Assume an m-way tensor $\mathcal{Z}$.


$\mathcal{Z}\_{p\_1 p\_2 ... p\_m} = 0$ if any different indices match
and $\mathcal{Z}\_{p\_1 p\_2 ... p\_m} = 1$ otherwise. 


It is a symmetric tensor. Now if it is 2-way tensor, i.e., a matrix, I can decompose it by diagonalization (of a symmetric hollow matrix). For a general tensor, probably I can do a numerical tensor decomposition (e.g., a symmetric tensor decomposition).


But I was wondering, since it is such a simple tensor (elements are either 1 or 0), is there an analytical decomposed expression for this tensor? 


I want to avoid storing the full tensor and then decompose it numerically.


I am a not a mathematician so I apologize if my terminologies are not correct.



Analytical decomposed form of a specific traceless symmetric tensor
Thanks to Zach Teitler for the comment that this tensor is associated with elementary symmetric polynomial (ESP). I searched for the decomposed form of ESP and found a paper ([Power Sum Decompositions of Elementary Symmetric Polynomials by Lee (2015)](https://arxiv.org/abs/1508.05235)).


I followed that and all I had to do was to translate the algebraic expressions in the context of tensors. There might be small errors in my part but this is basically what I found.


The $N$-way, $M$-dimensional tensor $\mathcal{Z}$ is equivalent to the ESP $S\_{N}^{M}$ -


\begin{align\*} 
\mathcal{Z}\_{p\_1 ... p\_N} 
&= \sum\_{\mu (I) = 1}^{R} \: \alpha\_{\mu} \: t\_{p\_1}^{\mu} ... t\_{p\_N}^{\mu}
\\
S\_{N}^{M}
&= \sum\_{1 \leq p\_1 \leq ... \leq p\_N \leq M} X\_{p\_1} ... X\_{p\_N}
\quad (2 \leq N \leq M)
\end{align\*}


where R is the rank.


The vector $\alpha$ and matrix $t$ are defined as -


\begin{align\*} 
t\_{p}^{\mu (I)} 
&= - 1, \quad \mbox{if} \: p \in \{ I \}
\\ 
&= + 1, \quad \mbox{otherwise}
\end{align\*}


\begin{align\*} 
\alpha\_{\mu (I)} 
&= (-1)^{n\_I} \: \frac{1}{2^{N-1}} \:
\begin{pmatrix}
M - k - n\_I - 1 \\
k - n\_I
\end{pmatrix}, \quad \mbox{if} \: k = \frac{N-1}{2} \: 
\mbox{(odd)} 
\\
&= (-1)^{n\_I} \: \frac{1}{2^N \: (M-N)} \:
\begin{pmatrix}
M - k - n\_I - 1 \\
k - n\_I
\end{pmatrix} \: (M - 2 \: n\_I), \quad \mbox{if} \: k = \frac{N}{2} \: \mbox{(even)}
\end{align\*}
where $ \{ I \} \subset \{ M \} $ and $ n\_I = | I | \leq k $. All the subsets are ordered and assigned an integer which belongs to $ \{ \mu (I) \} $. R has an upper bound ($ R \leq \sum\_{r=0}^{k} \: \begin{pmatrix}
M \\ r \end{pmatrix}$).


