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Given a finite set of real numbers $X_1, \ldots, X_n$, we can compute the first $n$ power sums of these numbers. From the power sums, the set $\{X_1, \ldots, X_n\}$ can be recovered. Essentially we lose the order by symmetrizing.

What about when $X_1, \ldots, X_n$ are vectors in $\mathbb R^m$? As many pointed out, my initial statement was not clear so I'll explain what I have in mind. I am hoping these functions $f_k(Y_1, \ldots, Y_k) = \sum_{1\le i_1, \ldots, i_k\le n}\prod_{j=1}^k \langle X_{i_j}, Y_j \rangle$ will be sufficient to represent the set $\{X_1, \ldots, X_n\}$. To "store" $f_k$, I can pick standard basis vectors for $Y_i$ and compute the result, but then there are $m^k$ possibilities. Symmetry should be able to reduce this number, probably by $O(m!)$. (I'm guessing again.) Anyway, I still don't know if $f_1, \ldots, f_n$ will be sufficient to recover the set $\{X_1, \ldots, X_n\}$.

I heard about multi-symmetric polynomials too, but I'm not sure about the set of generators to use. Does there exist an isometry-invariant set of generators? (Are my $f_k$ isometry-invariant?)

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What kind of product of vectors in $\real^n$ do you use? There is a concept of invariant polynomials of matrix argument, which is usually taken to be symmetric polynomials in the eigenvalues. Maybe thet coincide with the multi-symmetric polynomials you mention. –  kjetil b halvorsen Sep 10 '12 at 22:08
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Yes, you need to define some notion of what a moment is in $\mathbb{R}^n$ for $n>1$. –  gt6989b Sep 10 '12 at 22:55
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