P,Q])$. Hence
$(1,1)\notin\sigma([\varphi(P),\varphi(Q)]),(1,-1)\notin\sigma([\varphi(P),\varphi(Q)])$.
Thus $\varphi(P)+\varphi(Q)$ and $\varphi(P)-\varphi(Q)$ are both invertible.
Hence $\varphi(P)\vee\varphi(Q)=I,\varphi(P)\wedge\varphi(Q)=0$. ∎

In the following, we denote by $r(P)$ the rank of $P$ for any
$P\in\mathcal{P}(\mathcal{H})$.

######  Lemma 2.4.

Let $P,Q\in\mathcal{P}(\mathcal{H})$. If $r(P)=r(Q)$, then
$r(\varphi(P))=r(\varphi(Q))$. Moreover,
$\varphi(\mathcal{P}_{k}(\mathcal{H}))=\mathcal{P}_{k}(\mathcal{H}),\forall\
k\in\\{0,1,2,\dots,n\\}$.

###### Proof.

Notice $\varphi(I)=I,\varphi(0)=0$. We may assume that
$P,Q\in\mathcal{P}_{k}(\mathcal{H})$, where $k\in\\{1,2,\dots,n-1\\}$.

We first assume that $r(P\wedge Q)=k-1$. It follows that $r(P\vee Q)=k+1$.
Then there exist linearly independent vectors
$x_{1},x_{2},\dots,x_{k-1},\alpha,\beta\in\mathcal{H}$ such that $P$ is the
projection onto the subspace generated by $x_{1},x_{2},\dots,x_{k-1},\alpha$
and $Q$ is the projection onto the subspace generated by
$x_{1},x_{2},\dots,x_{k-1},\beta$. Take $R=P_{1}+(I-P\vee Q)$, where $P_{1}$
is the rank one projection onto $\mathbb{C}(\alpha+\beta)$. It follows that
$P\vee R=I,P\wedge R=0$ and $Q\vee R=I,Q\wedge R=0$. By Lemma 2.3 we obtain
that $\varphi(P)\vee\varphi(R)=I,\varphi(P)\wedge\varphi(R)=0$ and
$\varphi(Q)\vee\varphi(R)=I,\varphi(Q)\wedge\varphi(R)=0$. Hence

$r(\varphi(P))=n-r(\varphi(R))=r(\varphi(Q)).$

Now assume that $r(P\wedge Q)=k-r$, where $1\leq r\leq k$. Then there exist
linearly independent vectors
$x_{1},x_{2},\dots,x_{k-r},\alpha_{1},\alpha_{2},\dots,\alpha_{r},\beta_{1},\beta_{2},\dots,\beta_{r}$
such that $P$ is the projection onto the subspace generated by
$x_{1},x_{2},\dots,x_{k-r},\alpha_{1},\alpha_{2},\dots,\alpha_{r}$ and $Q$ is
the projection onto the subspace generated by
$x_{1},x_{2},\dots,x_{k-r},\beta_{1},\beta_{2},\dots,\beta_{r}$. Take
$Q_{0}=P,Q_{r}=Q$. For each $i\in\\{1,2,\dots,r-1\\}$, let $Q_{i}$ be the
projection onto the subspace generated by
$x_{1},x_{2},\dots,x_{k-r},\beta_{1},\dots,\beta_{i},\alpha_{i+1},\dots,\alpha_{r}$.
It follows that $Q_{0},Q_{1},\dots,Q_{r}\in\mathcal{P}_{k}(H)$ and
$r(Q_{i}\wedge Q_{i+1})=k-1$ for every $i\in\\{0,1,\dots,r-1\\}$. Then the
result of the previous paragraph implies that

$r(\varphi(P))=r(\varphi(Q)).$

Hence there exists a map $g:\\{0,1,2,\dots,n\\}\to\\{0,1,2,\dots,n\\}$ such
that
$\varphi(\mathcal{P}_{k}(\mathcal{H}))\subseteq\mathcal{P}_{g(k)}(\mathcal{H}),\forall\
k\in\\{0,1,2,\dots,n\\}$. By the fact that $\varphi$ is surjective, we obtain
that $g$ is a bijection and
$\varphi(\mathcal{P}_{k}(\mathcal{H}))=\mathcal{P}_{g(k)}(\mathcal{H}),\forall\
k\in\\{0,1,2,\dots,n\\}$. In particular, by Lemma 2.3 we have
$g(n-k)=n-g(k),\forall\ k\in\\{0,1,2,\dots,n\\}$.

Clearly, $g(0)=0$ and $g(n)=n$. Assume that $s=g(1)>1$. Then $g(n-1)=n-s<n-1$.
By the fact that $g$ is a bijection, there exists $l>1$ such that $g(l)=1$.
Take two projections
$P_{1}\in\mathcal{P}_{n-1}(\mathcal{H}),P_{2}\in\mathcal{P}_{l}(\mathcal{H})$
with $P_{1}\vee P_{2}=I$. It follows that
$\varphi(P_{1})\in\mathcal{P}_{n-s}(H),\varphi(P_{2})\in\mathcal{P}_{1}(H)$.
Then $r(\varphi(P_{1}))+r(\varphi(P_{2}))<n$. Therefore
$\varphi(P_{1})\vee\varphi(P_{2})\neq I$ and we obtain a contradiction
according to Lemma 2.2. Hence $g(1)=1,g(n-1)=n-1$. Continuing in this way, we
have
$\varphi(\mathcal{P}_{k}(\mathcal{H}))=\mathcal{P}_{k}(\mathcal{H}),\forall\
k\in\\{0,1,2,\dots,n\\}$. ∎

In the following we will show that the restriction of $\varphi$ on each
Grassmann space $\mathcal{P}_{k}(\mathcal{H})$ is a bijection and thus
$\varphi$ is a bijection. We present two necessary lemmas.

######  Lemma 2.5.

Let $Q\in\mathcal{P}_{n-1}(\mathcal{H}),P\in\mathcal{P}(\mathcal{H})$. If
$\varphi(P)\leq\varphi(Q)$, then $P\leq Q$. Moreover,
$\varphi|_{\mathcal{P}_{n-1}(\mathcal{H})}$ is a bijection.

###### Proof.

By Lemma 2.4, $\varphi(Q)\in\mathcal{P}_{n-1}(\mathcal{H})$. Since
$\varphi(P)\leq\varphi(Q)$, $\varphi(P)\vee\varphi(Q)\neq I$. By Lemma 2.2 and
the fact that $Q\in\mathcal{P}_{n-1}(H)$, we have $P\leq Q$. It is easy to
verify that $\varphi|_{\mathcal{P}_{n-1}(\mathcal{H})}$ is a bijection. ∎

For convenience, we denote by $\Phi=\phi|_{\mathcal{P}_{n-1}(\mathcal{H})}$ in
the following proposition.

######  Proposition 2.6.

Let $k\in\\{1,2,\dots,n\\}$ and $P\in\mathcal{P}_{n-k}(\mathcal{H})$. Assume
that $P^{\prime}\in\mathcal{P}_{n-k}(\mathcal{H})$ with
$\varphi(P^{\prime})=P$. Then for any $k$ projections
$Q_{1},Q_{2},\dots,Q_{k}\in\mathcal{P}_{n-1}(\mathcal{H})$ with $P=Q_{1}\wedge
Q_{2}\wedge\dots\wedge Q_{k}$, $P^{\prime}=\wedge_{1\leq i\leq
k}\Phi^{-1}(Q_{i})$. Moreover, $\varphi$ is a bijection.

###### Proof.

We prove the result by a mathematical induction on $k$. From Lemma 2.5, the
result is true when $k=1$. Assume that the result is true when $k=s$. Now let
$k=s+1$ and assume that
$Q_{1},Q_{2},\dots,Q_{s},Q_{s+1}\in\mathcal{P}_{n-1}(\mathcal{H})$ with
$P=Q_{1}\wedge Q_{2}\wedge\dots\wedge Q_{s}\wedge Q_{s+1}$.

Take $P_{1}=Q_{1}\wedge Q_{2}\wedge\dots\wedge Q_{s}$ and $P_{2}=Q_{1}\wedge
Q_{2}\wedge\dots\wedge Q_{s-1}\wedge Q_{s+1}$. Clearly $P_{1},P_{2}$ are two
different projections in $\mathcal{P}_{n-s}(\mathcal{H})$. By the assumption
that the result is true when $k=s$, we have
$\varphi|_{\mathcal{P}_{n-s}}(\mathcal{H})$ is a bijection and

$\displaystyle(\varphi|_{\mathcal{P}_{n-s}(\mathcal{H})})^{-1}(P_{1})=\Phi^{-1}(Q_{1})\wedge\Phi^{-1}(Q_{2})\wedge\dots\wedge\Phi^{-1}(Q_{s}),$
$\displaystyle(\varphi|_{\mathcal{P}_{n-s}(\mathcal{H})})^{-1}(P_{2})=\Phi^{-1}(Q_{1})\wedge\Phi^{-1}(Q_{2})\wedge\dots\wedge\Phi^{-1}(Q_{s-1})\wedge\Phi^{-1}(Q_{s}+1).$
(2.1)

By Lemma 2.5, $P^{\prime}\leq\Phi^{-1}(Q_{i})$ for each
$i\in\\{1,2,\dots,s+1\\}$. Hence
$P^{\prime}\leq(\varphi|_{\mathcal{P}_{n-s}(\mathcal{H})})^{-1}(P_{1})\wedge(\varphi|_{\mathcal{P}_{n-s}(\mathcal{H})})^{-1}(P_{2})$.
Since $\varphi|_{\mathcal{P}_{n-s}}$ is a bijection,
$(\varphi|_{\mathcal{P}_{n-s}(\mathcal{H})})^{-1}(P_{1})\neq(\varphi|_{\mathcal{P}_{n-s}(\mathcal{H})})^{-1}(P_{2})$
and thus
$r((\varphi|_{\mathcal{P}_{n-s}(\mathcal{H})})^{-1}(P_{1})\wedge(\varphi|_{\mathcal{P}_{n-s}(\mathcal{H})})^{-1}(P_{2}))\leq
n-s-1=r(P^{\prime})$. Therefore
$P^{\prime}=(\varphi|_{\mathcal{P}_{n-s}(\mathcal{H})})^{-1}(P_{1})\wedge(\varphi|_{\mathcal{P}_{n-s}(\mathcal{H})})^{-1}(P_{2})=\wedge_{1\leq
i\leq s+1}\Phi^{-1}(Q_{i})$ from (2.1). Hence
$\varphi|_{\mathcal{P}_{n-s-1}}(\mathcal{H})$ is also a bijection. Moreover,
$\varphi$ is a bijection. ∎

According to Proposition 2.6, we have the following corollary.

######  Corollary 2.7.

If $P,Q\in\mathcal{P}(H)$, then we have the following results.

  * (1)

If $P\neq I$ and
$\\{Q_{\lambda}:\lambda\in\Omega\\}\subseteq\mathcal{P}_{n-1}(\mathcal{H})$
with $\wedge_{\lambda\in\Lambda}Q_{\lambda}=P$, then
$\varphi^{-1}(P)=\wedge_{\lambda\in\Omega}\varphi^{-1}(Q_{\lambda})$;

  * (2)

If $P\leq Q$, then $\varphi^{-1}(P)\leq\varphi^{-1}(Q)$;

  * (3)

$\varphi^{-1}(P\wedge Q)=\varphi^{-1}(P)\wedge\varphi^{-1}(Q)$,
$\varphi^{-1}(P\vee Q)=\varphi^{-1}(P)\vee\varphi^{-1}(Q)$;

###### Proof.

(1) Notice that $P\leq Q_{\lambda}$ for every $\lambda\in\Omega$. It follows
from Lemma 2.5 that $\varphi^{-1}(P)\leq\varphi^{-1}(Q_{\lambda}),\forall\
\lambda\in\Omega$ and hence
$\varphi^{-1}(P)\leq\wedge_{\lambda\in\Omega}\varphi^{-1}(Q_{\lambda})$.
Assume that $P\in\mathcal{P}_{n-k}(\mathcal{H})$. Then there exist
$\lambda_{1},\lambda_{2},\dots,\lambda_{k}\in\Omega$ such that
$P=\wedge_{1\leq i\leq k}Q_{\lambda_{i}}$. It follows from Proposition 2.6
that $\varphi^{-1}(P)=\wedge_{1\leq i\leq k}\varphi^{-1}(Q_{\lambda_{i}})$.
Hence

$\varphi^{-1}(P)=\wedge_{\lambda\in\Omega}\varphi^{-1}(Q_{\lambda}).$

(2) This is clear from (1).

(3) From (1), it is obvious that $\varphi^{-1}(P\wedge
Q)=\varphi^{-1}(P)\wedge\varphi^{-1}(Q)$. By (2), we have
$\varphi^{-1}(P)\vee\varphi^{-1}(Q)\leq\varphi^{-1}(P\vee Q)$. Since
$r(P)=r(\varphi^{-1}(P)),r(Q)=r(\varphi^{-1}(Q))$ and
$r(\varphi^{-1}(P)\wedge\varphi^{-1}(Q))=r(\varphi^{-1}(P\wedge Q))=r(P\wedge
Q)$, we have $r(\varphi^{-1}(P)\vee\varphi^{-1}(Q))=r(\varphi^{-1}(P\vee Q))$
and therefore $\varphi^{-1}(P\vee Q)=\varphi^{-1}(P)\vee\varphi^{-1}(Q)$. ∎

Now we are able to prove the following result, which states that a surjective
map $\varphi:\mathcal{P}(\mathcal{H})\to\mathcal{P}(\mathcal{H})$ that shrinks
the joint spectrum of any two projections will preserve the joint spectrum of
any two projections.

######  Theorem 2.8.

If $\varphi:\mathcal{P}(\mathcal{H})\to\mathcal{P}(\mathcal{H})$ is a
surjective map such that
$\sigma([\varphi(P),\varphi(Q)])\subseteq\sigma([P,Q]),\ \forall\
P,Q\in\mathcal{P}(\mathcal{H})$, then

$\sigma([\varphi(P),\varphi(Q)])=\sigma([P,Q]),\ \forall\
P,Q\in\mathcal{P}(\mathcal{H}).$

###### Proof.

Note that $\varphi(I)=I,\varphi(0)=0$. By Corollary 2.7, we obtain that $P\vee
Q=I$ if and only if $\varphi(P)\vee\varphi(Q)=I$ and $P\wedge Q=0$ if and only
if $\varphi(P)\wedge\varphi(Q)=0$. It follows from the result in [16] that
$\sigma([\varphi(P),\varphi(Q)])=\sigma([P,Q])$ for any
$P,Q\in\mathcal{P}(\mathcal{H})$. ∎

######  Remark 2.9.

By the previous discussions, we obtain that $\varphi$ is bijective and
$\varphi^{-1}$ also preserves the joint spectrum of any two projections in
$\mathcal{P}(\mathcal{H})$. Applying the same argument to
$\varphi=(\varphi^{-1})^{-1}$, it follows from Corollary 2.7 that
$\varphi(P)\vee\varphi(Q)=\varphi(P\vee
Q),\varphi(P)\wedge\varphi(Q)=\varphi(P\wedge Q)$ and
$\varphi(P)\leq\varphi(Q)$ if and only if $P\leq Q$ for every
$P,Q\in\mathcal{P}(H)$. If $\mathcal{H}$ has dimension $2$, then any two
different rank one projections $P,Q$ satisfy $P\vee Q=I,P\wedge Q=0$ and
therefore
$\sigma([P,Q])=\\{(c,0):c\in\mathbb{C}\\}\cup\\{(0,c):c\in\mathbb{C}\\}$.
Hence any bijection on $\mathcal{P}(\mathcal{H})$ which fixes $0$ and $I$
preserves the joint spectrum of any two projections. In the following of this
section we focus on the case that $3\leq n<+\infty$. Fix an orthonormal basis
$\alpha_{1},\alpha_{2},\dots,\alpha_{n}$ for $\mathcal{H}$. For a nonzero
vector $\alpha\in\mathcal{H}$, we denote by $P_{\alpha}$ the projection onto
the one-dimensional subspace generated by $\alpha$. It follows that
$\varphi(P_{\alpha_{i}})$ is a rank one projection for each
$i\in\\{1,2,\dots,n\\}$ and there exist nonzero vectors
$\beta_{1},\beta_{2},\dots,\beta_{n}$ such that
$P_{\beta_{i}}=\varphi(P_{\alpha_{i}})$. By the fact that
$\varphi(P)\vee\varphi(Q)=\varphi(P\vee Q)$ for every $P,Q\in\mathcal{P}(H)$,
we have $\vee_{i}P_{\beta_{i}}=\varphi(\vee_{i}P_{\alpha_{i}})=I$ and
therefore $\beta_{1},\beta_{2},\dots,\beta_{n}$ are linearly independent.
Notice that for every $i\in\\{2,\dots,n\\}$, $P_{\alpha_{1}+\alpha_{i}}$ is a
rank one projection and $P_{\alpha_{1}+\alpha_{i}}\leq P_{\alpha_{1}}\vee
P_{\alpha_{i}}$. We have $\varphi(P_{\alpha_{1}+\alpha_{i}})\leq
P_{\beta_{1}}\vee P_{\beta_{i}}$ and thus there exists a constant $0\neq
t_{i}\in\mathbb{C}$ such that
$\varphi(P_{\alpha_{1}+\alpha_{i}})=P_{\beta_{1}+t_{i}\beta_{i}}$.

######  Lemma 2.10.

With notations as in Remark 2.9, we have that

$\varphi(P_{\alpha_{i}+\alpha_{j}})=P_{t_{i}\beta_{i}+t_{j}\beta_{j}},\
\forall\ i,j\in\\{2,3,\dots,n\\}.$

###### Proof.

Fix $i,j\in\\{2,3,\dots,n\\}$. We only need to consider the case when $i\neq
j$. Let $\alpha=\alpha_{1}+\alpha_{i}+\alpha_{j}$. It follows that
$P_{\alpha}\leq P_{\alpha_{1}+\alpha_{i}}\vee P_{\alpha_{j}},P_{\alpha}\leq
P_{\alpha_{1}+\alpha_{j}}\vee P_{\alpha_{i}}$. Therefore

$\varphi(P_{\alpha})\leq P_{\beta_{1}+t_{i}\beta_{i}}\vee
P_{\beta_{j}},\varphi(P_{\alpha})\leq P_{\beta_{1}+t_{j}\beta_{j}}\vee
P_{\beta_{i}}.$

Hence there exists constants $c_{1},c_{2}$ with
$\beta_{1}+t_{1}\beta_{i}+c_{1}\beta_{j},\beta_{1}+t_{j}\beta_{j}+c_{2}\beta_{i}$
are both in the range of $\varphi(P_{\alpha})$. Since
$\beta_{1},\beta_{2},\dots,\beta_{n}$ are linearly independent, we have
$c_{1}=t_{j},c_{2}=t_{i}$ and

$\displaystyle\varphi(P_{\alpha_{1}+\alpha_{i}+\alpha_{j}})=P_{\beta_{1}+t_{i}\beta_{i}+t_{j}\beta_{j}}.$
(2.2)

On the other hand, $\alpha=\alpha_{1}+(\alpha_{i}+\alpha_{j})$. It follows
that

$\displaystyle\varphi(P_{\alpha})\leq\varphi(P_{\alpha_{1}})\vee\varphi(P_{\alpha_{i}+\alpha_{j}}).$
(2.3)

Clearly, $\varphi(P_{\alpha_{i}+\alpha_{j}})\leq P_{\beta_{i}}\vee
P_{\beta_{j}}$. It follows from (2.2) and (2.3) that
$\varphi(P_{\alpha_{i}+\alpha_{j}})=P_{t_{i}\beta_{i}+t_{j}\beta_{j}}$. ∎

By Remark 2.9 and Lemma 2.10, replacing $\beta_{i}$ by $t_{i}\beta_{i}$ for
every $i\in\\{2,3,\dots,n\\}$ if necessary, in the following of this section
we may assume that

$\varphi(P_{\alpha_{i}+\alpha_{j}})=P_{\beta_{i}+\beta_{j}},\ \forall\
i,j\in\\{1,2,\dots,n\\}.$

######  Remark 2.11.

Notice that for every $z\in\mathbb{C}$, $P_{\alpha_{1}+z\alpha_{2}}\leq
P_{\alpha_{1}}\vee P_{\alpha_{2}}$. Hence
$\varphi(P_{\alpha_{1}+z\alpha_{2}})\leq P_{\beta_{1}}\vee P_{\beta_{2}}$.
Thus we can define a map $f:\mathbb{C}\to\mathbb{C}$ with
$\varphi(P_{\alpha_{1}+z\alpha_{2}})=P_{\beta_{1}+f(z)\beta_{2}}$. It is easy
to verify that $f$ is a bijection with $f(1)=1,f(0)=0$.

######  Lemma 2.12.

For each $i\in\\{2,3,\dots,n\\}$ and $z\in\mathbb{C}$,
$\varphi(P_{\alpha_{1}+z\alpha_{i}})=P_{\beta_{1}+f(z)\beta_{i}}$.

###### Proof.

We only need to prove the result for $i>2$. Let
$\alpha=\alpha_{1}+z\alpha_{2}+z\alpha_{i}$. It follows that $P_{\alpha}\leq
P_{\alpha_{1}}\vee P_{\alpha_{2}+\alpha_{i}}$ and $P_{\alpha}\leq
P_{\alpha_{1}+z\alpha_{2}}\vee P_{\alpha_{i}}$. Therefore
$\varphi(P_{\alpha})\leq P_{\beta_{1}}\vee
P_{\beta_{2}+\beta_{i}},\varphi(P_{\alpha})\leq
P_{\beta_{1}+f(z)\beta_{2}}\vee P_{\beta_{i}}$ and there exist two constants
$c_{1},c_{2}$ such that $\beta_{1}+c_{1}(\beta_{2}+\beta_{i})$ and
$\beta_{1}+f(z)\beta_{2}+c_{2}\beta_{i}$ are both in the range of
$\varphi(P_{\alpha})$. Hence $c_{1}=c_{2}=f(z)$ and

$\displaystyle\varphi(P_{\alpha})=P_{\beta_{1}+f(z)\beta_{2}+f(z)\beta_{i}}.$
(2.4)

On the other hand, $\alpha=(\alpha_{1}+z\alpha_{i})+\alpha_{2}$ and thus
$\varphi(P_{\alpha})\leq\varphi(P_{\alpha_{1}+z\alpha_{i}})\vee
P_{\beta_{2}}$. By the fact that $\varphi(P_{\alpha_{1}+z\alpha_{i}})\leq
P_{\beta_{1}}\vee P_{\beta_{i}}$ and (2.4) we obtain that
$\varphi(P_{\alpha_{1}+z\alpha_{i}})=P_{\beta_{1}+f(z)\beta_{i}}$. ∎

######  Lemma 2.13.

The map $f$ given in Remark 2.11 is a ring automorphism of $\mathbb{C}$. That
is, $f$ is a bijection and
$f(z_{1}+z_{2})=f(z_{1})+f(z_{2}),f(z_{1}z_{2})=f(z_{1})f(z_{2})$ for any
$z_{1},z_{2}\in\mathbb{C}$.

###### Proof.

Clearly $f$ is a bijection and $f(0)=0$. We only need to prove the result for
the case that $z_{1},z_{2}$ are both nonzero.

Take
$\alpha=(\alpha_{1}+(z_{1}+z_{2})\alpha_{2})+\alpha_{3}=(\alpha_{1}+z_{1}\alpha_{2})+(\alpha_{3}+z_{2}\alpha_{2})=(\alpha_{1}+\alpha_{3})+(z_{1}+z_{2})\alpha_{2}$.
Then $\varphi(P_{\alpha})\leq P_{\beta_{1}+f(z_{1}+z_{2})\beta_{2}}\vee
P_{\beta_{3}}$, $\varphi(P_{\alpha})\leq P_{\beta_{1}+f(z_{1})\beta_{2}}\vee
P_{\beta_{3}+f(z_{2})\beta_{2}}$ and $\varphi(P_{\alpha})\leq
P_{\beta_{1}+\beta_{3}}\vee P_{\beta_{2}}$. A similar calculation as in the
proof of Lemma 2.10 gives that $f(z_{1}+z_{2})=f(z_{1})+f(z_{2})$.

Take $\xi=\alpha_{1}+z_{1}z_{2}\alpha_{2}+z_{1}\alpha_{3}$. It follows that
$P_{\xi}\leq P_{\alpha_{1}+z_{1}z_{2}\alpha_{2}}\vee
P_{\alpha_{3}},P_{\xi}\leq P_{\alpha_{1}}\vee P_{\alpha_{3}+z_{2}\alpha_{2}}$
and $P_{\xi}\leq P_{\alpha_{1}+z_{1}\alpha_{3}}\vee P_{\alpha_{2}}$. Hence we
have

$\varphi(P_{\xi})\leq P_{\beta_{1}+f(z_{1}z_{2})\beta_{2}}\vee
P_{\beta_{3}},\varphi(P_{\xi})\leq P_{\beta_{1}}\vee
P_{\beta_{3}+f(z_{2})\beta_{2}},\varphi(P_{\xi})\leq
P_{\beta_{1}+f(z_{1})\beta_{3}}\vee P_{\beta_{2}}.$

It follows that there exist constants $c_{1},c_{2},c_{3}\in\mathbb{C}$ such
that

$\beta_{1}+f(z_{1}z_{2})\beta_{2}+c_{1}\beta_{3},\beta_{1}+c_{2}(\beta_{3}+f(z_{2})\beta_{2}),\beta_{1}+f(z_{1})\beta_{3}+c_{3}\beta_{2}$

are vectors in the range of $\varphi(P_{\xi})$. By the fact that
$\beta_{1},\beta_{2},\beta_{3}$ are linearly independent, we have

$\beta_{1}+f(z_{1}z_{2})\beta_{2}+c_{1}\beta_{3}=\beta_{1}+c_{2}(\beta_{3}+f(z_{2})\beta_{2})=\beta_{1}+f(z_{1})\beta_{3}+c_{3}\beta_{2},$

which implies that $f(z_{1}z_{2})=f(z_{1})f(z_{2})$. ∎

######  Corollary 2.14.

For any two distinguished numbers $i,j\in\\{1,2,\dots,n\\}$,

$\varphi(P_{\alpha_{i}+z\alpha_{j}})=P_{\beta_{i}+f(z)\beta_{j}}.$

###### Proof.

We may assume that $i\neq 1$ and $z\neq 0$. If $j=1$, then
$\varphi(P_{\alpha_{i}+z\alpha_{j}})=\varphi(P_{\alpha_{1}+\frac{1}{z}\alpha_{i}})=P_{\beta_{1}+f(\frac{1}{z})\beta_{i}}$.
By Lemma 2.13 we obtain that $f(\frac{1}{z})=\frac{1}{f(z)}$. Hence
$P_{\beta_{1}+f(\frac{1}{z})\beta_{i}}=P_{\beta_{i}+z\beta_{1}}$.

Assume that $i,j$ are distinguished numbers in $\\{2,3,\dots,n\\}$. Clearly
$\varphi(P_{\alpha_{i}+z\alpha_{j}})\leq P_{\beta_{i}}\vee P_{\beta_{j}}$.
Take
$\alpha=(\alpha_{1}+z\alpha_{j})+\alpha_{i}=(\alpha_{1}+\alpha_{i})+z\alpha_{j}=(\alpha_{1}+\alpha_{i}+z\alpha_{j})$.
Combining $\varphi(P_{\alpha})\leq P_{\beta_{1}+f(z)\beta_{j}}\vee
P_{\beta_{i}},\varphi(P_{\alpha})\leq P_{\beta_{1}+\beta_{i}}\vee
P_{\beta_{j}}$ with $\varphi(P_{\alpha})\leq
P_{\beta_{1}}\vee\varphi(P_{\alpha_{i}+z\alpha_{j}})$, we obtain the required
result. ∎

By Corollary 2.14, a similar argument as in the proof of Lemma 2.10 and Lemma
2.12 yields the following result(a mathematical induction is needed). We omit
its proof.

######  Lemma 2.15.

For any $z_{2},z_{3},\dots,z_{n}\in\mathbb{C}$ and any perturbation
$i_{1},i_{2},\dots,i_{n}$ of $1,2,\dots,n$,

$\varphi(P_{\alpha_{i_{1}}+z_{2}\alpha_{i_{2}}+\dots+z_{n}\alpha_{i_{n}}})=P_{\beta_{i_{1}}+f(z_{2})\beta_{i_{2}}+\dots+f(z_{n})\beta_{i_{n}}}.$

By Lemma 2.13 and Lemma 2.15, we have the following corollary directly.

######  Corollary 2.16.

For any $z_{1},z_{2},z_{3},\dots,z_{n}\in\mathbb{C}$ which are not all zero,

$\varphi(P_{z_{1}\alpha_{1}+z_{2}\alpha_{2}+\dots+z_{n}\alpha_{n}})=P_{f(z_{1})\beta_{1}+f(z_{2})\beta_{2}+\dots+f(z_{n})\beta_{n}}.$

######  Definition 2.17.

Let $\alpha_{1},\alpha_{2},\dots,\alpha_{n}$ be an orthonormal basis and
$\beta_{1},\beta_{2},\dots,\beta_{n}$ a basis for $\mathcal{H}$. For any ring
automorphism $f:\mathbb{C}\to\mathbb{C}$, define a map
$\hat{f}:\mathcal{H}\to\mathcal{H}$ (depending on
$\alpha_{1},\alpha_{2},\dots,\alpha_{n}$ and
$\beta_{1},\beta_{2},\dots,\beta_{n}$) as follows:

$\hat{f}(z_{1}\alpha_{1}+z_{2}\alpha_{2}+\dots+z_{n}\alpha_{n})=f(z_{1})\beta_{1}+f(z_{2})\beta_{2}+\dots+f(z_{n})\beta_{n}.$

######  Remark 2.18.

Given an orthonormal basis $\alpha_{1},\alpha_{2},\dots,\alpha_{n}$ and a
basis $\beta_{1},\beta_{2},\dots,\beta_{n}$ for $\mathcal{H}$, we define a map
$\psi:\mathcal{P}(\mathcal{H})\to\mathcal{P}(\mathcal{H})$ by
$\psi(P)(\mathcal{H})=\\{\hat{f}(P\xi):\xi\in\mathcal{H}\\}$, where $\hat{f}$
is defined in Definition 2.17. It is clear that $\psi$ is bijective with

$\psi(P\vee Q)=\psi(P)\vee\psi(Q),\psi(P\wedge Q)=\psi(P)\wedge\psi(Q)$

for any $P,Q\in\mathcal{P}(\mathcal{H})$. Notice that for any positive integer
$k$ and any matrix $(a_{ij})_{k\times k}\in M_{k}(\mathbb{C})$,
$det((f(a_{ij})))=f(det((a_{ij})))$. Therefore $det(f(a_{ij}))\neq 0$ if and
only if $det(a_{ij})\neq 0$. Hence $r(\psi(P))=r(P)$ for any
$P\in\mathcal{P}(\mathcal{H})$. Therefore $\psi(P)\vee\psi(Q)=I$ if and only
if $P\vee Q=I$ and, $\psi(P)\wedge\psi(Q)=0$ if and only if $P\wedge Q=0$. We
obtain that $\psi$ is joint spectrum preserving for any two projections.

Proof of Theorem 1.1. Theorem 2.8 gives that $(1)$ and $(2)$ are equivalent.
$(3)\Rightarrow(2)$ follows directly from Remark 2.18. Now we only need to
show that $(2)\Rightarrow(3)$. Assume that $\varphi$ preserves the joint
spectrum of any two projections. It follows from Corollary 2.16 and the
definition of $\hat{f}$ that
$\varphi(P)=\\{\hat{f}(P\xi):\xi\in\mathcal{H}\\},\forall\
P\in\mathcal{P}_{1}(\mathcal{H})$. By the fact that
$\hat{f}(\eta_{1}+\eta_{2})=\hat{f}(\eta_{1})+\hat{f}(\eta_{2})$ for any
$\eta_{1},\eta_{2}\in\mathcal{H}$ and $\varphi(P\vee
Q)=\varphi(P)\vee\varphi(Q)$, we obtain that
$\varphi(P)=\\{\hat{f}(P\xi):\xi\in\mathcal{H}\\},\forall\
P\in\mathcal{P}(\mathcal{H})$.

##  3\. Maps shrinking the joint spectrum of more than two projections

Assume that $n\geq 3$. In this section we look at a surjective map $\varphi$
on $\mathcal{P}(\mathcal{H})$ which shrinks the joint spectrum of any $k$
projections, where $k\geq 3$. It is easy to verify that $\varphi$ also shrinks
the joint spectrum of any $2$ projections and thus $\varphi$ is also induced
by a ring automorphism $f$ on $\mathbb{C}$ as in Theorem 1.1.

######  Lemma 3.1.

Assume that $\varphi:\mathcal{P}(\mathcal{H})\to\mathcal{P}(\mathcal{H})$ is a
surjective map which shrinks the joint spectrum of any $3$ projections. Then
$\varphi$ preserves the orthogonality, i.e., if $PQ=0$, then
$\varphi(P)\varphi(Q)=0$. In particular, the vectors
$\beta_{1},\beta_{2},\ldots,\beta_{n}$ in Theorem 1.1 are mutually orthogonal
in $\mathcal{H}$.

###### Proof.

By way of contradiction, we assume that there exist two nonzero projections
$P,Q$ on $\mathcal{H}$ with $PQ=0$ and $\varphi(P)\varphi(Q)\neq 0$. By the
arguments in Section 2 we have $\varphi$ preserves the order of projections.
Replacing $P$ by $I-Q$ if necessary, we may assume that $P+Q=I$ and
$\varphi(P)\varphi(Q)\neq 0$. It follows that there exist a unit vector
$\xi\in\varphi(P)(\mathcal{H})$ such that $\varphi(Q)\xi\neq 0$. Let
$c=\|\varphi(Q)\xi\|>0$. It is easy to verify that

$\|\xi+\varphi(Q)\xi\|^{2}=1+3c^{2},\langle\xi,\xi+\varphi(Q)\xi\rangle=1+c^{2}.$

Take a rank one projection $R$ with $\xi+\varphi(Q)\xi$ in its range. It
follows that $R\xi=\frac{1+c^{2}}{1+3c^{2}}(\xi+\varphi(Q)\xi)$. Hence we
obtain that $(\varphi(P)+\varphi(Q)-\frac{1+3c^{2}}{1+c^{2}}R)\xi=0$ and thus
$\varphi(P)+\varphi(Q)-\frac{1+3c^{2}}{1+c^{2}}R$ is not invertible. Therefore
$(1,1,-\frac{1+3c^{2}}{1+c^{2}})\in\sigma([\varphi(P),\varphi(Q),R])$.
However, $\frac{1+3c^{2}}{1+c^{2}}>1$,
$P+Q-\frac{1+3c^{2}}{1+c^{2}}\varphi^{-1}(R)=I-\frac{1+3c^{2}}{1+c^{2}}\varphi^{-1}(R)$
is invertible and
$(1,1,-\frac{1+3c^{2}}{1+c^{2}})\notin\sigma([P,Q,\varphi^{-1}(R)])$. We
obtain a contradiction.

It is easy to verify that $\beta_{1},\beta_{2},\ldots,\beta_{n}$ in Theorem
1.1 are mutually orthogonal in $\mathcal{H}$. ∎

######  Remark 3.2.

Notice that for each $i\in\\{2,\ldots,n\\}$, the projections
$P_{\alpha_{1}+\alpha_{i}}$ and $P_{\alpha_{1}-\alpha_{i}}$ are orthogonal.
Thus $\varphi(P_{\alpha_{1}+\alpha_{i}})$ and
$\varphi(P_{\alpha_{1}-\alpha_{i}})$ are orthogonal. Then we have
$\langle\beta_{1}+\beta_{i},\beta_{1}-\beta_{i}\rangle=0$, which implies that
$\|\beta_{1}\|=\|\beta_{i}\|$ for any $i\in\\{2,\ldots,n\\}$. Thus replacing
each $\beta_{i}$ by $\frac{\beta_{i}}{\|\beta_{1}\|}$ for $i=1,2,\ldots,n$, we
may assume that $\beta_{1},\beta_{2},\ldots,\beta_{n}$ form an orthonormal
basis for $\mathcal{H}$.

######  Lemma 3.3.

Assume that $\varphi:\mathcal{P}(\mathcal{H})\to\mathcal{P}(\mathcal{H})$ is a
surjective map which shrinks the joint spectrum of any $3$ projections. Then
either

$f(z)=z,\ \forall\ z\in\mathbb{C}$

or

$f(z)=\bar{z},\ \forall\ z\in\mathbb{C},$

where $f$ is the map defined in Section 2.

###### Proof.

Notice that $f(i)^{2}=f(i^{2})=f(-1)=-1$. We have either $f(i)=i$ or
$f(i)=-i$. In the following we only need to show that $f(t)=t$ for all
$t\in\mathbb{R}$. We first assume that $t\geq 0$. Let $P_{1},P_{2}$ be rank
one projections such that $t\alpha_{1}+\sqrt{t}\alpha_{2}\in
P_{1}(\mathcal{H})$ and $\alpha_{1}-\sqrt{t}\alpha_{2}\in P_{2}(\mathcal{H})$.
By Theorem 1.1, we have
$f(t)\beta_{1}+f(\sqrt{t})\beta_{2}\in\varphi(P_{1})(\mathcal{H})$ and
$\beta_{1}-f(\sqrt{t})\beta_{2}\in\varphi(P_{2})(\mathcal{H})$. Notice that
$P_{1}P_{2}=0$. By Lemma 3.1, $\varphi(P)\varphi(Q)=0$. Hence $\langle
f(t)\beta_{1}+f(\sqrt{t})\beta_{2},\beta_{1}-f(\sqrt{t})\beta_{2}\rangle=0$.
Therefore

$f(t)=|f(\sqrt{t})|^{2}\geq 0.$

Assume that $s_{1}<t<s_{2}$, where $s_{1},s_{2}$ are rational numbers. Note
that it is easy to verify that $f(s)=s$ for every rational number
$s\in\mathbb{R}$. Then

$s_{2}-f(t)=f(s_{2})-f(t)=f(s_{2}-t)\geq
0,f(t)-s_{1}=f(t)-f(s_{1})=f(t-s_{1})\geq 0$

and therefore $s_{1}\leq f(t)\leq s_{2}$. Hence we have $f(t)=t$ for all
$t\in\mathbb{R}$ and we obtain that (1) if $f(i)=i$, then $f(z)=z$ for all
$z\in\mathbb{C}$ and (2) if $f(i)=-i$, then $f(z)=\bar{z}$ for all
$z\in\mathbb{C}$. ∎

Proof of Theorem 1.2. It is clear that
$(5)\Rightarrow(4)\Rightarrow(3)\Rightarrow(1)$ and
$(4)\Rightarrow(2)\Rightarrow(1)$. We only need to show that
$(1)\Rightarrow(5)$. It follows from $(1)$ that $\varphi$ shrinks the joint
spectrum of any $3$ projections. Hence we may assume
$\alpha_{1},\alpha_{2},\dots,\alpha_{n}$ and
$\beta_{1},\beta_{2},\dots,\beta_{n}$ in Theorem 1.1 are two orthonormal basis
for $\mathcal{H}$. By Lemma 3.3 we have either

$f(z)=z,\ \forall\ z\in\mathbb{C}$

or

$f(z)=\bar{z},\ \forall\ z\in\mathbb{C},$

where $f$ is the map defined in Remark 2.11. If $f(z)=z$ for all
$z\in\mathbb{C}$, then we define a unitary $U$ by
$U(z_{1}\beta_{1}+z_{2}\beta_{2}+\dots+z_{n}\beta_{n})=z_{1}\alpha_{1}+z_{2}\alpha_{2}+\dots+z_{n}\alpha_{n}$.
If $f(z)=\bar{z}$ for all $z\in\mathbb{C}$, then we define an anti-unitary $U$
by
$U(z_{1}\beta_{1}+z_{2}\beta_{2}+\dots+z_{n}\beta_{n})=\bar{z_{1}}\alpha_{1}+\bar{z_{2}}\alpha_{2}+\dots+\bar{z_{n}}\alpha_{n}$.
It follows from Theorem 1.1 that $\varphi(P)=U^{*}PU$ in both cases.

##  4\. Joint spectrum shrinking maps on rank one projections

Assume that $n\geq 3$. In this section we assume that
$\phi:\mathcal{P}_{1}(\mathcal{H})\to\mathcal{P}_{1}(\mathcal{H})$ is a
surjective map. It is easy to verify that for any positive integer $m<n$, the
joint spectrum of any $m$ rank one projections $P_{1},P_{2},\dots,P_{m}$ is
$\mathbb{C}^{m}$. Therefore every map on $\mathcal{P}_{1}(\mathcal{H})$
preserves the joint spectrum of any $m$ rank one projections if $m<n$.

###  4.1. Maps preserving the joint spectrum of any $n$ rank one projections

We start with a description of the joint spectrum of $n$ rank one projections.

######  Lemma 4.1.

Let $P_{1},P_{2},\dots,P_{n}\in\mathcal{P}_{1}(\mathcal{H})$. Then

  1. (1)

if $P_{1}\vee P_{2}\vee\dots\vee P_{n}\neq I$ , then
$\sigma([P_{1},P_{2},\dots,P_{n}])=\mathbb{C}^{n}$;

  2. (2)

if $P_{1}\vee P_{2}\vee\dots\vee P_{n}=I$, then
$\sigma([P_{1},P_{2},\dots,P_{n}])=\\{(c_{1},c_{2},\dots,c_{n})\in\mathbb{C}^{n}:c_{1}c_{2}\dots
c_{n}=0\\}$.

###### Proof.

If $P_{1}\vee P_{2}\vee\dots\vee P_{n}\neq I$, then the range of any linear
combination of $P_{1},P_{2},\dots,P_{n}$ is contained in the range of
$P_{1}\vee P_{2}\vee\dots\vee P_{n}$ and thus any linear combination of
$P_{1},P_{2},\dots,P_{n}$ is not invertible. Therefore
$\sigma([P_{1},P_{2},\dots,P_{n}])=\mathbb{C}^{n}$.

On the other hand, assume that $P_{1}\vee P_{2}\vee\dots\vee P_{n}=I$ and
$c_{1}P_{1}+c_{2}P_{2}+\dots+c_{n}P_{n}$ is not invertible. Then there exists
a nonzero vector $\beta\in\mathcal{H}$ such that
$c_{1}P_{1}\beta+c_{2}P_{2}\beta+\dots+c_{n}P_{n}\beta=0$. Hence
$c_{i}P_{i}\beta=-c_{1}P_{1}\beta-\dots-c_{i-1}P_{i-1}\beta-
c_{i+1}P_{i+1}\beta-\dots-c_{n}P_{n}\beta=0$. By the fact that $P_{1}\vee
P_{2}\vee\dots\vee P_{n}=I$ we have $P_{i}\wedge(P_{1}\vee\dots\vee
P_{i-1}\vee P_{i+1}\vee\dots\vee P_{n})=0$ for each $i\in\\{1,2,\dots,n\\}$.
If $c_{1}c_{2}\dots c_{n}\neq 0$, then
$P_{1}\beta=P_{2}\beta=\dots=P_{n}\beta=0$, which is a contradiction to that
$P_{1}\vee P_{2}\vee\dots\vee P_{n}=I$ and $\beta\neq 0$. Therefore
$\sigma([P_{1},P_{2},\dots,P_{n}])\subseteq\\{(c_{1},c_{2},\dots,c_{n})\in\mathbb{C}^{n}:c_{1}c_{2}\dots
c_{n}=0\\}$. It is obvious that
$\\{(c_{1},c_{2},\dots,c_{n})\in\mathbb{C}^{n}:c_{1}c_{2}\dots
c_{n}=0\\}\subseteq\sigma([P_{1},P_{2},\dots,P_{n}])$. ∎

In order to prove the main result of this subsection, we will extend $\phi$ to
a bijective map on $\mathcal{P}(\mathcal{H})$ which preserves the joint
spectrum of any two projections. We give some necessary lemmas.

######  Lemma 4.2.

Assume that $\phi:\mathcal{P}_{1}(\mathcal{H})\to\mathcal{P}_{1}(\mathcal{H})$
is a surjective map which preserves the joint spectrum of any $n$ rank one
projections. Then for any $1\leq k\leq n$ and
$P_{1},P_{2},\dots,P_{k}\in\mathcal{P}_{1}(\mathcal{H})$, $r(P_{1}\vee
P_{2}\vee\dots\vee P_{k})=k$ if and only if
$r(\phi(P_{1})\vee\phi(P_{2})\vee\dots\vee\phi(P_{k}))=k$. In particular,
$\phi$ is injective.

###### Proof.

We first prove the “only if” part. Assume that $r(P_{1}\vee P_{2}\vee\dots\vee
P_{k})=k$. Then there exist $n-k$ rank one projections
$P_{k+1},P_{k+2},\dots,P_{n}$ with $P_{1}\vee P_{2}\vee\dots\vee P_{n}=I$.
Since $\phi$ preserves the joint spectrum of any $n$ rank one projections, it
follows from Lemma 4.1 that
$r(\phi(P_{1})\vee\phi(P_{2})\vee\dots\vee\phi(P_{n}))=n$. Hence
$r(\phi(P_{1})\vee\phi(P_{2})\vee\dots\vee\phi(P_{k}))=k$. If $P,Q$ are two
distinguished rank one projections, then $r(P\vee Q)=2$. It follows that
$r(\phi(P)\vee\phi(Q))=2$ and hence $\phi$ is injective. The other direction
follows directly from the fact that $\phi^{-1}$ also preserves the joint
spectrum of any rank one projections. ∎

######  Lemma 4.3.

Assume that $\phi:\mathcal{P}_{1}(\mathcal{H})\to\mathcal{P}_{1}(\mathcal{H})$
is a surjective map which preserves the joint spectrum of any $n$ rank one
projections. Then for any $1\leq k\leq n$ and
$P_{1},P_{2},\dots,P_{k}\in\mathcal{P}_{1}(\mathcal{H})$,
$r(\phi(P_{1})\vee\phi(P_{2})\dots\vee\phi(P_{k}))=r(P_{1}\vee P_{2}\dots\vee
P_{k})$.

###### Proof.

Denote $s=r(P_{1}\vee P_{2}\vee\dots\vee P_{k})$. It follows easily from Lemma
4.2 that $\phi$ is invertible on $\mathcal{P}_{1}(H)$ and $\phi^{-1}$ also
preserves the joint spectrum of any tuple of rank one projections. Hence we
only need to show that
$r(\phi(P_{1})\vee\phi(P_{2})\vee\dots\vee\phi(P_{k}))\geq s$. Since
$s=r(P_{1}\vee P_{2}\vee\dots\vee P_{k})$, there exist
$i_{1},i_{2},\dots,i_{s}\in\\{1,2,\dots,k\\}$ such that $r(P_{i_{1}}\vee
P_{i_{2}}\vee\dots\vee P_{i_{s}})=s$. By Lemma 4.2,
$r(\phi(P_{i_{1}})\vee\phi(P_{i_{2}})\vee\dots\vee\phi(P_{i_{s}}))=s$.
Therefore $r(\phi(P_{1})\vee\phi(P_{2})\vee\dots\vee\phi(P_{k}))\geq s$. ∎

######  Lemma 4.4.

Assume that $\phi:\mathcal{P}_{1}(\mathcal{H})\to\mathcal{P}_{1}(\mathcal{H})$
is a surjective map which preserves the joint spectrum of any $n$ rank one
projections. For any $P,Q_{1},Q_{2},\dots,Q_{k}$ in
$\mathcal{P}_{1}(\mathcal{H})$, $P\leq Q_{1}\vee Q_{2}\vee\dots\vee Q_{k}$ if
and only if $\phi(P)\leq\phi(Q_{1})\vee\phi(Q_{2})\vee\dots\vee\phi(Q_{k})$.
In particular, if $E_{1},E_{2},\dots,E_{l};F_{1},F_{2},\dots
F_{m}\in\mathcal{P}_{1}(\mathcal{H})$, then $E_{1}\vee E_{2}\vee\dots\vee
E_{l}=F_{1}\vee F_{2}\vee\dots\vee F_{m}$ if and only if
$\phi(E_{1})\vee\phi(E_{2})\vee\dots\vee\phi(E_{l})=\phi(F_{1})\vee\phi(F_{2})\vee\dots\vee\phi(F_{m})$.

###### Proof.

Assume that $P\leq Q_{1}\vee Q_{2}\vee\dots\vee Q_{k}$. Then $r(P\vee
Q_{1}\vee\dots\vee Q_{k})=r(Q_{1}\vee\dots\vee Q_{k})$. It follows from Lemma
4.3 that

$r(\phi(P)\vee\phi(Q_{1})\vee\dots\vee\phi(Q_{k}))=r(\phi(Q_{1})\vee\dots\vee\phi(Q_{k})),$

which implies that
$\phi(P)\leq\phi(Q_{1})\vee\phi(Q_{2})\vee\dots\vee\phi(Q_{k})$. The other
direction is similar from Lemma 4.3.

If $E_{1}\vee E_{2}\vee\dots\vee E_{l}=F_{1}\vee F_{2}\vee\dots\vee F_{m}$,
then the previous argument implies that
$\phi(E_{i})\leq\phi(F_{1})\vee\phi(F_{2})\vee\dots\vee\phi(F_{m})$ for each
$i\in\\{1,2,\dots,m\\}$ and
$\phi(F_{j})\leq\phi(E_{1})\vee\phi(E_{2})\vee\dots\vee\phi(E_{l})$ for each
$j\in\\{1,2,\dots,m\\}$. Hence
$\phi(E_{1})\vee\phi(E_{2})\vee\dots\vee\phi(E_{l})=\phi(F_{1})\vee\phi(F_{2})\vee\dots\vee\phi(F_{m})$.
The other direction is also similar. ∎

Proof of Theorem 1.3. $(2)\Rightarrow(1):$ We naturally extend $\phi$ to a
bijective map $\psi:\mathcal{P}(\mathcal{H})\to\mathcal{P}(\mathcal{H})$ by

$\psi(P)(\mathcal{H})=\\{\hat{f}(P\xi):\xi\in\mathcal{H}\\},\forall\
P\in\mathcal{P}(\mathcal{H}).$

Clearly $\psi$ is bijective and it follows from Theorem 1.1 that $\psi$ is
joint spectrum preserving for any two projections. Hence by the arguments in
Section 2 we have that $\psi(P\vee Q)=\psi(P)\vee\psi(Q)$ for any
$P,Q\in\mathcal{P}(\mathcal{H})$. Clearly $\psi(I)=I$. It follows that for any
$n$ rank one projections $P_{1},P_{2},\dots,P_{n}\in\mathcal{P}(\mathcal{H})$,
$P_{1}\vee P_{2}\vee\dots\vee P_{n}=I$ if and only if
$\psi(P_{1})\vee\psi(P_{2})\vee\dots\vee\psi(P_{n})=I$. By Lemma 4.1, $\phi$
preserves the joint spectrum of any $n$ rank one projections.

$(1)\Rightarrow(2):$ We define a map
$\Psi:\mathcal{P}(\mathcal{H})\to\mathcal{P}(\mathcal{H})$ as follows: if
$P=P_{1}\vee P_{2}\vee\dots\vee P_{k}$ with
$P_{1},P_{2},\dots,P_{k}\in\mathcal{P}_{1}(\mathcal{H})$, then define

$\Psi(P)=\phi(P_{1})\vee\phi(P_{2})\vee\dots\vee\phi(P_{k})$

and let $\Psi(0)=0$. By Lemma 4.4, $\Psi$ is well-defined and injective. Fix
$R\in\mathcal{P}(\mathcal{H})$ and let $s=r(R)$. Then there exist $s$ rank one
projections $R_{1},R_{2},\dots,R_{s}\in\mathcal{P}_{1}(\mathcal{H})$ such that
$R=R_{1}\vee R_{2}\vee\dots\vee R_{s}$. Note that $\phi$ is bijective. Take
$R^{\prime}=\phi^{-1}(R_{1})\vee\phi^{-1}(R_{2})\vee\dots\vee\phi^{-1}(R_{s})$
and it follows that $\Psi(R^{\prime})=R$. Therefore $\Psi$ is bijective. To
show $\Psi$ is joint spectrum preserving for any two projections, we only need
to prove (1) for any $P,Q\in\mathcal{P}(\mathcal{H})$, $P\vee Q=I$ if and only
if $\Psi(P)\vee\Psi(Q)=I$ and (2) for any $P,Q\in\mathcal{P}(\mathcal{H})$,
$P\vee Q=I,P\wedge Q=0$ if and only if
$\Psi(P)\vee\Psi(Q)=I,\Psi(P)\wedge\Psi(Q)=0$. By Lemma 4.2 and the definition
of $\Psi$, the proof of (1) is obvious. Now assume $P\vee Q=I,P\wedge Q=0$. It
follows that $r(P)+r(Q)=n$ and $\Psi(P)\vee\Psi(Q)=I$. By Lemma 4.3 we have
$r(\Psi(P))=r(P),r(\Psi(Q))=r(Q)$. Hence $r(\Psi(P))+r(\Psi(Q))=n$. Therefore
$\Psi(P)\wedge\Psi(Q)=0$. The rest follows directly from Theorem 1.1.

###  4.2. Maps shrinking the joint spectrum of more than $n$ rank one
projections

Now we assume that
$\phi:\mathcal{P}_{1}(\mathcal{H})\to\mathcal{P}_{1}(\mathcal{H})$ is a
surjective map which shrinks the joint spectrum of $n+1$ projections. Notice
that $\phi$ also shrinks the joint spectrum of any $n$ rank one projections. A
similar argument as in Lemma 4.2 gives that $\phi$ is bijective on
$\mathcal{P}_{1}(\mathcal{H})$. We follow a similar line as in Section 3 to
show that $\phi$ preserves the orthogonality.

######  Lemma 4.5.

Assume that $\phi:\mathcal{P}_{1}(\mathcal{H})\to\mathcal{P}_{1}(\mathcal{H})$
is a surjective map which shrinks the joint spectrum of any $n+1$ projections.
Then $\phi$ preserves the orthogonality.

###### Proof.

By way of contradiction, assume that $P,Q\in\mathcal{P}_{1}(\mathcal{H})$ such
that $PQ=0$ and $\phi(P)\phi(Q)\neq 0$. Take a unit vector
$\xi\in\phi(P)\mathcal{H}$ such that $\phi(Q)\xi\neq 0$. Take a rank one
projection $R$ with $\xi+\phi(Q)\xi$ in its range. Let $c=\|\phi(Q)\xi\|>0$.
It follows that

$\displaystyle(\phi(P)+\phi(Q)-\frac{1+3c^{2}}{1+c^{2}}R)\xi=0.$ (4.1)

Notice that $R\leq\phi(P)\vee\phi(Q)$. We have that
$Ran(\phi(P)+\phi(Q)-\frac{1+3c^{2}}{1+c^{2}}R)\leq\phi(P)\vee\phi(Q)$, where
$Ran(\phi(P)+\phi(Q)-\frac{1+3c^{2}}{1+c^{2}}R)$ denotes the range projection
of $\phi(P)+\phi(Q)-\frac{1+3c^{2}}{1+c^{2}}R$. It follows from (4.1) that
$r(Ran(\phi(P)+\phi(Q)-\frac{1+3c^{2}}{1+c^{2}}R))=1$. Take
$P_{3},P_{4},\dots,P_{n}\in\mathcal{P}_{1}(\mathcal{H})$ such that
$P+Q+P_{3}+\dots+P_{n}=I$. Since $\frac{1+3c^{2}}{1+c^{2}}>1$,
$P+Q-\frac{1+3c^{2}}{1+c^{2}}\phi^{-1}(R)+P_{3}+\dots+P_{n}=I-\frac{1+3c^{2}}{1+c^{2}}\phi^{-1}(R)$
is invertible. Since $r(Ran(\phi(P)+\phi(Q)-\frac{1+3c^{2}}{1+c^{2}}R))=1$,
$r(Ran(\phi(P)+\phi(Q)-\frac{1+3c^{2}}{1+c^{2}}R)+\phi(P_{3})+\dots+\phi(P_{n})))\leq
n-1$ and thus
$\phi(P)+\phi(Q)-\frac{1+3c^{2}}{1+c^{2}}R+\phi(P_{3})+\dots+\phi(P_{n})$ is
not invertible. We obtain a contradiction. ∎

######  Lemma 4.6.

Assume that $Q_{1},Q_{2},\dots,Q_{k}$ are mutually orthogonal projections in
$\mathcal{P}_{1}(\mathcal{H})$ and $P\in\mathcal{P}_{1}(\mathcal{H})$ with
$P\leq Q_{1}+Q_{2}+\dots+Q_{k}$. Then
$\phi(P)\leq\phi(Q_{1})+\phi(Q_{2})+\dots+\phi(Q_{k})$. Moreover, if
$R_{1},R_{2},\dots,R_{k}$ are mutually orthogonal with
$R_{1}+R_{2}+\dots+R_{k}=Q_{1}+Q_{2}+\dots+Q_{k}$, then
$\phi(R_{1})+\phi(R_{2})+\dots+\phi(R_{k})=\phi(Q_{1})+\phi(Q_{2})+\dots+\phi(Q_{k})$.

###### Proof.

Take $Q_{k+1},Q_{k+2},\dots,Q_{n}\in\mathcal{P}_{1}(\mathcal{H})$ with
$Q_{1}+Q_{2}+\dots+Q_{n}=I$. It follows from Lemma 4.5 that
$\phi(Q_{1}),\dots,\phi(Q_{k}),\phi(Q_{k+1}),\dots,\phi(Q_{n})$ are mutually
orthogonal with sum $I$. Notice that for each $i\in\\{k+1,k+2,\dots,n\\}$,
$PQ_{i}=0$. By Lemma 4.5 again, $\phi(P)\phi(Q_{i})=0,\ \forall\
i\in\\{k+1,k+2,\dots,n\\}$. Hence
$\phi(P)\leq\phi(Q_{1})+\phi(Q_{2})+\dots+\phi(Q_{k})$. If
$R_{1},R_{2},\dots,R_{k}$ are mutually orthogonal with
$R_{1}+R_{2}+\dots+R_{k}=Q_{1}+Q_{2}+\dots+Q_{k}$, then we get
$\phi(R_{i})\leq\phi(Q_{1})+\phi(Q_{2})+\dots+\phi(Q_{k})$ and
$\phi(Q_{i})\leq\phi(R_{1})+\phi(R_{2})+\dots+\phi(R_{k})$ for each
$i\in\\{1,2,\dots,k\\}$. Hence
$\phi(R_{1})+\phi(R_{2})+\dots+\phi(R_{k})=\phi(Q_{1})+\phi(Q_{2})+\dots+\phi(Q_{k})$.
∎

Now we can get the main result of this subsection.

Proof of Theorem 1.4. It is clear that
$(5)\Rightarrow(4)\Rightarrow(3)\Rightarrow(1)$ and
$(4)\Rightarrow(2)\Rightarrow(1)$. In the following we only need to verify
$(1)\Rightarrow(5)$.

We define a map $\Psi:\mathcal{P}(\mathcal{H})\to\mathcal{P}(\mathcal{H})$ as
follows, $\Psi(0)=0$ and $\Psi(P)=\phi(P_{1})+\phi(P_{2})+\cdots+\phi(P_{k})$
when $P\in\mathcal{P}(\mathcal{H})$ and $P=P_{1}+P_{2}+\dots+P_{k}$ with
$P_{1},P_{2},\dots,P_{k}\in\mathcal{P}_{1}(\mathcal{H})$. By Lemma 4.5 and
Lemma 4.6, $\Psi$ is well defined such that $r(\Psi(P))=r(P)$ for any
$P\in\mathcal{P}(\mathcal{H})$. Since $\phi$ shrinks the joint spectrum of any
$n+1$ rank one projections, it also shrinks the joint spectrum of any $n$ rank
one projections. A similar argument as in the proofs of Lemma 4.2 and Lemma
4.3 can also imply $r(\phi(Q_{1})\vee\phi(Q_{2})\vee\dots\vee\phi(Q_{s}))\geq
r(Q_{1}\vee Q_{2}\vee\dots\vee Q_{s}),\ \forall\
Q_{1},Q_{2},\dots,Q_{s}\in\mathcal{P}_{1}(\mathcal{H})$. Hence if $P=E_{1}\vee
E_{2}\vee\dots\vee E_{l}$ with
$E_{1},E_{2},\dots,E_{l}\in\mathcal{P}_{1}(\mathcal{H})$, then

$\Psi(P)=\phi(E_{1})\vee\phi(E_{2})\vee\dots\vee\phi(E_{l}),$

which implies that $\Psi(P\vee Q)=\Psi(P)\vee\Psi(Q)$. A similar argument as
in the proof of Theorem 1.3 gives that $\Psi$ is bijective and preserves the
joint spectrum of any two projections. The rest follows from the same line as
in the proofs of Lemma 3.1 and Theorem 1.2.

######  Remark 4.7.

Notice that the method used in this subsection can not be applied to a
surjective map on $\mathcal{P}_{1}(\mathcal{H})$ which is joint spectrum
shrinking for any $n$ rank one projections. Indeed, even if a surjective map
on $\mathcal{P}_{1}(\mathcal{H})$ is joint spectrum preserving for any $n$
rank one projections, we can take $\beta_{1},\beta_{2},\dots,\beta_{n}$ in
Theorem 1.3 to be non-orthonormal so that $\phi$ will not preserve the
orthogonality. A further question is, for a surjective map on
$\mathcal{P}_{1}(\mathcal{H})$, whether joint spectrum shrinking for any $n$
rank one projections implies joint spectrum preserving for any $n$ rank one
projections.

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