A question on the dual relationship between the regressive product and the exterior product I am trying to understand the following sentence, which I came across in a book:

The underlying beauty of the Ausdehnungslehre is due to this symmetry [the duality between the regressive and exterior product], which in turn is due to the fact that linear spaces of $m$-elements and linear spaces of $(n-m)$-elements have the same dimension...

According to the book, this is related to the dual nature of the regressive and exterior products:

For example, the exterior product of $m$ 1-elements is an $m$-element. The dual to this is that the regressive product of $m$ $(n-1)$-elements is an $(n-m)$-element.

What exactly is confusing about the above?
1) How do linear spaces of $m$-elements and linear spaces of $(n-m)$-elements have the same dimension? Actually, what is meant by dimension here? I understand dimension to be the number of $m$-elements necessary to build an object of a particular space - i.e. the grade of the object. Is this correct? With this view, $(n-m) \neq m$.
2) I don't see how the fact that the result of the regressive product when applied to $m$ $n-1$ elements is an $(n-m)$-element makes it the dual of the exterior product...
3) How is duality akin to symmetry? I am unable to appreciate the beauty in the duality between two operators.  
 A: I'm just going to answer this in the language of geometric algebra (because I can't seem to translate it into Grassmann algebra right).

A $k$-blade is the exterior product of $k$ linearly independent vectors.  In geometric algebra, the dual $A^*$ is the $(n-k)$-blade such that $A \wedge A^* = |A|^2I$ (but we're not particularly interested in that constant, $|A|^2$, here), where $I$ is called the unit pseudoscalar -- the highest grade element of our $n$-space.  Thus we can consider this "dual" to be the part of $I$ which doesn't "contain" any of $A$.
The "regressive product" seems to be what I've heard called the "meet" of blades.  The meet $M=A \vee B$ (which if you learn Geometric Algebra has the simple formula $A \vee B = A^* \cdot B$) is the blade of largest grade such that $A=M \wedge A'$ and $B=B' \wedge M$.  Thus the regressive product is the part of $A$ and $B$ "contained" by both $A$ and $B$.
You'll want to note that in $\Bbb G^n$, there are $\binom{n}{k}$ orthogonal unit $k$-blades. Thus each $k$-space (the subspace of elements of grade $k$) has the same dimension as $(n-k)$-space.  Where dimension here means the same thing it does in linear algebra: it's the maximum number of linearly independent "basis blades" the any $k$-space can hold -- this is distinct from the grade of a $k$-space/ $k$-blade.  For example, take our space to be $\Bbb G^4$.  This is the space of scalars, vectors, 2-blades, 3-blades, 4-blades, and the sums of these objects.  In this space there are $\binom 42=6$ basis 2-blades -- usually they are called $e_1 \wedge e_2, e_1 \wedge e_3, e_1 \wedge e_4, e_2 \wedge e_3, e_2 \wedge e_4, $ and $e_3 \wedge e_4$ -- and thus the dimension of the subspace formed by the span of these basis blade is $6$.  However each element of the subspace is of grade $2$.
The fact that each $k$-space then has the same dimension as $(n-k)$-space is really the important part for understanding what your book is saying:
What this section of your book is saying is that: $ (\wedge^m\ V)^* = \vee^m\ V^*$ and (by the fact that the dual of the dual of an element is that same element up to sign) likewise $(\vee^m\ V)^* = \wedge^m\ V^*$.  That is $(A \wedge B)^*=A^* \vee B^*$.  So the smallest blade "containing" $A$ & $B$ is dual to the largest blade "contained" by both the dual of $A$ and the dual of $B$.
Because in geometric algebra we have a very simple formula for finding the dual of a blade (for reference it's $A^* = AI^{-1}$), we could theoretically represent any exterior product by regressive products (AKA meets) and vice versa.  I think this rather unexpected and potentially powerful result is what this author is calling the "beauty" of this relation.

I don't know if any of this has made any sense to you, so just comment below if you still have questions.
