[proofplan] We use the reduced tensor decomposition of full Fock space with respect to the orthogonal family $(H_i)_{i \in I}$. The key point is that, relative to a fixed colour $i$, every operator in $B_i$ acts only on the initial block of $H_i$-letters. Therefore a centered operator in $B_i$ sends any vector whose first colour is different from $i$ into vectors whose first colour is $i$, with no vacuum component. Applying this observation from right to left to an alternating centered product shows that the final vector is orthogonal to the vacuum, so its vacuum expectation is zero. [/proofplan] [step:Decompose Fock space into initial colour blocks] Let $K$ denote the orthogonal complement in $H$ of the closed linear span of $\bigcup_{i \in I} H_i$. We regard $K$ as an additional colour, distinct from all elements of $I$. A reduced colour word is a finite sequence $\alpha = (\alpha_1,\dots,\alpha_m)$ with entries in $I \cup \{0\}$ such that consecutive entries are distinct, where colour $0$ denotes $K$. For each colour $\alpha_r$, define $E_{\alpha_r} := H_{\alpha_r}$ if $\alpha_r \in I$ and $E_0 := K$. For a reduced colour word $\alpha = (\alpha_1,\dots,\alpha_m)$, define \begin{align*} \mathcal{K}_\alpha := \mathcal{F}(E_{\alpha_1})^\circ \otimes \cdots \otimes \mathcal{F}(E_{\alpha_m})^\circ, \end{align*} where $\mathcal{F}(E)^\circ := \bigoplus_{r=1}^{\infty} E^{\otimes r}$ is the non-vacuum part of the full Fock space over a closed subspace $E \subset H$. The multiplication map that concatenates tensor blocks gives a unitary identification \begin{align*} \mathcal{F}(H) \cong \mathbb{C}\Omega \oplus \bigoplus_{\alpha} \mathcal{K}_\alpha, \end{align*} where the direct sum is over all reduced colour words $\alpha$. [guided] The point of this decomposition is to group consecutive tensor factors coming from the same orthogonal subspace. A vector such as \begin{align*} h_1 \otimes h_2 \otimes h_3 \otimes h_4 \end{align*} with $h_1,h_2 \in H_i$ and $h_3,h_4 \in H_j$, $i \neq j$, belongs to the block \begin{align*} \mathcal{F}(H_i)^\circ \otimes \mathcal{F}(H_j)^\circ. \end{align*} This grouping is orthogonal because the subspaces $H_i$ are pairwise orthogonal, and tensor products of orthogonal one-particle subspaces are orthogonal in full Fock space. The extra colour $0$ accounts for the part of $H$ not generated by the $H_i$; it will not be produced by the operators in the proof, but including it makes the decomposition a decomposition of all of $\mathcal{F}(H)$. Thus every non-vacuum vector in the algebraic tensor subspace generated by the colour decomposition has a well-defined first colour. The vacuum vector $\Omega$ is kept separate, since it has no first colour. [/guided] [/step] [step:Identify how one colour algebra acts on the initial block] Fix $i \in I$. Let $\mathcal{R}_i$ be the closed subspace of $\mathcal{F}(H)$ spanned by $\Omega$ and all reduced blocks whose first colour is not $i$. Define the unitary map \begin{align*} U_i: \mathcal{F}(H_i) \otimes \mathcal{R}_i \to \mathcal{F}(H) \end{align*} by concatenating the $H_i$-block with the remaining reduced block and then reducing adjacent blocks of the same colour. Under this unitary identification, for every $f \in H_i$, \begin{align*} U_i^{-1}\ell(f)U_i = \ell_i(f) \otimes I_{\mathcal{R}_i}, \end{align*} where $\ell_i(f) \in \mathcal{L}(\mathcal{F}(H_i))$ is the left creation operator on $\mathcal{F}(H_i)$ and $I_{\mathcal{R}_i}$ is the identity operator on $\mathcal{R}_i$. Taking [Hilbert space](/page/Hilbert%20Space) adjoints gives \begin{align*} U_i^{-1}\ell(f)^*U_i = \ell_i(f)^* \otimes I_{\mathcal{R}_i}. \end{align*} Therefore every $T \in B_i$ has the form \begin{align*} U_i^{-1}TU_i = A \otimes I_{\mathcal{R}_i} \end{align*} for some $A$ in the unital algebra of $\mathcal{L}(\mathcal{F}(H_i))$ generated by $\{\ell_i(f)+\ell_i(f)^*: f \in H_i\}$. [/step] [step:Centered one colour operators create a non-vacuum initial block] [claim:Centered colour operators send nonmatching first colours to matching first colours] Let $i \in I$ and let $T \in B_i$ satisfy $\varphi(T)=0$. If $\xi \in \mathcal{R}_i$, then $T\xi$ belongs to the closed span of reduced blocks whose first colour is $i$. [/claim] [proof] By the previous step, there exists an operator $A \in \mathcal{L}(\mathcal{F}(H_i))$ in the corresponding one-colour algebra such that \begin{align*} U_i^{-1}TU_i = A \otimes I_{\mathcal{R}_i}. \end{align*} Let $\Omega_i$ denote the vacuum vector in $\mathcal{F}(H_i)$. Under the unitary $U_i$, every vector $\xi \in \mathcal{R}_i$ is represented as $\Omega_i \otimes \xi$. Hence \begin{align*} U_i^{-1}T\xi = A\Omega_i \otimes \xi. \end{align*} Since $U_i(\Omega_i \otimes \Omega)=\Omega$, the vacuum expectation of $T$ is \begin{align*} \varphi(T) = (T\Omega,\Omega)_{\mathcal{F}(H)} = (A\Omega_i,\Omega_i)_{\mathcal{F}(H_i)}. \end{align*} The hypothesis $\varphi(T)=0$ therefore implies $A\Omega_i \perp \Omega_i$, so \begin{align*} A\Omega_i \in \mathcal{F}(H_i)^\circ. \end{align*} Consequently $A\Omega_i \otimes \xi$ has a non-vacuum initial block in colour $i$. Applying $U_i$ shows that $T\xi$ lies in the closed span of reduced blocks whose first colour is $i$. [/proof] [/step] [step:Apply the centered colour action to an alternating product] Let $n \in \mathbb{N}$, let $i_1,\dots,i_n \in I$ satisfy $i_k \neq i_{k+1}$ for $1 \leq k < n$, and let $T_k \in B_{i_k}$ satisfy $\varphi(T_k)=0$ for each $k$. We prove by backward induction on $k$ that \begin{align*} T_kT_{k+1}\cdots T_n\Omega \end{align*} belongs to the closed span of reduced blocks whose first colour is $i_k$. For $k=n$, the vector $\Omega$ belongs to $\mathcal{R}_{i_n}$. By the claim applied to $T_n \in B_{i_n}$, the vector $T_n\Omega$ has first colour $i_n$. Assume the assertion has been proved for $k+1$. Then $T_{k+1}\cdots T_n\Omega$ lies in the closed span of reduced blocks whose first colour is $i_{k+1}$. Since $i_k \neq i_{k+1}$, this vector belongs to $\mathcal{R}_{i_k}$. Applying the claim to $T_k \in B_{i_k}$ gives that \begin{align*} T_kT_{k+1}\cdots T_n\Omega \end{align*} lies in the closed span of reduced blocks whose first colour is $i_k$. The induction is complete. [/step] [step:Take the vacuum coefficient and conclude freeness] The vector $T_1T_2\cdots T_n\Omega$ lies in the closed span of reduced blocks whose first colour is $i_1$. Every such reduced block is orthogonal to the vacuum vector $\Omega$. Hence \begin{align*} \varphi(T_1T_2\cdots T_n) = (T_1T_2\cdots T_n\Omega,\Omega)_{\mathcal{F}(H)} = 0. \end{align*} This is exactly the centered alternating product condition for free independence of the family $(B_i)_{i \in I}$ with respect to the vacuum state $\varphi$. [/step]