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$?

  • 2
    $\begingroup$ The isomorphism is called the Hodge dual en.wikipedia.org/wiki/Hodge_dual but I only know it in dimension 2 (where it rotates by 90 degrees using the formula for perpendicular slope) and dimension 3 (where it is called the cross product more or less). $\endgroup$ – Jack Schmidt Oct 20 '11 at 2:21

Here are the steps to proving the (non-canonical) isomorphism you ask about:

  • Prove that $\wedge^k F$ is free of rank $\binom nk$. (This is just the same as in the context of vector spaces over a field; if you choose a free basis $x_i$, $i = 1,\ldots, n$, for $F$, then the $k$th exterior power is freely generated by the products $x_{i_1}\wedge x_{i_2} \wedge \cdots \wedge x_{i_k}$, for sequences $1 \leq i_1 < i_2 < \cdots < i_k \leq n$.)

  • Observe that $\binom nk=\binom{n}{n-k}$, and that free modules of the same free rank are isomorphisc.

Here is a related, but more canonical, isomorphism:

Wedge product induces a bilinear pairing $\wedge^k F \times \wedge^{n-k}F \to \wedge^n F$, which one checks to be non-degenerate.

The target is a line (i.e. one-dimensional), and so from this one deduces a canonical isomorphism
$\wedge^{n-k}F \cong (\wedge^k F)^*\otimes \wedge^n F.$

(Here ${}^*$ denotes the dual of a free $R$-module.)


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