When I discuss a question with my colleague, he said he has a feeling that the symmetrization of functions should be unique up to a coefficient. However, I keep being critical at the point.

Specifically, symmetrization(standard) of a multivariable function is

$$Sf(x_1,\dots,x_n)=\frac{1}{n!}\sum_{\sigma\in S_n}f(\sigma(x_1,...,x_n))$$ Here, $S_n$ is the symmetric group of n elements with order $|n!|$. $\sigma(x_1,...,x_n)$ is a short notation for $(x_{\sigma(1)},\dots,x_{\sigma(n)})$ since $\sigma$ is a permutation acting on these subscripts.

Then the question can be rephrased as the following:

Let $S$ be the symmetrization operation for $n-$variable functions as the above formula, if $T$ is another symmetrization operation which is not trivial (say the constant operator, $Tf=c$, where $c$ is some constant) for $n-$variable functions, which satisfies $T(Tf)=Tf$, then $Sf-Tf=0$. (For this case, there cannot be extra coefficient since the operation should be projective ($T\circ T=T$)). Or if $T$ is a projection operator of $n-$variable function such that for all permutation $\sigma\in S_n$, $\sigma(Tf):=(Tf)(\sigma(x_1,...,x_n))=Tf$. Then $Tf=Sf$.

Can someone show this thing, find a proof in any books, or just find a counter example? Appreciated.

  • $\begingroup$ The null operator $T(f)=0$ (0 is trivially a symmetric function) is a trivial counterexample, i.e. $T^2f = Tf$ but $Sf \neq 0$ in general. Why instead non try to proving that Sf is a metric projector on the subspace of symmetric function for some metric? (Just an idea I have no clue if it is true) $\endgroup$ – Warlock of Firetop Mountain May 4 '18 at 15:38
  • $\begingroup$ I may exclude this trivial one, but this is indeed a good counterexample. $\endgroup$ – Hamio Jiang May 4 '18 at 15:41

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