The following is a question that comes from a counterexample to a problem about exterior algebras of modules and the notes mentions somehow this all relates to the fact that symplectic geometry is done over even dimension.
Let $F$ be a free $R$-module of rank $ n\geq 3$ where $R$ is a commutative ring with identity.
Let $T(F) = \oplus_{k=0}^{\infty} T^{k}(F)$ where $T^k(F) = F \otimes F \otimes \ldots \otimes F$ is tensor product of $k$ modules.
Let $\wedge F $ denote the exterior algebra of the $R$-module $F$, that is the quotient of the tensor algebra $T(F)$ by the ideal $A(F)$ generated by all $x \otimes x$ for $x \in F$.
Suppose $1 \leq k \leq n-1$.
How do we show $ \wedge ^k F \cong \wedge^{n-k} F$?