Let $A$ be a C*-algebra and $\pi:A\to B(H)$ a non-degenerate *-representation. We also denote $M_{\pi}$ by the von Neumann algebra generated by $\pi(A)$ in $B(H)$.
The predual of the von Neumann algebra $M_{\pi}$ may be considered as a closed subspace of $A^*$. To see this, one may check that the following map is an isometric: $$\gamma:(M_{\pi})_*\to A^* : \gamma(f)(a)=\langle f,\pi(a)\rangle$$ Moreover $(M_{\pi})_*$ forms an invariant subspace of $A^{*}$, that is $$(M_{\pi})_*=A^{**}(M_{\pi})_*A^{**}$$ Therefore $(M_{\pi})_*^{\perp}$ forms a $w^*$-closed two sided ideal in $A^{**}$. It means that there is a central projection $z_{\pi}$ (which is called the support of $\pi$) such that $$(M_{\pi})_*^{\perp}=A^{**}(1-z_{\pi})\Longrightarrow (M_{\pi})_*=A^*z_{\pi}\Longrightarrow M_{\pi}\simeq A^{**}z_{\pi}$$
To sum up: for given a non-degenerate representation $\pi$ of $A$ there is a central projection with $\pi(A)''=M_{\pi}\simeq A^{**}z_{\pi}$.
Now, let us consider the representation $$\rho_{z_{\pi}}:A\to B(z_{\pi}H_{u})~:~a\to az_{\pi}$$ where $H_u$ is the Hilbert space obtained by the universal representation.
Question: It seems that $\pi$ is not necessairly unitarily equivalent to $\rho_{z_{\pi}}$. For example, let us consider $A=K(H)$. If it were true then there exists just one non-degenerate representation on $K(H)$!
I would like to know that when $\pi$ is uitarily equivalent to $\rho_{z_{\pi}}$ ?