Unitization of (non-unital) subalgebras of a unital $C^*$-algebra

Question

Let $\mathfrak{D}$ be a unital $C^*$-algebra, we'll denote its unit as $1_D$. Let $\mathfrak{A}$ be a finite-dimensional $C^*$-subalgebra of $\mathfrak{D}$ and let $\mathfrak{B}$ be a non-unital $C^*$-subalgebra of $\mathfrak{D}$.

Since $\mathfrak{A}$ is finite-dimensional, it's unital and we can denote this unit by $1_A$.

Since every non-unital $C^*$-algebra is contained in a unital $C^*$-algebra as a maximal ideal of codimension 1, we can unitize $\mathfrak{B}$ by setting $\mathfrak{B}^+ := \mathfrak{B} \oplus \mathbb{C}$. Then the unit of $\mathfrak{B}^+$ is $1_{B+} = (0,1)$.

But what is the relationship (if any) between all these different units? Is there any way we can think of $1_A$ as "equivalent" to $1_D$? Can we just unitize $\mathfrak{B}$ in a different way so that its unit is $1_D$ by doing something like $\mathfrak{B}^{++} := \mathfrak{B} \oplus \mathbb{C}1_D$?

Answer 1

In general, if $B\subset A$ is an inclusion of $C^*$-algebras and $p\in A$ is a projection (i.e. $p=p^*=p^2$) such that $p\not\in B$ and $pb=bp=b$ for all $b\in B$, then $\tilde{B}\cong B+\mathbb{C}p$.

First $B+\mathbb{C}p$ is a $C^*$-subalgebra of $A$ (why?) and define $\phi:\tilde{B}\to B+\mathbb{C}p$ by $\phi(b,\lambda)=b+\lambda p$.

The map $\phi$ is obviously well-defined, linear, $*$-preserving and surjecitve. It is also multiplicative:

$$\phi((b,\lambda)(c,\mu))=\phi(bc+\lambda c+\mu b,\lambda\mu)=bc+\lambda c+\mu b+\lambda\mu p=bc+\lambda pc+\mu bp+\lambda\mu p=(b+\lambda p)(c+\mu p)=\phi(b,\lambda)\cdot\phi(c,\mu)$$ and finally it is injective: if $\phi(b,\lambda)=0$, then $\lambda p=-b$, so, if $\lambda\neq0$ we have $p\in B$, a contradiction. Hence $\lambda=0$ and $b=0$ follows. So $\phi$ is a bijective $*$-homomorphism, i.e. a $*$-isomorphism.

The above shows that, whenever you have an inclusion of $C^*$-algebras and there is a projection in the large $C^*$-algebra that is not already included in the small $C^*$-algebra and this projection acts as a unit for the small $C^*$-algebra, then the unitization is the same thing as thinking the direct sum of the small $C^*$-algebra and a copy of complex scalars of this projection. See for example Martin Argerami's answer: he provides an example where there are many ways that you can unitize; all these ways fall in the above situation with a projection that behaves as a unit.

Answer 2

You can take $1_B=1_D$. But you can also take it smaller. For instance you could have $$D=\mathbb C\oplus c\oplus M_3(\mathbb C),$$ $$ A=\mathbb C\oplus 0\oplus 0,\qquad B=0\oplus c_0\oplus 0. $$ So $$ 1_A=(1,0,0),\qquad 1_D=(1,1, I_3). $$ And you can unitize $B$ with any of $$ (0,1,0),\ (1,1,0),\ (0,1,E_{11}),\ (0,1,E_{11}+E_{22}),\ (1,1,E_{11}+E_{22}),\ 1_D $$ and a couple others.