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 $V$.

Feel free to add/edit tags!

Answer: 
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.
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 $F\_2=c\_{x}^{4}$.

---

**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.