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The special linear group $\text{SL}_2(\mathbb{Z})$ of $2\times 2$ invertible matrices in $\mathbb{Z}$ acts on binary cubic forms $\{ax^3 + bx^2y + cxy^2 + dy^3\}$ by acting on the vector $(x,y)^T$. This action descends to an action of the group $\Gamma_\infty := \left\{ \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix} : n \in \mathbb{Z}\right\}$ on binary cubic forms. It happens to be that $\Gamma_\infty$ is generated by $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$. So understanding the action of this one element allows us to understand the whole $\Gamma_\infty$ action. (This is the action that sends $(x,y) \to (x+y, y)$)

I am interested in the invariants of binary forms under this action. In this (the binary cubic forms) case, the invariants form a 3-dimensional algebra, and I know the invariants. This action makes sense on other binary forms as well. It happens to be that the invariants of the binary quadratic case form a 2-dimensional algebra, which I also know. But I found these through the "brute force" method.

I am not really familiar with invariant theory at all, so I don't know if this is an easy or a hard question. But I'm wondering if there's some nice form for the invariants of binary n-degree forms under $\Gamma_\infty$? Alternately, perhaps there's a nice form for the invariants under $\text{SL}_2(\mathbb{Z})$ that one might perhaps be able to modify?

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Let $f(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_0$ (I'm not sure how to frame the following with homogeneized polynomials, so I will stick with this together with the action induced by translations)

The translation by the opposite of the unique root of $f^{(n-1)}(x)$ gives you a canonical way to pick a representative in each equivalence class :

$I_n = a_n$ is always an invariant of the action induced by translation, and if $x$ is the root of $f^{(n-1)}$, then $I_0 = f(x), I_1 = f'(x), I_2 = f''(x),\ldots, I_{n-2} = f^{(n-2)}(x)$ are all invariant. This gives you $n-1$ invariants.

Conversely, given $I_0, I_1, \ldots I_{n-2}$ and $I_n \in \Bbb K$ (a field with characteristic $0$) , the corresponding unique class of polynomials modulo translation by an element of $\Bbb K$ with those invariants is the equivalence class of $I_n x^n + I_{n-2} \frac {x^{n-2}}{(n-2)!} + \ldots + I_1 x + I_0$

For example with $f(x) = ax^3+bx^2+cx+d$, The only root of $f''$ is the inflexion point of $f$, $x = - \frac b{3a}$, and by evaluating $f$ and $f'$ at $x$, you get the other two invariants $3ac-b^2$ and $2b^3-9abc+27a^2d$.


Note that if a rational function $g \in \Bbb K(a_0,a_1,\ldots,a_n)$ is invariant by translation-by-one, then it is invariant by all the translations with parameter in $\Bbb K$ (I need this because I'm translating by $-a_{n-1}/na_n)$ :

Here the action on the degree $1$ part of $\Bbb K[a_0,a_1,\ldots,a_n]$ is a linear transformation. In fact there is a particular matrix $T \in GL_{n+1}(\Bbb Z[X])$ such that the action of the translation by $q$ is the linear transformation given by $T(q)$.

Therefore, for each $g$ there is an element $\tilde{g} \in \Bbb K(a_0,a_1,\ldots,a_n,X)$ such that forall $q \in \Bbb K$, $g^{T(q)}(a_0,a_1,\ldots,a_n) = g(T(q)(a_0,a_1,\ldots,a_n)) = \tilde{g}(a_0,a_1,\ldots,a_n,q)$.

Now, if $g$ is invariant by $T(1)$, then forall $q \in \Bbb N$, $\tilde{g}(a_0,a_1,\ldots,a_n,q) = \tilde{g}(a_0,a_1,\ldots,a_n,0)$, which implies that $g(a_0,a_1,\ldots,a_n,X) - g(a_0,a_1,\ldots,a_n)$ is a rational function of $X$ with infinitely many roots, so it is zero, and so $g$ is invariant by every translation $T(q)$ for $q \in \Bbb K$

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Really, I could not have hoped for a simpler process. This is great! Do you know a proof (or a reference) of why $I_0, \ldots, I_{n-2}$ actually are invariants? I checked a few, and it works, which is a bit miraculous. But now I have to wonder why. –  mixedmath Dec 14 '12 at 7:05
    
@mixedmath: It's because we have a way of selecting a unique representative.The rational function $q(f) = -a_{n-1}/na_n$ is such that $q(f^{T(q')}) = q(f)-q'$ so the operation $f \mapsto f^{T(q(f))}$ is an invariant. Then I'm just looking at its coefficients. Basically you can always translate a polynomial until its second coefficient is $0$, and there is only one way to do it. –  mercio Dec 14 '12 at 8:43
    
I'm sorry, but I'm not quite there yet. I see that $q(f^{T(q')}) = q(f)-q'$, so that $q(f^{T(q(f))}) = 0$. I also see that the coefficients of $f^{T(q(f))}$ are the things we want to see are invariant under translation. And I see that there is a unique translation that removes the second coefficient of a polynomial. But I don't yet see what allows us to conclude that these coefficients are invariants. I suspect this has something to do with the fact that q is rational, or that there is something obvious that I don't see. –  mixedmath Dec 14 '12 at 22:22
    
@mixedmath : By "$\Phi : f \mapsto f^{T(q(f))}$ is invariant" I was thinking $\Phi(f^{T(r)}) = (f^{T(r)})^{T(q(f^{T(r)}))} = (f^{T(r)})^{T(q(f)-r)} = f^{T(r+q(f)-r)} = f^{T(q(f))} = \Phi(f)$, which is probably illegible, so $\Phi$ corresponds to a map $\Bbb A^{n+1} \to \Bbb A^{n+1}$ where all the coordinates are invariant by translation. And also yes, it is nice that the acting group and the action itself are algebraic and that we have nice rational functions everywhere we look. –  mercio Dec 15 '12 at 0:49
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