Given $\cal{A},\cal{B}$ two dense $*$-algebras of two $C^*$-algebras $A$ and $B$ respectively, together with a $*$-algebra homomorphism $f:\cal{A} \to \cal{B}$, is it clear that $f$ extends to a bounded linear operator $f:A \to B$?
$\begingroup$
$\endgroup$
2
-
$\begingroup$ By "$*$-map" do you mean "$*$-homomorphism"? $\endgroup$– Matthew DawsCommented Jun 20, 2018 at 15:23
-
$\begingroup$ yes, I have edited $\endgroup$– Max SchattmanCommented Jun 20, 2018 at 15:25
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
7
I believe this is a counter-example.
Let $\newcommand{\mc}{\mathcal}\mc A$ be the algebra of complex polynomials restricted to $[0,1]$, with closure $A=C[0,1]$. Let $X\in\mc A$ be the coordinate function, $X(t)=t$ for $t\in[0,1]$, so $X$ generates $\mc A$ as a $*$-algebra.
Let $H$ be a Hilbert space and $x\in\mc B(H)$ be any non-zero self-adjoint operator. Let $\mc B$ be the $*$-algebra generated by $x$ and let $B$ be the closure, a $C^*$-algebra.
Let $f:\mc A\rightarrow\mc B$ be the unique $*$-homomorphism with $f(X)=x$. Suppose that $f$ extends by continuity to $A$, say $\pi:A\rightarrow B$ with $\pi$ bounded. By continuity, $\pi$ is a $*$-homomorphism, and so is contractive. But then $\|x\| = \|\pi(X)\| \leq \|X\|=1$. As not every self-adjoint operator is a contraction, this provides the required contradiction.
answered Jun 20, 2018 at 15:48
-
-
$\begingroup$ Can one exclude such examples by asking more of $f$? $\endgroup$ Commented Jun 20, 2018 at 15:54
-
1$\begingroup$ Well, tautologically, yes! ($f$ by assumption extends to $A$...) Some years ago, I came across similar problems when working with Hopf $*$-algebras: it is rather subtle as to when you can extend such an $f$, and I don't recall having seen an "abstract" criteria: rather, there are domain specific arguments. So without knowing a lot more about $\mathcal A, \mathcal B$ and $f$ I doubt one can say more. Of course, I could be wrong... $\endgroup$ Commented Jun 20, 2018 at 15:58
-
2$\begingroup$ It would be enough to know that the unitization of $\mathcal{A}$ is inverse closed ... $\endgroup$ Commented Jun 20, 2018 at 15:59
-
1$\begingroup$ Even though the answer is false in general, there are many special situations in which the extension does exist. This is discussed in Section 4 of [R. Exel, T. Giordano and D. Gonçalves, Enveloping algebras of partial actions as groupoid C*-algebras, J. Operator Theory, 65 (2011), 197-210]. $\endgroup$– RuyCommented Jun 21, 2018 at 0:54