To describe an invariant trivector in dimension 8 geometrically

Question

$\newcommand\Alt{\bigwedge\nolimits}$Let $G=\operatorname{SL}(2,\Bbb C)$, and let $R$ denote the natural 2-dimensional representation of $G$ in ${\Bbb C}^2$. For an integer $p\ge 0$, write $R_p=S^p R$; then $R_1=R$ and $\dim R_p=p+1$.

Using Table 5 in the book of Onishchik and Vinberg, I computed that the representation $$ R_2\otimes\Alt^2 R_4 $$ contains the trivial representation with multiplicity one. I used the table as a black box.

Question. Let $V\subset R_2\otimes\Alt^2 R_4$ denote the corresponding one-dimensional subspace. How can one describe $V$ as a subspace geometrically?

Motivation: I want to consider a $\operatorname{PGL}(2,k)$-fixed trivector $$v\in V\subset R_2\otimes\Alt^2 R_4\subset \Alt^3(R_2\oplus R_4)$$ of the 8-dimensional vector space $W=R_2\oplus R_4$ over a field $k$ of characteristic 0, and then to twist all this using a Galois-cocycle of $\operatorname{PGL}(2,k)$. For this end I need a geometric description of $V$.

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Answer 1

Here's another very nice (but still algebraic) interpretation that explains some of the geometry: Recall that $\operatorname{SL}(2,\mathbb{C})$ has a $2$-to-$1$ representation into $\operatorname{SL}(3,\mathbb{C})$ so that the Lie algebra splits as $$ {\frak{sl}}(3,\mathbb{C}) = {\frak{sl}}(2,\mathbb{C})\oplus {\frak{m}} $$ where ${\frak{m}}$ is the ($5$-dimensional) orthogonal complement of ${\frak{sl}}(2,\mathbb{C})$ using the Killing form of ${\frak{sl}}(3,\mathbb{C})$. Note that ${\frak{m}}$ is an irreducible ${\frak{sl}}(2,\mathbb{C})$-module, and that every element $x\in {\frak{sl}}(3,\mathbb{C})$ can be written uniquely as $x = x_0 + x_1$ with $x_0\in {\frak{sl}}(2,\mathbb{C})$ and $x_1\in{\frak{m}}$. Note also that $[{\frak{m}},{\frak{m}}]= {\frak{sl}}(2,\mathbb{C})$.

This defines the desired pairing ${\frak{sl}}(2,\mathbb{C})\times \bigwedge\nolimits^2({\frak{m}})\to\mathbb{C}$: Send $(x_0,y_1,z_1)$ to $\operatorname{tr}(x_0[y_1,z_1])$. Of course, this makes the $\operatorname{SL}(2,\mathbb{C})$-invariance of the pairing obvious.

Answer 2

For a purely geometric construction, see further below, after the following algebraic considerations.

There is a Wronskian isomorphism which as a particular case says that the second exterior power of $R_4$ is isometric to the second symmetric power of $R_3$. So the invariant in question is $I(Q,C)$, a joint invariant in a binary quadratic $Q$ and a binary cubic $C$, which is linear in $Q$ and quadratic in $C$. This is indeed unique up to scale and is given in classical symbolic notation (see, e.g., Grace and Young) by $$ (ab)(ac)(bc)^2 $$ where $Q=a_{x}^{2}$ and $C=b_{x}^{3}=c_{x}^{3}$.

Another construction is to start from the binary discriminant, and polarize it to get a bilinear form (the unique invariant one on $R_2$), and apply this bilinear form to $Q$ and the Hessian of $C$.

If one does not want to use the Wronskian isomorphism then the invariant would be $J(Q,F_1,F_2)$, trilinear in the quadratic $Q$ and the two binary quartics $F_1,F_2$. It would satisy the antisymmetry $J(Q,F_2,F_1)=-J(Q,F_1,F_2)$ and would be given in symbolic form by $$ (ab)(ac)(bc)^3 $$ where now $Q=a_{x}^{2}$, $F_1=b_{x}^{4}$, and $F_2=c_{x}^{4}$.

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Geometric construction:

Consider $\mathbb{P}^1$ embedded by Veronese as a conic $\mathscr{C}$ in $\mathbb{P}^2$. A binary quadratic $Q$ corresponds to a point in $\mathbb{P}^2$. A binary cubic $C$ corresponds to a divisor or an unordered collection of three points $\{P_1,P_2,P_3\}$ on $\mathscr{C}$. Let $T_1, T_2, T_3$ be the tangents to the conic at $P_1,P_2,P_3$. Consider the points of intersection $T_1\cap P_2P_3$, $T_2\cap P_1P_3$, $T_3\cap P_1P_2$. They are aligned and thus define a line $L$. The vanishing of the invariant $I(Q,C)$ detects the situation where the point $Q$ is on the line $L$. I don't remember if the collinearity result I mentioned has a name, but it is a degenerate case of Pascal's Theorem.