let's say I have a tensor:
A = [3, 5, 7]
If I do a Kronecker product of A with A I would get the tensor [9,15,21,15,25,35,21,35,49]
One can notice that 15 appear twice since the 1st 3x2nd 5 are the same as 1st 5x2nd 3 (also 21 & 35).
I want to know if there is a name for an equivalent operation to square the tensor without repeated elements so the result tensor would be [9, 15, 21, 25, 35, 49].
AFAIK there's not a special name for this, but it amounts to dropping the $(4,7,8)^{th}$ rows from the vector. And there is an algebraic expression for such an operation $$\eqalign{ &b = M(a\otimes a) \\ &M=\left[\matrix{ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 }\right] }$$ where each row of $M$ contains a single ${\tt1}\,$ while its $(4,7,8)^{th}$ columns are zeros.
More generally, if $a\in{\mathbb R}^{n}$, then $M\in{\mathbb R}^{(n^2+n)/2\,\times n^2}\,$ and the zero columns are at $$\eqalign{ &1n+1 \\ &2n+1,\quad 2n+2 \\ &3n+1,\quad 3n+2,\quad 3n+3 \\ &4n+1,\quad 4n+2,\quad \ldots \\ &\ldots\\ &(n-1)n+1,\;(n-1)n+2,\quad \ldots,\quad &n^2-1 }$$