Isomorphism between $\bigwedge^2\mathbb R^3$ and the dual of $\mathfrak{so}(3)$

Question

I need to show that the map $L:\bigwedge^2\mathbb{R}^3\to \mathfrak{so}(3)^*$ defined by $L(x\wedge y)\Omega=\langle \Omega x,y\rangle$ for $\Omega\in \mathfrak{so}(3)$ is an isomorphism. This map is clearly skewsymmetric, but I cannot show that it is injective. Clearly, if $L(x_1\wedge x_2)=L(y_1\wedge y_2)$, then we have $\langle \Omega(x_1-y_1),(x_2-y_2)\rangle=0$, no-degeneracy of the inner product should to the trick, but I cannot justify this explicitly.

Answer 1

Assuming you are identifying $\mathbb{R}^3$ with $\mathfrak{so}(3)$ and the action is given by the Lie bracket you have:$$L(e_i \wedge e_j)(e_k)=\langle[e_k,e_i],e_j\rangle=\langle[e_i,e_j],e_k\rangle$$ for $i,j,k\in \{1,2,3\}$. Then \begin{eqnarray*} L(e_1\wedge e_2)&=&e_3^*,\\ L(e_2\wedge e_3)&=&e_1^*,\\ L(e_3\wedge e_1)&=&e_2^*.\\ \end{eqnarray*}

Thus you have an isomorphism of vector spaces, as your map takes one basis to another.