Let $V$ a vector space of dimension $2$ over a field $k$ of characteristic different from $2$ and $3$. Let $S^{3}V$ the third symmetric power of $V$.  

How to construct a symplectic form on $S^{3}V$ such that elements coming from the linear group of $V$ are similitudes for this form?  

I believe it's some miraculous formula that I can't write down.



Symplectic form on the third symmetric power of a plane
There's nothing miraculous about this: Here is an explicit formula: Let $x$ and $y$ be a basis for $V$. Then $A,B\in S^3(V)$ can be written in the form
$$
A = a\_{-3}\,x^3+3a\_{-1}\,x^2y+3a\_1\,xy^2+a\_3\,y^3
\quad\text{and}\quad
B = b\_{-3}\,x^3+3b\_{-1}\,x^2y+3b\_1\,xy^2+b\_3\,y^3
$$
where $a\_i$ and $b\_i$ are in $k$. Set
$$
\langle A,B\rangle = a\_{-3}b\_3 - 3a\_{-1}b\_1 + 3a\_1b\_{-1} - a\_3b\_{-3}\,.
$$
This anti-symmetric pairing is then preserved (up to a multiple) under basis change. I.e., if $\phi:V\to V$ is an isomorphism, it induces a map $\phi\_3:S^3V\to S^3V$, and one has
$$
\langle \phi\_3(A),\phi\_3(B)\rangle = \det(\phi)^{-3}\langle A,B\rangle.
$$ 


Note that this generalizes to give a non-degenerate skew-symmetric pairing on $S^{2m+1}V$ when $V$ has dimension $2$ and the characteristic $p$ of $k$ does not divide any of the binomial coefficients ${2m{+}1}\choose{i}$ for $0\le i\le 2m{+}1$.


