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?


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).

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  • $\begingroup$ Thanks, Marty. I am familiar with this construction. However, it still strikes me as peculiar that these four forms are exactly those that one needs to make $\mathfrak{g}_2$. I am voting up your answer because it offers some good perspective/context, but I cannot yet accept this as the whole story. How does $\mathfrak{sl}_2$ "know" that its 4-dimensional representation lives inside $\mathfrak{g}_2$? $\endgroup$ – David Richter Dec 23 '10 at 12:11
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    $\begingroup$ Can I ask, how does $\mathfrak{gl}_n$ "know" that its representation $Sym^2(C^n)$ lives inside of $\mathfrak{sp}_{2n}$? If you can give an answer to that question, perhaps I'll understand better what kind of answer you're seeking for $\mathfrak{g}_2$. $\endgroup$ – Marty Dec 23 '10 at 17:21

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