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

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.

• What is your action of $\mathfrak{so}(3)$ on $\mathbb R^n$? Or should $n=3$? – Aaron Jul 16 '20 at 1:20
• @Aaron yes, n=3. – JerryCastilla Jul 16 '20 at 1:32

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 $$\begin{eqnarray*} L(e_1\wedge e_2)&=&e_3^*,\\ L(e_2\wedge e_3)&=&e_1^*,\\ L(e_3\wedge e_1)&=&e_2^*.\\ \end{eqnarray*}$$
• In hindsight the action was probably supposed to be induced from the standard representation of $SO(3)$ on $\mathbb{R}^3$. However, this action may be identified with the one I used, so the answer is the same. – tkf Jul 16 '20 at 18:21