Is there a standard normal form for homogeneous cubic polynomials in $\mathbb{R}[x\_1, x\_2, x\_3]$? Or, put another way, is there a nice way to describe the orbit space of the natural (diagonal) action of $GL(3, \mathbb{R})$ on $Sym^3(\mathbb{R}^3)$?


(This is related to my previous question [Normal form for trace-free real cubic forms in 3 variables under SO(3)-action?](https://mathoverflow.net/questions/104628/normal-form-for-trace-free-real-cubic-forms-in-3-variables-under-so3-action), but now I'm interested in the full $GL(3)$ action and not just the $SO(3)$ action, and I also don't want to impose the trace-free condition.)



Normal forms for homogeneous cubic polynomials in $\mathbb{R}[x_1, x_2, x_3]$
Hi, Jeanne! It may help to have some geometric explanation of the normal forms over $\mathbb{R}$. The standard account is this:


If the projective cubic curve $F(x\_1,x\_2,x\_3)=0$ is nonsingular (the 'generic' case), it has exactly three real flexes, they are distinct and lie on a line. One can make a linear change to make them lie on the line $x\_1{+}x\_2{+}x\_3=0$ and have this line intersect the curve at the three points where some $x\_i=0$ (so that these are the three flexes). From this, one sees that one can make a real linear change of variables so that 
$$
F = {x\_1}^3 + {x\_2}^3 + {x\_3}^3 + 6\sigma\ x\_1x\_2x\_3\ ,
$$  

where $\sigma\not=-\tfrac12$ is a real number. (When $\sigma=-\tfrac12$, the above cubic factors as a line $x\_1{+}x\_2{+}x\_3=0$ and an irreducible quadratic form.)


Enumerating the singular cases over the reals gets a little messy, but the final result is that when the curve is irreducible and singular there is exactly one singular point, which is necessarily real, and it is either a hyperbolic node, elliptic node, or a cusp. These have the normal forms 
$$
F = {x\_2}^2x\_3 - \epsilon\ {x\_1}^2 x\_3 - {x\_1}^3
$$
where $\epsilon$ is $1$, $-1$, or $0$, respectively.


If the curve is the union of a line and a smooth quadric, i.e., $F = LQ$, where $L$ is linear and $Q$ is nonsingular (possibly without real points), you can put the quadric in normal form, $Q = {x\_1}^2+{x\_2}^2\pm{x\_3}^2$ and then use the stabilizer group of the quadric to normalize $L$. In the $+$ case, you can always rotate, using $\mathrm{SO}(3)$, to make $L=x\_1{+}x\_2{+}x\_3$ (say). In the $-$ case, there are three cases, and you can rotate, using $\mathrm{SO}(2,1)$ to make $L$ be one of $x\_1$, $x\_1{+}x\_3$, or $x\_3$.


Finally, if the curve is the union of three (complex) lines, it depends on whether the lines are all distinct or not and whether they are concurrent or not. In the distinct case, you get that $F$ lies on the orbit of either $F=x\_1x\_2x\_3$ (all real, not concurrent), $x\_1x\_2(x\_1{+}x\_2)$ (all real, concurrent), $x\_1({x\_2}^2{+}{x\_3}^2)$ (two complex conjugate, not concurrent), or $x\_1({x\_1}^2{+}{x\_2}^2)$ (two complex, concurrent). If they are not all distinct, you get either $x\_1{x\_2}^2$ (two distinct) or ${x\_1}^3$ (all the same).


These are the normal forms for all the nonzero cubics in $3$ variables.


*Remark:* You also asked about the 'orbit space' of $\mathrm{GL}(3,\mathbb{R})$ acting on $\mathrm{Sym}^3(\mathbb{R}^3)$. The naïve quotient, endowed with the quotient topology, is not very nice because it is highly non-Hausdorf, as every open neighborhood of the fixed point $0$ meets every orbit. Even when you remove $0$, so that the remainder can be thought of as $\mathrm{PGL}(3,\mathbb{R})\simeq \mathrm{SL}(3,\mathbb{R})$ acting on $\mathbb{P}\bigl(\mathrm{Sym}^3(\mathbb{R}^3)\bigr)\simeq \mathbb{RP}^9$, the orbit space with the quotient topology is *still* non-Hausdorf. The whole point of developing Geometric Invariant Theory (aka GIT) is to figure out the right way to remove the bad points so that the remaining 'good' points (aka, the 'semi-stable' points) will make a nice quotient space when you divide out by the action of the group. I don't know how good this theory is in the real case, though. In any case, the nonsingular cubics are all 'good' points for the moduli space.


