I feel a little confused about the conclusion of unitisation of unial $C^*$algebras.There are two different statements . 1. If $A$ is unital,then the only unitisation is $A$ itself. 2.If $A$ is unital,then $\tilde{A}$ is $*$-isomorphic to $A\oplus \Bbb C$. The above conclusions are from different reference books. Which conclusion is correct?
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2Presumably the two books use different definitions of "unitisation". Check their definitions. – Eric Wofsey Jul 05 '19 at 16:26
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I assume the definition of unitization from your first reference is something like:
A unitization of a $C^*$-algebra $A$ is an embedding of $A$ as an essential ideal in a unital $C^*$-algebra.
In this case, if $A$ is unital then the only unitization of $A$ is itself. Indeed, if $A,B$ are unital and $A\subset B$ is an essential ideal, then $(1_B-1_A)a=0$ for all $a\in A$, so $1_B=1_A$ and it follows that $B=A$.
The definition from your second reference is probably more standard. In this case, the unitization of $A$ is $\tilde A=A\oplus\mathbb C$ as a vector space, with multiplication and involution $$(a,\lambda)(b,\mu)=(ab+\mu a+\lambda b,\lambda\mu),\qquad (a,\lambda)^*=(a^*,\overline\lambda),$$ and the norm of $(a,\lambda)$ is the maximum of $|\lambda|$ and $\sup\{\|ab+\lambda b\|:\|b\|\leq1\}$. In this case, if $A$ is unital, then the algebra homomorphism $\tilde A\to A\oplus\mathbb C$ given by $(a,\lambda)\mapsto(a+\lambda 1_A,\lambda)$, is a $*$-isomorphism.
To summarize, both statements are correct, but the references use different definitions.
Aweygan
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