The question is about the following part of the paper "A Note on essentially Normal Operators" by Arveson.
First we start with the dilation theorem of Naimark which is $\phi(f) = P\pi(f)|_{\mathcal{H}} = P\pi(f)P$. I think the second equality is the same as the restriction?
Then following the argument I have $\phi(fg)-\phi(f)\phi(g) = P\pi(fg)P - P\pi(f)P^2\pi(g)P = PSTP - PSP^2TP = PSTP - PSPTP$.
We take $S=T^{*}$ hence $PT^{*}TP - PT^{*}PTP$. Now since $P=P^{\perp}$ we can write $PT^{*}TP - PT^{*}P^{\perp}TP$. From here I am a bit stuck. The paper claims $PT^{*}P^{\perp}TP$ is compact and furthermore $P^{\perp}TP$ and $PTP^{\perp}$ are compact.
From here it is also a mystery to me how we can see from that that we can write down any operator in $\pi(C(X))$ a 2x2 block matrix and the rest of the proof?
I am kind of stuck here for quite a bit to understand the reasoning. Thanks for any help!
In spirit, yes. In reality, no. The functions $P\pi(f)|_{\mathcal{H}}$ and $ P\pi(f)P$ have different domains.
No, not at all. $P^\perp=I-P$. The equality $P=P^\perp$ is impossible. What happens is that you have $K=PT^*TP-PT^*PTP$ is compact. This you can write as $$ K=PT^*P^\perp TP. $$ Next what Arveson uses is that $K=(P^\perp TP)^*P^\perp TP$, and that if $X^*X$ is compact, then so is $X$. There are probably several ways to see this. One is to notice that $X^*X=|X|^2$, and that the square root of a compact positive operator is compact; hence $|X|$ is compact and then $X$ is compact by the polar decomposition.
So now $P^\perp TP$ is compact for any $T\in \pi(C(X))$. In particular for $T^*$, so $PTP^\perp=(P^\perp T^*P)^*$ is compact. Then $PT-TP$ is compact.
As for the $2\times 2$ matrix, given any projection, like $P$, you can write $H=PH\oplus P^\perp H$, and any operator can be seen like a $2\times 2$ block matrix. Namely, you associate $T$ with $$ \begin{bmatrix} PTP& PTP^\perp\\ P^\perp TP&P^\perp TP^\perp\end{bmatrix}. $$ Such a decomposition behaves like the usual matrices, particularly in the sense that multiplication is "row times column". In this picture, $$ P=\begin{bmatrix} 1&0\\0&0\end{bmatrix},\qquad\qquad P^\perp=\begin{bmatrix} 0&0\\0&1\end{bmatrix}, $$ so it is trivial to check that $$ PT-TP=\begin{bmatrix} 0&PTP^\perp\\ -P^\perp TP&0\end{bmatrix}. $$
As for $B$, $N$ is normal and $C$ is compact. So $\def\abajo{\\[0.2cm]}$ \begin{align} A^*A-AA^*\oplus B^*B-BB^*& = (N+C)^*(N+C)-(N+C)(N+C)^*\abajo &=N^*N-NN^*+C\,(\text{stuff})+(\text{stuff})\,C\abajo &=C\,(\text{stuff})+(\text{stuff})\,C, \end{align} which is compact. So $$ B^*B-BB^*=P^\perp(A^*A-AA^*\oplus B^*B-BB^*)P^\perp $$ is compact.