$ \newcommand\Ext\bigwedge \newcommand\pair[1]{\langle#1\rangle} \newcommand\tensor\otimes \newcommand\Iso\mathscr \newcommand\intrsym\lrcorner \newcommand\intr{{\,\intrsym\,}} \DeclareMathOperator\span{span} $
My actual question is towards the bottom, bolded. The rest of this is context.
Geometry-Preserving Isomorphisms
Let $V$ be an $n$-dimensional vector space, $V^*$ it dual, and $\Ext V$ its exterior algebra. I am primarily interested in $V = \mathbb R^n$, but I don't think that makes much difference here.
We identify $\Ext V^*$ with $(\Ext V)^*$ in the usual way: for $v_1,\dotsc,v_k \in V$ and $\alpha_1,\dotsc,\alpha_k \in V^*$ define $$ \pair{\alpha_1\wedge\cdots\wedge\alpha_k, v_1\wedge\cdots\wedge b_k} = \det(M),\quad M_{ij} = \pair{\alpha_i, v_j}, $$ where $\pair{\alpha_i, v_j}$ is the natural pairing, and we extend by linearity to the entirety of $\Ext V^*\tensor\Ext V$, defining $\pair{\cdot,\cdot}$ to be zero when the grades don't match. We will not distinguish between $\Ext V^*$ and $(\Ext V)^*$.
We can identify (up to a scalar multiple) simple $k$-vectors $A \in \Ext^k V$ with subspaces of $V$ by $$ A^\wedge := \{v \in V \;:\; v\wedge A = 0\}, $$ and we can show that (up to a scalar) there is a unique simple $\alpha \in \Ext^{n-k}V^*$ such that $$ \alpha^\intrsym := \{v \in V \;:\; v\intr\alpha)\} = A^\wedge, \tag{$*$} $$ where $\intrsym$ is the interior product, which we'll define for both multivectors and dual multivectors by $$ \pair{\xi, X\wedge Y} = \pair{X\intr\xi, Y},\quad \pair{\xi\wedge\zeta, X} = \pair{\zeta, \xi\intr X} $$ for all $\xi, \zeta \in \Ext V^*$ and $X, Y \in \Ext V$. The exterior products then correspond to spans and intersections of subspaces of $V$: $$\begin{aligned} (A\wedge B)^\wedge &= \begin{cases} \{0\} &\text{if }\,\exists v\in V\setminus\{0\}.\, v \in A^\wedge\text{ and }v \in B^\wedge, \\ \span(A^\wedge\cup B^\wedge) &\text{otherwise}, \end{cases} \\ (\alpha\wedge\beta)^\intrsym &= \begin{cases} \{0\} &\text{if }\,\exists H\subset V.\, H\text{ is a hyperplane and } \alpha^\intrsym \subseteq H\text{ and }\beta^\intrsym \subseteq H, \\ \alpha^\intr\cap B^\intr &\text{otherwise} \end{cases} \end{aligned}$$ for any simple $A, B \in \Ext V$ and simple $\alpha, \beta \in \Ext V^*$. Call these equations ($**$).
A geometry-preserving isomorphism $\Iso I : \Ext V \to \Ext V^*$ is such that (1) if $A$ is simple then $\Iso I(A)$ is simple, and (2) $A^\wedge = (\Iso I(A))^\intrsym$ for simple $A$. We will abbreviate "geometry-preserving isomorphism" to "GP-isomorphism". (I do acknowledge the possibility that there may be a better definition of "geometry-preserving" that captures the same spirit, and would not mind that as a means of answering my question.)
Two particular GP-isomorphisms of interest are the Poincaré isomorphisms. Choose $I \in \Ext^n V$. Since $\Ext^n V$ and $\Ext^n V^*$ are 1-dimensional, there is a unique $I^* \in \Ext^n V^*$ such that $\pair{I^*, I} = 1$. We then define $I^\# : \Ext V \to \Ext V^*$ and $I_\# : \Ext V^* \to \Ext V$ by $$ I^\#(X) = X\intr I^*,\quad I_\#(\xi) = \xi\intr I. $$ We can show that $$ (I^\#)^{-1} = I_\#^*,\quad I_\#^{-1} = (I^\#)^*,\quad (xI)^\# = \frac1xI^\#,\quad (xI)_\# = xI_\#, $$ where here $*$ denotes the dual of a transformation and $x$ is a scalar.
Why Should We Prefer $I^\#$ and $I_\#$?
My main interest is in defining the regressive product $\vee : \Ext V\tensor\Ext V \to \Ext V$, which is usually done by $$ X\vee Y := (I^\#)^{-1}(I^\#(X)\wedge I^\#(Y)). $$ Because $I^\#$ is geometry-preserving, $\vee$ corresponds to the intersection of subspaces just like $\wedge$ does over $\Ext V^*$. We can show that using $I_\#^{-1}$ instead of $I^\#$ results in the same product.
But let's take a step back and work ourselves towards a definition of $\vee$. Given any three GP-isomorphisms $\Iso M, \Iso L, \Iso R : \Ext V \to \Ext V^*$ we could define $$ X\vee_{\Iso M,\Iso L,\Iso R} Y := \Iso M^{-1}(\Iso L(X)\wedge\Iso R(Y)). $$ However, upon requiring that $\vee_{\Iso M,\Iso L,\Iso R}$ have an identity $E$, $$ X\vee_{\Iso M,\Iso L,\Iso R} E = E\vee_{\Iso M,\Iso L,\Iso R} X = X, $$ we can show that it suffices to take $\Iso M = \Iso L = \Iso R$ (and so we will simply write $\vee_{\Iso M}$). Note that this also automatically gives us associativity of $\vee_{\Iso M}$.
So we need only find a single suitable GP-isomorphism.
The correspondence ($*$) above tells us that a GP-isomorphism must be grade reversing, i.e. grade $k$ maps to $n-k$. Not only that, but since all Poincaré isomorphisms are multiples of each other (and every multiple of a Poincaré isomorphism is Poincaré) this correspondence also tells us that the restriction of any GP-isomorphism to a space of simple $k$-vectors must be Poincaré; i.e. for any GP-isomorphism $\Iso I$ and $S \subseteq \Ext^k V$ a one-dimensional subspace of simple $k$-vectors, there is $I \in \Ext^n V$ such that $\Iso I|_S = I^\#|_S$. (It is interesting to note that if our field is $\mathbb F_2$, then the only GP-isomorphism is the unique Poincaré isomorphism.) But it appears to me that this is the only restriction. For example, define $\Iso I = I^\#$ on $k$-vectors for $k \not= 2$ and $\Iso I = xI^\#$ on $2$-vectors for some $x \not= 0, 1$. This acts like a Poincaré isomorphism on each grade individually, but isn't one when considered on the entirety of $\Ext V$, Additionally, $X\vee_{\Iso I} Y \not= X\vee Y$ in general: If $n = 4$, $A \in \Ext^2 V$, and $B \in \Ext^3 V$ then $$ A\vee_{\Iso I} B = \Iso I^{-1}(\Iso I(A)\wedge\Iso I(B)) = (I^\#)^{-1}(xI^\#(A)\wedge I^\#(B)) = xA\vee B, $$ noting that $I^\#(A)\wedge I^\#(B) \in \Ext^3 V^*$ and $3 \not= 4-2$.
So this is my question: solely with an eye towards defining $\vee$, is there any reason to prefer the Poincaré isomorphisms (or perhaps some other isomorphism) over any other geometry-preserving isomorphism? I.e., do the Poincaré isomorphisms give $\vee$ some desirable property that other geometry-preserving isomorphisms do not?
I am aware that $$ X\wedge Y = \pair{I^\#(X), Y}I,\quad \xi\wedge\zeta = \pair{\zeta, I_\#(\xi)}I^*. $$ which give $I^\#$ and $I_\#$ another interpretation as coming from bilinear forms induced by the exterior product. Fixing a non-degenerate quadratic form $Q$ on $V$ (and now assuming characteristic $\not=2$) endows $\Ext V$ with the corresponding Clifford product, and the isomorphism $\Iso Q : V \to V^*$ from $Q$ extends by the universal property of Clifford algebras to a Clifford algebra isomorphism $\Ext V \to \Ext V^*$. I am then also aware that $$ (\Iso Q^{-1}\circ I^\#)(X) = I^{-1}X,\quad (I_\#\circ\Iso Q)(X) = \widetilde XI, $$ where $I^{-1}$ is the Clifford inverse of $I$, $\widetilde X$ is the reversal of $X$ (i.e. main anti-involution), and $I^{-1}X$ and $\widetilde XI$ are Clifford products.
These properties are very, very nice and make the Poincaré isomorphisms desirable; however, I am specifically curious as to whether they are special for defining $\vee$.