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In CFT, a singular vector $|\chi\rangle$ indicated the original Verma module $V(c,h)$ was reducible, where itself corresponding to a full-fledged Virasoro algebra.(Francesco section 7.1.3)

It may happen that the representation of the Virasoro algebra comprising all the states(7.7) is irreducible. By this, we mean that there is a subspace(or submodule) that itself a full-fledged representation of the Virasoro algebra. The state of this submodule transform amongst themself under any conformal transformation. Such a submodule is also generated from a highest-weight states$|\chi\rangle$, such that $L_n|\chi\rangle=0(n>0)$. although this state is also of the form (7.7). $$L_{-k_1} L_{-k_2}...L_{-k_n}|h\rangle(1\leq k_1\leq ...\leq k_n) \tag{Eq. 7.7}$$

But the Virasoro algebra was the generator relation, $$[L_m,L_n]=(m-n)L_{m+m} +\frac{c}{12}(m^3-m)\delta_{m+n,0}$$ What does it mean by the singular vector generated a full-fledged Virasoro algebra? I guessed that it meant the $L_n$ mode could act on $|\chi\rangle$ just like a primary states, but didn't this hold for any states? i.e. $L_{-2}|h\rangle$. Why $L_{-2}|h\rangle$ did not correspondence to a full-fledged Virasoro algebra?

From ACuriousMind's answer https://physics.stackexchange.com/a/130415/209383 $L_{-2}|h\rangle$ might not be the highest weight states, but I don't know if there's any proof that connects the highest weight states with the null vector, and, still, arbitrary $L_n$ modes could be applied in front of $L_{-2}|h\rangle$ just like that of the $|\chi\rangle$.

Why does it mean by the singular vector generated the full-fledged Virasoro algebra?

To explain, the definition read:

  1. the highest weight states: $L_n|\chi\rangle=0,(n>0)$. If the Hermitian conjugation could be assumed, this implied that $|\chi\rangle$ a singular vector.

  2. the singular vector: $\langle\chi|\chi\rangle=0$, i.e. $|\chi\rangle$ was a singular vector.

  3. the subspace generated from a state $|h\rangle$: a subspace which the operators $L_n$ acts on a state, say $|h\rangle$ which, and it's closed under the conformal transformation, i.e. the states transform among themselves.

Could there be a highest weight states other than primary $|h\rangle$ such that it's not singular?(which means the hermitian conjugate probably would not hold) How to show that 1 or 2 implied 3, and vise versa?

ShoutOutAndCalculate
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    It is not clear to me what this question is asking without going and reading di Francesco's section 7.1.3: A vector is not an algebra, so I don't know how a vector is supposed to "correspond" to a "full-fledged" algebra. Could you quote the relevant passage where di Francesco makes the claim you're asking about? – ACuriousMind Jul 10 '22 at 19:43
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  • Is there a "not" missing in the first sentence of the quote? Because what it gives is the definition of the representation being reducible/not irreducible. 2. I feel you've misunderstood what the quote is trying to say, but I don't know what the misunderstanding is. All di Francesco is doing is saying the same thing I say in my answer: It can happen that the Verma module is not irreduclble - that's my " if one of the descendants of our initial highest weight vector is itself of highest weight", just differently phrased. Where does this $\lvert 0\rangle$ you're talking about come from?
    • – ACuriousMind Jul 10 '22 at 19:56
  • @ACuriousMind I think it's the part (the additional highest weight states inside the verma module $\leftrightarrow$ singular vector). Could there be a highest weight states in $V(c,h)$ that's non singular? – ShoutOutAndCalculate Jul 10 '22 at 20:03
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  • There's always one highest weight vector that's not singular, namely the one you started the construction of the Verma module with. 2. If there is a second highest weight vector, then it forms a subrepresentation. It doesn't matter at all whether it's singular or not - if it isn't singular, then the subrepresentation it forms is unitary and the subrep is isomorphic to one of the other unitary $V(h,c)$ Verma quotients.
    • – ACuriousMind Jul 10 '22 at 20:10
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    I don't really see what your comments have to do with what you've written in your question: What is $\lvert 0\rangle$? How did you go from Francesco's "this is a rep of the Virasoro algebra" to your "this vector corresponds to the Virasoro algebra"? You're being very imprecise with terminology and this makes the question very confusing. – ACuriousMind Jul 10 '22 at 20:10
  • @ACuriousMind Corrected. There were three different things mixed together, basically the confusion was from there. – ShoutOutAndCalculate Jul 10 '22 at 20:42