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Let $A$ and $B$ be two $\ast$-algebras and let $\varphi:A\to B$ be a $\ast$-algebra morphism.

I am interested in all the ways that $\varphi$ could fail to be unital (that is, if $A$ has a unit $1_A$, how could it be that $\varphi(1_A)\neq1_B$?).

Using the same proof as in group theory, it seems like if both algebras contain only invertible elements, then $\varphi$ is always unital: $$ \varphi(1_A)=\varphi(a\cdot a^{-1})=\varphi(a)\cdot\varphi(a^{-1})=\varphi(a)\cdot\varphi(a)^{-1}=1_B$$where in the first equality we have used the fact that $A$ contains an invertible element $a$ and in the third equality the fact that a group morphism preserves inverses.

So my question is:

How else could $\varphi$ fail to be unital except for these cases:

1) $A$ does not have a unit.

2) $A$ does not have any invertible elements.

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    $\begingroup$ $A$ will never be a group with respect to multiplication (except in the trivial case $A=\{0\}$) because $0$ is not invertible. If $B$ has non-invertible elements, your proof breaks down as you cannot assure that invertible elements are mapped to invertible elements. Indeed, $a\mapsto 0$ is a non-unital $\ast$-homomorphism. $\endgroup$ – MaoWao May 19 '16 at 8:43
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    $\begingroup$ How can you say that $\phi(a^{-1})=\phi(a)^{-1}$ if you do not assume that $\phi(1_A)=1_B$? $\endgroup$ – Mathematician 42 May 19 '16 at 8:47
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Let $A=\left\{\begin{pmatrix} a&0\\0&b \end{pmatrix}\mid a,b\in \mathbb{C}\right\}$ and $B=\left\{\begin{pmatrix}a&0&0\\0&b&0\\0&0&c\end{pmatrix}\mid a,b,c\in \mathbb{C}\right\}.$ Then $A$ and $B$ are two unital $*$-algebras in the usual way. Consider the map

$$\phi:A\rightarrow B:\begin{pmatrix} a&0\\0&b \end{pmatrix}\mapsto \begin{pmatrix} a&0&0\\0&b&0\\0&0&0 \end{pmatrix}.$$ Then $\phi$ is a $*$-algebra morphism that is not unital.

Edit: Notice however that $\phi(1_A)$ is the unit of $\text{Im}(\phi)$.

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  • $\begingroup$ More generally, if $A$ is a unital $\ast$-algebra and $A\to B$ is a $\ast$-homomorphism, then $\phi(A)$ is a unital $\ast$-algebra and $\phi$ is unital when viewed as map between $A$ and $B$. $\endgroup$ – MaoWao May 19 '16 at 13:38
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    $\begingroup$ As a map between $A$ and $\phi(A)$. $\endgroup$ – Martin Argerami May 19 '16 at 16:24

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