Kac Determinant and Verma Module Null Vectors

Hello! Welcome to our fourth lesson on 2D Conformal Field Theory.

In our last session, we constructed the fundamental representations of the Virasoro algebra: Verma modules. We learned that each module $V_{c, h}$ is built upon a primary state $∣ h ⟩$ and its descendants. We also discovered that for specific values of the central charge $c$ and conformal dimension $h$ , a Verma module can be reducible, meaning it contains a null vector—a descendant that is also a primary state. We found by direct calculation that for $h = 0$ , a null vector appears at level 1.

While direct calculation works for low levels, it quickly becomes intractable. Today, our goal is to introduce a powerful and elegant tool that systematically tells us precisely when null vectors appear. This will allow us to identify null vectors in Verma modules using the Kac determinant formula. This formula is one of the cornerstones of 2D CFT, providing a complete classification of the reducible representations of the Virasoro algebra.

Unitarity, the Gram Matrix, and Null Vectors

Before introducing the formula, let's establish the physical and mathematical context. In a unitary quantum theory, the Hilbert space must have a positive-definite inner product. This means there can be no states with negative norm, and any state with zero norm is physically identified with the zero vector.

In our CFT, we can define a natural inner product on the Verma module $V_{c, h}$ by postulating that the adjoint of a Virasoro generator is:

L_{n}^{†} = L_{- n}

The primary state is normalized as $⟨ h ∣ h ⟩ = 1$ . With this, we can compute the inner product of any two states in the module. For example, for two level-1 states:

⟨ h ∣ L_{1} \cdot L_{- 1} ∣ h ⟩ = ⟨ h ∣ ([L_{1}, L_{- 1}] + L_{- 1} L_{1}) ∣ h ⟩ = ⟨ h ∣2 L_{0} ∣ h ⟩ = 2 h ⟨ h ∣ h ⟩ = 2 h

For this state to have a non-negative norm, we need $h \geq 0$ .

The inner products between states at different levels are always zero. For example, $⟨ L_{- 1} h ∣ L_{- 2} h ⟩ = ⟨ h ∣ L_{1} L_{- 2} ∣ h ⟩ = ⟨ h ∣3 L_{- 1} ∣ h ⟩ = 0$ . This means the matrix of inner products, called the Gram matrix $M$ , is block-diagonal, with one block $M^{(N)}$ for each level $N$ .

The condition of unitarity requires that all eigenvalues of each $M^{(N)}$ must be non-negative. If a block $M^{(N)}$ has a zero eigenvalue, its determinant is zero, and the module contains a zero-norm state.

This is the crucial connection:
A null vector at level $N$ is a state $∣ χ ⟩$ with zero norm, $⟨ χ ∣ χ ⟩ = 0$ . Its existence implies that the Gram matrix at that level is singular, i.e., $det M^{(N)} = 0$ .

These zero-norm states are precisely the null vectors we are looking for. They are orthogonal to every state in the module (including themselves), and to build a unitary, irreducible representation, we must quotient them out.

The Kac Determinant Formula

The determinant of the Gram matrix at level $N$ , $det M^{(N)} (c, h)$ , is known as the Kac determinant. In 1978, Victor Kac computed a beautiful closed-form expression for it.

CFT notes

The following sections from the provided notes by Jakob Stern and the anonymous Rutgers author introduce the Gram matrix and the celebrated Kac determinant formula. This formula provides the complete answer to the question of when a Verma module is reducible.

First, please read the first three paragraphs of section 4.3, 'Kac determinant and restrictions imposed by unitarity' in the 'CFT notes' by the anonymous author. This provides a clear motivation for why the determinant of the level-N Gram matrix block is the key object of study.

Now that you've seen the motivation, let's look at the formula itself. The formula is most elegant when the central charge $c$ is parameterized by a single number $m$ :

c (m) = 1 - \frac{6}{m ( m + 1 )}

For any given $c$ , this equation can be solved for $m$ . The formula for the Kac determinant at level $N$ is then given by:

det M^{(N)} (c, h) = α_{N} r, s \in Z^{+} r s \leq N \prod (h - h_{r, s} (m))^{p (N - r s)}

where:

$α_{N}$ is a positive constant that depends only on $N$ .
$p (k)$ is the integer partition function, which counts the number of basis states at level $k$ .
$h_{r, s} (m)$ are special values of the conformal dimension given by the Kac formula: $h_{r, s} (m) = \frac{(( m + 1 ) r - m s ) ^{2} - 1}{4 m ( m + 1 )}$

This formula is a remarkable, non-perturbative result. It tells us that the determinant, which is a complicated polynomial in $h$ and $c$ , has zeros only at the specific values $h = h_{r, s} (m)$ .

Conformal Field Theory

The notes by Jakob Stern provide a very compact presentation of this formula. Let's read it to solidify our understanding.

Please read the section 3.3, 'Kac Determinants', in Stern's notes. Focus on the formula for det M^(N)(c, h) and the associated expressions for c(m) and h_{r,s} under 'Conjecture (Kac)' (this conjecture was later proven by Feigin and Fuchs).

Using the Formula to Identify Null Vectors

The Kac determinant formula provides a complete recipe for finding null vectors.
A null vector appears in the Verma module $V_{c, h}$ if and only if its determinant vanishes. This happens when $h$ is equal to one of the $h_{r, s} (m)$ for some positive integers $r, s$ .

The product in the formula runs over pairs $(r, s)$ such that their product $r s$ is less than or equal to the level $N$ . This means that a new zero, corresponding to $h_{r, s}$ , first appears in the determinant at level $N = r s$ . This signals the emergence of a new null vector at that level.

Let's see this in action with the examples we've already encountered.

Level 1: $N = 1$

The condition on the product is $r s \leq 1$ . The only pair of positive integers that satisfies this is $(r, s) = (1, 1)$ .
The Kac formula gives:

h_{1, 1} (m) = \frac{(( m + 1 ) \cdot 1 - m \cdot 1 ) ^{2} - 1}{4 m ( m + 1 )} = \frac{( 1 ) ^{2} - 1}{4 m ( m + 1 )} = 0

So, $det M^{(1)}$ has a zero at $h = 0$ . This confirms our result from the last lesson: a null vector appears at level 1 if and only if $h = 0$ . The null vector is $∣ χ ⟩ = L_{- 1} ∣ h = 0 ⟩$ .

Level 2: $N = 2$

The condition is $r s \leq 2$ . The pairs of positive integers are $(1, 1), (1, 2), (2, 1)$ .
The determinant is:

det M^{(2)} (c, h) = α_{2} (h - h_{1, 1})^{p (1)} (h - h_{1, 2})^{p (0)} (h - h_{2, 1})^{p (0)}

Using $p (1) = 1$ and $p (0) = 1$ , this simplifies to:

det M^{(2)} (c, h) = α_{2} (h - 0) (h - h_{1, 2}) (h - h_{2, 1})

The new zeros appear at $h = h_{1, 2} (m)$ and $h = h_{2, 1} (m)$ . Let's calculate them:

h_{1, 2} (m) = \frac{(( m + 1 ) - 2 m ) ^{2} - 1}{4 m ( m + 1 )} = \frac{( 1 - m ) ^{2} - 1}{4 m ( m + 1 )} = \frac{m ^{2} - 2 m}{4 m ( m + 1 )} = \frac{m - 2}{4 ( m + 1 )}

h_{2, 1} (m) = \frac{( 2 ( m + 1 ) - m ) ^{2} - 1}{4 m ( m + 1 )} = \frac{( m + 2 ) ^{2} - 1}{4 m ( m + 1 )} = \frac{m ^{2} + 4 m + 3}{4 m ( m + 1 )} = \frac{( m + 1 ) ( m + 3 )}{4 m ( m + 1 )} = \frac{m + 3}{4 m}

So, if $h$ takes either of these two values (for a given $c$ , which fixes $m$ ), a new null vector appears at level 2. This is precisely the condition that one finds by laboriously solving $L_{1} ∣ χ ⟩ = 0$ and $L_{2} ∣ χ ⟩ = 0$ for a generic level-2 state $∣ χ ⟩ = (L_{- 2} + a L_{- 1}^{2}) ∣ h ⟩$ . The Kac formula gives us the answer immediately.

Test your understanding!

At what level $N$ does a null vector corresponding to $h_{2, 2} (m)$ first appear? What is the value of $h_{2, 2} (m)$ ?

Show answer

A null vector corresponding to $h_{r, s}$ first appears at level $N = r s$ . For $(r, s) = (2, 2)$ , this is level $N = 4$ .

The value of $h_{2, 2} (m)$ is:

h_{2, 2} (m) = \frac{(( m + 1 ) 2 - m \cdot 2 ) ^{2} - 1}{4 m ( m + 1 )} = \frac{( 2 m + 2 - 2 m ) ^{2} - 1}{4 m ( m + 1 )} = \frac{2 ^{2} - 1}{4 m ( m + 1 )} = \frac{3}{4 m ( m + 1 )}

The Kac Table: A Map of Reducible Representations

The pairs $(c (m), h_{r, s} (m))$ for which null vectors exist form a grid-like pattern in the $(c, h)$ plane. When $m$ is a positive integer, these correspond to a special class of theories called minimal models, which have a finite number of primary fields. The plot of these $(h_{r, s}, c)$ values is often called the Kac table.

The image below shows this structure. Each curve represents $h_{r, s}$ as a function of $c$ . A Verma module $V_{c, h}$ is reducible if its $(c, h)$ coordinates fall on one of these lines.

Kac table of h_r,s versus c

This plot shows the conformal dimensions h_{r,s} where null vectors appear, plotted as a function of the central charge c. Each line corresponds to a specific pair (r,s). A Verma module characterized by a point (c,h) is reducible if that point lies on one of these curves. The regions between the curves are where Verma modules are irreducible. The special values of c < 1 marked for m=2, 3, 4 correspond to the unitary minimal models.

Conclusion

Today we have unveiled a central tool in the study of 2D CFT. The Kac determinant formula provides a complete and elegant answer to the question of when representations of the Virasoro algebra are reducible.

Key Takeaways:

Unitarity requires the Gram matrix of inner products to be positive semi-definite.
A null vector is a state of zero norm, and its existence corresponds to a zero of the Gram matrix determinant.
The Kac determinant formula gives a closed-form expression for $det M^{(N)} (c, h)$ , the determinant of the Gram matrix at level $N$ .
A null vector appears if and only if the conformal dimension $h$ matches one of the special values $h_{r, s} (m)$ given by the Kac formula.
The first null vector corresponding to the pair $(r, s)$ appears at level $N = r s$ .

With this powerful result in hand, we are now equipped to classify entire conformal field theories. In our next lesson, we will focus on a particularly important set of theories: the unitary minimal models. We will see how combining the Kac formula with the additional constraint of unitarity restricts $c$ and $h$ to a discrete set of values, leading to theories with finite, solvable spectra.

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