# proof of quotients in $C^{*}$-algebras

Proof: We have that $\mathcal{I}$ is self-adjoint (http://planetmath.org/InvolutaryRing), since it is a closed ideal of a $C^{*}$-algebra (http://planetmath.org/CAlgebra) (see this entry (http://planetmath.org/ClosedIdealsInCAlgebrasAreSelfAdjoint)). Hence, the involution in $\mathcal{A}$ induces a well-defined involution in $\mathcal{A}/\mathcal{I}$ by $(x+\mathcal{I})^{*}:=x^{*}+\mathcal{I}$.

Recall that, since $\mathcal{I}$ is closed, the quotient norm is indeed a norm in $\mathcal{A}/\mathcal{I}$ that makes $\mathcal{A}/\mathcal{I}$ a Banach algebra (see this entry (http://planetmath.org/QuotientsOfBanachAlgebras)). Thus we only have to prove the $C^{*}$ to prove that $\mathcal{A}/\mathcal{I}$ is a $C^{*}$-algebra.

Recall that $C^{*}$-algebras have approximate identities (http://planetmath.org/CAlgebrasHaveApproximateIdentities). Notice that $\mathcal{I}$ itself is a $C^{*}$-algebra and pick an approximate identity $(e_{\lambda})$ in $\mathcal{I}$ such that

• each $e_{\lambda}$ is positive.

• $\|e_{\lambda}\|\leq 1$

We will only prove the case when $\mathcal{A}$ has an identity element $e$. For the non-unital case, one can consider $\mathcal{A}$ as a $C^{*}$-subalgebra of its minimal unitization and the same proof will still work.

Let $\|\cdot\|_{q}$ denote the quotient norm in $\mathcal{A}/\mathcal{I}$. We claim that for every $x\in\mathcal{A}$:

 $\displaystyle\|x+\mathcal{I}\|_{q}=\lim_{\lambda}\|x(e-e_{\lambda})\|$ (1)

We will prove the above equality as a lemma at the end of the entry. Assuming this result, it follows that for every $a\in\mathcal{A}$

 $\displaystyle\|x+\mathcal{I}\|_{q}^{2}=\lim\|x(e-e_{\lambda})\|^{2}=\lim\|(e-e% _{\lambda})x^{*}x(e-e_{\lambda})\|\leq\lim\|(e-e_{\lambda})\|\|x^{*}x(e-e_{% \lambda})\|$

Since each $e_{\lambda}$ is positive and $\|e_{\lambda}\|\leq 1$ we know that its spectrum lies on the interval $[0,1]$. Hence $e-e_{\lambda}$ is also positive and its spectrum also lies on the interval $[0,1]$. Thus, $\|e-e_{\lambda}\|\leq 1$. Therefore:

 $\displaystyle\|x+\mathcal{I}\|_{q}^{2}\leq\lim\|(e-e_{\lambda})\|\|x^{*}x(e-e_% {\lambda})\|\leq\lim\|x^{*}x(e-e_{\lambda})\|=\|x^{*}x+\mathcal{I}\|_{q}$

Since $\mathcal{A}/\mathcal{I}$ is a Banach algebra, we also have $\|x^{*}x+\mathcal{I}\|_{q}\leq\|x+\mathcal{I}\|_{q}^{2}$ and so

 $\|x+\mathcal{I}\|_{q}^{2}=\|x^{*}x+\mathcal{I}\|_{q}$

which proves that $\mathcal{A}/\mathcal{I}$ is a $C^{*}$-algebra. $\square$

$\;$

We now prove equality (1) as a lemma.

Lemma - Suppose $\mathcal{A}$ is a $C^{*}$-algebra with identity element $e$. Let $\mathcal{I}\subset\mathcal{A}$ be a closed ideal and $(e_{\lambda})$ be an approximate identity in $\mathcal{I}$ such that each $e_{\lambda}$ is positive and $\|e_{\lambda}\|\leq 1$. Then

 $\|x+\mathcal{I}\|_{q}=\lim_{\lambda}\|x(e-e_{\lambda})\|$

for every $x$ in $\mathcal{A}$.

Proof: Since $y(e-e_{\lambda})\longrightarrow 0$ for every $y\in\mathcal{I}$ it follows that

 $\displaystyle\limsup\|x(e-e_{\lambda})\|$ $\displaystyle=$ $\displaystyle\limsup\|x-xe_{\lambda}-y+ye_{\lambda}\|$ $\displaystyle=$ $\displaystyle\limsup\|(x-y)(e-e_{\lambda})\|$ $\displaystyle\leq$ $\displaystyle\|x-y\|$

Therefore, taking the infimum over all $y\in\mathcal{I}$ we obtain:

 $\limsup\|x(e-e_{\lambda})\|\leq\inf_{y\in\mathcal{I}}\|x-y\|=\|x+\mathcal{I}\|% _{q}$

Also, since $xe_{\lambda}\in\mathcal{I}$,

 $\liminf\|x(e-e_{\lambda})\|\geq\inf_{y\in\mathcal{I}}\|x-y\|=\|x+\mathcal{I}\|% _{q}$

and this proves the lemma. $\square$

Title proof of quotients in $C^{*}$-algebras ProofOfQuotientsInCalgebras 2013-03-22 17:41:56 2013-03-22 17:41:56 asteroid (17536) asteroid (17536) 7 asteroid (17536) Proof msc 46L05