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A little context for the following question, first. As Noah Stein notes in a comment below, the present question is closely related to the free semialgebraic geometry studied by Helton and his colleagues. In fact, working on the free nonabelian monoid rather than the free abelian group as I do below, something very close to an affirmative answer to the present question holds (see Proposition 2.1 here), but the proof of this strongly uses the fact that in a free monoid there can be no "multiplicative cancellation of terms". Roughly, the present question asks what happens if we allow multiplicative inverses of elements. A related question of mine is here.

Suppose $A\in M_n(\mathbb{R})$ is a symmetric matrix, and let $G$ be the free abelian group on $n$ generators $a_{1},...,a_{n}$. Let $\mathbb{R}G$ be the real group algebra of $G$ with involution $(\sum \lambda_{g}g)^{*}=\sum \lambda_{g}g^{-1}$. Suppose that the quadratic *-polynomial $\sum_{i,j=1}^{n}A_{ij}a_{i}^{*}a_{j}=\sum_{k=1}^{N}h_{k}^{*}h_{k}$ for some $h_{1},...,h_{N}\in \mathbb{R}G$, i.e. is a sum of hermitian squares in the group algebra. Does this imply $A=B^{T}B+D$ for some $B\in M_n(\mathbb{R})$ and $D\in M_n(\mathbb{R})$ a diagonal matrix with zero trace?

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    $\begingroup$ The setup reminds me a bit of Bill Helton's work (with various coauthors) on polynomials in non-commuting variables and the positivity and convexity thereof. However, in that work it is the algebra on the free monoid that is involved (there are no inverses) and the involution is reversal of order of the strings. So I'm not sure if any of the techniques there would be useful, but I'm leaving this comment just in case. $\endgroup$
    – Noah Stein
    May 15, 2016 at 14:51
  • $\begingroup$ Thanks, Noah! This is precisely what I want to know: What happens when we add these inverses? $\endgroup$
    – Jon Bannon
    May 15, 2016 at 17:10
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    $\begingroup$ This feels like it should have something to do with classical Fourier analysis and positive-definite functions... (which, I believe, is part of what Helton and al are trying to generalize) $\endgroup$
    – Yemon Choi
    May 15, 2016 at 19:55
  • $\begingroup$ @Yemon: My calculations suggest you may be right about this. Can you think of any good references that may discuss this sort of thing? $\endgroup$
    – Jon Bannon
    May 15, 2016 at 20:02
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    $\begingroup$ @YemonChoi: Embarrassing blind spot at the top, but the modified question may actually be the modification needed, if you'd like to have a look: mathoverflow.net/questions/238952/… $\endgroup$
    – Jon Bannon
    May 16, 2016 at 11:12

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