Annihilator for $\mu\in\Lambda^3(V)$ Let $V$ a 4-dimensional real vector space and let $\mu\in \Lambda^3(V)$. I have to characterize the elements in the annihilator of $\mu$, i.e., $\text{Ann }\mu=\left\lbrace  u\in V \vert u\wedge \mu=0 \right\rbrace$, I'm stuck with this.
My try: I know that $e_1\wedge e_2 \wedge e_3, e_1\wedge e_2\wedge e_4, e_1\wedge e_3\wedge e_4, e_2\wedge e_3\wedge e_4$ is a basis for $\Lambda^3(V)$ then I write $\mu$ in terms of the basis where the coefficients for $\mu$ are $v_1, v_2, v_3, v_4$. So, I write $u=\sum u_ie_i$ and then I calculate $u\wedge \mu$. Therefore I get that
\begin{equation}
0=u\wedge\mu= (u_1v_4-u_4v_1+u_3v_2-u_2v_3)(e_1\wedge e_2\wedge e_3 \wedge e_4)
\end{equation}
but I don't know how to made any conclusion for $u$.
 A: I think your solution is fine. Here's another approach. Rather than choosing a basis for $V$, let's define an inner product on $V$. If we do that, we have an isomorphism $\Psi\colon \Lambda^3 V \cong V^*$. Then $\Psi(\mu)=\phi\in V^*$, and the annihilator of $\mu$ is the annihilator of $\phi$, i.e., $\{v\in V: \phi(v)=0\}$.
How does this work out with your viewpoint? Start with an orthonormal basis $e_i$ and dual basis $\omega^i$. The natural volume element is then $\Omega = \omega^1\wedge\dots\wedge\omega^4$, and we define $\Psi$ by $\Psi(\xi) = -\iota_\xi\Omega\in V^*$ for $\xi\in\Lambda^3 V$. (This will give the same signs you have.) In particular, $\Psi(e_1\wedge e_2\wedge e_3) = \omega^4$, etc. This results, I believe, in
\begin{multline*}
\Psi(v_1e_1\wedge e_2\wedge e_3 + v_2e_1\wedge e_2\wedge e_4 + v_3e_1\wedge e_3\wedge e_4 + v_4 e_2\wedge e_3\wedge e_4) \\ = v_1\omega^4 - v_2\omega^3 + v_3\omega^2 - v_4\omega^1.
\end{multline*}
Now we're at the same place :) Perhaps (but not really) this approach makes it more obvious that the annihilator of $\mu$ is a hyperplane in $V$. But that's really clear, anyhow, since $\Lambda^4 V\cong\Bbb R$ and you have one linear equation.
