Does the Nakajima-Zwanzig equation preserve the trace of the projected density matrix? Looking at the Nakajima-Zwanzig equation,  wich gives the time evolution of a projection $\cal {P} \rho$  of a full density matrix $\rho$, I am wondering if the trace of $\cal P \rho$  is preserved under time evolution.
The terms like $LX \stackrel{def}= \dfrac{i}{\hbar} [X,H]$ have a null trace, but  it seems there is no reason why terms like $\mathcal {P} L X$ necessarily have a null trace, so it seems that the trace of the projected density matrix is not necessarily conserved.
Is it correct ?
 A: If you use the canonical choice $\mathcal{P}X = \mathrm{Tr}_B(X)\otimes\rho_B$, where $B$ denotes the irrelevant degrees of freedom in the total Hilbert space $\mathbb{H} = \mathbb{H}_A \otimes \mathbb{H}_B$, and $\rho_B$ is an arbitrary reference state on $\mathbb{H}_B$, then $$ \mathrm{Tr} (\mathcal{P} L X )= \mathrm{Tr} (\mathrm{Tr}_B(LX)\otimes\rho_B) = \mathrm{Tr}_A(\mathrm{Tr}_B(LX))\times\mathrm{Tr}_B(\rho_B) = \mathrm{Tr}(LX) = 0. $$
I expect similar reasoning will show that the full Nakajima-Zwanzig (NZ) equation is trace-preserving. This must be the case since it is formally equivalent to the von-Neumann equation. However, usually one makes some approximation to the NZ equation, such as truncating the time propagator for the irrelevant subspace $e^{\mathcal{Q}Lt}$ to zeroth or maybe lowest non-trivial order in $L$ (the first order term typically vanishes). In this case there is no guarantee of a physical (completely positive) evolution. However, it still looks like even such a truncated NZ equation must be trace-preserving by the same argument, since it is formulated in terms of commutators that vanish under the trace operation. 
I don't know how to prove it for other definitions of the projection $\mathcal{P}$, but the choice I have used is by far the most common (if not the only) one in use in the open quantum systems literature. 
A: OP links to the Wikipedia page for the Nakajima-Zwanzig equation.
Now, let ${\cal H}$ be the Hilbert space of the full system and ${\cal H}_{\rm red}$ be the Hilbert space of the reduced system. The density operator $\rho\in B({\cal H})$ should be semipositive and have trace one. 
What is perhaps not emphasized sufficiently on Wikipedia is 
that the reduced/projected density matrix $\rho_{\rm red}={\cal P}\rho\in B({\cal H}_{\rm red})$ should also have the properties of a density matrix, i.e. be a semipositive operator $\rho_{\rm red}:{\cal H}_{\rm red}\to {\cal H}_{\rm red}$ with trace one. 
The above conditions lead to severe constraints on the type of projection ${\cal P}$ that we can allow in the first place. In fact, generically we can only suggest the canonical reduction construction via tensor product splitting: 
$$\text{system } ~=~ \text{ environment } \otimes \text{ reduced system}, $$
i.e.  ${\cal H}={\cal H}_{\rm env}\otimes {\cal H}_{\rm red}$, so that 
${\cal P}\rho={\rm tr}_{{\cal H}_{\rm env}} \rho,$ 
cf. Mark Mitchison's answer.
Example (which ultimately doesn't work): Consider the Hilbert spaces ${\cal H}=\mathbb{C}^2$ and ${\cal H}_{\rm red}=\mathbb{C}$. Let the Hamiltonian be $H=\frac{1}{2}\hbar\omega\sigma_{1}$, Here $\sigma_1$, $\sigma_2$, and $\sigma_3$ are the Pauli matrices. Define
$$\sigma_{\pm}~:=~\frac{1}{2}({\bf 1}_{2 \times 2}\pm \sigma_3).$$ 
Let 
$$\rho(t)~=~\frac{1}{2}({\bf 1}_{2 \times 2} +\sigma(t))$$ 
belong to the Bloch ball. Let $\sigma(t=0)=\sigma_2$ be a initial condition. Then 
$$\sigma(t)~=~e^{Lt}\sigma_2~=~\sin(\omega t)\sigma_3 +\cos(\omega t)\sigma_2,$$
which happens to correspond to pure states. The projection ${\cal P}$ to spin up ($\uparrow$) is
$$\rho_{\rm red}(t)~:=~{\cal P}\rho(t)~:=~\sigma_{+}\rho(t)\sigma_{+}~=~\frac{1}{2}(1 +\sin(\omega t)),$$
which doesn't have trace one.
