Geometric interpretation of intersection product

Question

Let $E,E^*$ be a pair of dual $n$-dimensional real vector spaces and let $e,e^*$ be a pair of dual basis vectors of $\bigwedge^n E$ and $\bigwedge^n E^*$, respectively. For $u,v\in\bigwedge E$, we can define an "intersection product" by $$u\cap v=D^e[(D^e)^{-1}u\wedge(D^e)^{-1}v]\tag{1}$$ where $D^e:\bigwedge E^*\to\bigwedge E$ is the Poincaré duality map given by $D^e u^*=i(u^*)e$ where $i$ is the insertion operator. This product makes $\bigwedge E$ into an associative, unital, anti-commutative algebra (the "intersection algebra").

Question: Is there a simple geometric interpretation of the product in (1), analogous to the geometric interpretation of the wedge product in terms of volume? By "simple" I mean using basic concepts of linear and exterior algebra and not more advanced theories (manifolds, homology, cohomology, etc.).

My thoughts so far:

• When I first saw the name "intersection product" I thought maybe when $u$ is a $p$-blade representing a subspace $U$ of $E$ and $v$ is a $q$-blade representing a subspace $V$ of $E$, then $u\cap v$ would represent the intersection $U\cap V$, but that's clearly not right since $u\cap v$ has degree $p+q-n$. If $p+q=n$, then $u\cap v$ is a scalar and $u\cap v\ne 0$ if and only if $E=U\oplus V$, with the sign of $u\cap v$ describing the relative orientations of $U$, $V$, and $E$.

• The intersection algebra in $\bigwedge E$ is isomorphic to the exterior algebra in $\bigwedge E^*$, so maybe this leads to a simple geometric description.

• If $E$ is an oriented Euclidean space with oriented unit volume $e$, then we can take $E^*=E$ under the inner product and Poincaré duality reduces to Hodge duality, which may also lead to a simple description.

Any pointers are appreciated.

Answer 1

$ \newcommand\Ext{{\bigwedge}} $

Yes. Let $A, B \in \Ext E$ be simple multivectors, i.e. products of vectors. We can associate each such simple multivector with a subspace of $E$ by $$ [A] := \{x \in E \;:\; x\wedge A = 0\}. $$ Now let $H$ be the set of all hyperplanes of $E$, which could be expressed as $$ H := \{\ker\phi \;:\; \phi \in E^*\}. $$ We may then derive that $$ [A\wedge B] = \begin{cases} \{0\} &\text{if }\exists x\in E.\:x \in [A]\text{ and }x \in [B], \\ [A]\oplus [B] &\text{otherwise}, \end{cases} $$ and $$ [A\vee B] = \begin{cases} \{0\} &\text{if }\exists U\in H.\:[A]\subseteq U\text{ and }[B]\subseteq U, \\ [A]\cap [B] &\text{otherwise}. \end{cases} $$ By $[A]\oplus[B]$ I specifically mean the span of $[A]\cup[B]$.