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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.

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

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  • $\begingroup$ 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. $\endgroup$ – tkf Jul 16 '20 at 18:21

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