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The standard form of a positive semidefinite quadratic polynomial of $n$ variables is a sum of n squares. Or in the language of linear algebra, a positive semidefinite symmetric matrix can be diagonalized by congruence transformation such that the diagonal elements are either 0 or 1. What about positive semidefinite polynomial of higher degree (of course the degree must be even)? Or in a different language, the standard form of positive semidefinite higher rank symmetric tensors?

liyiontheway
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    suspect there is no standard form; two books, Squares by A. R. Rajwade. Positive Polynomials by Prestel and Delzell. Example of Choi, Lam, Resnick: $w^4 + x^2 y^2 + y^2 z^2 + z^2 x^2 - 4wxyz$ – Will Jagy Jul 12 '24 at 17:21
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    https://www.math.ucdavis.edu/~deloera/MISC/LA-BIBLIO/trunk/ReznickBruce/Reznick3.pdf – Will Jagy Jul 12 '24 at 17:35
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    https://www.researchgate.net/profile/Man-Duen-Choi/publication/248657584_An_old_question_of_Hilbert/links/559092cd08ae1e1f9bae203c/An-old-question-of-Hilbert.pdf – Will Jagy Jul 12 '24 at 17:48

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There's no hope for anything generalizing the quadratic case, and for a pretty simple reason: by a dimension count, $GL_n$ has dimension $n^2$ and the vector space of homogeneous polynomials in $n$ variables of degree $d$, or equivalently the vector space of symmetric $d$-tensors, has dimension ${n+d-1 \choose d}$. This is only less than $n^2$ when $d \le 2$ in general, which is why there's a hope for the orbits to be nice and admit nice descriptions. (A similar idea applies to antisymmetric tensors or arbitrary tensors, and can be used to explain heuristically why the orbits of $GL(V)$ acting on $\text{End}(V)$ are relatively nice, for example.)

When $d \ge 3$ this is larger than $n^2$; for example for $n = 3$ and $d = 3$ we have ${n+d-1 \choose d} = {5 \choose 3} = 10 > n^2 = 9$. This tells us that orbits now come in continuous families so there's no hope for a discrete classification; I think it's even true that the generic orbit is free, or equivalently the generic homogeneous polynomial / symmetric tensor has trivial automorphism group (edit: maybe it's finite, e.g. multiplication by a suitable root of unity). So things get much wilder immediately past the quadratic case.

Qiaochu Yuan
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  • Thank you for your insightful answer. I understand when $d\geq 3$ then $GL_n$ orbits become continuous. But can we have a standard form with continuous parameters? For example, for the case you gave $n=3,d=3$, something like $x_1^3+x_2^3+x_3^3 + \lambda(x_1^2x_2+x_2^2x_3+x_3^2x_1)$ with $\lambda$ being a continuous parameter. I know it's perhaps not well-defined what the ''simplest'' standard form is if continuous parameters are allowed, but just take the face value. – liyiontheway Jul 13 '24 at 03:16
  • @liyiontheway: I don't think there's any hope for anything like this either but it would require a stronger argument to rule out. – Qiaochu Yuan Jul 13 '24 at 04:26