Show that Quotient of a C-algebra is a C-algebra in its own right

Question

Let $A$ be a commutative C*-algebra over $\mathbb{C}$ and let $J$ be an ideal of $A$ such that $J\in Closed(A)$ and that contains $x^*$ if it contains $x$, for all $x\in A$.

I know that $A/J$ is also a Banach algebra (Rudin FA 11.4) and also has an involution of the form $[x]^*:=[x^*]$ which is well defined (if $[x]=[y]$, $x-y\in J$, so that $x^*-y^* \in J$ and so $[x]^*=[y]^*$).

I would like to prove that $A/J$ is in fact a C*-algebra, that is, that $$||[x][x]^*||=||[x]||^2$$ (the B*-condition is equivalent to the C*-condition).

However I got stuck:

For one direction it is easy:

$$ ||[x][x]^*|| \leq ||[x]||||[x]^*|| = ||[x]||^2 $$ using the fact that $||[x]^*|| \equiv \inf(\{||x^*-j|| | j\in J\}) = \inf(\{||x^*-j^*|| | j\in J\}) = \inf(\{||x-j|| | j\in J\}) \equiv ||[x]|| $ because if $j\in J$ then $j^* \in J$ and $A$ is a C*-algebra, so that $||x-j||=||x^*-j^*||$.

For the other direction, $$ ||[x]||^2 = ||[x]|| ||[x]^*|| \leq ||x-j_1|| ||x^*-j_2|| \stackrel{A\,is\,a\,C*}{=} ||(x-j_1)(x^*-j_2^*)|| = ||x x^*-j_1 x^*-xj_2^*-j_1j_2^* || $$ for all $(j_1,j_2)\in J^2$.

If $\varepsilon>0$, then $\exists j_\varepsilon\in J$ such that $$ ||x x^* -j_\varepsilon||<||x x^*+J||+\varepsilon $$ So now if I could find $(j_1,j_2)\in J^2$ such that $j_\varepsilon=-j_1 x^*-xj_2^*-j_1j_2^*$ then the proof would be complete.

However, I am not sure how to do that, if that is even possible.

Maybe not, but Rudin FA 11.4 applies to commutative algebras, so otherwise one would have to rely on something else. — user118719, Jul 03 '15 at 13:26
Do you have a little knowledge of approximate units? (It will be alot easier!) — C-star-W-star, Jul 03 '15 at 13:55
Unfortunately not, and this concept is not mentioned in the book I am reading, so I would prefer a direct as possible proof. In particular I wonder if my approach may be salvaged. — user118719, Jul 03 '15 at 13:56
See when you want to prove that $F(t):=(\sin(t),\cos(t))$ is continuous you can either directly prove it (hard!) or you exploit the characteristic property (easy!). That is the reason why most books follow with quotients only after approximate units. I will give a short intro if you're fine with that? — C-star-W-star, Jul 03 '15 at 14:05

score 1 · Accepted Answer · answered Jul 03 '15 at 14:56

1

Approximate Unit

Regard ball cone: $$\mathcal{B}_+:=\{A\in\mathcal{A}:\|A\|<1:A\geq0\}$$

Order elements: $$E,E'\in\mathcal{I}\cap\mathcal{B}_+:\quad E\leq E'$$

Then one has: $$I\in\mathcal{I}:\quad\|I-IE\|,\|I-EI\|\stackrel{E\to1}{\longrightarrow}0$$

(That is the hard part!)

Quotient Norm

Note that it holds: $$1-\sigma(E)\geq0\implies\|1-E\|\leq1$$

An estimate gives: $$\|A+\mathcal{I}\|\leq\|A-AE\|=\|A(1-E)\|\\ \leq\|A+I\|\cdot1+\|I-IE\|\to\|A+I\|$$

So the norm writes: $$\|A+\mathcal{I}\|=\lim_{E\to1}\|A-AE\|$$

That gives rise to: $$\|A+\mathcal{I}\|^2=\lim_{E\to1}\|A-AE\|^2=\lim_{E\to1}\|(1-E)A^*A(1-E)\|\\ \leq1\lim_{E\to1}\|A^*A-A^*AE\|=\|A^*A+\mathcal{I}\|=\|(A+\mathcal{I})^*(A+\mathcal{I})\|$$

Thus it satisfies: $$\|A+\mathcal{I}\|^2=\|(A+\mathcal{I})^*(A+\mathcal{I})\|$$

Concluding C*-algebra.

answered Jul 03 '15 at 14:56

C-star-W-star

16,275

Let me know if you're stuck somewhere. – C-star-W-star Jul 03 '15 at 15:06
Thanks! By $E\leq E'$ you mean $E'-E$ is a positive operator (positive means self-adjoint and with non-negative real spectrum)? – user118719 Jul 03 '15 at 15:14
Yes, exactly!! :) – C-star-W-star Jul 03 '15 at 15:15
And be careful not to drop selfadjoint as there are nontrivial nilpotents with positive spectrum. – C-star-W-star Jul 03 '15 at 15:17
I'm trying to understand why you order elements in $\mathcal{I}\cap\mathcal{B}+$, and it seems (from googling "approximate units") that you are trying to say $\mathcal{I}\cap\mathcal{B}+$ is a net, and then examine the limits of these nets. Is there a reason why you are looking at nets and not just good old sequences? I thought in metric spaces (of which Banach algebras are an example of) sequences are sufficient. All this machinery seems like a huge overkill to prove an inequality. – user118719 Jul 03 '15 at 20:22
@user118719: Yes, exactly: Nets are simple to handle. :) Of course, they admit a sequence but that is needless rather than overkill. ;) – C-star-W-star Jul 03 '15 at 22:46
@user118719: Besides, nets are EXTREMELY!!!!!!!!!! powerful: Arbitrary sums, integrals, direct sums, tensor products, direct integrals, etc. etc. etc. (ONB's for example become a one-liner.) – C-star-W-star Jul 03 '15 at 22:47
@user118719: For extracting sequence see: Cauchy Net For order of nets see: Directed Indices For equivalence see: Completeness – C-star-W-star Jul 03 '15 at 22:58
Translating this to sequences, you're basically saying there's a sequence in $\mathcal{I}$ intersected with the open unit ball, whose elements are positive, and which basically converges to $1$. But how do you know $1$ is a limit point of this thing? – user118719 Jul 03 '15 at 23:47
Let us continue this discussion in chat. – C-star-W-star Jul 03 '15 at 23:52

Show that Quotient of a C*-algebra is a C*-algebra in its own right

1 Answers1

Show that Quotient of a C-algebra is a C-algebra in its own right