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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?

ShoutOutAndCalculate
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  • The first paragraph makes no sense (and I can't quite tell what you meant to say, either): A group cannot be "split" into representations. A representation is a vector space on which the group acts, the group itself is not a representation of itself because it's (usually) not a vector space. 2. I don't understand the rest of the question, either: Any CFT Hilbert space is a sum of irreps of Vir. alg., and since it's a Hilbert space there are no states of zero norm. The irreps are enumerated by the quotients of Verma modules $V(h,c)$. Why would you want to "use" $\chi$ and how?
    • – ACuriousMind Jul 10 '22 at 19:51
  • @ACuriousMind But the non unitary theory(contain the negative norm states) are still useful in the condensed matter(and the negative norm states wasn't always quotiented out and do show up in the characters) such as the open system, why the zero norm states could not be used? – ShoutOutAndCalculate Jul 10 '22 at 19:56
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    Sure, if you drop the unitarity requirement on your theory you might use them. But I think many people (in particular those giving intros to CFT) usually talk about the unitary case unless they explicitly say otherwise. – ACuriousMind Jul 10 '22 at 19:57
  • @ACuriousMind I was guessing that just like the usage of the negative norm states in the non unitary CFT, the quotient of the singular vector states might be of some physical reasons. But I wasn't sure what's the reason behind it. – ShoutOutAndCalculate Jul 10 '22 at 20:00