In the conformal field theory book by Francesco, Mathieu, Senechal, the Verma module is built from a primary field $|\phi\rangle$, and if one of the descendants is a singular vector $|\chi\rangle$, the singular vector can also build a Verma module.

However, the definition of $|\phi\rangle$ and $|\chi\rangle$ is almost identical, that

$$L_n|\chi\rangle = 0, n>0$$


$$L_n|\phi\rangle = 0, n>0$$


$$\langle h| L_{-1}...L_{-i}|h\rangle= \langle h| L_1...L_i|h\rangle^*=0$$

Though $\langle h|h'\rangle=\delta_{h,h'}$

But what's the point then to distinguish the Virasoro primary fields and the reducible Verma modules? Where in the computation of the characters those Verma module generated by the singular vectors were to be subtracted?

What is the physical interpretation of the singular vector? Does it corresponding to a gauge freedom that should be subtracted? Or does this corresponding to the procedure of "choosing the physical states"?

What does the Verma module generated by the singular vector in the reducible Virasoro algebra represent? Also, could the descendant of the primary fields that's nonsingular also generate a Verma module?

  • 1
    $\begingroup$ The distinction may mostly be one of terminology. The notion of a Verma module arose in the pure mathematical exploration of the representation theory of Lie groups—not in physics. $\endgroup$
    – Buzz
    Apr 11 at 4:50
  • $\begingroup$ I edited your question to fix several typos and the non-idiomatic use of the past tense, but I still cannot figure out what the paragraph starting with "But what's the point then[...]" is supposed to mean. Who is "distinguishing" "the Virasoro primary fields" and "the reducible Verma modules"? A primary field and a module are completely different things, why would one need to "distinguish" them? I suspect you wanted something other than "distinguish" here, but I don't know what. $\endgroup$
    – ACuriousMind
    Apr 11 at 9:51
  • $\begingroup$ @ACuriousMind From the answer it seemed that there's no distinction between the minimal singular vector and the highest weight states, but in the case of a "physical model" such as the 2d Ising mdel, the singular vector subtracted in the character was not added back in. It's became an irredeemable model but the resulting character was not a tensor sum. If $|\chi\rangle$ was negative norm states, then it should be removed, but why should a zero norm simply be removed and forgot? Also, from the state operator correspondence weren't they still a well behaved zero norm descendant field? $\endgroup$ Apr 11 at 16:19

1 Answer 1


The general idea of a Verma module is that you start with a single highest weight vector - in this case probably $\lvert \phi\rangle$ - and see what you get by applying the ladder operators of your algebra to it. I discuss a more elementary application of this technique to $\mathrm{SU}(2)$ in this answer of mine - let us here just note that if one of the descendants of our initial highest weight vector is itself of highest weight, then our Verma module is not an irreducible representation of the algebra, since the descendants of this second vector form a subrepresentation (this is the $\lvert \chi\rangle$ in your case).

So, since you're actually looking for irreducible representations, the Verma module itself has clearly failed to produce one (and it is not evident whether the subrepresentation is irreducible or not, either). However, it turns out that the quotient of the Verma module by this subrepresentation always yields an irreducible representation of the algebra.

There is no physical meaning to this procedure in particular. What you're doing is trying to determine the irreducible representations of the Virasoro algebra because you know every 2d CFT Hilbert space must be constructed from these representations, the method of Verma modules is just a convenient way to do so.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.