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.

  • 2
    $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.