It is convenient at this point to define the restriction of the vector
\(\textbf{A}\) to a subset of components, i.e., to the subspace spanned by a subset of indices. E.g.,
$$\begin{aligned} \textbf{A}^{(i_1 i_2)\,(j_1 j_2)\,(k_1 k_2)} = \{a_{i j k}\,a_{i' j' k'}\}\ \ \text {with}\ \ i,i'\in \{i_1, i_2\},\, j,j'\in \{j_1, j_2\},\, k,k'\in \{k_1, k_2\}.\nonumber \\ \end{aligned}$$
(62)
Note that the action of
\(P_i, P_j\) or
\(P_k\) on this restricted vector keeps it inside the same subspace. In this notation, the previous result for qutrits reads as follows. Given a state such that
\((\mathbb {1}-P_i)(\mathbb {1}-P_j)\textbf{A}= 0\) and
$$\begin{aligned} (\mathbb {1}-P_j)\textbf{A}^{(i_1 i_2)\,(j_1 j_2)\,(k_1 k_2)}\ne 0 \end{aligned}$$
(63)
(for some component(s)), then
$$\begin{aligned} (\mathbb {1}-P_i)\textbf{A}^{(i_1 i_2 i_3)\,(j_1 j_2 j_3)\,(k_1 k_2 k_3)}= 0. \end{aligned}$$
(64)
To extend this result to higher dimensions, let us consider the multipartite Hilbert space
$$\begin{aligned} {\mathcal {H}} = {\mathcal {H}}_{1}\otimes {\mathcal {H}}_{2}\otimes {\mathcal {H}}_{3}, \end{aligned}$$
(65)
with
\(\text {dim}\, {\mathcal {H}}_n=d_n\ge 4\). Our hypothesis is again Eqs. (
57) and (
63) in some subspace corresponding to the indices
\(\{(i_1 i_2)\,(j_1 j_2)\,(k_1 k_2)\}\), i.e., some subspace of three qubits. Let us consider the addition of a fourth set of indices
\(\{i_4 j_4 k_4\}\). Our goal is to show that
$$\begin{aligned} (\mathbb {1}-P_i)\textbf{A}^{(i_1 i_2 i_3 i_4)\,(j_1 j_2 j_3 j_4)\,(k_1 k_2 k_3 k_4)}= 0. \end{aligned}$$
(66)
Since we have shown that we can go from Eqs. (
63) and (
64) for
any new indices
\(i_3, j_3, k_3\), it follows that Eq. (
66) is satisfied by many subsets of components, namely those corresponding to extend the three qubits of Eq. (
63) to any three qutrits:
$$\begin{aligned} (\mathbb {1}-P_i)\textbf{A}^{(i_1 i_2 i_3)\,(j_1 j_2 j_3)\,(k_1 k_2 k_3)}= & {} 0, \nonumber \\ (\mathbb {1}-P_i)\textbf{A}^{(i_1 i_2 i_3)\,(j_1 j_2 j_3 )\,(k_1 k_2 k_4)}= & {} 0, \nonumber \\ (\mathbb {1}-P_i)\textbf{A}^{(i_1 i_2 i_3)\,(j_1 j_2 j_4)\,(k_1 k_2 k_4)}= & {} 0, \\ \text {etc.}\hspace{3cm}{} & {} \nonumber \end{aligned}$$
(67)
Hence, the only components to check are those
\(A_{i j k ;\, i' j' k'}=(a_{i j k})\, (a_{i' j' k'})\) where
\(i=i_3, i'=i_4\) (or vice versa) and/or
\(j=j_3, j'=j_4\) (or vice versa) and/or
\(k=k_3, k'=k_4\) (or vice versa). Let
\(A_{pqr; p'q'r'}= (a_{p q r})\, (a_{p' q' r'})\) be one of such components. We have to show that
\((\mathbb {1}-P_i)A_{pqr; p'q'r'}=0\), i.e.,
$$\begin{aligned} (a_{p q r})\, (a_{p' q' r'})=(a_{p' q r})\, (a_{p q' r'}). \end{aligned}$$
(68)
From Eq. (
63), there must be some component of
\(\textbf{A}^{(i_1 i_2)\,(j_1 j_2)\,(k_1 k_2)}\) different from zero. Suppose
\(A_{i_1 j_1 k_1 ;\, i_2 j_2 k_2}=(a_{i_1 j_1 k_1})\, (a_{i_2 j_2 k_2})\ne 0\) (the argument goes the same for any other component). Then the following chain of equalities follow from Eq. (
67):
$$\begin{aligned} (a_{i_1 j_1 k_1})\, (a_{i_2 j_2 k_2})\, (a_{p q r})\, (a_{p' q' r'})= & {} (a_{p j_1 k_1})\, (a_{p' j_2 k_2})\, (a_{i_1 q r })\, (a_{i_2 p' r'})\nonumber \\ {}= & {} (a_{i_2 j_1 k_1})\, (a_{i_1 j_2 k_2})\, (a_{p' q r})\, (a_{p q' r' }) \nonumber \\ {}= & {} (a_{i_1 j_1 k_1})\, (a_{i_2 j_2 k_2})\, (a_{p' q r})\, (a_{p q' r'}). \end{aligned}$$
(69)
Canceling the common factor in the first and latter expression, which by hypothesis is different from zero, we obtain
$$\begin{aligned} (a_{p q r})\, (a_{p' q' r'})=(a_{p' q r})\, (a_{p q' r'}) \end{aligned}$$
(70)
as desired. Consequently, Eq. (
66) holds.
The extension of the previous result to any other set of indices is straightforward. E.g., if we enlarge the dimension of
\({\mathcal {H}}_1\) up to 6, so that there two new possibilities:
\(i_5, i_6\), the whole argument holds for
$$\begin{aligned} (\mathbb {1}-P_i)\textbf{A}^{(i_1 i_2\ i'\ i'')\,(j_1 j_2 j_3 j_4)\,(k_1 k_2 k_3 k_4)}= 0, \end{aligned}$$
(71)
with
\((i'\ i')\in \{(i_3 i_5), (i_3 i_6), (i_4 i_5), (i_4 i_6), (i_5 i_6) \}\). This includes all the components of
\(\textbf{A}^{(i_1 i_2 i_3 i_4 i_5 i_6)\,(j_1 j_2 j_3 j_4)\,(k_1 k_2 k_3 k_4)}\), thus
$$\begin{aligned} (\mathbb {1}-P_i)\textbf{A}^{(i_1 i_2 i_3 i_4 i_5 i_6)\,(j_1 j_2 j_3 j_4)\,(k_1 k_2 k_3 k_4)}= 0. \end{aligned}$$
(72)
In a similar fashion the result is extended to arbitrary dimensions of the three subsystems.