I am a post graduate student who is currently studying some optimization for quadratic form. From the class lecture, I know that we can always turn any quadratic functions into theirs corresponding canonical form using some changes of variable.

For example: $f\left( {x,y} \right) = x \times y$ can be rewritten as $f\left( {X,Y} \right) = \frac{1}{4}{X^2} - \frac{1}{4}{Y^2}$ through the following change of variable $\left\{ {\begin{array}{*{20}{c}} {X = x + y}\\ {Y = x - y} \end{array}} \right.$.

Out of pure curiosity, my question is:

Does "canonical cubic form" and "canonical quartic form" for function of the form $f\left( {x,y,z} \right) = xyz$ and $g\left( {x,y,z,t} \right) = xyzt$ respectively even existed ?

If these canonical form existed then what would be a systematic way to find them ?

Thank you for your enthusiasm !

  • 1
    $\begingroup$ A cubic surface has 27 straight lines, real or imaginary. $\endgroup$
    – Narasimham
    Commented Apr 5 at 17:39
  • $\begingroup$ Are you working over $\Bbb C$? $\Bbb R$? In either case, not all quadratic forms can be written in the form $X^2 - Y^2$ for some choice of basis, e.g., the zero quadratic form. One can likewise ask about normal forms for symmetric trilinear forms on $\Bbb R^2$, i.e., in $2$ variables. $\endgroup$ Commented Apr 5 at 20:17
  • $\begingroup$ @TravisWillse sorry for not being clear enough, I am working with the real number since this is an engineering problem. But if complex number does reveal some interesting insight, I am willing to take a look at it too. $\endgroup$ Commented Apr 6 at 12:20
  • $\begingroup$ There's always the polynomial analogue of the Waring Problems and e.g. that the form defining a cubic surface can be written as as sum of five cubes of linear forms. $\endgroup$ Commented Apr 6 at 14:16
  • 1
    $\begingroup$ @TuongNguyenMinh It sounds like you're essentially asking for a cubic/quartic version of Sylvester's Law of Inertia. Is that correct? $\endgroup$ Commented Apr 7 at 1:33

1 Answer 1


$2\cdot 2! xy=(x+y)^2-(x-y)^2$

$2^2\cdot 3! xyz=(x+y+z)^3-(x+y-z)^3-(x-y+z)^3+(x-y-z)^3 $

$2^3\cdot 4! xyzt=(x+y+z+t)^4-(x+y+z-t)^4-(x+y-z+t)^4-(x-y+z+t)^4 +(x+y-z-t)^4+(x-y+z-t)^4+(x-y-z+t)^4-(x-y-z-t)^4$



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