Representation of a C* algebra

Question

Let $A$ be a $C^*$- algebra. Let $\pi$ be a representation of $A$ on a Hilbert space $H$. Let $p \in A$ be a projection. Consider the $C^*$- algebra $pAp$. Let $P=\pi(p)$. I want to represent $pAp$ in $PH$. Is it correct if I define $\tilde \pi: pAp \to B(PH)$ by $\tilde \pi(pap)=P\pi(a)P$?

Answer 1

Yes, but written like that you make things unnecessarily complicated as you need to look whether your $\tilde\pi$ is well-defined. You simply need to define $$ \tilde\pi(pap)=\pi(pap)|_{PH}. $$

Answer 2

Martin's answer is impecable but it might be interesting to analyze the meaning of the "restriction to $PH$" mentioned there.

By $π(pap)|_{PH}$ he certainly means the operator from $PH$ to itself given by $ξ↦π(pap)ξ$, for all $ξ$ in $PH$.

In this interpretation it is implicit that the co-domain has also been restricted, since otherwise $π(pap)|_{PH}$ might be seen as being an operator from $PH$ to $H$, rather than to $PH$.

The goal here is simply to describe a slick algebraic way to deal with the restriction question. The crucial ingredient is to consider the inclusion of $P(H)$ into $H$, which we will denote by. $$ j:P(H) \to H. $$

The adjoint of $j$ then turns out to be the orthogonal projection from $H$ to $P(H)$ but, technically speaking, $P\neq j^*$, because the co-domain of $P$ is $H$, while the co-domain of $j^*$ is $P(H)$. In other words, $P\in B(H,H)$, while $j^*\in B(H,P(H))$.

Moreover $j^*j=I_{P(H)}$ (because $j$ is an isometry), while $jj^*$ coincides with $P$ in all respects, co-domain included!

This said, the desired representation $\tilde \pi $ may be defined by $$ \tilde \pi (b) = j^*\pi (b)j, \quad \forall b\in pAp. $$

It is a lot of fun to prove that $\tilde \pi $ is multiplicative: for $b,c\in pAp$, we have $$ \tilde \pi (b) \tilde \pi (c) = j^*\pi (b)j j^*\pi (c)j = j^*\pi (b)P\pi (c)j {\buildrel (*) \over =} $$$$ = j^*P\pi (b)\pi (c)j = j^*jj^*\pi (bc)j = j^*\pi (bc)j = \tilde \pi (bc). $$

The main fact supporting this calculation is that $P$ commutes with $\pi (b)$ (see (*) above), so the same argument would still yield a representation should we just assume that $P$ lies in the commutant of our representation.