TITLE: The exceptional Lie algebra $\mathfrak{g}_2$ and binary cubics
QUESTION [8 upvotes]: How is the exceptional 14-dimensional Lie algebra $\mathfrak{g}_2$ related to the covariant algebra for the binary cubic?
Here are some details on this question.  This algebra is generated by 4 forms, these being the form itself $Q$, the discriminant $\Delta$, the Hessian $H$, and a transvectant $T$ of $Q$ and $H$.  These four forms ${Q,H,T,\Delta}$ obey a syzygy because $Q^2\Delta$ is a linear combination of $H^3$ and $T^2$.  (This is well-known.  One may find an exposition in Peter Olver's book on classical invariant theory, for example.)
Let $(d,w)$ denote the degree and weight of a form.  Then the degrees and weights of ${1,Q,H,T,\Delta}$ are respectively ${(0,0),(1,3),(2,2),(3,3),(4,0)}$.  This sequence is peculiar because these nearly give the branching weights of the adjoint representation of $\mathfrak{g}_2$ with respect to a Lie subalgebra generated by a pair of opposite short root vectors.  The only one "missing" is a pair of the form $(2,0)$, which cannot occur.
Is this just a happy coincidence, or is there some "deep" reason reason for this?

REPLY [5 votes]: Well, I don't know about "deep", but here is another perspective.  My point of view is much different than the classical invariant point of view -- I use the word "syzygy" about as often as I use the word "zeugma".  
A good reference for everything I'm about to write is the Duke paper of Wee Teck Gan, Benedict Gross, and Gordan Savin, "Fourier coefficients of modular forms on $G_2$".
In the Lie algebra $\mathfrak{g}_2$, there is a maximal parabolic subalgebra $\mathfrak{p} = \mathfrak{l} \oplus \mathfrak{u}$, where $\mathfrak{u}$ is a two-step nilpotent Lie algebra with one-dimensional center $\mathfrak{z}$.  Correspondingly, there is a parabolic subgroup $P = LU$ of the simple complex Lie group $G_2$.  The Levi subalgebra $\mathfrak{l}$ contains root spaces for a short root and its negative.  The derived subalgebra $[\mathfrak{l}, \mathfrak{l}]$ is generated by a pair of opposite short root vectors.
The associated Levi subgroup $L$ is isomorphic to $GL_2$, and $L$ acts by the adjoint action on $\mathfrak{g}_2$.  Since $L$ normalizes $U$, we get an action of $L$ on the 5-dimensional vector space $\mathfrak{u}$, and this action stabilizes the one-dimensional $\mathfrak{z} \subset \mathfrak{u}$ (note also $\mathfrak{z} = [\mathfrak{u}, \mathfrak{u}]$).  The resulting action of $L$ on $\mathfrak{u} / \mathfrak{z}$ is a 4-dimensional representation of $GL_2$.
This representation of $GL_2$ is isomorphic to the representation of $GL_2$ on the space of binary cubic forms $C(x,y) = ax^3 + bx^2 y + c x y^2 + d y^3$, in which $g \in GL_2$ acts on a cubic form via:
$$ \[ gC \](x,y) = det(g)^{-1} \cdot C( (x,y) \cdot g).$$
See Proposition 3.1 of Gan-Gross-Savin for more.
One could go the other way as well, and construct the Lie algebra $\mathfrak{g}_2$ beginning with the space of cubic forms (which has some additional structure).  I think that the machinery of "structurable algebras" developed by Bruce Allison would work for that (and general construction of Lie algebras with 5-term grading).