Surjectivity of unitization functor of C*-algebras

Question

Let $A$ be a unital $C^*$-algebra.

Question: Is $A$ isomorphic to the unitization of a $C^*$-algebra $A_0$?

Clarification: Here, I am using the functorial definition of unitization. The unitization of $A_0$ is the algebra with underlying vector space $A_0\oplus\mathbb C$ and norm $$||(a,\lambda)\|=\max\Big\{|\lambda|,\sup_{\|b\|=1}\{\|ab+\lambda b\|\}\Big\}.$$

Thoughts: Suppose $A$ is commutative. Then by Gelfand duality, $A$ is isomorphic to $C(X)$ for some compact Hausdorff space $X$. So in the commutative case it would suffice that any such $X$ is the one-point compactification of some space $X_0$; but I'm not even sure if this is true.

The answer depends on the definition see – Ryszard Szwarc Mar 05 '23 at 08:42 — Ryszard Szwarc, Mar 05 '23 at 08:42

score 4 · Accepted Answer · answered Mar 05 '23 at 15:23

No. The algebra $A_0$ in question would necessarily be an ideal -- the multiplication you have on the unitization is $$(a,\mu)(b,\nu) = (ab + \mu b + \nu a, \mu \nu).$$ Thus if $\mu$ or $\nu$ is 0, then the right component will be 0, i.e., $A_0$ is an ideal. Are there C*-algebras without proper ideals? (try $M_n$)

As for the commutative case, see Which spaces can be seen as to be the one-point compactification of some other space.

Of course, the unitization you describe is a "minimal" unitization. However every unital C*-algebra $A$ is in fact the multiplier algebra of itself (which is in some sense a "maximal unitization" -- it corresponds to the Stone–Čech compactification in the commutative case).

J. De Ro · Answer 2 · 2023-03-06T09:01:21.343

2

No, but if $A_0$ is unital, then $A\cong A_0 \oplus_\infty \mathbb{C}$ as $C^*$-algebras. The map $$A_0\oplus_\infty \mathbb{C}\to A: (a, \lambda) \mapsto (a-\lambda 1_{A_0})+ \lambda 1_{A}$$ is an isomorphism of $C^*$-algebras.

edited Mar 06 '23 at 09:01

answered Mar 05 '23 at 18:57

J. De Ro

21,438

What does $\oplus_\infty$ mean? – geometricK Mar 06 '23 at 05:20
Max norm of the two Banach spaces – J. De Ro Mar 06 '23 at 07:31
I'm not sure I understand: if $A=\mathbb C$, then you're saying $A_0$ is $\mathbb C\oplus\mathbb C$ as a vector space? That doesn't fit inside $A$. – geometricK Mar 06 '23 at 08:03
@geometricK Ah sorry, I misunderstood your notation. I thought $A_0$ is the $C^*$-algebra and $A$ is the unitisation. Edited. – J. De Ro Mar 06 '23 at 09:01

score 2 · Answer 3 · answered Jun 26 '23 at 15:55

The first time I came up with this question is when I learned K-theory. I realized that $B(H)$ cannot be such a C*-algebra.

If $A$ is the unitization of $B$, then we have a short exact sequence which is necessarily split: $$0\rightarrow B\rightarrow A\rightarrow \mathbb{C}\rightarrow0.$$ So by the split exactness of $K_0$, we have a split exact sequence: $$0\rightarrow K_0(B)\rightarrow K_0(A)\rightarrow K_0(\mathbb{C})\rightarrow 0.$$ Or equivalently, $K_0(A)\cong K_0(B)\oplus\mathbb{Z}$. The conclusion is: if $A$ is a unital C*-algebra whose $K_0$ group cannot admit $\mathbb{Z}$ as a direct summand, then $A$ cannot be a unitalization of another one.

For example, the Cuntz algebras (but they are also simple), $B(H)$, and more generally, $M(A\otimes\mathbb{K})$, where $M$ denotes the multiplier algebra, $\mathbb{K}$ is the algebra of compact operators on a infinite-dimensional separable Hilbert space.

Surjectivity of unitization functor of C*-algebras

3 Answers3