The proof of a tensor cross product identity

Question

If $l=a\times b,l_i=\epsilon_{ijk}a^jb^k$. then $a^ib^j+a^jb^i=l^il^j-l^2g^{ij}$. $g$ is the metric tensor. I tried to dot product $g_{ij}$ to two sides, then I found it became $2(a\cdot b)^2=2(a\cdot b)^2$, but it seems only proved necessarity. I don't know how to prove it completely.

Answer 1

I assume you have a typo and the right identity should be: $$ l^il^j-l^2g^{ij}=(a^ib^j+a^jb^i)a^kb_k-(a^2b^ib^j+b^2a^ia^j). $$

One can deduct this identity by expanding the product $\epsilon_{ikl}\epsilon_{jpq}$: $$ l^il^j-l^2g^{ij}=\begin{vmatrix}g^{ij}&g^{ip}&g^{iq}\\g^{kj}&g^{kp}&g^{kq}\\g^{lj}&g^{lp}&g^{lq}\end{vmatrix}a_kb_la_pb_q-g^{ij}\begin{vmatrix}g^{kp}&g^{kq}\\g^{lp}&g^{lq}\end{vmatrix}a_kb_la_pb_q = \\ \left(-g^{ip}\begin{vmatrix}g^{kj}&g^{kq}\\g^{lj}&g^{lq}\end{vmatrix}+g^{iq}\begin{vmatrix}g^{kj}&g^{kp}\\g^{lj}&g^{lp}\end{vmatrix}\right)a_kb_la_pb_q = \\ -g^{ip}(a^ja_pb^lb_l-a^qb_qb^ja_p)+g^{iq}(a^ja^lb_lb_q-a^pa_pb^jb_q)=\\ a^kb_k(a^ib^j+a^jb^i)-(a^ia^jb^2+b^ib^ja^2). $$