Can we extend a multiplicative linear functional of a closed left ideal on whole of the algebra?

Question

Let B be a closed left ideal of a Banach algebra A. Also, B has a right approximate identity (in B).

If g is a nonzero multiplicative linear functional on B, can we always extend g to a multiplicative linear functional on A ?

Answer 1

Let $H$ be a complex Hilbert space with $\dim(H)\geq 2$. Denote by $B(H)$ the Banach algebra of all bounded linear operators on $H$. Since there is no closed ideal of codimension $1$ in $B(H)$ we see that

(1) there is no nonzero multiplicative linear functional on $B(H)$.

Choose and fix a vector $e\in H$, $\| e\|=1$. Let $K=\{ x\in H;\quad \langle x,e\rangle=0\}$. Then $K^\perp ={\mathbb C}\, e$ and $H={\mathbb C}\, e \oplus K$.

Let $$ {\mathcal I}=\{ T\in B(H);\quad K\subseteq \ker T\}. $$ It is easy to see that this is a closed left ideal in $B(H)$. If $P\in B(H)$ is the orthogonal projection onto ${\mathbb C}\, e$ along $K$, then, of course, $P\in {\mathcal I}$. Since any $x\in H$ has a unique decomposition $x=\alpha e+y$ with $\alpha \in {\mathbb C}$ and $y\in K$ we have $$ TPx=\alpha Te=Tx $$ for any $T\in {\mathcal I}$. Hence $P$ is a right identity in ${\mathcal I}$.

Define $\varphi: {\mathcal I} \to {\mathbb C}$ by $$ \varphi(T)=\langle Te,e\rangle \qquad (T\in {\mathcal I}). $$ It is obvious that $\varphi $ is a linear functional and because of $\varphi(P)=\langle Pe,e\rangle=1$ it is nonzero.

Let $S, T\in {\mathcal I}$ be arbitrary. Let $Te=\lambda_T e+y_T$ with $\lambda_T\in {\mathbb C}$ and $y_T\in K$. Then $\varphi(T)=\langle Te,e\rangle=\lambda_T$. Hence $$ \varphi(ST)=\langle STe,e\rangle=\langle S(\varphi(T)e+y_T),e\rangle=\varphi(S)\varphi(T). $$ Thus, $\varphi$ is a nonzero multiplicative linear functional on ${\mathcal I}$. However it cannot be extended to a multiplicative linear functional on $B(H)$ because of (1).