Vertex Modules, Rationality, and Logarithmic Regimes: Examples and Counterexamples

In the ordered basis $e_0,e_1$, $N$ has matrix $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ and square zero. Thus $(I+N{)}^2\mathrm{/}2=I\mathrm{/}2+N$, giving the two displayed matrices by [F1]. In particular $(L_F(0)-1\mathrm{/}2)e_1=e_0\mathrm{\ne }0$ and $(L_F(0)-1\mathrm{/}2)e_0=0$. These are an actual length-two conformal Jordan chain, not just a nonsemisimple action of an unrelated algebra element.

On a weighted-degree-$d$ monomial $p$, $L_F(0)(p\otimes e_0)=(d+1\mathrm{/}2)p\otimes e_0$ and $L_F(0)(p\otimes e_1)=(d+1\mathrm{/}2)p\otimes e_1+p\otimes e_0$. The generalized weights are exactly $d+1\mathrm{/}2$, with finite spaces $P_d\otimes E$ and lower bound $1\mathrm{/}2$, by [F1]. Therefore the module meets every grading restriction and is logarithmic by [F2]. Formally on this pair, exponentiating the nilpotent part terminates and gives $z^{L_F(0)}(p\otimes e_1)=z^{d+1\mathrm{/}2}(p\otimes e_1+(\log z)p\otimes e_0)$; this explains the terminology while the module fields themselves remain the ordinary Laurent fields already proved in [F1].

Every state is a scalar multiple of $\mathbf{1}$. The identity field is lower truncated by [F1], and [F2] gives the vacuum and creation identities. Translation covariance holds because both sides are zero by [F3]; locality holds because the identity commutes with itself by [F4]. Thus $V$ satisfies the vertex-algebra axioms.

Every state is $a=\lambda \mathbf{1}$. Linearity and the vacuum property in [F6] give $Y_M(a,z)=\lambda {\mathrm{id}}_M$. Hence every graded subspace of $M$ is a $V$-submodule, and each $M(n)$ in [F7] is a submodule.

Since $Y(\omega ,z)=0$, all $L(n)=0$ and the Virasoro relations hold with central charge $0$; also $L(-1)=0=T$. The grading is $V_0=V=\mathbb{C}\mathbf{1}$ and $V_n=0$ for $n>0$, so $V$ is conformal of CFT type by [F5].

If $M=0$, it is the empty direct sum of simple modules. Otherwise, for each $n$ with $M(n)\mathrm{\ne }0$, [F9] gives a nonempty set of bases of $M(n)$; [A1] chooses one basis in every such degree. Each basis line is a one-dimensional admissible submodule concentrated in degree $n$, hence is simple, and the direct sum of all these lines is $M$. Thus every admissible module is completely reducible, which is rationality by [F8].

Since $A$ is unital, commutative, and associative, the constant field assignment $Y(a,z)b=ab$ has only the mode $a_{(-1)}b=ab$ and all other modes zero. It is lower truncated; the vacuum and creation identities hold because $1b=b$; translation covariance holds because $T=0$ and the field is constant; and locality holds because multiplication operators commute. Thus $V$ is a vertex algebra by [F1].

For any finite-dimensional unital $A$-module $(M,\rho )$, set $Y_M(a,z)=\rho (a)$, a constant field. It is lower truncated and satisfies the vacuum axiom. In its module Jacobi identity, associativity gives $\rho (a)\rho (b)=\rho (ab)=\rho (b)\rho (a)$, so the left side is $(\alpha -\beta )\rho (ab)$, where $\alpha$ and $\beta$ are the first two displayed delta terms; the right side is $\gamma \rho (ab)$, where $\gamma$ is the third term. Identity [F3] gives $\alpha -\beta =\gamma$, so this is a $V$-module by [F2]. Conversely, by the definition of $\mathcal{C}$, its objects are exactly these unital $A$-modules.

For a constant-field object, the only potentially nonzero mode is $a_{(-1)}^M=\rho (a)$, so its $V$-submodules are exactly its $A$-submodules. Let $S$ be a nonzero simple object of $\mathcal{C}$ in the sense of [F5]. The subspace $\epsilon S$ is an $A$-submodule. It cannot equal $S$, since then ${\epsilon }^{2}S=\epsilon S=S$ would contradict ${\epsilon }^2=0$ and $S\mathrm{\ne }0$. Simplicity therefore gives $\epsilon S=0$, so the action factors through $A\mathrm{/}(\epsilon )\cong \mathbb{C}$. Every one-dimensional subspace is then a submodule; simplicity forces $\dim S=1$. Conversely, the one-dimensional module with $\epsilon$ acting by zero is simple. Hence $\mathcal{C}$ has exactly one simple isomorphism class.

In the regular module $A$, the subspace $\epsilon A=\mathbb{C}\epsilon$ and quotient $A\mathrm{/}\epsilon A\cong \mathbb{C}$ are both the simple module just described, giving $0\to \mathbb{C}\epsilon \to A\to \mathbb{C}\to 0$. If it split as $A$-modules, $A$ would be a direct sum of two modules annihilated by $\epsilon$, forcing multiplication by $\epsilon$ on $A$ to be zero; but $\epsilon \cdot 1=\epsilon \mathrm{\ne }0$. Thus this extension does not split, and $\mathcal{C}$ is not semisimple by [F5].

For every candidate conformal vector $\omega \in V$, the field $Y(\omega ,z)$ is constant, so in $Y(\omega ,z)={\sum }_{n}L(n)z^{-n-2}$ the only possibly nonzero mode is $L(-2)$ and $L(0)=0$. Its zero eigenspace is all of $V$, which has dimension two, not $\mathbb{C}\mathbf{1}$; hence no conformal vector makes this a CFT-type algebra under [F4]. The example is therefore an algebraic module-category counterexample, not a counterexample to a rationality theorem for conformal vertex operator algebras.