Highest-Weight Modules, Null States, and the Kac Determinant

The Virasoro algebra is not just a useful way of organizing two-dimensional conformal transformations. It is the engine that makes two-dimensional CFT unusually solvable. Once we know the central charge $c$ and the spectrum of primary fields, the descendants of each primary are generated algebraically by the negative Virasoro modes.

This page develops that representation theory carefully. The main ideas are:

• a primary state generates a Verma module by acting with $L_{-n}$;
• radial quantization gives a natural inner product and hence Gram matrices at each level;
• special modules contain null states, descendants with zero norm that are themselves highest-weight;
• after quotienting by null descendants, the irreducible representation can be much smaller;
• the Kac determinant tells us exactly when null states occur.

In string theory this representation theory has two roles. First, it gives the CFT language for vertex operators. Second, it explains why zero-norm states are not merely an annoyance: they encode gauge redundancies.

Verma modules

Work holomorphically for the moment. The Virasoro algebra is

$$[L_m,L_n] = (m-n)L_{m+n} + \frac{c}{12}m(m^2-1)\delta_{m+n,0}.$$

A highest-weight state $|h\rangle$ obeys

$$L_0|h\rangle=h|h\rangle, \qquad L_n|h\rangle=0\quad n>0.$$

The Verma module $V(c,h)$ is the vector space obtained by acting with all products of lowering operators $L_{-n}$, $n>0$:

$$V(c,h) = \text{span}\left\{ L_{-n_1}L_{-n_2}\cdots L_{-n_k}|h\rangle \right\}.$$

We usually order the modes as

$$n_1\geq n_2\geq\cdots\geq n_k\geq1.$$

The level of such a descendant is

$$N=n_1+n_2+\cdots+n_k.$$

Because

$$[L_0,L_{-n}]=nL_{-n},$$

a level-$N$ descendant has $L_0$ eigenvalue $h+N$:

$$L_0\left(L_{-n_1}\cdots L_{-n_k}|h\rangle\right) = (h+N)L_{-n_1}\cdots L_{-n_k}|h\rangle.$$

Thus the module is graded:

$$V(c,h)=\bigoplus_{N=0}^{\infty}V_N(c,h).$$

The dimension of the level-$N$ subspace is the partition number $p(N)$, at least before null states are removed. For example,

$$\begin{aligned} V_0&:\quad |h\rangle,\\ V_1&:\quad L_{-1}|h\rangle,\\ V_2&:\quad L_{-2}|h\rangle,\quad L_{-1}^2|h\rangle,\\ V_3&:\quad L_{-3}|h\rangle,\quad L_{-2}L_{-1}|h\rangle,\quad L_{-1}^3|h\rangle. \end{aligned}$$

A primary state $|h\rangle$ generates a Verma module. At level $N$, the number of naive descendants is the partition number $p(N)$.

This is already reminiscent of the string oscillator Fock space. There, too, a level is built by partitions of an integer. The difference is that Virasoro descendants are generated by stress-tensor modes, not by target-space oscillator modes.

Inner products and descendant norms

In radial quantization, Hermitian conjugation is defined by inversion on the complex plane. For the Virasoro modes,

$$L_n^{\dagger}=L_{-n}.$$

This gives an inner product on each level of a Verma module. The Gram matrix at level $N$ is the matrix of inner products among the $p(N)$ level-$N$ descendants.

At level one there is only one descendant:

$$L_{-1}|h\rangle.$$

Its norm is

$$\langle h|L_1L_{-1}|h\rangle = \langle h|[L_1,L_{-1}]|h\rangle = 2h.$$

So unitarity requires

$$h\geq0.$$

If $h=0$, this level-one descendant has zero norm. In the vacuum module, the state $L_{-1}|0\rangle$ corresponds to $\partial\mathbf 1$, which must vanish.

At level two choose the ordered basis

$$|1\rangle=L_{-2}|h\rangle, \qquad |2\rangle=L_{-1}^{2}|h\rangle.$$

The Gram matrix is

$$M_2(c,h) = \begin{pmatrix} \langle 1|1\rangle & \langle 1|2\rangle\\ \langle 2|1\rangle & \langle 2|2\rangle \end{pmatrix} = \begin{pmatrix} 4h+\frac c2 & 6h\\[2pt] 6h & 4h(2h+1) \end{pmatrix}.$$

The entries follow directly from the Virasoro algebra. For instance,

$$[L_2,L_{-2}]=4L_0+\frac c2,$$

so

$$\langle h|L_2L_{-2}|h\rangle=4h+\frac c2.$$

The determinant is

$$\det M_2(c,h) = 2h\left(16h^2+2(c-5)h+c\right).$$

The level-two Gram determinant vanishes when the two descendants become linearly dependent after quotienting by a zero-norm state.

The determinant detects zero-norm combinations. If $\det M_2=0$, there is a nonzero level-two descendant orthogonal to every level-two state. In a unitary representation, such a state must be null and must be removed from the physical Hilbert space.

Level-two null states

Let us find the null vector explicitly. Try

$$|\mathcal N_2\rangle = \left(L_{-2}+aL_{-1}^{2}\right)|h\rangle.$$

A null descendant that generates a submodule must itself be highest-weight:

$$L_n|\mathcal N_2\rangle=0, \qquad n>0.$$

It is enough to impose $L_1$ and $L_2$, because higher positive modes follow from commutators. First,

$$L_1L_{-2}|h\rangle=3L_{-1}|h\rangle,$$

and

$$L_1L_{-1}^{2}|h\rangle=2(2h+1)L_{-1}|h\rangle.$$

Therefore

$$3+2a(2h+1)=0,$$

or

$$a=-\frac{3}{2(2h+1)}.$$

The condition $L_2|\mathcal N_2\rangle=0$ gives

$$4h+\frac c2+6ah=0.$$

Substituting the value of $a$ gives

$$16h^2+2(c-5)h+c=0.$$

Thus the standard level-two null vector is

$$|\mathcal N_2\rangle = \left( L_{-2}-\frac{3}{2(2h+1)}L_{-1}^{2} \right)|h\rangle,$$

provided

$$16h^2+2(c-5)h+c=0.$$

Equivalently,

$$h=h_{\pm}(c) = \frac{5-c\pm\sqrt{(c-1)(c-25)}}{16}.$$

These two branches are the first glimpse of the Kac table.

Null submodules and irreducible modules

A null vector has zero norm, but its importance is larger than that one number. If $|\mathcal N\rangle$ is both null and highest-weight, then its descendants

$$L_{-n_1}\cdots L_{-n_k}|\mathcal N\rangle$$

form an entire submodule. In a unitary theory, every state in this null submodule is orthogonal to every state in the full module.

The irreducible representation is obtained by quotienting:

$$\mathcal H_{c,h} = \frac{V(c,h)}{\text{null submodules}}.$$

This quotienting is not a technical afterthought. It is part of the definition of the physical state space. In string theory, zero-norm descendants similarly encode gauge redundancies. For example, the massless vector polarization

$$\epsilon_\mu\sim \epsilon_\mu+\lambda k_\mu$$

is the spacetime shadow of a null state on the worldsheet.

A null highest-weight descendant generates its own Verma submodule. The irreducible module is obtained by quotienting out that entire null submodule.

The Kac determinant

The level-two determinant is the first nontrivial case of a general theorem. At level $N$, the Gram matrix $M_N(c,h)$ has size $p(N)\times p(N)$. Its determinant factorizes as

$$\det M_N(c,h) = C_N \prod_{\substack{r,s\geq1\\rs\leq N}} \left(h-h_{r,s}(c)\right)^{p(N-rs)},$$

where $C_N$ is a nonzero constant independent of $h$.

The interpretation is direct:

$$h=h_{r,s}(c) \quad\Longrightarrow\quad \text{a null vector first appears at level }rs.$$

Zeros of the Kac determinant are labeled by integer pairs $(r,s)$. The first null vector appears at level $rs$, and its descendants produce zeros at higher levels.

A convenient parameterization is

$$c=1-6\frac{(p-q)^2}{pq},$$

with

$$h_{r,s}^{(p,q)} = \frac{(qr-ps)^2-(p-q)^2}{4pq}.$$

For general complex $p/q$, this is just a way to parametrize $c$ and the zeros of the determinant. When $p$ and $q$ are coprime positive integers, the formula becomes the entry point to minimal models.

Unitarity and why null states matter

A unitary CFT must have a positive-semidefinite inner product. The Kac determinant gives strong constraints on which $(c,h)$ are allowed.

For $c\geq1$, the familiar unitary highest-weight representations have

$$h\geq0.$$

Null states can occur at special values, but generic modules are irreducible.

For $0<c<1$, unitarity is far more restrictive. The allowed central charges are the discrete series

$$c_m=1-\frac{6}{m(m+1)}, \qquad m=3,4,5,\ldots,$$

and the allowed weights are Kac-table weights

$$h_{r,s} = \frac{\left((m+1)r-ms\right)^2-1}{4m(m+1)},$$

with

$$1\leq r\leq m-1, \qquad 1\leq s\leq m.$$

These are the unitary minimal models. We will examine them on the next page.

The lesson is worth emphasizing: two-dimensional conformal symmetry is infinite-dimensional, so representation theory is a dynamical constraint. A few algebraic data, plus null-vector consistency, can determine entire families of correlation functions.

Exercises

Exercise 1. Count descendants at low levels

List a basis of descendants at levels $N=1,2,3,4$ in a generic Verma module and check that the number of states is $p(N)$.

Solution

The basis states are labeled by partitions of $N$:

$$\begin{aligned} N=1 &: \quad L_{-1}|h\rangle,\\ N=2 &: \quad L_{-2}|h\rangle,\quad L_{-1}^2|h\rangle,\\ N=3 &: \quad L_{-3}|h\rangle,\quad L_{-2}L_{-1}|h\rangle,\quad L_{-1}^3|h\rangle,\\ N=4 &: \quad L_{-4}|h\rangle,\quad L_{-3}L_{-1}|h\rangle,\quad L_{-2}^2|h\rangle,\quad L_{-2}L_{-1}^2|h\rangle,\quad L_{-1}^4|h\rangle. \end{aligned}$$

Thus the counts are

$$1, \quad 2, \quad 3, \quad 5,$$

which are $p(1),p(2),p(3),p(4)$.

Exercise 2. Derive the level-two Gram matrix

Using the Virasoro algebra, show that in the basis $L_{-2}|h\rangle$, $L_{-1}^{2}|h\rangle$,

$$M_2(c,h) = \begin{pmatrix} 4h+\frac c2 & 6h\\ 6h & 4h(2h+1) \end{pmatrix}.$$

Solution

The first entry is

$$\langle h|L_2L_{-2}|h\rangle = \langle h|[L_2,L_{-2}]|h\rangle =4h+\frac c2.$$

For the off-diagonal entry,

$$\langle h|L_2L_{-1}^2|h\rangle = \langle h|[L_2,L_{-1}]L_{-1}+L_{-1}[L_2,L_{-1}]|h\rangle.$$

Since $[L_2,L_{-1}]=3L_1$ and $L_1|h\rangle=0$, this reduces to

$$3\langle h|L_1L_{-1}|h\rangle=6h.$$

Finally,

$$\langle h|L_1^2L_{-1}^2|h\rangle =4h(2h+1),$$

obtained by commuting the two $L_1$ operators through the two $L_{-1}$ operators until they annihilate $|h\rangle$.

Exercise 3. Find the level-two null vector

Let

$$|\mathcal N_2\rangle=(L_{-2}+aL_{-1}^{2})|h\rangle.$$

Impose $L_1|\mathcal N_2\rangle=L_2|\mathcal N_2\rangle=0$.

Solution

The $L_1$ equation gives

$$L_1L_{-2}|h\rangle=3L_{-1}|h\rangle,$$

and

$$L_1L_{-1}^2|h\rangle=2(2h+1)L_{-1}|h\rangle.$$

Therefore

$$3+2a(2h+1)=0, \qquad a=-\frac{3}{2(2h+1)}.$$

The $L_2$ equation gives

$$4h+\frac c2+6ah=0.$$

Substitution yields

$$16h^2+2(c-5)h+c=0.$$

Exercise 4. Orthogonality of a null submodule

Suppose $|\mathcal N\rangle$ is a null highest-weight descendant. Show that all descendants of $|\mathcal N\rangle$ are orthogonal to all states in the Verma module.

Solution

A general descendant of $|\mathcal N\rangle$ has the form

$$L_{-n_1}\cdots L_{-n_k}|\mathcal N\rangle.$$

Pair it with a general descendant of $|h\rangle$. Moving all lowering operators in the bra to raising operators using $L_n^{\dagger}=L_{-n}$, the inner product reduces to a linear combination of terms of the form

$$\langle \mathcal N|L_{m_1}\cdots L_{m_\ell}|\psi\rangle$$

or, equivalently after commuting, to matrix elements involving positive modes acting on $|\mathcal N\rangle$. Since $|\mathcal N\rangle$ is highest-weight, all positive modes annihilate it. The remaining term is proportional to $\langle\mathcal N|\mathcal N\rangle=0$. Hence the whole null submodule is orthogonal to the full module.

Exercise 5. Locate the first few Kac zeros

Using the Kac determinant formula, explain why a zero at $h=h_{2,1}$ first occurs at level $2$, while a zero at $h=h_{2,2}$ first occurs at level $4$.

Solution

The determinant at level $N$ contains factors with $rs\leq N$:

$$\det M_N\propto \prod_{rs\leq N}\left(h-h_{r,s}\right)^{p(N-rs)}.$$

For $(r,s)=(2,1)$, the product $rs=2$, so the factor $h-h_{2,1}$ first appears at $N=2$.

For $(r,s)=(2,2)$, the product is $rs=4$, so the factor $h-h_{2,2}$ first appears at $N=4$. At higher levels it reappears with multiplicity $p(N-4)$ because the null state has descendants.