Is it possible to build a trilinear map $\tau$ such that

Question

$\tau: E^3 \to V$ (where $E,V$ are vector spaces with $char\ne 2$) and such that for all $(x,y,z)\in E^3$, $\tau(x,y,z)=-\tau(y,x,z)$ and $\tau(x,y,z)=\tau(x,z,y)$.

Thanks in advance !

Answer 1

$$ \tau(x,y,z) = -\tau(y,x,z) = - \tau(y,z,x) = \tau(z,y,x) $$

$$ \tau(x,y,z) = \tau(x,z,y) = - \tau(z,x,y) = -\tau(z,y,x) $$

So $\tau\equiv 0$.