Representation of a C* algebra 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 $\tilde \pi(pap)=P\pi(a)P$?
 A: 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
$$
\tilde\pi(pap)=\pi(pap)|_{PH}.
$$
A: 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.
