Let $(M,g)$ be a Riemannian manifold, differential forms are defined using tensors, could we define a tensor using a differential form? For example, if $\omega$ is a two-form on $M$ which is expressed locally as $$F(p)=F_{ij}(p)e^i\wedge e^j,$$ where $e_i$ is an o.n. basis for $T_pM$ and $F_{ji}=-F_{ij}$, is $T=(F_{ij}-F_{ji})e^i\otimes e^j$ a well-defined two-tensor on $M$?
Added: is $F(p)=\sigma_{ij}F_{ij}(p)e^i\wedge e^j$ well-defined? where $\sigma_{ij}=1$ or $-1$.
Edited according to Hurkyl's answer.
Nope! For example, $e^1 \wedge e^2$ and $-(e^2 \wedge e^1)$ are the same 2-form, however $e^1 \otimes e^2$ and $-(e^2 \otimes e^1)$ are quite different tensors.
However, this can be repaired, as the map $x \wedge y \mapsto \frac{1}{2} (x \otimes y - y \otimes x)$ does induces a homomorphism from 2-forms to tensors. Furthermore, this map splits the projection map induced by $x \otimes y \mapsto x \wedge y$. (i.e. doing the former then the latter gives you the 2-form you started with)
(alternatively, you could insist on $F_{ij}$ being anti-symmetric)