On the Nonexistence of Nontrivial Involutive n-Homomorphisms of C*-algebras

TL;DR

This paper proves that *-preserving n-homomorphisms between C*-algebras are necessarily trivial or decomposable into simpler *-homomorphisms, resolving a question about their continuity and structure.

Contribution

It establishes that all *-preserving n-homomorphisms of C*-algebras are either ordinary *-homomorphisms or differences of two orthogonal *-homomorphisms, showing nonexistence of nontrivial cases.

Findings

All *-preserving n-homomorphisms are continuous.

For even n > 2, such maps are *-homomorphisms.

For odd n >= 3, such maps are differences of two orthogonal *-homomorphisms.

Abstract

An n-homomorphism between algebras is a linear map $\phi : A \to B$ such that $\phi(a_1 ... a_n) = \phi(a_1)... \phi(a_n)$ for all elements $a_1, >..., a_n \in A.$ Every homomorphism is an n-homomorphism, for all n >= 2, but the converse is false, in general. Hejazian et al. [7] ask: Is every *-preserving n-homomorphism between C*-algebras continuous? We answer their question in the affirmative, but the even and odd n arguments are surprisingly disjoint. We then use these results to prove stronger ones: If n >2 is even, then $\phi$ is just an ordinary *-homomorphism. If n >= 3 is odd, then $\phi$ is a difference of two orthogonal *-homomorphisms. Thus, there are no nontrivial *-linear n-homomorphisms between C*-algebras.

Equations

$$\varphi (a_1{a}_2\cdots a_n)=\varphi (a_1)\varphi (a_2)\cdots \varphi (a_n).$$

$$\sum _{i,j=1}^m⟨\varphi (a_i^{\ast }{a}_j)v_j,v_i⟩\ge 0.$$

$$\sum _{i,j=1}^m⟨\varphi (a_i^{\ast }{a}_j)v_j,v_i⟩$$

$$=\sum _{i,j=1}^m⟨\varphi (a_{ik}{)}^{\ast }\cdots \varphi (a_{i1}{)}^{\ast }\varphi (a_{j1})\cdots \varphi (a_{jk})v_j,v_i⟩$$

$$=⟨\sum _{j=1}^m\varphi (a_{j1})\cdots \varphi (a_{jk})v_j,\sum _{i=1}^m\varphi (a_{i1})\cdots \varphi (a_{ik})v_j⟩$$

$$=⟨x,x⟩\ge 0,$$

$$a_{ij}=\sum _{m_1,\cdots ,m_{n-1}=1}^k{a}_{i{m}_1}^1{a}_{m_1{m}_2}^2\cdots a_{m_{n-1}j}^n.$$

$$\varphi (a_{ij})$$

$$=\sum _{m_1,\cdots ,m_{n-1}=1}^k\varphi (a_{i{m}_1}^1)\varphi (a_{m_1{m}_2}^2)\cdots \varphi (a_{m_{n-1}j}^n)$$

$$=[{\varphi }_k(a^1){\varphi }_k(a^2)\cdots {\varphi }_k(a^n){]}_{ij},$$

$${\varphi }_k(a^{\ast })={\varphi }_k((a_{ji}^{\ast }))=(\varphi (a_{ji}^{\ast }))=(\varphi (a_{ji}{)}^{\ast })={\varphi }_k(a{)}^{\ast }$$

$$\varphi (a^{\ast }a)$$

$$=\varphi (a_k{)}^{\ast }\cdots \varphi (a_1{)}^{\ast }\varphi (a_1)\cdots \varphi (a_k)$$

$$=(\varphi (a_1)\cdots \varphi (a_k){)}^{\ast }(\varphi (a_1)\cdots \varphi (a_k))$$

$$=b^{\ast }b\ge 0,$$

$$\Bigg{\{}\begin{aligned} &\|x\|^{2k}=\|(x^{*}x)^{k}\|\\ &\|x\|^{2k+1}=\|x(x^{*}x)^{k}\|\end{aligned}$$

$$\Vert x{\Vert }^{2k}=(\Vert x{\Vert }^2{)}^k=\Vert x^{\ast }x{\Vert }^k=\Vert (x^{\ast }x{)}^k\Vert$$

$$\Vert x(x^{\ast }x{)}^k{\Vert }^2$$

$$=\Vert (x^{\ast }x{)}^k{x}^{\ast }x(x^{\ast }x{)}^k\Vert$$

$$=\Vert (x^{\ast }x{)}^{2k+1}\Vert =\Vert (x^{\ast }x){\Vert }^{2k+1}$$

$$=(\Vert x{\Vert }^2{)}^{2k+1}=(\Vert x{\Vert }^{2k+1}{)}^2;$$

$$\phi\big{(}(x^{*}x)^{k}\big{)}=\phi(x^{*}x\cdots x^{*}x)=\big{(}\phi(x^{*})\phi(x)\big{)}^{k}=\big{(}\phi(x)^{*}\phi(x)\big{)}^{k}.$$

$$\Vert \varphi (x){\Vert }^n$$

$$=\Vert (\varphi (x{)}^{\ast }\varphi (x){)}^k\Vert =\Vert \varphi ((x^{\ast }x{)}^k)\Vert$$

$$\le \Vert \varphi \Vert \Vert (x^{\ast }x{)}^k\Vert =\Vert \varphi \Vert \Vert x{\Vert }^{2k}=\Vert \varphi \Vert \Vert x{\Vert }^n,$$

$$c(\lambda -(a^{\ast }a{)}^k)=a.$$

$$c(\lambda -(a^{\ast }a{)}^k)=a(\lambda -(a^{\ast }a{)}^k{)}^{-1}(\lambda -(a^{\ast }a{)}^k)=a.$$

$$a^{\ast }(a{a}^{\ast }{)}^{k-1}(a{b}_m)=(a^{\ast }a{)}^k{b}_m=\lambda b_m-d_m\nrightarrow 0,$$

$$a{b}_m=c(\lambda -(a^{\ast }a{)}^k)b_m\to 0,$$

$$a=c(\lambda -(a^{\ast }a{)}^k)=(\lambda c-c(a^{\ast }a{)}^k).$$

$$\varphi (a)$$

$$=\lambda \varphi (c)-\varphi (c)(\varphi (a{)}^{\ast }\varphi (a){)}^k=\varphi (c)(\lambda -(\varphi (a{)}^{\ast }\varphi (a){)}^k)$$

$$\varphi (c)(\lambda -(\varphi (a{)}^{\ast }\varphi (a){)}^k)=\varphi (a).$$

$$\sigma ((\varphi (a{)}^{\ast }\varphi (a){)}^k)\subseteq \sigma ((a^{\ast }a{)}^k)\cup \{0\}.$$

$$\Vert \varphi (a){\Vert }^{2k}$$

$$=r((\varphi (a{)}^{\ast }\varphi (a){)}^k)\le r((a^{\ast }a{)}^k)$$

$$=\Vert (a^{\ast }a{)}^k\Vert =\Vert a{\Vert }^{2k},$$

$$e=\varphi (1)=\varphi (1^{\ast })=\varphi (1{)}^{\ast }=e^{\ast }.$$

Full text

1 Introduction

Let $A$ and $B$ be algebras and $n\geq 2$ an integer. A linear map $\phi:A\to B$ is an $n$-homomorphism if for all $a_{1},a_{2},\dots,a_{n}\in A$,

[TABLE]

A $2$-homomorphism is then just a homomorphism, in the usual sense, between algebras. Furthermore, every homomorphism is clearly also an $n$-homomorphism for all $n\geq 2$, but the converse is false, in general. The concept of $n$-homomorphism was studied for complex algebras by Hejazian, Mirzavaziri, and Moslehian [7]. This concept also makes sense for rings and (semi)groups. For example, an $AE_{n}$-ring is a ring $R$ such that every additive endomorphism $\phi:R\to R$ is an $n$-homomorphism; Feigelstock [4, 5] classified all unital $AE_{n}$-rings.

In [7], Hejazian et al. ask: Is every $*$-preserving $n$-homomorphism between ${C^{\star}}$-algebras continuous? We answer in the affirmative by proving that every involutive $n$-homomorphism $\phi:A\to B$ between ${C^{\star}}$-algebras is in fact norm contractive: $\|\phi\|\leq 1$. Surprisingly, the arguments for the even and odd $n$ cases are disjoint and, thus, are discussed in different sections. When $n=3$, automatic continuity is reported by Bračič and Moslehian [2], but note that the proof of their Theorem 2.1 does not extend to the nonunital case since the unitization of a $3$-homomorphism is not a $3$-homomorphism, in general.

Using these automatic continuity results, we prove the following stronger results: If $n>2$ is even, every $*$-linear $n$-homomorphism $\phi:A\to B$ between ${C^{\star}}$-algbras is in fact a $*$-homomorphism. If $n\geq 3$ is odd, every $*$-linear $n$-homomorphism $\phi:A\to B$ is a difference $\phi(a)=\psi_{1}(a)-\psi_{2}(a)$ of two orthogonal $*$-homomorphisms $\psi_{1}\perp\psi_{2}$. Regardless, for all integers $n\geq 3$, every positive linear $n$-homomorphism is a $*$-homomorphism. Note that if $\psi$ is a $*$-homomorphism, then $-\psi=0-\psi$ is a norm contractive $*$-preserving $3$-homomorphism that is not positive linear.

There is also a dichotomy between the unital and nonunital cases. When the domain algebra $A$ is unital, there is a simple representation of an $n$-homomorphism as a certain $n$-potent multiple of a homomorphism (discussed in the Appendix.) The nonunital case is more subtle. For example, if $A$ and $B$ are nonunital (Banach) algebras such that $A^{n}=B^{n}=\{0\}$, then every linear map $L:A\to B$ (bounded or unbounded) is, trivially, an $n$-homomorphism (see Examples 2.5 and 4.3 of [7]).

The outline of the paper is as follows: In Section $2$, we prove automatic continuity for the even case and in Section $3$ for the odd case. In Section $4$, we prove our nonexistence results. A key fact in many of our proofs is the Cohen Factorization Theorem [3] of ${C^{\star}}$-algebras. (See Proposition 2.33 [8] for an elementary proof of this important result.) Finally, in Appendix A, we collect some facts about $n$-potents that we need.

The authors would like to thank Dana Williams and Tom Shemanske for their helpful comments and suggestions.

2 Automatic Continuity: The Even Case

In this section, we prove that when $n>2$ is even, every involutive (i.e., $*$-linear) $n$-homomorphism between ${C^{\star}}$-algebras is completely positive and norm contractive, which generalizes the well-known result for $*$-homomorphisms ($n=2$). Recall that a linear map $\theta:A\to B$ between ${C^{\star}}$-algebras is positive if $a\geq 0$ implies $\theta(a)\geq 0$ or, equivalently, for every $a\in A$ there is a $b\in B$ such that $\theta(a^{*}a)=b^{*}b$. We say that $\theta$ is completely positive if, for all $k\geq 1$, the induced map $\theta_{k}:M_{k}(A)\to M_{k}(B)$, $\theta_{k}((a_{ij}))=(\theta(a_{ij}))$, on $k\times k$ matrices is positive.

Theorem 2.1**.**

Let $\mathcal{H}$ be a Hilbert space. If $n\geq 2$ is even, then every involutive $n$-homomorphism from a C-algebra $A$ into $\mathcal{B}(\mathcal{H})$ is completely positive.*

Proof. Let $\phi:A\to\mathcal{B}(\mathcal{H})$ be an involutive $n$-homomorphism. We may assume $n=2k>2$. Let $\langle\cdot,\cdot\rangle$ denote the inner product on $\mathcal{H}$. By Stinespring’s Theorem [9] (see Prop. II.6.6 [1]), $\phi$ is completely positive if and only for any $m>1$ and elements $a_{1},\dots,a_{m}\in A$ and vectors $v_{1},\dots,v_{m}\in\mathcal{H}$ we have

[TABLE]

We proceed as follows: for each $1\leq i\leq m$ use the Cohen Factorization Theorem [3] to factor $a_{i}=a_{i1}\cdots a_{ik}$ into a product of $k$ elements. Thus, their adjoints factor as $a_{i}^{*}=a_{ik}^{*}\cdots a_{i1}^{*}$. Since $n=2k$, we compute

[TABLE]

where $x=\sum_{i=1}^{m}\phi(a_{i1})\cdots\phi(a_{ik})v_{i}\in\mathcal{H}$. The result now follows. $\Box$

Even though the previous result is a corollary of the more general theorem below, we have included it because the proof technique is different.

Lemma 2.2**.**

Let $\phi:A\to B$ be an $n$-homomorphism. Then, for all $k\geq 1$, the induced maps $\phi_{k}:M_{k}(A)\to M_{k}(B)$ on $k\times k$ matrices are $n$-homomorphisms. Moreover, if $\phi$ is involutive $(\phi(a^{*})=\phi(a)^{*})$, then each $\phi_{k}$ is also involutive.

Proof. Given $n$ matrices $a^{1}=(a^{1}_{ij}),\dots,a^{n}=(a^{n}_{ij})$ in $M_{k}(A)$, we can express their product $a^{1}a^{2}\cdots a^{n}=(a_{ij})$, where the $(i,j)$-th entry $a_{ij}$ is given by the formula

[TABLE]

Since $\phi_{k}(a^{1}a^{2}\cdots a^{n})=(\phi(a_{ij}))$ by definition and

[TABLE]

it follows that $\phi_{k}:M_{k}(A)\to M_{k}(B)$ is an $n$-homomorphism. Now suppose that $\phi$ is involutive. We compute for all $a=(a_{ij})\in M_{k}(A)$:

[TABLE]

and hence each $\phi_{k}:M_{k}(A)\to M_{k}(B)$ is involutive. $\Box$

Theorem 2.3**.**

Let $\phi:A\to B$ be an involutive $n$-homomorphism between C-algebras. If $n\geq 2$ is even, then $\phi$ is completely positive. Thus, $\phi$ is bounded.*

Proof. We may assume $n=2k>2$. Since $\phi$ is linear, we want to show that for every $a\in A$ we have $\phi(a^{*}a)\geq 0$. By the Cohen Factorization Theorem, for any $a\in A$ we can find $a_{1},...,a_{k}\in A$ such that the factorization $a=a_{1}\cdots a_{k}$ holds. Thus, the adjoint factors as $a^{*}=a_{k}^{*}\cdots a_{1}^{*}$. Since $n=2k$ and $\phi$ is $n$-multiplicative and $*$-preserving,

[TABLE]

where $b=\phi(a_{1})\cdots\phi(a_{k})\in B$. Thus, $\phi$ is a positive linear map. By the previous lemma, all of the induced maps $\phi_{k}:M_{k}(A)\to M_{k}(B)$ on $k\times k$ matrices are involutive $n$-homomorphisms and are positive. Hence, $\phi$ is completely positive and therefore bounded [1]. $\Box$

We now wish to show that if $n\geq 2$ is even, then an involutive $n$-homomorphism is actually norm-contractive. First, we will need generalizations of the familiar ${C^{\star}}$-identity appropriate for $n$-homomorphisms.

Lemma 2.4**.**

Let $A$ be a ${C^{\star}}$-algebra. For all $k\geq 1$, we have that

[TABLE]

for all $x\in A$.

Proof. In the even case, we have easily that

[TABLE]

by the functional calculus since $x^{*}x\geq 0$. In the odd case, we compute again using the ${C^{\star}}$-identity and functional calculus:

[TABLE]

the result follows by taking square roots. $\Box$

Theorem 2.5**.**

Let $\phi:A\to B$ be an involutive $n$-homomorphism of ${C^{\star}}$-algebras. If $\phi$ is bounded, then $\phi$ is norm contractive ($$\|\phi\|\leq 1$$).

Proof. Suppose $n=2k$ is even. Then for all $x\in A$ we have

[TABLE]

Thus by the previous lemma,

[TABLE]

which implies that $\|\phi\|\leq 1$ by taking $n$-th roots.

The proof for the odd case $n=2k+1$ is similar. $\Box$

3 Automatic Continuity: The Odd Case

The positivity methods above do not work when $n$ is odd, since the negation of a $*$-homomorphism defines an involutive $3$-homomorphism that is (completely) bounded, but not positive. We need the following slight generalization of Lemma 3.5 of Harris [6].

Lemma 3.1**.**

Let $A$ be a ${C^{\star}}$-algebra and let $\lambda\neq 0$ and $k\geq 1$. If $a\in A$ then $\lambda\in\sigma((a^{*}a)^{k})$ if and only if there does not exist an element $c\in A$ with

[TABLE]

Proof. If $\lambda\not\in\sigma((a^{*}a)^{k})$, then $c=a(\lambda-(a^{*}a)^{k})^{-1}\in A$ satisfies

[TABLE]

and so (1) holds.

On the other hand, if $\lambda\in\sigma((a^{*}a)^{k})$ then, by the commutative functional calculus, there is a sequence $\{b_{m}\}_{1}^{\infty}$ in the unitization $A^{+}$ with $b_{m}\not\to 0$ but $d_{m}=_{\text{def}}(\lambda-(a^{*}a)^{k})b_{m}\to 0$. Since $\lambda\neq 0$ we must have

[TABLE]

which implies $ab_{m}\not\to 0$. Hence, there does not exist an element $c\in A$ that can satisfy equation (1), since this would imply that

[TABLE]

which is a contradiction. This proves the lemma. $\Box$

We now prove automatic continuity for involutive $n$-homomorphisms of ${C^{\star}}$-algebras for all odd values of $n$. Note that we do not assume that $A$ is unital, nor do we appeal to the unitization $\phi^{+}:A^{+}\to B^{+}$ of $\phi$, which is not an $n$-homomorphism, in general.

Theorem 3.2**.**

Let $\phi:A\to B$ be an involutive $n$-homomorphism between ${C^{\star}}$-algebras. If $n\geq 3$ is odd, then $\|\phi\|\leq 1$, i.e., $\phi$ is norm contractive.

Proof. Let $n=2k+1$ where $k\geq 1$. Given any $a\in A$ and $\lambda>0$ such that that $\lambda\not\in\sigma((a^{*}a)^{k})$, there is, by the previous lemma, an element $c\in A$ such that

[TABLE]

Noting that $c(a^{*}a)^{k}$ is a product of $2k+1=n$ elements in $A$, and $\phi$ is a $*$-linear $n$-homomorphism, we compute:

[TABLE]

which yields that there is an element $\phi(c)\in B$ with:

[TABLE]

By the previous lemma, we conclude that $\lambda\not\in\sigma((\phi(a)^{*}\phi(a))^{k})$. Thus, we have shown the following inclusion of spectra:

[TABLE]

Therefore, by the spectral radius formula [1, II.1.6.3] and the generalization of the ${C^{\star}}$-identity in Lemma 2.4, we must deduce that:

[TABLE]

which implies that $\|\phi(a)\|\leq\|a\|$ for all $a\in A$, as desired. $\Box$

Note that the argument in the previous proof does not work for $n=2k$ even, since we would need to employ $(a^{*}a)^{k-1}a$ which is a product of $2k-1=n-1$ elements as needed, but not self-adjoint, in general. Thus, we could not appeal to the spectral radius formula for self-adjoint elements and Lemma 3.1 would not apply. Hence, the even and odd $n$ arguments are essentially disjoint.

4 Nonexistence of Nontrival Involutive $n$-homomorphisms of

${C^{\star}}$-algebras

Our first main result is the nonexistence of nontrivial $n$-homomorphisms on unital ${C^{\star}}$-algebras for all $n\geq 3$. We do the unital case first since it is much simpler to prove and helps to frame the argument for the nonunital case.

Theorem 4.1**.**

Let $\phi:A\to B$ be an involutive $n$-homomorphism between the ${C^{\star}}$-algebras $A$ and $B$, where $A$ is unital. If $n\geq 2$ is even, then $\phi$ is a $*$-homomorphism. If $n\geq 3$ is odd, then $\phi$ is the difference $\phi(a)=\psi_{1}(a)-\psi_{2}(a)$ of two orthogonal $*$-homomorphisms $\psi_{1}\perp\psi_{2}:A\to B$.

Proof. In either case, by Proposition A.1, the element $e=\phi(1)\in B$ is an $n$-potent ($e^{n}=e$) and is self-adjoint, because

[TABLE]

Also, there is an associated algebra homomorphism $\psi:A\to B$ defined for all $a\in A$ by the formula

[TABLE]

such that $\phi(a)=e\psi(a)=\psi(a)e$. In either case, $\psi$ is $*$-linear since $\phi$ is $*$-linear and $e$ is self-adjoint and commutes with the range of $\phi$:

[TABLE]

Now, if $n=2k$ is even, $e=e^{n}=(e^{k})^{*}e^{k}\geq 0$ and so $e=p$ is a projection. Thus, $\phi(a)=p\psi(a)=\psi(a)p=p\psi(a)p$ is a $*$-homomorphism. If $n\geq 3$ is odd, then by Lemma A.8, $e$ is the difference of two orthogonal projections $e=p_{1}-p_{2}$ which must commute with both $\psi$ and $\phi$ by the functional calculus. Define $\psi_{1},\psi_{2}:A\to B$ by $\psi_{i}(a)=p_{i}\psi(a)p_{i}$ for all $a\in A$ and $i=1,2$. Then $\psi_{2}\perp\psi_{2}$ are orthogonal $*$-homomorphisms, and

[TABLE]

for all $a\in A$, from which the desired result follows. $\Box$

Corollary 4.2**.**

Let $\phi:A\to B$ be a linear map between ${C^{\star}}$-algebras. If $A$ is unital, the following are equivalent for all integers $n\geq 2$:

• a.)

$\phi$* is a $*$-homomorphism.*

• b.)

$\phi$* is a positive $n$-homomorphism.*

• c.)

$\phi$* is an involutive $n$-homomorphism and $\phi(1)\geq 0$.*

Proof. Clearly (a) $\implies$ (b) $\implies$ (c). If $n\geq 2$ is even, then (c) $\implies$ (a) by the previous result. If $n\geq 3$ is odd, then by the previous result, we only need to show that $\phi$ is positive. Let $n=2k+1$. Given any $a\in A$, by the Cohen Factorization Theorem, we can write $a=a_{1}\cdots a_{k}$. Since $\phi(1)\geq 0$, by hypothesis, and $n=2k+1$, we compute:

[TABLE]

where $b=\phi(a_{1})\cdots\phi(a_{k})\in B$. Thus, $\phi$ is positive linear and therefore a $*$-homomorphism. $\Box$

Next, we extend our nonexistence results to the nonunital case, by appealing to approximate unit arguments (which require continuity!) and the following important factorization property of $*$-preserving $n$-homomorphisms.

Lemma 4.3** (Coherent Factorization Lemma).**

Let $\phi:A\to B$ be an involutive $n$-homomorphism of ${C^{\star}}$-algebras. For any $1\leq k\leq n$ and any $a\in A$, if $a=a_{1}\cdots a_{k}=b_{1}\cdots b_{k}$ in $A$, then

[TABLE]

Note that, in general, $\phi(a)\neq\phi(a_{1})\cdots\phi(a_{k})$ when $1<k<n$.

Proof. Clearly, we may assume $1<k<n$. Since $\phi$ is $*$-linear, the range $\phi(A)\subset B$ is a self-adjoint linear subspace of $B$ (but not necessarily a subalgebra, in general). Given any $d=\phi(c)\in\phi(A)$, using the Cohen Factorization Theorem, write $d=d_{1}\cdots d_{n}=\phi(c_{1})\cdots\phi(c_{n})$ where $d_{i}=\phi(c_{i})$ for $1\leq i\leq n$. Consider the following computation:

[TABLE]

Let $f=\phi(a_{1})\cdots\phi(a_{k})-\phi(b_{1})\cdots\phi(b_{k})$. Then $fd=0$ for all $d\in\phi(A)\subset B$, and thus $fd=0$ for all $d$ in the $*$-subalgebra $A_{\phi}$ of $B$ generated by $\phi(A)$. In particular, for the element

[TABLE]

Hence, $ff^{*}=fd_{a}=0$ and so $\|f\|^{2}=\|ff^{*}\|=0$ by the ${C^{\star}}$-identity. Therefore,

[TABLE]

and the result is proven. $\Box$

Definition 4.4**.**

An approximate unit for a $($nonunital$)$ ${C^{\star}}$-algebra $A$ is a net $\{e_{\lambda}\}_{\lambda\in\Lambda}$of elements in $A$ indexed by a directed set $\Lambda$ such that

• a.)

$0\leq e_{\lambda}$* and $\|e_{\lambda}\|\leq 1$ for all $\lambda\in\Lambda$;*

• b.)

$e_{\lambda}\leq e_{\mu}$* if $\lambda\leq\mu$ in $\Lambda$;*

• c.)

For all $a\in A$,

[TABLE]

Every ${C^{\star}}$-algebra has an approximate unit, which is countable ($\Lambda=\mathbb{N}$) if $A$ is separable (see Section II.4 of Blackadar [1].)

Theorem 4.5**.**

Suppose $\phi:A\to B$ is an involutive $n$-homomorphism of ${C^{\star}}$-algebras, where $A$ is nonunital. Then, for all $a\in A$, the limit

[TABLE]

exists, independently of the choice of the approximate unit $\{e_{\lambda}\}$ of $A$, and defines a $*$-homomorphism $\psi:A\to B$ such that

[TABLE]

for all $a\in A$.

Proof. We may assume $n\geq 3$. Given $a\in A$, use the Cohen Factorization Theorem to factor $a=a_{1}a_{2}\cdots a_{n}$. Define a map $\psi:A\to B$ by

[TABLE]

which is well-defined by the Coherent Factorization Lemma. The continuity of $\phi$ implies that

[TABLE]

It follows that we can write:

[TABLE]

and so $\psi:A\to B$ is linear since $\phi$ is linear. Moreover, since $\phi$ is $*$-linear, it follows that $\psi$ is also $*$-linear:

[TABLE]

In the computation above, we factored $a_{n}=a_{n2}a_{n1}$ and set $a_{12}=a_{1}a_{2}$ to obtain the factorization $a^{*}=a_{n}^{*}\cdots a_{1}^{*}=(a_{n1}^{*}a_{n2}^{*})a_{n-1}^{*}\cdots a_{12}^{*}$ into $n$ elements. Given $a,b\in A$ with factorizations $a=a_{1}\cdots a_{n}$ and $b=b_{1}\cdots b_{n}$, the fact that $\phi$ is an $n$-homomorphism implies:

[TABLE]

note that $ab=(ab_{1})b_{2}b_{3}\cdots b_{n}$ is a factorization of $ab$ into $n$ elements. A second proof of multiplicativity goes as follows:

[TABLE]

Thus, $\psi$ is a well-defined $*$-homomorphism. Finally, we compute:

[TABLE]

$\Box$

Using similar factorizations, the fact that $\{e_{\lambda}^{n}\}$ is also an approximate unit for $A$, and the fact that the strict completion of the ${C^{\star}}$-algebra ${C^{\star}}(\phi(A))$ generated by the range $\phi(A)$ is the multiplier algebra $M({C^{\star}}(\psi(A)))$, we obtain the nonunital version of Proposition A.1.

Corollary 4.6**.**

Suppose that $A$ and $B$ are ${C^{\star}}$-algebras with $A$ nonunital, and let $\phi:A\to B$ be an involutive $n$-homomorphism with associated $*$-homomorphism $\psi:A\to B$. Then there is a self-adjoint $n$-potent $e=e^{*}=e^{n}\in M(C^{*}(\phi(A)))$ such that $\phi(e_{\lambda})\to e$ strictly for any approximate unit $\{e_{\lambda}\}$ of $A$, and with the property that

[TABLE]

for all $a\in A$.

Proof. By the previous proof, we can define $e\in M({C^{\star}}(\phi(A)))$ on generators $\phi(a)$ by

[TABLE]

for any $a=a_{1}\cdots a_{n}\in A$. It follows that:

[TABLE]

which implies $e\in M({C^{\star}}(\phi(A)))$ is $n$-potent. The fact that $e=e^{*}$ follows from $\phi(e_{\lambda})^{*}=\phi(e_{\lambda}^{*})=\phi(e_{\lambda})$. The other statements follow from the previous proof. $\Box$

The dichotomy between the unital and nonunital cases is now clear. If $A$ is unital, then ${C^{\star}}(\phi(A))\subset B$ is a unital ${C^{\star}}$-subalgebra of $B$ with unit $\psi(1)=\phi(1)^{n-1}\in B$ (which is a projection!) and so

[TABLE]

However, for $A$ nonunital, we cannot identify the multiplier algebra $M({C^{\star}}(\phi(A)))$ as a subalgebra of $B$, or even $M(B)$, unless $\phi$ is surjective. In general, we only have inclusions $\psi(A)\subset{C^{\star}}(\phi(A))\subset B.$

Now that we know, as in the unital case, every involutive $n$-homomorphism is an $n$-potent multiple of a $*$-homomorphism, we can prove the following general version of Theorem 4.1 and its corollary in a similar manner using Lemma A.8.

Theorem 4.7**.**

Let $\phi:A\to B$ be an involutive $n$-homomorphism of ${C^{\star}}$-algebras. If $n\geq 2$ is even, then $\phi$ is a $*$-homomorphism. If $n\geq 3$ is odd, then $\phi$ is the difference $\phi(a)=\psi_{1}(a)-\psi_{2}(a)$ of two orthogonal $*$-homomorphisms $\psi_{1}\perp\psi_{2}:A\to B$.

Corollary 4.8**.**

For all $n\geq 2$ and ${C^{\star}}$-algebras $A$ and $B$, $\phi:A\to B$ is a positive $n$-homomorphism if and only if $\phi$ is a $*$-homomorphism.

Appendix A On $n$-homomorphisms and $n$-potents

An element $x\in A$ is called an $n$-potent if $x^{n}=x$. Note that if $\phi:A\to B$ is an $n$-homomorphism, then $\phi(x)=\phi(x^{n})=\phi(x)^{n}\in B$ is also an $n$-potent. The following important result is Proposition 2.2 [7], whose proof is included for completeness.

Proposition A.1**.**

If $A$ is a unital algebra $($or ring$)$ and $\phi:A\to B$ is an $n$-homomorphism, then there is a homomorphism $\psi:A\to B$ and an $n$-potent $e=e^{n}\in B$ such that $\phi(a)=e\psi(a)=\psi(a)e$ for all $a\in A$. Also, $e$ commutes with the range111Note that the range $\phi(A)$ is not a subalgebra of $B$ in general. of $\phi$, i.e., $e\phi(a)=\phi(a)e$ for all $a\in A$.

Proof. Note that $e=\phi(1)=\phi(1^{n})=\phi(1)^{n}=e^{n}\in B$ is an $n$-potent. Define a linear map $\psi:A\to B$ by $\psi(a)=e^{n-1}\phi(a)$ for all $a\in A$. For all $a,b\in A$,

[TABLE]

and so $\psi$ is an algebra homomorphism. Furthermore,

[TABLE]

Similarly, $\psi(a)e=\phi(a)$ for all $a\in A$. The final statement is a consequence of the fact that for all $a\in A$,

[TABLE]

$\Box$

The following computation will be more significant when we consider the nonunital case (see the proof of Theorem 4.5.)

Corollary A.2**.**

Let $\phi$ and $\psi$ be as in Proposition A.1 and $n\geq 3$. Then for all $a\in A$, if $a=a_{1}a_{2}\cdots a_{n}$ with $a_{1},\dots,a_{n}\in A$,

[TABLE]

Proof. We compute as follows:

[TABLE]

Definition A.3**.**

Let $A$ be a unital algebra. An $n$-partition of unity is an ordered $n$-tuple $(e_{0},e_{1},\dots,e_{n-1})$ of idempotents $(e_{k}^{2}=e_{k})$ that sum to the identity $e_{0}+e_{1}+\cdots+e_{n-1}=1$ and are pairwise mutually orthogonal, i.e., $e_{j}e_{k}=\delta_{jk}1$ for all $0\leq j,k\leq n-1$, where $\delta_{jk}$ is the Kronecker delta.

Note that $e_{0}=1-(e_{1}+\cdots+e_{n-1})$ is completely determined by $e_{1},e_{2},\dots,e_{n-1}$ and is thus redundant in the notation for an $n$-partition of unity.

Definition A.4**.**

Let $\omega_{0}=0$ and $\omega_{k}=e^{2\pi i(k-1)/(n-1)}$ for $1\leq k\leq n-1$. Note that $\omega_{1}=1$ and $\omega_{1},\dots,\omega_{n-1}$ are the $(n-1)$-th roots of unity and $\Sigma_{n}=\{\omega_{0},\omega_{1},\dots,\omega_{n-1}\}$ are the $n$ roots of the polynomial equation $x^{n}-x=x(x^{n-1}-1)=0$.

If $A$ is a complex algebra, we let $\tilde{A}$ denote $A$, if $A$ is unital, or the unitization $A^{+}=A\oplus\mathbb{C}$, if $A$ is nonunital.

Theorem A.5**.**

Let $A$ be a complex algebra. If $e\in A$ is an $n$-potent, there is a unique $n$-partition of unity $(e_{0},e_{1},\dots,e_{n-1})$ in $\tilde{A}$ such that

[TABLE]

If $A$ is nonunital, then $e_{1},\dots,e_{n-1}\in A$.

Proof. Define the $n$ polynomials $p_{0},p_{1},\dots,p_{n-1}$ by

[TABLE]

In particular, $p_{0}(x)=1-x^{n-1}$. Each polynomial $p_{k}$ has degree $n-1$ and satisfies $p_{k}(\omega_{k})=1$ and $p_{k}(\omega_{j})=0$ for all $j\neq k$. It follows that $p_{j}(x)p_{k}(x)=0$ for all $x\in\Sigma_{n}$. We also claim that for all $x\in\mathbb{C}$ that

[TABLE]

[TABLE]

Indeed, these identities follow from the fact that these polynomial equations have degree $n-1$ but are satisfied by the $n$ distinct points in $\Sigma_{n}$.

Now, given any $x^{n}=x$ in $\mathbb{C}$ it follows that $p_{k}(x)^{2}=p_{k}(x)$. Hence, for any $n$-potent $e\in A$, if we define $e_{k}=p_{k}(e)$ then $(e_{0},e_{1},\dots,e_{n-1})$ consists of idempotents $e_{k}^{2}=p_{k}(e)^{2}=p_{k}(e)=e_{k}$ and satisfy, by (2),

[TABLE]

They are pairwise orthogonal, because $e_{j}e_{k}=p_{j}(e)p_{k}(e)=0$ for $j\neq k$. Moreover,

[TABLE]

by Equation (3). For $1\leq k\leq n-1$, note that $p_{k}(x)=xq_{k}(x)$ for some polynomial $q_{k}(x)$. Hence, if $A$ is nonunital and $1\leq k\leq n-1$, we have $e_{k}=p_{k}(e)=eq_{k}(e)\in A$, since $A$ is an ideal in $\tilde{A}$. $\Box$

The following result is the $n$-homomorphism version of the previous $n$-potent result. Recall say that two linear maps $\psi_{i},\psi_{j}:A\to B$ are orthogonal ($\psi_{i}\perp\psi_{j}$) if

[TABLE]

for all $a,b\in A$.222Note that the zero homomorphism is orthogonal to every homomorphism.

Proposition A.6**.**

Let $A$ and $B$ be complex algebras. If $A$ is unital then a linear map $\phi:A\to B$ is an $n$-homomorphism if and only if there exist $n-1$ mutually orthogonal homomorphisms $\psi_{1},\dots,\psi_{n-1}:A\to B$ such that for all $a\in A$,

[TABLE]

Proof. $(\Rightarrow)$ Let $\phi:A\to B$ be an $n$-homomorphism. By Proposition A.1, there is an $n$-potent $e\in B$ and a homomorphism $\psi:A\to B$ such that $\phi(a)=e\psi(a)=\psi(a)e$. Using the previous result, write $e=\sum_{k=1}^{n-1}\omega_{k}e_{k}$, where $(e_{0},e_{1},\dots,e_{n-1})$ is the associated $n$-partition of unity in $\tilde{A}$ defined by the polynomials $p_{k}$. Since $e_{k}=p_{k}(e)$, we have that $e_{k}\psi(a)=\psi(a)e_{k}$ for $1\leq k\leq n-1$. Define $\psi_{k}:A\to B$ by

[TABLE]

Then $\psi_{1},\dots,\psi_{n-1}$ are orthogonal homomorphisms and, for all $a\in A$,

[TABLE]

$(\Leftarrow)$ Follows from the fact that $\omega_{k}^{n}=\omega_{k}$ for all $k=1,\dots,n-1$. $\Box$

Remark A.7**.**

If $A$ is nonunital, the above result does not hold. One reason is that the unitization $\phi^{+}:A^{+}\to B^{+}$ of an $n$-homomorphism is not, in general, an $n$-homomorphism. Also, if $A^{n}=B^{n}=\{0\}$, then every linear map $L:A\to B$ is an $n$-homomorphism $($See Examples 2.5 and 4.3 of Hejazian et al [7]$)$.

Let $\Sigma_{n}$ be the $n$ roots of the polynomial equation $x=x^{n}$ from Definition A.4. If $A$ is a ${C^{\star}}$-algebra, it follows that a normal $n$-potent $e=e^{n}$ must have spectrum $\sigma(e)\subseteq\Sigma_{n}$. Recall that a projection is an element $p=p^{*}=p^{2}\in A$. Two projections $p_{1}$ and $p_{2}$ are orthogonal if $p_{1}p_{2}=0$. A tripotent is a $3$-potent element $e^{3}=e\in A$.

The following characterization of self-adjoint $n$-potents in ${C^{\star}}$-algebras is important for our nonexistence results on $n$-homomorphisms.

Lemma A.8**.**

Let $A$ be a ${C^{\star}}$-algebra.

• a.)

If $n\geq 2$ is an even integer, the following are equivalent:

• i.)

$e$* is a projection.*

• ii.)

$e$* is a positive $n$-potent.*

• iii.)

$e$* is a self-adjoint $n$-potent.*

• b.)

If $n\geq 3$ is an odd integer, the following are equivalent:

• i.)

$e$* is a self-adjoint tripotent.*

• ii.)

$e=p_{1}-p_{2}$* is a difference of two orthogonal projections.*

• iii.)

$e$* is a self-adjoint $n$-potent.*

Proof. In both the even and odd cases, (i) $\implies$ (ii) $\implies$ (iii) (See Theorem A.5). Suppose (iii) holds. If $n=2k$ is even,

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and so the spectrum of $e$ satisfies $\sigma(e)\subset\Sigma_{n}\cap[0,\infty]=\{0,1\}$. Thus, $e$ is a projection. If $n\geq 3$ is odd, then since $e=e^{*}$ we must have $\sigma(e)\subset\Sigma_{n}\cap\mathbb{R}=\{-1,0,1\}$. Thus, $\lambda=\lambda^{3}$ for all $\lambda\in\sigma(e)$, which implies $e=e^{3}$ is tripotent. $\Box$