[proofplan]
We prove both directions using the [moment-cumulant formula for free cumulants](/theorems/7107). The main combinatorial tool is the cumulant-of-products formula for an interval partition, which we prove directly by comparing moment-cumulant expansions. For the implication from freeness to vanishing mixed cumulants, we induct on the order of the cumulant, first removing scalar parts and then grouping consecutive entries from the same subalgebra. For the converse, we expand an alternating centered mixed moment over noncrossing partitions; mixed blocks vanish by hypothesis, while color-respecting noncrossing partitions contain a centered singleton factor.
[/proofplan]
[step:Fix the noncrossing notation and prove the cumulant-of-products formula]
For $n \in \mathbb N$, let $NC(n)$ denote the lattice of noncrossing partitions of $\{1,\dots,n\}$, with maximal element $1_n=\{\{1,\dots,n\}\}$. If $\pi \in NC(n)$ and $a_1,\dots,a_n \in \mathcal A$, write
\begin{align*}
\kappa_\pi[a_1,\dots,a_n] := \prod_{V \in \pi}\kappa_{|V|}(a_{v_1},\dots,a_{v_{|V|}})
\end{align*}
where $V=\{v_1<\cdots<v_{|V|}\}$.
Let $\sigma=\{I_1,\dots,I_m\}$ be an interval partition of $\{1,\dots,n\}$, listed in increasing order. For each $1 \leq \ell \leq m$, define the product
\begin{align*}
b_\ell := \prod_{j \in I_\ell}^{\nearrow} a_j \in \mathcal A
\end{align*}
where the arrow means that the factors are multiplied in increasing order of the index $j$. Define
\begin{align*}
S(\sigma):=\{\pi \in NC(n): \pi \vee \sigma = 1_n\}.
\end{align*}
We claim that
\begin{align*}
\kappa_m(b_1,\dots,b_m)=\sum_{\pi \in S(\sigma)}\kappa_\pi[a_1,\dots,a_n].
\end{align*}
To prove this, first apply the moment-cumulant formula to the word $a_1\cdots a_n$:
\begin{align*}
\varphi(a_1\cdots a_n)=\sum_{\pi \in NC(n)}\kappa_\pi[a_1,\dots,a_n].
\end{align*}
Since $b_1\cdots b_m=a_1\cdots a_n$, the same moment is also
\begin{align*}
\varphi(b_1\cdots b_m)=\sum_{\rho \in NC(m)}\kappa_\rho[b_1,\dots,b_m].
\end{align*}
For a fixed $\rho \in NC(m)$, expanding each cumulant over the products indexed by the interval blocks of $\sigma$ gives exactly the sum of those $\kappa_\pi[a_1,\dots,a_n]$ for which the partition of interval blocks induced by $\pi \vee \sigma$ is $\rho$. Thus the contribution corresponding to $\rho=1_m$ is precisely the sum over $\pi \in NC(n)$ satisfying $\pi \vee \sigma=1_n$. This is the displayed formula.
[guided]
The point of this step is to make precise how a cumulant of products decomposes into cumulants of the original individual factors. We are given an interval partition
\begin{align*}
\sigma=\{I_1,\dots,I_m\}
\end{align*}
of $\{1,\dots,n\}$, and each interval $I_\ell$ is collapsed into a single product
\begin{align*}
b_\ell := \prod_{j \in I_\ell}^{\nearrow} a_j.
\end{align*}
The product $b_1\cdots b_m$ is the original word $a_1\cdots a_n$, only with parentheses inserted according to the intervals.
The free cumulants are defined by the moment-cumulant formula over noncrossing partitions. Applied to the original word, this gives
\begin{align*}
\varphi(a_1\cdots a_n)=\sum_{\pi \in NC(n)}\kappa_\pi[a_1,\dots,a_n].
\end{align*}
Applied to the shorter word $b_1,\dots,b_m$, it gives
\begin{align*}
\varphi(b_1\cdots b_m)=\sum_{\rho \in NC(m)}\kappa_\rho[b_1,\dots,b_m].
\end{align*}
These two moments are equal because $b_1\cdots b_m=a_1\cdots a_n$.
Now ask which terms in the expansion of the original word contribute to the top cumulant $\kappa_m(b_1,\dots,b_m)$. A partition $\pi \in NC(n)$ connects the original positions, while $\sigma$ records which positions were grouped into the same product $b_\ell$. After collapsing each interval of $\sigma$ to one point, the partition $\pi$ induces the partition of the collapsed points determined by $\pi \vee \sigma$. Therefore the terms contributing to the one-block cumulant of the collapsed variables are exactly those for which the collapse is one block:
\begin{align*}
\pi \vee \sigma = 1_n.
\end{align*}
Hence
\begin{align*}
\kappa_m(b_1,\dots,b_m)=\sum_{\pi \in S(\sigma)}\kappa_\pi[a_1,\dots,a_n],
\end{align*}
where
\begin{align*}
S(\sigma):=\{\pi \in NC(n): \pi \vee \sigma=1_n\}.
\end{align*}
[/guided]
[/step]
[step:Show that cumulants with a unit entry vanish in order at least two]
We record the unit-entry consequence of the moment-cumulant formula. For every $r \geq 2$, every $1 \leq q \leq r$, and every $x_1,\dots,x_{q-1},x_{q+1},\dots,x_r \in \mathcal A$,
\begin{align*}
\kappa_r(x_1,\dots,x_{q-1},1_{\mathcal A},x_{q+1},\dots,x_r)=0.
\end{align*}
Indeed, use induction on $r$. The case $r=2$ follows by expanding
\begin{align*}
\varphi(x1_{\mathcal A})=\kappa_2(x,1_{\mathcal A})+\kappa_1(x)\kappa_1(1_{\mathcal A})
\end{align*}
and using $\kappa_1(y)=\varphi(y)$ and $\varphi(1_{\mathcal A})=1$. For the induction step, expand the moment containing the unit entry. Terms in which the unit is a singleton reproduce the moment-cumulant expansion of the word with that unit removed. Terms in which the unit belongs to a block of size at least two contain a lower-order cumulant with a unit entry, except for the possible one-block term. The induction hypothesis cancels all lower-order such terms, and comparing both expansions leaves only the one-block cumulant, which must therefore be zero.
[/step]
[step:Use freeness to show mixed cumulants vanish by induction]
Assume that the unital subalgebras $(\mathcal A_i)_{i \in I}$ are freely independent. We prove by induction on $n \geq 2$ that every $n$-th cumulant with entries from not-all-equal specified subalgebras vanishes.
Fix $n \geq 2$, assume the assertion is known for all orders $<n$, and choose indices $i_1,\dots,i_n \in I$ not all equal with elements $a_j \in \mathcal A_{i_j}$. By multilinearity and the unit-entry vanishing from the previous step, replacing each $a_j$ by
\begin{align*}
a_j^\circ := a_j-\varphi(a_j)1_{\mathcal A}
\end{align*}
does not change $\kappa_n(a_1,\dots,a_n)$, because $n \geq 2$. Thus it suffices to prove the result when every $a_j$ is centered:
\begin{align*}
\varphi(a_j)=0
\end{align*}
for $1 \leq j \leq n$.
Group the sequence $1,\dots,n$ into maximal consecutive intervals
\begin{align*}
\sigma=\{I_1,\dots,I_m\}
\end{align*}
on which the index $i_j$ is constant. Since the indices are not all equal, $m \geq 2$, and by maximality adjacent intervals correspond to different subalgebras. For each $1 \leq \ell \leq m$, let $r_\ell \in I$ be the unique index such that $a_j \in \mathcal A_{r_\ell}$ for $j \in I_\ell$, and define
\begin{align*}
b_\ell := \prod_{j \in I_\ell}^{\nearrow} a_j \in \mathcal A_{r_\ell}.
\end{align*}
Now define the centered block variable
\begin{align*}
c_\ell := b_\ell-\varphi(b_\ell)1_{\mathcal A} \in \mathcal A_{r_\ell}.
\end{align*}
The variables $c_1,\dots,c_m$ are centered and lie in alternating free subalgebras, so freeness gives
\begin{align*}
\varphi(c_1\cdots c_m)=0.
\end{align*}
Expanding this moment over $NC(m)$,
\begin{align*}
0=\sum_{\rho \in NC(m)}\kappa_\rho[c_1,\dots,c_m].
\end{align*}
For $\rho \neq 1_m$, either some block of $\rho$ contains positions whose associated indices $r_\ell$ are not all equal, in which case the corresponding lower-order mixed cumulant vanishes by the induction hypothesis, or every block is contained in a single index class. In the latter case, since adjacent indices $r_\ell$ are distinct, the standard interval-block property for noncrossing partitions gives a singleton block; its factor is $\kappa_1(c_\ell)=\varphi(c_\ell)=0$. Hence every term with $\rho \neq 1_m$ vanishes, and therefore
\begin{align*}
\kappa_m(c_1,\dots,c_m)=0.
\end{align*}
Since $m \geq 2$, replacing $c_\ell$ by $b_\ell=c_\ell+\varphi(b_\ell)1_{\mathcal A}$ does not change the cumulant, again by the unit-entry vanishing. Thus
\begin{align*}
\kappa_m(b_1,\dots,b_m)=0.
\end{align*}
[/step]
[step:Isolate the original cumulant from the grouped cumulant]
Apply the cumulant-of-products formula from the first step to the interval partition $\sigma=\{I_1,\dots,I_m\}$. It gives
\begin{align*}
0=\kappa_m(b_1,\dots,b_m)=\sum_{\pi \in S(\sigma)}\kappa_\pi[a_1,\dots,a_n].
\end{align*}
The partition $1_n$ belongs to $S(\sigma)$, and its contribution is the desired cumulant
\begin{align*}
\kappa_{1_n}[a_1,\dots,a_n]=\kappa_n(a_1,\dots,a_n).
\end{align*}
Let $\pi \in S(\sigma)$ with $\pi \neq 1_n$. Because $\pi \vee \sigma=1_n$, the partition $\pi$ must contain at least one block meeting two of the maximal intervals $I_\ell$ with different associated subalgebras. The corresponding cumulant factor is a lower-order mixed cumulant, since $\pi \neq 1_n$, and it vanishes by the induction hypothesis. Hence every term in the sum except the $1_n$ term vanishes. Therefore
\begin{align*}
\kappa_n(a_1,\dots,a_n)=0.
\end{align*}
This completes the induction and proves that freeness implies vanishing of all mixed free cumulants.
[/step]
[step:Use vanishing mixed cumulants to recover alternating centered mixed moments]
Conversely, assume that every mixed free cumulant vanishes. To prove freeness, let $n \in \mathbb N$, choose indices $i_1,\dots,i_n \in I$ satisfying
\begin{align*}
i_j \neq i_{j+1}
\end{align*}
for $1 \leq j < n$, and choose centered elements $a_j \in \mathcal A_{i_j}$ with
\begin{align*}
\varphi(a_j)=0.
\end{align*}
We must prove
\begin{align*}
\varphi(a_1\cdots a_n)=0.
\end{align*}
By the moment-cumulant formula,
\begin{align*}
\varphi(a_1\cdots a_n)=\sum_{\pi \in NC(n)}\kappa_\pi[a_1,\dots,a_n].
\end{align*}
Fix $\pi \in NC(n)$. If some block $V=\{v_1<\cdots<v_s\}$ of $\pi$ contains positions whose indices $i_{v_1},\dots,i_{v_s}$ are not all equal, then the corresponding factor
\begin{align*}
\kappa_s(a_{v_1},\dots,a_{v_s})
\end{align*}
is a mixed cumulant and is zero by hypothesis. If every block of $\pi$ contains only positions from a single subalgebra, then $\pi$ has a singleton block. Indeed, if no singleton block existed, the noncrossing interval-block argument would produce two adjacent positions in the same block; but adjacent positions have different indices, contradicting the assumption that every block is contained in one index class. For such a singleton block $\{q\}$, the corresponding factor is
\begin{align*}
\kappa_1(a_q)=\varphi(a_q)=0.
\end{align*}
Thus every summand in the moment-cumulant expansion is zero, and consequently
\begin{align*}
\varphi(a_1\cdots a_n)=0.
\end{align*}
[guided]
We now prove the reverse implication directly from the definition of free independence. Freeness asks us to start with an alternating word
\begin{align*}
a_1\cdots a_n
\end{align*}
where $a_j \in \mathcal A_{i_j}$, adjacent indices are different, and every $a_j$ is centered:
\begin{align*}
\varphi(a_j)=0.
\end{align*}
The goal is to prove that the mixed moment is zero.
The moment-cumulant formula expresses this moment as
\begin{align*}
\varphi(a_1\cdots a_n)=\sum_{\pi \in NC(n)}\kappa_\pi[a_1,\dots,a_n].
\end{align*}
So it is enough to show that each summand indexed by $\pi \in NC(n)$ vanishes.
Take one partition $\pi$. There are two possibilities. First, some block
\begin{align*}
V=\{v_1<\cdots<v_s\}
\end{align*}
contains entries from at least two different subalgebras. Then the factor
\begin{align*}
\kappa_s(a_{v_1},\dots,a_{v_s})
\end{align*}
is a mixed free cumulant, so it is zero by the hypothesis of this direction. Since $\kappa_\pi[a_1,\dots,a_n]$ is a product over the blocks of $\pi$, the entire product is zero.
Second, every block of $\pi$ is contained inside a single index class. We claim that $\pi$ must then have a singleton block. Suppose instead that every block had size at least two. A standard elementary property of noncrossing partitions says that such a partition has a block containing two adjacent positions. One way to see this is to choose a block whose convex hull has minimal positive length; if it contains a gap, noncrossing forces another non-singleton block strictly inside that gap, contradicting minimality. Thus some adjacent positions $q$ and $q+1$ lie in the same block. But the word is alternating, so $i_q \neq i_{q+1}$, which contradicts the assumption that each block stays inside one index class. Hence a singleton block $\{q\}$ exists. Its cumulant factor is
\begin{align*}
\kappa_1(a_q)=\varphi(a_q)=0.
\end{align*}
Again the whole product $\kappa_\pi[a_1,\dots,a_n]$ is zero.
Since every noncrossing partition falls into one of these two cases, every term in the expansion of $\varphi(a_1\cdots a_n)$ vanishes. Therefore
\begin{align*}
\varphi(a_1\cdots a_n)=0.
\end{align*}
This is exactly the alternating centered mixed moment condition defining free independence.
[/guided]
[/step]
[step:Conclude the equivalence]
The previous step proves that the assumed vanishing of mixed free cumulants forces all alternating centered mixed moments to vanish, which is precisely free independence of the family $(\mathcal A_i)_{i \in I}$. Together with the induction proving the forward implication, this establishes the equivalence.
[/step]
