Bounded linear functionals and representations - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T13:52:06Z http://mathoverflow.net/feeds/question/98955 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/98955/bounded-linear-functionals-and-representations Bounded linear functionals and representations Martin Wanvik 2012-06-06T14:44:43Z 2012-06-06T17:39:49Z <p>Suppose that $A$ is a unital C$^*$-algebra and that $\varphi: A \to \mathbb{C}$ is a bounded linear functional. Then there exists a Hilbert space $H$, a representation $\pi: A \to B(H)$ and vectors $\psi, \eta \in H$ such that $$\varphi(a) = \langle \pi(a)\psi, \eta \rangle$$ for all $a \in A$ (this can be proved by decomposing the functional as a linear combination of four states and considering the direct sum of the representation spaces associated to the GNS-construction for each state). </p> <p>My question is: </p> <blockquote> <p>Assuming further that $\| \varphi \| \leq 1$, can we choose $H$, $\pi$ and $\psi, \eta$ as above, <strong>satisfying the additional requirement that $\| \psi \| \leq 1$ and $\| \eta \| \leq 1$</strong>, such that (again) $\varphi(a) = \langle \pi(a) \psi, \eta \rangle$ for all $a \in A$?</p> </blockquote> <p>Note that this is clearly true for a positive functional - simply write $\varphi(a) = \langle \pi(a)\xi, \xi \rangle$ (using the GNS-construction) and note that $$1 \geq \| \varphi \| = \sup_{\| a \| \leq 1} |\langle \pi(a) \xi, \xi \rangle | \geq |\langle \pi(1) \xi, \xi \rangle| = \| \xi \|^2$$</p> http://mathoverflow.net/questions/98955/bounded-linear-functionals-and-representations/98963#98963 Answer by Nik Weaver for Bounded linear functionals and representations Nik Weaver 2012-06-06T15:58:34Z 2012-06-06T17:39:49Z <p>Yes, you can do this using polar decomposition. We can also consider <code>$\phi$</code> to be a normal linear functional on <code>$A^{**}$</code>, and there is a positive <code>$\omega \in A^*$</code> and a partial isometry <code>$v \in A^{**}$</code> such that <code>$\phi(a) = \omega(va)$</code> for all <code>$x \in A$</code>. (I'm sure this is in volume 1 of Takesaki, probably also in Pedersen.) We have <code>$\|\omega\| = \|\phi\| \leq 1$</code>, so we can apply GNS to <code>$\omega$</code> and get <code>$\phi(a) = \omega(va) = \langle \pi(va)\xi,\xi\rangle = \langle \pi(a)\psi,\eta\rangle$</code> with <code>$\psi = \xi$</code> and <code>$\eta = \pi(v^*)\xi$</code>. As you note, <code>$\|\xi\|^2 = \langle \pi(1)\xi,\xi\rangle = \omega(\xi) \leq 1$</code>.</p> http://mathoverflow.net/questions/98955/bounded-linear-functionals-and-representations/98968#98968 Answer by Mike Jury for Bounded linear functionals and representations Mike Jury 2012-06-06T16:25:28Z 2012-06-06T16:25:28Z <p>Another point of view is that this result can be seen as a special case of its "completely contractive" version (in the same way that Stinespring's dilation theorem for completely positive maps generalizes the GNS representation of states). In particular, the following is true: if $\Phi:A\to B(H)$ is any completely contractive map, then there exists another Hilbert space $K$, a representation $\pi:A\to B(K)$, and operators $V_1, V_2:H\to K$ with $\|V_i\|\leq 1$ such that $$\Phi(a) =V_2^*\pi(a)V_1$$ for all $a\in A$. This result may be found e.g. in Chapter 8 of Paulsen's <em>Completely Bounded Maps and Operator Algebras</em>. Essentially the idea is to use Arveson's extension theorem to embed the original $\Phi$ as the "upper right corner" of a completely positive map into the $2\times 2$ matrices over $B(H)$, and then use the Stinespring theorem.</p>