1
$\begingroup$

Given a two-form $\omega\in \Lambda^2V$ for some (say finite dimensional) vector space $V$ we may associate with $\omega$ a skew map $f_{\omega}:V\rightarrow V^*$ given by $X\mapsto \iota_X\omega$, where $V^*$ is the dual space to $V$ and $\iota_x\omega$ is the linear functional $\omega(X,\cdot)$. My question is how do we go the other way, i.e., given a skew map $f:V\rightarrow V^*$ (by skew I mean $f^*:(V^*)^*=V\rightarrow V^*$ satisfies $f^*=-f$) how do we explicitly associate with $f$ a two-form $\omega_f\in \Lambda^2V^*$ ?

$\endgroup$

1 Answer 1

2
$\begingroup$

You can define $\omega_f$ by $$\omega_f(u, v) = f(u)(v),$$ i.e. you take the action of $f(u) \in V^\ast$ on the vector $v \in V$. Since $f$ is a skew map, we have that $$\omega_f(v, u) = f(v)(u) = f^\ast(u)(v) = -f(u)(v) = -\omega_f(u,v),$$ so $\omega_f$ is skew-symmetric. Presumably $f$ is also linear, from which the bilinearity of $\omega_f$ follows. Hence $\omega_f$ is a $2$-form on $V$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .