# Sub $C^*$-algebra + Ideal of $C^*$ algebra is complete space

Let $A$ be a $C^*$ Algebra. Let $J$ be a closed ideal in A . Let $B$ be $C^*$-sub-algebra of $A$. Prove $B+J$ is complete space (i.e. every cauchy sequence in $B+J$ converges to an element of $B+J$).

Denote by $x_n + y_n$ a Cauchy sequence in $B+J$ ($x_n \in B$, $y_n \in J$). We need to show $x_n$ is Cauchy sequence in $B$ and $y_n$ is Cauchy sequence in $J$. It seems we need to use the fact $J$ is ideal, But how ?

Any hints ?

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In general $\{x_n\}$ and $\{y_n\}$ need not be Cauchy. Let $A = B = J = \mathbb{C}$ and take $x_n = n$, $y_n = -n$. Then $\{x_n + y_n\}$ is Cauchy but neither $\{x_n\}$ nor $\{y_n\}$ is. So that can't be the right approach. –  Nate Eldredge May 24 '12 at 23:36

I assume that "ideal" means "two-sided ideal".

You are getting stuck with the approach you mention for a very good reason. It cannot work! If you arbitrarily decompose a Cauchy sequence in $B + J$ as a sum of a sequence in $B$ and a sequence in $J$, it need not be the case that these two sequences must individually be Cauchy. e.g. let $A$ consist of all $4 \times 4$ complex matrices that are $2 \times 2$ block diagonal, let $B$ be the subalgebra of $A$ consisting of those matrices with the lower block a scalar multiple of the identity, and let $J$ be those matrices in $A$ whose first block is $0$. For any $n$, the identity element of $A$ is the sum of the digonal matrices with diagonals $(1,1,n+1,n+1)$ (in $B$) and $(0,0,-n,-n)$ (in $J$) and the sequences in $B$ and $J$ thus defined clearly do not converge.

So you need to do something else.

Here is one way to solve the problem, making rather intense use of nontrivial but fundamental facts about quotients of $C^*$ algebras and $*$-homomorphisms from one $C^*$-algebra to another.

Since $J$ is a closed ideal in $A$, we have a quotient $C^*$ algebra $A/J$. Let $p: A \to A/J$ denote the canonical quotient map. Let $p': B \to A/J$ denote the restriction of $p$ to $B$. Then $p'$ is a $*$-homomorphism from one $C^*$-algebra to another, and so its range $R = p'(B)$ is a $C^*$-subalgebra of $A/J$. Let $A' = p^{-1}(R)$. Being the inverse image of a $C^*$-algebra under a $*$-homomorphism, $A'$ is a $C^*$-algebra. I claim that $A' = B+J$.

To see that $B+J \subseteq A'$, fix any $a \in B + J$. We can choose some $b \in B$ and $j \in J$ with $a = b + j$. We then have $$p(a) = p(b + j) = p(b) + p(j) = p(b) = p'(b)$$ showing that $p(a) \in p'(B) = R$ and hence that $a$ is in $p^{-1}(R) = A'$ by definition. As $a$ was an arbitrary element of $B+J$, we deduce that $B + J \subseteq A'$.

Conversely, fix any $a \in A' = p^{-1}(R)$. By definition of the set $A'$, we know that $p(a) \in R = p'(B)$, and hence that there is $b \in B$ with $p(a) = p'(b)$. It follows that $p(a - b) = p(a) - p(b) = p(a) - p'(b) = 0$ so that $a - b$ is in the kernel of $p$, which is $J$. Letting $j = a - b$ we deduce that $a = b + (a - b) = b + j$ is in $B + J$ as desired. As $a$ was an arbitrary element of $B+J$ we deduce that $A' \subseteq B + J$.

It follows that $B + J$ is not only complete, but a $C^*$-subalgebra of $A$. (The argument also shows that the quotient $C^*$-algebras $B/(B \cap J)$ and $(B + J)/J$ are isomorphic, which might remind you of theorems in other parts of algebra.)

The main nontrivial facts used here are that the range of a $C^*$-algebra under a $*$-homomorphism from one $C^*$-algebra to another is a $C^*$ algebra (ie, that a $*$-homomorphism from one $C^*$-algbera to another has closed range), and that the quotient $A/J$ has a natural $C^*$-algebra structure. Neither of these facts are immediately obvious from the definitions. But if you look at their proofs, you will get a sense of how they allow us to completely circumvent the issues one runs into when one tries to decompose a sequence in $B + J$ as a sum of sequences in $B$ and $J$.

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You start by saying I assume that "ideal" means "two-sided ideal". I think you rather assume that $J$ is a $\ast$-ideal in order to be able to conclude that $A/J$ is a $C^\ast$-algebra. So you assume a little more than what you're announcing since a $\ast$-ideal is in particular a two-sided ideal. –  t.b. May 24 '12 at 23:47
@t.b. Every closed two-sided ideal in a $C^*$-algebra is a $*$-ideal, so Leslie's assumption seems fine to me... –  user16299 May 24 '12 at 23:57
Oh, right. Thanks! –  t.b. May 25 '12 at 0:00
@t.b. Oops, yeah. I'm also assuming the standard (but nontrivial) characterizations of two-sided ideals in $C^*$-algebras. (After a while, it is very difficult to remember how much of this fundamental stuff is nontrivial!) –  leslie townes May 25 '12 at 0:18
Thank you for the detailed answer, Leslie ! –  colge May 25 '12 at 15:42