The structure of the Hilbert space of 2d CFT In many textbooks, I found similar statements that in 2d CFT (which I hope I'm not misunderstanding), one can decompose the space of states into primaries and their Virasoro descendants, or into quas-primaries and their $SL(2,\mathbb{C})$ descendants.
And I'm confused about the above statement.
Consider the $bc$ ghost (or similarly, $\beta \gamma$ ghost), with the simplest stress tensor $T = (b \partial c)$. Also consider the current $j = (bc)$. We know that
$$
T(z)j(w) = \frac{ 1 }{ (z-w)^3 } + \frac{ j(w) }{ (z-w)^2 } + \frac{ \partial j(w) }{ z-w }\ ,
$$
which tells us that $j$ is not primary. In fact, it's not quasi-primary either, since under infinitesimal transformation $z \to z' = z + \epsilon(z)$,
$$
j(z) \to j'(z') = \text{standard terms} + \partial^2 \epsilon(z) \ ,
$$
where $\partial^2\epsilon(z)$ is nonvanishing for special conformal transformations.
And I think $j(z)$ is not a Virasoro descendant of any primary either (perhaps I'm wrong about this?). So it seems that $j(z)$ is something beyond the textbook statement.
I wonder how to understand this issue?
==== some more simple calculations (maybe not useful) ====
Starting from the $j \equiv j(z)|0\rangle = b_{-1}c_0|0\rangle$, one can do
$$
L_{+1} j =|0\rangle \ .
$$
Or alternatively
$$
T(w)j(z) \sim \frac{ one(w) }{ (z-w)^3 } + ... \Rightarrow (L_{+1}j)(w) = \{Tj\}_3(w) = one(w)
$$
So $j$ can reach the $SL(2,\mathbb{C})$-inv. vacuum by special conformal transformation "$K$". However, one cannot come back to $j$ by doing Virasoro descendant, say, $L_{-1}|0\rangle = 0$.
==== some other thoughts ====
I'm having the feeling that the Hilbert space of $bc$, or other more general VOA, is not a "direct sum" of "Virasoro conformal families", but instead it should be some reducible but non-decomposible structure (as in Ribault's comment).
 A: It's no accident that you invoked ghosts to find a counter-example. Ghosts are non-unitary and the standard proof of the primary / descendant classification uses unitarity. In particular, given a local operator which is not a quasi-primary, the state $\mathcal{O}(0) \left | 0 \right >$ has some scaling dimension $\Delta$. We can then show using the conformal algebra that
\begin{equation}
K_a \mathcal{O}(0) \left | 0 \right >, \,\, K_b K_a \mathcal{O}(0) \left | 0 \right >, \,\, K_c K_b K_a \mathcal{O}(0) \left | 0 \right >
\end{equation}
and so on have dimensions $\Delta - 1$, $\Delta - 2$ and $\Delta - 3$. If we are given that there's a unique ground state, then we must eventually get to an operator which is annihilated by special conformal translations in order to prevent $\Delta$ from decreasing forever. This state is the quasi-primary and then we can work backwards to find out that the module of states we had to hit with $K_a$ to arrive there is spanned by its $SL(2, \mathbb{C})$ descendants.
We can demonstrate the non-unitarity of the ghost theory by observing that the states $c_1 \left | 0 \right >$ and $c_0 c_1 \left | 0 \right >$ both have lower energy than the $SL(2, \mathbb{C})$ vacuum. However, because $bc$ ghosts anti-commute, it is not possible to get more states like this by acting with additional copies of $c_0$ and $c_1$. Therefore, they are nicer than $\beta \gamma$ ghosts. As a result, we get to say there is something like a Hilbert space except the inner product is not positive-definite and the ground state is degenerate. Everything is built on top of
\begin{equation}
\left | \uparrow \right > = c_0 c_1 \left | 0 \right >, \,\, \left | \downarrow \right > = c_1 \left | 0 \right >
\end{equation}
which are Hermitian conjugates of each other.
A: Primaries, quasi-primaries, and their descendants, are all eigenvectors of the dilation operator. In logarithmic CFTs, the dilation operator is not diagonalizable, so you have states that are not eigenvectors. Therefore, logarithmic CFTs provide other examples of CFTs where not every state is primary or descendant.
