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Note: Geometric algebra here refers to the Clifford algebra of Euclidean space, and thus is distinct from algebraic geometry, although both are related to differential forms (to the best of my knowledge).

Question: The tensor product of two homomorphisms of abelian groups, $f:A \to A'$, $g:B\to B'$, satisfies for all $a\in A$, $b\in B$, the identity: $$f\otimes g ( a\otimes b)=f(a) \otimes g(b).$$

Does this mean that we can interpret differential 2-forms as linear functionals of bivectors, and in general differential $k$-forms as linear functionals of $k$-vectors?

Or am I oversimplifying? (Or should I say $k$-blades instead of $k$-vectors? I always confuse the two.)

Comments: Such an interpretation might make Hodge duality easier to understand in terms of geometric algebra.

David Hestenes has almost certainly addressed this in one of his numerous works on the subject, so a reference (please don't forget the page number!) to one of his or someone else's works on the subject will suffice for an answer.

Chill2Macht
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    The answer is yes. I'll see if I can find a reference on it. But, I should mention that in the GA literature (or at least Hestenes's recent works), $k$-forms are simply $k$-blades in the tangent space of a vector manifold. BTW, because $k$-forms are linear, saying they take $k$-blades or $k$-vectors as arguments is equivalent (because $k$-vectors are just linear combinations of $k$-blades). –  Dec 04 '16 at 04:03
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    Note that if you're talking about the regular geometric algebra -- $\mathcal G^n(\Bbb R)$ with either the Euclidean or Lorentzian inner product (which honestly are the only Clifford algebras I really study) -- then it's a trivial matter to prove that $dx^i\wedge dx^j(e_k,e_l) = e^k\wedge e^l(e_k \wedge e_l)$ for all $i,j,k,l$ (where on the left is the antisymmetric tensor wedge and the on the right is just one notation that I use for a linear functional of bivectors). And because both are (multi-)linear functions this proves that it is equivalent to work with these bivector ... –  Feb 15 '17 at 14:32
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    linear functionals instead of the usual covectors. Note however that the tensor-defined exterior product doesn't require your vector space be equipped with a metric and thus can be applied more generally. But, at least in this situation, and probably more generally for Clifford algebras (though I can't seem to find a reference), the two ideas are equivalent. –  Feb 15 '17 at 14:34

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Consider exercise 4.(d), p.349, Chapter 9 "Riemannian Metrics", of Spivak's A Comprehensive Introduction to Differential Geometry Volume I, 3rd edition, 1999.

Using the isomorphisms $\bigotimes^k V^* \approx (\bigotimes^k V)^*$ and $\Lambda^k(V^*) \approx (\Lambda^k V)^*$, define inner products on $\otimes^K V$ and $\Lambda^k V$ by using the isomorphism $V \to V^*$ given by the inner product on $V$.

Assuming that the implicit claim $\Lambda^k(V^*) \cong (\Lambda^k V)^*$ is correct, we have as a special case that: $$\Lambda^k(V^{**}) \cong (\Lambda^k (V^*))^* \,. $$ By definition of the space of differential $k$-forms, $\Omega^k (V) = \Lambda^k(V^*)$, thus we have: $$\Lambda^k(V^{**}) \cong (\Omega^k(V))^* \,. $$ If $V$ is finite-dimensional, we will also have the identification $V^{**} \cong V$, so that $$\Lambda^K(V) \cong (\Omega^k(V))^* \iff (\Lambda^k(V))^* \cong (\Omega^k(V))^{**} \,, $$ and then using the fact that $\Omega^k(V)$ itself is finite-dimensional if and only if $V$ is, thus $(\Omega^k(V))^{**} \cong \Omega^k(V)$, therefore from the above we have: $$(\Lambda^k(V))^* \cong \Omega^k(V)\,. $$ Thus linear functionals on $k$-vectors are up to vector space isomorphism equivalent to differential $k$-forms. (I imagine the isomorphism even extends up to wedge product algebras, but that would still have to be verified based on the above, so I am not really sure.)

Likewise, $\Lambda^k(V) \cong (\Omega^k(V))^*$, i.e. $k$-vectors are equivalent up to vector space isomorphism to linear functionals of differential $k$-forms.

Thus it does seem that (at least for finite-dimensional spaces) the operations of forming the exterior algebra and the dual space "commute" with one another.

Since the exercise was from a chapter about (pseudo-)Riemannian metrics (i.e. non-degenerate bilinear forms), I imagine that this commutativity property also extends to the formation of the Clifford algebra and the dual vector space (at least this seems to be what Spivak is implying).

This should hold at least when considered as vector spaces, since I know that the exterior and Clifford (i.e. geometric) algebras are isomorphic as vector spaces. I do not know if it extends to the algebra structure, since I don't even know if the exterior and Clifford algebras are isomorphic as wedge product algebras, much less if their duals are.

Spivak's mention of inner product on $\Lambda^kV$ seems like an oblique hint at Hodge duality.

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Chill2Macht
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