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Artin's solution to Hilbert's seventeenth problem tells us that a multivariate polynomial $f$ takes only non-negative values over the reals if and only if it is a sum of squares of rational functions.

Now suppose that, in addition to taking only non-negative values over the reals, $f$ is a symmetric polynomial; i.e., it is invariant under any permutation of the variables. Is there anything more we can say about $f$ beyond the fact that it is a sum of squares of rational functions? For example, can we say anything about what $f$ looks like when expanded in terms of various well-known bases for symmetric functions?

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    $\begingroup$ A trivial observation is that a non-negative symmetric polynomial can be written as a sum of squares of rational functions in such a way that permutation of the variables preserves the expression (i.e. the set of terms in the sum is preserved). $\endgroup$
    – Brendan McKay
    Mar 24 at 10:41
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    $\begingroup$ As Per's answer sort of highlights, in the transition between the last two questions you are eliding the distinction between rational functions and polynomials, which is important here. You can't necessarily write a nonnegative polynomial as a sum of squares of polynomials, so talking about bases of symmetric functions (which all consist of polynomials, not rational functions) is a little strange... $\endgroup$
    – Sam Hopkins
    Mar 24 at 13:39
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    $\begingroup$ @SamHopkins Fair comment. I just thought that "anything more we can say" might strike people as being overly vague, so I was groping for an example of something that one might imagine could be said. But really, anything nontrivial would interest me. $\endgroup$
    – Timothy Chow
    Mar 24 at 14:16

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This is just a long comment, but note that a symmetric polynomial in 2 variables, might not even be expressible as a linear combination of squares of symmetric polynomials. For example, $x^2+xy+y^2$ is not a linear combination of $(x+y)^2$ (the only possible option).

However, if $f(x_1,\dotsc,x_n)$ is a shift-invariant polynomial, i.e. $f(x_1,\dotsc,x_n) = f(x_1+t,\dotsc,x_n+t)$ for all $t$ (the discriminant is a natural example of such a polynomial), then (with the obvious constraint that it is of homgeneous even degree) it is a linear combination of squares of shift-invariant symmetric polynomials.

(The proof boils down to finding an explicit basis the vector space consisting of squares.)

So, perhaps in the symmetric and shift-invariant case, it might be possible that the rational functions themselves can be taken to be symmetric and shift-invariant...

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    $\begingroup$ Well, it is also the case that a polynomial which takes on only non-negative values for real numbers cannot necessarily be expressed as a sum of squares of polynomials: using rational functions is really crucial. $\endgroup$
    – Sam Hopkins
    Mar 24 at 13:37
  • $\begingroup$ @SamHopkins Absolutely! I would very much want to see one example of a shift-invariant symmetric polynomial, which is non-negative, but requires rational functions in the sum-of-squares-representation. In fact, ill donate $10 to OEIS for such an example :) $\endgroup$
    – Per Alexandersson
    Mar 24 at 15:53
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There is paper by Georges Glaeser [MR0143058], published by the Annals of Mathematics in 1963, in which he proves that

Every symmetric smooth fonction on $\mathbb R^n$ is equal to $g(\sigma_1,⋯,\sigma_n)$ for a $g$ smooth on $\mathbb R^n$, where $\sigma_1, \dots, \sigma_n$ are the elementary symmetric polynomials.

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    $\begingroup$ Does the result include anything about non-negativity and squares? Otherwise, how does it apply to the question? $\endgroup$
    – Federico Poloni
    Mar 25 at 10:33
  • $\begingroup$ @FedericoPoloni Right. I guess a specific question you could ask is: is it possible to express a nonnegative, symmetric polynomial as a sum of squares of symmetric rational functions? Maybe this is obvious? But I don't quite see it... $\endgroup$
    – Sam Hopkins
    Mar 25 at 13:21

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