Modules and fusion for the one-dimensional vertex algebra

Only the $(-1)$st mode is nonzero. Truncation, creation and vacuum hold immediately, translation has both sides zero, and scalar fields commute, so locality holds with exponent zero. The zero conformal field has $L(n)=0$ for every $n$ and satisfies Virasoro with $c=0$; $L(-1)=T$, $V_0=\mathbb{C}\mathbf{1}$ and all other graded pieces vanish. Thus $V$ has exactly the stipulated conformal structure.

On any module, linearity and vacuum force $Y_E(a\mathbf{1},x)=a{1}_E$, hence $Y_E(\omega ,x)=0$ and $L_E(n)=0$. Conversely this action satisfies Jacobi because each term is a scalar multiple of the basic formal identity $x_0^{-1}\delta \left(\frac{x_1-x_2}{x_0}\right)-x_0^{-1}\delta \left(\frac{x_2-x_1}{-x_0}\right)=x_2^{-1}\delta \left(\frac{x_1-x_0}{x_2}\right)\mathrm{.}$ This identity follows coefficientwise by expanding each integer power with the binomial convention. The only possible generalized eigenvalue of the zero $L_E(0)$ is zero. Thus grading restriction is exactly finite-dimensionality of $E$; its real lower weight bound is zero. Every linear map between these spaces commutes with all fields.

Since $L_E(-1)=0$, the derivative identity forces ${\mathrm{\partial }}_x\mathcal{Y}(e,x)f=0$. Write each coefficient as $x^a{P}_a(\log x)$ with $P_a$ a finite polynomial. Its derivative is $x^{a-1}(a{P}_a+P_a^{\mathrm{\prime }})$. For $a\mathrm{\ne }0$, its top coefficient forces $P_a=0$; for $a=0$ one has $P_a^{\mathrm{\prime }}=0$. Distinct exponents remain distinct after differentiation. The only remaining term is therefore a constant $B(e,f)$, bilinear in $e,f$. Conversely any such $B$ has the required truncation, zero derivative and zero Möbius brackets. Its Jacobi identity is the scalar delta identity in step 2.1 multiplied by $aB(e,f)$ for the state $a\mathbf{1}$. Thus the assignment $B\mapsto {\mathcal{Y}}_B$ is a linear bijection, and bases of $E,F,G$ show its dimension is the displayed product.

The completion of $G$ is $G$, since it has only weight zero. With the specified trivial auxiliary grading, every linear map preserves auxiliary degree and its two input degrees add to zero. A P(z)-map is consequently just a linear map $E\otimes F\to G$: weight compatibility is automatic and lower truncation is automatic outside that single weight. Its Jacobi identity reduces to $x_0^{-1}\delta \left(\frac{x_1-z}{x_0}\right)-z^{-1}\delta \left(\frac{x_1-x_0}{z}\right)=x_0^{-1}\delta \left(\frac{z-x_1}{-x_0}\right),$ multiplied by its value and the scalar state; this is the identity in step 2.1 with $x_2=z$, with the prescribed formal expansions. All three brackets vanish. Every linear map is therefore admitted. Composition with the identity of $E\otimes F$ gives precisely the required natural bijection of Hom spaces. This proves the P(z)-universal property directly for every $z\mathrm{\ne }0$.

Because $V_{+}=0$, $C_1(E)=0$ and its quotient has dimension $\dim E$. Each strict weight cutoff is either zero or $E$, hence finite dimensional. The opposite transform fixes $\mathbf{1}$, so its transposed scalar field on $E^{\ast }$ is again scalar multiplication, and evaluation $E\to E^{\ast \ast }$ is a module isomorphism by finite-dimensional linear algebra. Every finite product or iterate of the operators from step 3.1 is an actual finite composition of bilinear maps, independent of all insertion variables. There is a single nonzero weight projection, so the projected series are finite. Matrix coefficients converge absolutely, extend as single-valued constant functions, and have a regular expansion with exponent zero and logarithmic degree zero at every boundary divisor.

The associator sends $(e\otimes f)\otimes g$ to $e\otimes (f\otimes g)$, unit maps are scalar multiplication, and the braiding sends $e\otimes f$ to $f\otimes e$. Their inverses are the reverse rebracketing, insertion of $1$, and the reverse flip. They are linear module maps and natural because their formulas commute with arbitrary linear maps on the factors. Both routes through the associativity pentagon send $((e\otimes f)\otimes g)\otimes h$ to $e\otimes (f\otimes (g\otimes h))$; both triangle routes send $(e\otimes a)\otimes f$ to $e\otimes af$. In the first hexagon both routes send $(e\otimes f)\otimes g$ to $f\otimes (g\otimes e)$, and in the second both send $e\otimes (f\otimes g)$ to $(g\otimes e)\otimes f$. Pure tensors span, so these verify all coherence equations. Choosing a basis $e_i$ of $E$ with dual $e_i^{\ast }$, evaluation $e_i^{\ast }\otimes e_j\mapsto {\delta }_{ij}$ and coevaluation $1\mapsto {\sum }_i{e}_i\otimes e_i^{\ast }$ have triangle composites $e\mapsto {\sum }_i{e}_i{e}_i^{\ast }(e)=e$ and $\lambda \mapsto {\sum }_i\lambda (e_i)e_i^{\ast }=\lambda$. These formulas also cover dimension zero by the empty sum. Hence this particular example has duality by direct verification.