# Symmetric version of Hilbert's seventeenth problem?

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?

• 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). Commented Mar 24, 2023 at 10:41
• 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... Commented Mar 24, 2023 at 13:39
• @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. Commented Mar 24, 2023 at 14:16

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.

• Does the result include anything about non-negativity and squares? Otherwise, how does it apply to the question? Commented Mar 25, 2023 at 10:33
• @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... Commented Mar 25, 2023 at 13:21
• @SamHopkins is $(x-y)^2$ a sum of squares of symmetric rational functions? I doubt, since any such function must be divisible by $x-y$, thus (being symmetric) by $(x-y)^2$, thus its square is divisible by $(x-y)^4$ Commented Jul 30, 2023 at 13:14

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...

• 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. Commented Mar 24, 2023 at 13:37
• @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 :) Commented Mar 24, 2023 at 15:53

Let $$f\in\mathbb{R}[X_1,\ldots,X_n]$$ be symmetric (or more generally invariant by a compact group). Then, there exists $$n$$ symmetric polynomials (more generally, $$m$$ $$G$$-invariant polynomial) - for example, the elementary symmetric polynomials $$\sigma_1,\ldots,\sigma_n$$ as mentioned by Bazin above, such that every symmetric polynomial can be (uniquely in the case of $$G=S_n$$) written as $$f=g(\sigma_1,\ldots,\sigma_n)$$. Now, the fact that $$f$$ is non-negative amounts to the same as saying that $$g$$ is non-negative on the image of the map $$x\mapsto (\sigma_1,\ldots, \sigma_n)$$. Procesi and Schwarz 1 -or in fact already earlier Procesi 2 in the case of symmetric polynomials, make the link to Hilbert's 17th problem: The image can be described as a basic semi-algebraic set, given by explicit inequalities, say $$h_1\geq 0,\ldots, h_k\geq 0$$. So one gets directly (see for example Theorem 2.1 in Procesi's paper) that $$g=\sum_{1\leq i_1, where each $$s_{i_1\cdots i_t}$$ is a sum of squares of elementary symmetric functions. This is maybe the most general statement one can hope for, in the case of symmetric polynomials - and generally. Let me also add, that one can use general representation theory of linear groups to get a good handle over the cones of symmetric ($$G$$-invariant) sums of squares of a given degree - in particular in the case of the symmetric group and more generally finite reflection groups, where the representation in terms of the generators is unique. Let me shamelessly also advertise the papers on invariant sums of squares decompositions - in the case of the symmetric group 3 and in the (slightly) more general case 4.

• Hi Cordian, welcome to MO. You didn't include links to papers 3 and 4. Commented Jul 30, 2023 at 11:04
• Thanks for pointing this out - had posted this in a hurry. I fixed it! Commented Jul 30, 2023 at 12:05