= Solution
The <weight diagram> has <weights> $6,4,2,0,-2,-4,-6$ with multiplicities $1,1,2,2,2,1,1$. An <Irreducible Lie algebra representation> of the <sl2 Lie algebra> with <highest weight> $n$ has <weights> $n,n-2,\ldots,-n$, all of multiplicity one. The top <weight> therefore forces one $L(6)$ summand. Subtracting its <formal character> leaves <weights> $2,0,-2$, each once, which is $L(2)$. By <Weyl complete reducibility theorem>,
$$
\boxed{\Lambda^2(\operatorname{Sym}^4V)\cong L(6)\oplus L(2)
\cong\operatorname{Sym}^6V\oplus\operatorname{Sym}^2V.}
$$
Their <dimensions> $7+3=10$ account for the entire module.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/iii/paper-2-sl2-weights.png]
{title=The exterior-square weight diagram and its two irreducible summands; basis labels are defined in part iii}
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