Let \(f\) be any skew-symmetric bilinear form on \(\RR^3\). Prove that there are linear functionals \(L_1, L_2\) such that \[ f(\alpha, \beta) = L_1(\alpha)L_2(\beta) - L_1(\beta)L_2(\alpha). \]
Solution
Let \(e_1,e_2,e_3\) be the standard basis and put \[ a=f(e_1,e_2),\qquad b=f(e_1,e_3),\qquad c=f(e_2,e_3). \] Skew-symmetry gives, for \(\alpha=(x_1,x_2,x_3)\) and \(\beta=(y_1,y_2,y_3)\), \[ \begin{aligned} f(\alpha,\beta) ={}&a(x_1y_2-x_2y_1) +b(x_1y_3-x_3y_1)\\ &+c(x_2y_3-x_3y_2). \end{aligned} \]
1. If \(a\ne0\), construct the required functionals.
Define \[ L_1(x_1,x_2,x_3)=x_1-\frac ca x_3, \] and \[ L_2(x_1,x_2,x_3)=a x_2+b x_3. \] Then \[ \begin{aligned} &L_1(\alpha)L_2(\beta)-L_1(\beta)L_2(\alpha)\\ &=a(x_1y_2-x_2y_1) +b(x_1y_3-x_3y_1) +c(x_2y_3-x_3y_2)\\ &=f(\alpha,\beta). \end{aligned} \]
2. If \(a=0\) but \(b\ne0\), construct the required functionals.
Define \[ L_1(x_1,x_2,x_3)=x_1+\frac cb x_2, \qquad L_2(x_1,x_2,x_3)=b x_3. \] Then \[ L_1(\alpha)L_2(\beta)-L_1(\beta)L_2(\alpha) =b(x_1y_3-x_3y_1)+c(x_2y_3-x_3y_2), \] which is \(f(\alpha,\beta)\) because \(a=0\).
3. If \(a=b=0\), factor the remaining term.
If \(c\ne0\), take \[ L_1(x_1,x_2,x_3)=x_2, \qquad L_2(x_1,x_2,x_3)=c x_3. \] Then their alternating product is \[ c(x_2y_3-x_3y_2)=f(\alpha,\beta). \] If also \(c=0\), then \(f=0\) and one may take \(L_1=0\) and \(L_2=0\).
Thus in every case there exist \(L_1,L_2\in(\mathbb R^3)^*\) with \[ \boxed{ f(\alpha,\beta) =L_1(\alpha)L_2(\beta)-L_1(\beta)L_2(\alpha).} \]