In the group theory, a group $G$ could be split into the irreducible representations, i.e. $$G=G_1\oplus G_2\oplus...(\oplus G_N \text{ if it was finite many})$$ Once the irreducible representation was obtained, they were "added back in", i.e. the $G_2\oplus ...\oplus G_N$ was added back to $G_1$ to reconstruct the full group $G$.
However, in CFT, although the operator-correspondence associated the singular vector $|\chi\rangle$ with a null primary field(also secondary), the subspace generated by the $|\chi\rangle$ was simply subtracted from the vector space, and never added back in, i.e. the character used in the partition function was only from the irreducible of the non zero norm states $\chi_\text{irreducible}$ of $$\chi_h=\chi_\text{irreducible}\oplus \chi_{|\chi_1\rangle}\oplus \chi_{|\chi_2\rangle}\oplus...$$ , but does not contain the verma module from the singular states $|\chi\rangle$. *It's obvious that each verma module $V(c,h)$ correspondence to a unique (non zero norm subspace) irreducible character.
In the post, ACuriousMind explained that
every 2d CFT Hilbert space must be constructed from these (irreducible) representations
and
There is no physical meaning to this procedure in particular
But why the subspace generated by the singular vector wasn't used? Though the negative norm states wasn't useful unless it's a non unitary theory, the zero norm states, i.e. singular vector didn't affect the unitary, and could they be viewed as the degeneracy in the vacuum states or multiple fixed points(such as the two fixed points in the orbifold boson)?
Why wasn't the singular vector character used in the partition function?
