The definition of a non-degenerate skew symmetric $\omega : H \otimes V \to H^{*} \otimes V^{*} $, where H and V are finite dimensional vector spaces, is that for each $v \in V$ non-zero, $\omega : H \otimes \langle v \rangle \to H^{*} \otimes V^{*} $ is injective.
I am not able to find an example of a form that is not ''globally injective'', e.g. $ker \omega \neq 0$ but $\omega$ is nondegenerated.
I would appreciate any example of such $\omega$.
Thank you!
According to the definition stated in the post $$\omega : H \otimes V \to H^{*} \otimes V^{*}$$ is non-degenerate when for any non-zero $v\in V$, the restriction $$\omega|_v : H \otimes \langle v \rangle \to H^{*} \otimes V^{*}$$ is injective.
This is equivalent to saying that $\ker\omega|_v=0$ or that $\omega_v$ is an isomorphism onto its image. However, this implies that $\omega$ is itself injective.
To see that, let us suppose that $\omega$ is non-degenerate (according to your definition) but $\ker\omega\neq 0$. Let $h,v\neq 0$ be such that $h\otimes v\in\ker\omega$, thus $\omega(h\otimes v)=0$. But this means that for this particu-lar $v\neq 0$, we will have $\omega|_v(h\otimes v)=\omega(h\otimes v)=0$, which implies that $h\otimes v\in\ker\omega|_v\neq 0$, thus $\omega|_v$ would not be injective.
I think that your definition of non-degeneracy seems too restrictive. Usually, non-degeneracy in such multilinear maps means that for any non-zero $v\in V$, the restriction $\omega|_v$ is not identically zero, i.e. there exists at least one non-zero $h\in H$ such that $\omega|_v(h\otimes v)\neq 0$.
So, you ask for a form $\omega$ which is non-degenerate $\Leftrightarrow\ker\omega=0$ and at the same time $\ker\omega\neq0$ ?
– KonKan Feb 20 '16 at 02:44