Isomorphism between $sl_{4}$ and the orthogonal group of $6$ variables Let V be the irreducible $sl_{4}$-module with highest weight $\pi_{2}=\lambda_{1}+\lambda_{2}$ (i.e if $H=\left(\lambda_{1},\dots,\lambda_{4}\right)$ is a diagonal matrix in $sl_{4}$ with values $\lambda_{1},\dots,\lambda_{4}$ on it's diagonal then $\pi_{2}\left(H\right)=\lambda_{1}+\lambda_{2}$). Show that $\dim V=6$ and that there is a non-degenrate invariant quadratic form on $V$. Use this to create an Isomorphism of $sl_{4}$ onto the Lie-algebra of the orthogonal group in $6$ variables.
I know how to show that $V\simeq\wedge^{2}k^{4}$ and that it's dimension is $6$. I don't know how to find this quadratic form however. Thanks!
 A: Let $D$ be the "determinant" isomorphism $\wedge^4 k^4 \to k$. Define the $k$-bilinear form $h:V\times V\to k$ by $h\left(v,w\right)=D\left(v\wedge w\right)$ (where $v\wedge w$ denotes the wedge of $v$ and $w$ in the exterior algebra $\wedge k^4$). Show that this is symmetric (note that elements of $V=\wedge^2 k^4$ commute - not anticommute - in the exterior algebra $\wedge k^4$) and nondegenerate - and (by the multiplicativity of determinants) $\mathfrak{sl}_4$-invariant. So you have an invariant nondegenerate symmetric bilinear form, thus also an invariant nondegenerate quadratic form.
More generally (but still with the same proof), for every $n\in\mathbb N$, the irreducible $\mathfrak{sl}_{2n}$-module with highest weight $\lambda_1+\lambda_2+...+\lambda_n$ is the module $\wedge^n k^{2n}$ and has an $\mathfrak{sl}_{2n}$-invariant nondegenerate bilinear form. It is symmetric if $n$ is even and skew-symmetric if $n$ is odd.
Sacrificing the symmetry/skew-symmetry claim, we can generalize this even further: The dual of the $\mathfrak{sl}_m$-module with highest weight $p_1\lambda_1+p_2\lambda_2+...+p_m\lambda_m$ is the $\mathfrak{sl}_m$-module with highest weight $-p_m\lambda_1-p_{m-1}\lambda_2-...-p_1\lambda_m$. Since the only relation between the $\lambda_i$ is $\lambda_1+\lambda_2+...+\lambda_m=0$, you can use this to see which $\mathfrak{sl}_m$-modules are isomorphic to their own dual. Now, an $\mathfrak{sl}_m$-module is isomorphic to its own dual if and only if it has an invariant nondegenerate binary form. Such a form is always necessarily either symmetric or skew-symmetric (more or less by Schur's lemma). How to find out whether it is symmetric or skew-symmetric? This is something I don't know in general.
