Universal Property of the Multiplier Algerba

We call an injective $*$-isomorphism $i:A\to B$ between two C*-algebra an *ideal homomorphism* if $i(A)$ is an closed ideal of $B$. The universal property of the multiplier algebra $M(I)$ of a C*-algebra $I$ is: for any C*-algebra $A$ and an ideal homomorphism $i:I\to A$ there is a unique $*$-homomorphism $f$ from $A$ to $M(I)$ such that the following diagram is commutative:
\begin{xy}
\xymatrix{
A\ar[r]^f & M(I)\\
I\ar[u]^i\ar[ru]_c & ,
}
\end{xy}
where $c$ is the canonical ideal homomorphism.

*An application*. [If $I$ is a unital C*-algebra, then it cannot be an essential ideal of any other C*-algebra.][1]

[1]: https://math.stackexchange.com/q/3046342

left maximal modular ideal in $M_n$

Each pure state on $M_n=B(\mathbb{C}^n)$ is of form $$w_x:=\langle \cdot ~ x, x\rangle,$$ thus each left maximal modular ideal is of form $$\{u|w_x(u^*u)=0\}=\{u|\|ux\|=0\}.$$
Particularly, corresponding to $e_1=(1,0)$ in $\mathbb{C}^2$, the left maximal modular ideal is is of form
$$\left( \begin{array} & 0 & a\\ 0 & b \end{array}\right).$$

a Positive Bilinear Map is Necessarily Bounded

Suppose $A$ and $B$ are Banach spaces, $C$ is a normed space, and $\sigma:A\times B\to C$ is a bilinear form. For each $a\in A$, denote the map $b\mapsto \sigma(a,b)$ by $\hat{a}:B\to C$; and for each $b\in B$ denote the map $a\mapsto \sigma(a,b)$ by $\hat{b}:A\to C$.

**Lemma**. The bilinear form $\sigma$ is bounded is equivalent to $\hat{a}$ and $\hat{b}$ are both bounded for all $a\in A$ and $b\in B$.

*Proof*. Since $\hat{b}$ is a bounded map on $A$ for all elment $b$ in the closed unit ball of $B$, $\sup_{a\in A_1}\|\hat{b}(a)\|\leq \|\hat{b}\|$, i.e., $\sup_{a\in A_1}\|\hat{a}(b)\|\leq \|\hat{b}\|$. By PUB, the family $\{\hat{a}|a\in A_1\}$ of bounded maps is bounded uniformly, that is, $\sup_{a\in A_1}\|\hat a\|\leq M$ for some $M>0$ . Hence, $\sup_{a\in A_1, b\in B_1}\|\sigma(a, b)\|=\sup_{a\in A_1, b\in B_1}\|\hat{a}(b)\|\leq M$.

**Application**. Let $A$, $B$ and $C$ be C*-algebras.$\newcommand{\tensor}{\otimes}$
By a *positive* bilinear form $\sigma$ from $A\times B$ to $C$, we mean $\sigma(a,b)\geq 0$ for all positive elements $a$ in $A$ and $b$ in $B$, or equivalently, $\hat{a}$ and $\hat{b}$ are positive linear maps on $B$ and $A$ respectively.
Since every positive linear map between two C*-algebra is bounded, by the above lemma,
>a positive bilinear form in C*-algebras must be bounded.

If $A$ and $B$ are C*-algebras, $\gamma$ is a C*-norm on $A\tensor B$, then the map $A\times B\to A\tensor_\gamma B$ is a positive bilinear form and thus is bounded, that is, $\gamma(a,b)\leq M\|a\|\|b\|$ for some positive number $M$.

the atomic part of $C(X)$

Suppose $\pi_\tau(a)=k\id_\tau$, then $k=\langle \pi_\tau(a)x_\tau,x_\tau\rangle=\tau(a)$, so $\pi_\tau(a)=\tau(a)\id_\tau$.

If $\tau$ and $\rho$ are two pure states of $A$, $(\pi_\tau, H_\tau)$ is unitarily equivalent to $(\pi_\rho, H_\rho)$. Then $\tau(a)\id_\tau=\pi_\tau(a)=u^*\pi_\rho(a) u=\rho(a)u^*\id_\rho u=\rho(a)\id_\tau$,where $u$ is a unitary from $H_\tau$ to $H_\rho$. Therefore, $\tau=\rho$. That is,
the atomic representation of $C(X)$ is $(\pi_{atomic}, H_{atomic})=(\oplus_{\tau\in PS(A)}\pi_\tau,\oplus_{\tau\in PS(A)} H_\tau)=(\oplus_{\tau\in \Sigma(A)}\tau(\cdot)\id_\tau, \oplus_{\tau\in \Sigma(A)}\C)=(\oplus_{\tau\in \Sigma(A)}\tau(\cdot)\id_\tau, \ell^2(\Sigma(A))),$ where $\Sigma(A)$ is the character space of $A$.

Since $\oplus_{\tau\in PS(A)}\tau(a)\id_\tau=\widehat{a}:\tau\to \tau(a)\id_\tau$, $\pi_{atomic}(A)=C(\Sigma(A))$. As $\pi_{atomic}$ is non-degenerate, operators in $B(\oplus_{\tau\in PS(A)}H_\tau)$ commuting with $\oplus_{\tau\in PS(A)}\pi_\tau(a)$ are of form $\oplus_{\tau\in PS(A)}u_\tau$. By $\|\oplus_{\tau\in PS(A)}u_\tau\|=\sup_{\tau\in PS(A)}\|u_\tau\|< \infty$ we know that $\pi_{atomic}(A)’=\ell^\infty(\Sigma(A))$ and thus $\pi_{atomic}(A)”=\ell^\infty(\Sigma(A))$.

An closed left ideal $L$ is the intersection of the maximal modular left ideals containing $L$

**Lemma 1.** Let $L_1$ and $L_2$ be closed ideals of a C* algebra $A$. Suppose that $L_1\subset L_2$ and that every pure state of $A$ that vanishes on $L_1$ vanishes on $L_2$. Then $L_1=L_2$.

*Proof.* It is a corollary of Corollary 5.1.9 and Theorem 5.3.2 in [1].

**Lemma 2.** Let $\tau$ be a state of a C*-algebra $A$ and $N_\tau:=\{a|\tau(a^*a)=0\}$. Then a closed left ideal $L$ of $A$ is contained in $N_\tau$ if and only if it is contained in $\ker\tau$.

*Proof.* For any element $a$ in $A$, $\tau(a^*a)=0$ implies $\tau(a)=0$ since $|\tau(a)|^2\leq \|\tau\|\tau(a^*a)$. Thus $N_\tau\subset \ker \tau$.

Suppose $l\in L$, then $l^*l\in L$ since $L$ is a left ideal. Hence if $L\subset\ker \tau$ then $l^*l \in \ker\tau$, i.e. $l\in N_\tau$. So $L\subset N_\tau$.

>**Theorem.** Suppose $L$ is a closed left ideal of a C*-algebra $A$. If the set
$$\vee L:=\{N_\tau|\tau \mbox{ is a pure state of } A\}$$
is non-empty, then $L=\cap\vee L$.

*Proof.* $L\subset \cap\vee L$ is obvious.

Let $\tau$ be a pure state of $A$ and suppose $L\subset \ker\tau$.

Since
\begin{align*}
& L\subset \ker \tau & \\
\Leftrightarrow & L\subset N_\tau & (\mbox{ Lemma 2 })\\
\Rightarrow & \cap\vee L\subset N_\tau & (\mbox{ the definition of } \vee L)\\
\Leftrightarrow & \cap \vee L\subset \ker \tau & (\mbox{ Lemma 2 }),
\end{align*}
$L=\cap\vee L$ by Lemma 1.

References

[1] Gerald J Murphy. C*-algebras and operator theory. Academic press, 2014.