One-dimensional constant-field model satisfying full local CFT data

With the Euclidean topology, $\mathbb{C}$ is Hausdorff and has a countable basis of rational balls; it is connected because line segments join any two points. Its standard smooth structure makes it an oriented real two-manifold without boundary, and its standard complex atlas consists of all compatible biholomorphic charts. Thus the specified worldsheet has the required Riemann-surface structure.

The normed space $\mathbb{C}$ is complete, Hausdorff and locally convex; addition and scalar multiplication are continuous. Since $\mathcal{D}=\mathcal{H}$, it is dense and is invariant under all declared zero modes and scalar fields. The one-sector direct sum over $\{\ast \}$ is already complete and has the stated norm topology.

For all $m,n\in \mathbb{Z}$, both Virasoro commutators and the mixed commutator have zero left side. Their right sides also vanish: every mode is zero and each central term is multiplied by $c_L=0$ or $c_R=0$. Hence these are representations of both copies with the required central charges.

Each of $L_0$ and ${\bar{L}}_0$ is the zero operator, diagonalizable with spectrum $\{0\}$, which is real and bounded below. The only joint generalized-weight space is ${\mathcal{H}}_{(0,0)}=\mathbb{C}$, and its nilpotent Jordan part is zero. The chosen direct-sum presentation therefore gives discrete spectrum and positive energy.

The vector $\mathrm{\Omega }=1$ is nonzero and every mode with index $n\ge -1$ annihilates it. The functional $ϵ(a)=a$ is continuous and linear in the norm topology and satisfies $ϵ(\mathrm{\Omega })=1$. Since every mode is zero, every vector lies in the common global-vacuum subspace; that subspace is $\mathbb{C}=\mathbb{C}\mathrm{\Omega }$, so the vacuum is unique up to scalar.

Identify $A=\mathbb{C}\mathbf{1}$ with $\mathbb{C}$ by $a\leftrightarrow a\mathbf{1}$, with product $(a\mathbf{1})(b\mathbf{1})=ab\mathbf{1}$, vacuum $\mathbf{1}$, translation $T=0$, conformal vector $\omega =0$, and state-field map $Y_A(a\mathbf{1},z)b\mathbf{1}=ab\mathbf{1}$. The field/truncation condition [F14] holds because only the $(-1)$st product is nonzero. The vacuum and creation conditions [F15] hold because $Y_A(\mathbf{1},z)=\mathrm{Id}$ and $Y_A(a\mathbf{1},z)\mathbf{1}=a\mathbf{1}=e^{zT}(a\mathbf{1})$. Translation covariance [F16] holds because $T$ and the field derivative vanish; locality [F17] holds because multiplication operators commute. The zero conformal vector has all Virasoro modes and central charge zero, with $L(-1)=T$ as required by [F18] and step 3.1. Also $A_0=A=\mathbb{C}\mathbf{1}$ and $A_n=0$ for $n>0$, so its grading is diagonalizable, nonnegative, and finite dimensional with the required degree-zero space; this verifies [F19]. Declare the category of finite-dimensional ordinary $A$-modules and their module homomorphisms as in [F20]. The vacuum axiom and linearity force every module field to be $Y_M(a\mathbf{1},z)=a{\mathrm{Id}}_M$; conversely these fields are lower-truncated and satisfy the vacuum axiom. For the module Jacobi identity, both operator products are $ab{\mathrm{Id}}_M$, and $Y_A(a\mathbf{1},z_0)b\mathbf{1}=ab\mathbf{1}$, so the required identity is $z_0^{-1}\delta ((z_1-z_2)\mathrm{/}{z}_0)-z_0^{-1}\delta ((z_2-z_1)\mathrm{/}(-z_0))=z_2^{-1}\delta ((z_1-z_0)\mathrm{/}{z}_2)$ with the expansions of [F20]. For $k<0$, the coefficient of $z_0^k$ on the left is the difference of the two binomial expansions of the polynomial $(z_1-z_2{)}^{-k-1}$; they coincide because $-k-1$ is a nonnegative integer, so the difference is zero. The right side has only nonnegative powers of $z_0$, hence its coefficient is also zero. For $k\ge 0$, the left coefficient is ${\sum }_{j\ge 0}\left(\genfrac{}{}{0px}{}{k+j}{j}\right)z_1^{-k-1-j}{z}_2^j+(-1{)}^k{\sum }_{j\ge 0}\left(\genfrac{}{}{0px}{}{k+j}{j}\right)z_1^j{z}_2^{-k-1-j}$ by the two prescribed binomial expansions. The right coefficient is ${\sum }_{m\in \mathbb{Z}}(-1{)}^k\left(\genfrac{}{}{0px}{}{m+k}{k}\right)z_1^m{z}_2^{-m-k-1}$; terms $m\ge 0$ give the second sum, terms $m=-k-1-j$ give the first because $\left(\genfrac{}{}{0px}{}{-j-1}{k}\right)=(-j-1)(-j-2)\cdots (-j-k)\mathrm{/}k!=(-1{)}^k\left(\genfrac{}{}{0px}{}{j+k}{k}\right)$, and intermediate exponents $-k\le m\le -1$ have $0\le m+k<k$, so $\left(\genfrac{}{}{0px}{}{m+k}{k}\right)=0$. Thus the scalar module fields satisfy the Jacobi identity. For $m=0$, the module is the zero object, the empty direct sum. For $m>0$, choose a basis $v_1,\dots ,v_m$. Each line $\mathbb{C}{v}_j$ is a submodule because the action is scalar, and it is simple because a one-dimensional space has no proper nonzero subspace; thus every finite-dimensional module is a finite direct sum of simple modules. If a nonzero simple module had dimension greater than one, the span of any nonzero vector would be a proper nonzero submodule, contradicting [F21]. Any two one-dimensional modules are isomorphic: a linear map sending one nonzero basis vector to another is bijective and intertwines scalar actions. Finally, for a short exact sequence in this finite-dimensional category, choose a basis $p_1,\dots ,p_r$ of $P$ and lifts $m_i\in M$. The subspace $S=\mathrm{span}\{m_1,\dots ,m_r\}$ maps isomorphically onto $P$ because $q(m_i)=p_i$, so $M=N\oplus S$. Every subspace is a submodule under scalar action, and $(q{\mathrm{\mid }}_S{)}^{-1}$ is a module homomorphism; hence the sequence splits in the declared category. There is exactly one irreducible isomorphism class, so [F11] gives the rational flag.

The identity map $\sigma :\mathbb{C}\to \mathbb{C}$ is a bijection and sends the identity field label $1$ to $\mathrm{\Omega }=1$. The pointwise field $Y(a;z,\bar{z})$ is scalar multiplication by $a$, so $Y(1;z,\bar{z})={\mathrm{Id}}_{\mathbb{C}}$ and every field preserves $\mathcal{D}$.

For $f\in C_c^{\mathrm{\infty }}(U)$, its coordinate Lebesgue integral is finite and $Y(a;f)$ is scalar multiplication on $\mathcal{D}$. If $\mathrm{supp}f\subseteq K⋐U$, then for every $u\in \mathcal{D}$ and $\lambda \in {\mathcal{H}}^{\mathrm{\prime }}$, $\mathrm{\mid }\lambda (Y(a;f)u)\mathrm{\mid }\le \mathrm{\mid }a\mathrm{\mid }\mathrm{\mid }\lambda (u)\mathrm{\mid }{m}_2(K){\sup }_K\mathrm{\mid }f\mathrm{\mid }$, where $m_2(K)<\mathrm{\infty }$ is coordinate Lebesgue measure. This order-zero bound by the zeroth uniform seminorm proves continuity on each compact-support test-function space, hence the required weak distribution property.

On the vacuum, the vector representative is $Y(a;z,\bar{z})\mathrm{\Omega }=a$ for every $z$. It is continuous near the origin, and along every radial approach $z=r{e}^{i\vartheta }\to 0$ its norm limit is $a=\sigma (a)$.

For every $n$, the covariance commutator on either side is zero because the corresponding Virasoro mode vanishes. Every transformed label ${\sigma }^{-1}(L_{j-1}\sigma (a))$ and ${\sigma }^{-1}({\bar{L}}_{j-1}\sigma (a))$ is also zero, so every summand on both right sides vanishes and each series converges as the constant zero series. The covariance identities therefore hold in every matrix coefficient.

For each ordered pair $a,b$, the product is $Y(a;z,\bar{z})Y(b;w,\bar{w})=ab{\mathrm{Id}}_{\mathbb{C}}$. Since $Y(1;w,\bar{w})={\mathrm{Id}}_{\mathbb{C}}$, the declared single term with coefficient $ab$ is exactly the product on the nonempty radial domain $0<\mathrm{\mid }z-w\mathrm{\mid }<1$.

Equip $\mathcal{H}=\mathbb{C}$ with the positive-definite Hermitian product $⟨u,v⟩=u\bar{v}$, linear in the first argument; indeed $⟨u,u⟩=\mathrm{\mid }u{\mathrm{\mid }}^2>0$ for $u\mathrm{\ne }0$. Multiplication by $a$ has adjoint multiplication by $\bar{a}$, since $⟨au,v⟩=au\bar{v}=⟨u,\bar{a}v⟩$. Every Virasoro mode has adjoint zero, so $L_n^{\ast }=0=L_{-n}$ and ${\bar{L}}_n^{\ast }=0={\bar{L}}_{-n}$. Thus $Y(a;z,\bar{z}{)}^{\ast }=Y(\bar{a};z,\bar{z})$ and coordinate smearing gives $Y(a;f{)}^{\ast }=Y(\bar{a};\bar{f})$. The unitary flag has its required inner product, mode identities and field adjoints.

Every scalar test density in $E_i(V)$ is elementary, namely $1\cdot s(z)d^{2}z$, so the elementary span is the whole test space and hence dense. Since $V_i(a,z)=a$ and ${\mathrm{\Omega }}_i=1$, the prescribed vacuum pairing is exactly $\lambda (\int s{d}^{2}z)=\int \lambda (s(z))d^{2}z$ for every continuous linear $\lambda :\mathbb{C}\to \mathbb{C}$. For an overlap $w=g(z)$, the real Jacobian is $\mathrm{\mid }{g}^{\mathrm{\prime }}(z){\mathrm{\mid }}^2>0$; change of variables gives $\int (P_{ji}s)(w)d^{2}w=\int s(z)d^{2}z$. Thus ${\mathcal{Y}}_i(s)={\mathcal{Y}}_j(P_{ji}s)$, the required finite field identity since $W=I$. A biholomorphism maps compact supports to compact supports. On each fixed compact set its derivatives, inverse derivatives and inverse Jacobian derivatives are bounded, so the chain and product rules bound each output test seminorm by finitely many input seminorms. The pushforward is continuous, with inverse the inverse-coordinate pushforward, and respects extension by zero. The derivative chain rule gives $\mathrm{\mid }(g_{kj}\circ g_{ji}{)}^{\mathrm{\prime }}{\mathrm{\mid }}^{-2}=\mathrm{\mid }{g}_{kj}^{\mathrm{\prime }}\circ g_{ji}{\mathrm{\mid }}^{-2}\mathrm{\mid }{g}_{ji}^{\mathrm{\prime }}{\mathrm{\mid }}^{-2}$, so $P_{ki}=P_{kj}{P}_{ji}$. The $R$ and $W$ cocycles and identity/inverse rules hold because all their factors are identities. The transported vacuum and functional equations hold because every chart has ${\mathrm{\Omega }}_i=1$, ${ϵ}_i(a)=a$ and $W=I$. Along a differentiable coordinate flow the point field and label are constant, so both the finite point-field covariance derivative and the mode covariance identity vanish; the derivative of density transport is exactly the ordinary test-density Jacobian change already accounted for by the change-of-variables identity.

For any $u\in \mathcal{D}$ and $\lambda \in {\mathcal{H}}^{\mathrm{\prime }}$, the OPE matrix coefficient is the single constant $ab\lambda (u)$. Its sum is finite, hence absolutely and locally uniformly convergent on every compact subset of the radial domain. Treating $\bar{z},\bar{w}$ as independent complex variables gives the same constant entire representative; restriction to $\bar{z},\bar{w}$ equal to complex conjugates recovers the Euclidean matrix coefficient. The constant extends across both excluded radial boundaries, although the declared OPE claim uses the open domain only; no branch choice is needed.

The only joint weight is $(h,\bar{h})=(0,0)$, so every field has integral spin $h-\bar{h}=0$. Every finite vacuum correlator is constant on the configuration space, hence single-valued. For disjointly supported test functions $f,g$, the smeared operators are scalar multiples of the identity, so $Y(a;f)Y(b;g)=Y(b;g)Y(a;f)$. Thus the package is bosonic and local.

Let $\theta (z)=\bar{z}$, $U_{+}=\{z:\mathrm{Im}z>0\}$ and $U_{-}=\theta (U_{+})$. For $f\in C_c^{\mathrm{\infty }}(U_{+})$ define $f^{\theta }(z)=\overline{f(\stackrel{ˉ}{z})}$ on $U_{-}$. Define $\mathrm{\Theta }$ antilinearly on the algebra of polynomials of fields smeared in $U_{+}\cup U_{-}$ by $\mathrm{\Theta }(Y(a;f))=Y(\bar{a};f^{\theta })$, $\mathrm{\Theta }(FG)=\mathrm{\Theta }(G)\mathrm{\Theta }(F)$ and $\mathrm{\Theta }(1)=1$. Applying this twice returns each generator because ${\theta }^2$ and complex conjugation square to the identity, and reversing products twice restores their order; hence $\mathrm{\Theta }$ is an involution. Every such polynomial acts as $q\mathrm{Id}$, and its reflection acts as $\bar{q}\mathrm{Id}$: for each generator the change of variable $y\mapsto -y$ gives ${\int }_{U_{-}}{f}^{\theta }(x+iy)dxdy=\overline{{\int }_{U_{+}}f(x+iy)dxdy}$, and reversal leaves scalar products unchanged. If $G$ acts by $rI$, then $B(F,G)=\bar{q}r$. This is conjugate-linear in $F$, linear in $G$, satisfies $B(F,G)=\overline{B(G,F)}$, and has $Q(F)=\mathrm{\mid }q{\mathrm{\mid }}^2\ge 0$; the domain contains the identity and all operators preserve $\mathbb{C}$. This verifies the full reflection-positive form.

In every centered chart the vacuum field is the jointly continuous map $(a,z)\mapsto a$, with limit $a={\sigma }_i(a)$. For two such charts the centered-state compatibility is $a=a$. Every single or iterated OPE is an equality of constant functions with one term at each fusion; transport by $R=I$ leaves labels and coefficients unchanged and transports each declared domain to its coordinate image. Change of variables in step 7.1 transports all smeared equalities, and constant entire continuations agree. Thus restrictions, finite transport, vacuum creation, OPE and continuation satisfy the entire finite coordinate interface, including non-affine coordinate changes.

For $n\ge 2$, let $T$ be any rooted full binary OPE tree with leaves labelled by the insertions at distinct coordinates $z_1,\dots ,z_n$. Define the representative coordinate recursively by $r(i)=z_i$ at a leaf and $r(v)=r(v_R)$ at an internal node $v$ with right child $v_R$; thus each representative is a leaf coordinate in that node's subtree, and sibling representatives are distinct. Let ${\mathcal{D}}_T$ be the set of pairwise-distinct configurations satisfying $0<\mathrm{\mid }r(v_L)-r(v_R)\mathrm{\mid }<1$ at every internal node. It is open, being the finite intersection of the configuration space with inverse images of the open annulus $0<\mathrm{\mid }z\mathrm{\mid }<1$. In postorder, each node fuses its two child fields inside the declared pair domain, and its parent then fuses the resulting fields using that parent's child representatives; these are the required nested domains. The set is nonempty: choosing $z_j=j\mathrm{/}(8(n+1))$ makes all leaves and representatives distinct and every internal distance less than $1\mathrm{/}8$. This same choice works for every tree, so any two channel domains have a common nonempty open neighborhood. Each substitution has one term, so step 7.2 gives absolute, locally uniform convergence at every stage.

For any finite labels $a_1,\dots ,a_n$ with $n\ge 1$, the vacuum correlator is $ϵ(Y(a_1)\cdots Y(a_n)\mathrm{\Omega })={\prod }_{j=1}^n{a}_j$; for the empty insertion set it is $ϵ(\mathrm{\Omega })=1$. Every OPE tree multiplies the same scalars; associativity and commutativity of complex multiplication make all tree outputs equal. The correlator is constant and entire in the independent complexified insertion variables, so analytic continuation along any path is unique and every crossed channel, including all three four-point pairings, agrees.

The established flags are unitary, reflection-positive, rational, bosonic and discrete-spectrum, while the continuous-spectrum flag is false by step 4.1. The construction verifies only the stated genus-zero data; no higher-genus sewing, path-integral measure, physical realization, or equivalence with another axiom system is a premise or conclusion.