2
$\begingroup$

Let $B$ be a $C^{*}$-subalgebra of a $C^{*}$-algebra $A$ with a faithful conditional expectation $P: A\rightarrow B$. Kumjian suggests on page 15 that since $P$ is faithful we have

$\left\Vert a\right\Vert =\text{sup}\left\{ \left\Vert P\left(c^{*}a^{*}ac\right)\right\Vert ^{1/2}\mid c\in A\text{, }P\left(c^{*}c\right)\leq1\right\} $.

I assume the easiest way to see this is to check that the right side actually gives a $C^{*}$-norm on $A$ and then make use of it's uniqueness. The faithfulness obviously gives the definiteness. What I don't see is why the right expression satisfies the triangular equation and the $*$-condition. How can I check this?

$\endgroup$

1 Answer 1

5
$\begingroup$

I think this can be answered with GNS representation theory. Note that all is needed is that $P$ is a faithful positive map between C*-algebras.

Let $S(B)$ be the state space of $B$. Consider the representation $$ \pi = \bigoplus\big\{ \pi_{\phi\circ P} \mid \phi\in S(B) \big\} $$ of $A$ on some (huge) Hilbert space, where each $\pi_{\phi\circ P}$ is the GNS representation associated to the state $\phi\circ P$ on $A$. As $P$ is faithful, it follows that so is $\pi$. Thus $\pi$ is isometric. Given $a\in A$ and $\delta>0$, we may therefore find some $\phi\in S(B)$ with $$ \|a\|^2 \leq \delta+\|\pi_{\phi\circ P}(a)\|^2 = \delta+\sup\big\{ \|\pi_{\phi\circ P}(a)\xi\|^2 \mid \xi\in H_{\phi\circ P},\ \|\xi\|\leq 1 \big\}. $$ But now each $\xi$ is approximated by an element of the form $\pi_{\phi\circ P}(c)\xi_{\phi\circ P}$, where $\xi_{\phi\circ P}$ is the cyclic vector, and this element having norm at most one means $(\phi\circ P)(c^*c)\leq 1$. Applying this to the above supremum yields $$ \begin{array}{ccl} \|a\|^2 &\leq& \delta+\sup\big\{ (\phi\circ P)(c^*a^*ac) \mid c\in A,\ (\phi\circ P)(c^*c)\leq 1 \big\} \\ &\leq& \delta+\sup\big\{ P(c^*a^*ac) \mid c\in A,\ P(c^*c)\leq 1 \big\}. \end{array} $$ As $\delta>0$ was an arbitrary parameter, this shows the $\leq$-part in your equation. The inequality in the other direction is trivial as $P$ is positive.

$\endgroup$
3
  • 4
    $\begingroup$ Another way to package all those GNS representations appearing in this very nice answer is using Hilbert $C^*$-modules. Let $X$ denote the (right) Hilbert $B$-module obtained by completing $A$ in the norm induced by the $B$-valued inner product $\langle a,b\rangle = P(a^*b)$. Then the norm in question is the operator norm on the $C^*$-algebra of adjointable operators on $X$ (evaluated on the operator of left-multiplication by $a$). $\endgroup$
    – user85913
    Jul 16, 2017 at 10:04
  • $\begingroup$ This is indeed the more elegant way to see it if one knows Hilbert modules. $\endgroup$ Jul 16, 2017 at 12:12
  • $\begingroup$ Thanks guys, great answers! I'm not familiar with Hilbert-C*-modules yet. The GNS proof works for me. I appreciate your help! $\endgroup$ Jul 16, 2017 at 19:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.