# Can we construct the exterior algebra just from simple multivectors?

$$\newcommand\K{\mathbb K} \newcommand\Ext{\mathop{\textstyle\bigwedge}} \newcommand\Lip{\mathrm{Lip}} \newcommand\ev{\mathrm{ev}} \newcommand\Gr{\mathrm{Gr}}$$Let $$V$$ be a finite-dimensional $$\K$$-vector space and $$\Ext V$$ its exterior algebra. The Lipschitz monoid $$\Lip(V)$$ is the monoid of all simple multivectors, or equivalently the monoid of all multiples of exterior products of vectors: $$\Lip(V) = \{\alpha v_1\wedge\dotsb\wedge v_k \;:\; \alpha\in\K,\; v_1,\dotsc,v_k \in V,\: k \geq 0\}.$$

Can we construct $$\Ext V$$ from $$\Lip(V)$$?

$$\Ext V$$ "feels" quite like the monoid algebra $$\K[\Lip(V)]$$ of $$\Lip(V)$$ but modulo the vector space structure of $$V$$. Indeed, note that $$V \subseteq \Lip(V)$$. Denoting the scalar product of $$\K[\Lip(V)]$$ by $$\cdot$$, the set $$\K\cdot V = \left\{\sum_{i=1}^k\alpha_i\cdot v_i \;:\; \alpha_1,\dotsc,\alpha_k\in\K,\: v_1,\dotsc,v_k\in V,\: k\geq 0\right\} \subseteq \Lip(V)$$ of all formal linear combinations of vectors has a natural evaluation map $$\ev : \K\cdot V \to V$$ taking formal linear combinations to actual linear combinations; in symbols, $$\ev\left(\sum_{i=1}^k\alpha_i\cdot v_i\right) = \sum_{i=1}^k\alpha_iv_i.$$

This map easily extends to a homomorphism $$\K[\Lip(V)] \to \Ext V$$. First, for any $$\alpha \in \K$$ and $$X \in \Lip(V)$$, define $$\ev(\alpha\cdot X) = \alpha X$$ which suffices to define $$\ev$$ on all of $$\K[\Lip(V)] = \bigoplus_{X\in\Lip(V)}\K\cdot X$$. Now we check that it is as homomorphism: if $$\beta \in \K$$ and $$Y \in \Lip(V)$$ then $$\ev([\alpha\cdot X][\beta\cdot Y]) = \ev([\alpha\beta]\cdot[X\wedge Y]) = \alpha\beta X\wedge Y,$$$$\ev(\alpha\cdot X)\wedge\ev(\beta\cdot Y) = (\alpha X)\wedge(\beta Y) = \alpha\beta X\wedge Y,$$ so $$\ev$$ is indeed a homomorphism.

This gives $$\Ext V \cong \K[\Lip(V)]/\ker(\ev)$$ as vector spaces.

The above suggests to me that, perhaps, there is some universal property that allows us to generate $$\Ext V$$ from $$\Lip(V)$$; but it is not clear to me what exactly this universal property is, nor what category it should take place in.

This is just the approach that I first thought up; if anyone has a completely different way of deriving $$\Ext V$$ from $$\Lip(V)$$ that would be most appreciated.

This question could be extended to Clifford algebras in general, but I think there is some more subtlety in the general case which I think makes it best to leave that to a future question.

As for why I am interested in doing this:

Consider the following relation on $$\Lip(V)$$: $$X \sim Y \iff \exists\alpha\in\K\setminus\{0\}.\: X = \alpha Y.$$ This relation is compatible with the monoid structure of $$\Lip(V)$$, so $$\Lip(V)/{\sim}$$ is a monoid; denote the equivalence classes of this quotient by $$[{-}]$$. But we have a bijection $$\begin{gathered} (\Lip(V)/{\sim})\setminus\{[0]\} \\ [X] \end{gathered} \mathrel{\begin{gathered} \longleftrightarrow \\ \longmapsto \end{gathered}} \begin{gathered} \Gr(V), \\ \{v \in V \;:\; v\wedge X = 0\}. \end{gathered}$$ where $$\Gr(V)$$ is the Grassmannian of $$V$$.

This makes $$\Gr(V)\sqcup\{\mathbf0\}$$ into a monoid (where $$\mathbf0$$ is a formal zero) and shows how we can construct $$\Lip(V)/{\sim}$$ geometrically. It's worth describing this product in terms of $$\Gr(V)$$: for any $$S, T \in \Gr(V)$$ it is defined by $$S\wedge T = \begin{cases} \mathbf0 &\text{if }S\cap T\ne\{0\}, \\ \mathrm{span}(S\cup T) &\text{otherwise} \end{cases}$$ and by requiring $$\mathbf0$$ to be a zero element. Note that $$\{0\}$$ is the identity. Put another way, this product is just the internal direct sum of subspaces returning the sigil $$\mathbf0$$ when such a direct sum is invalid.

We could then go from here to $$\Lip(V)$$ and then to $$\Ext V$$, giving a very geometric construction $$\Ext V$$. It's unclear to me how to identify $$V$$ as a subset of $$\Lip(V)$$ from this construction, but I will leave that for another question.

• Does Lip$(V)$ include all scalars, or just the scalar $1$ (and $0$)? Commented May 2 at 4:00
• Good catch. I guess the two definitions I gave aren't equivalent. I think, morally for this specific question, I want it to include all scalars: just like all (nonzero) multiples of a vector or bivector represent a line or plane, all multiples of $1$ represent the origin. This is also how Micali and Helmstetter define it in Quadratic Mappings and Clifford algebras (see proposition 5.3.2); it seems that one of their goals in defining $\mathrm{Lip}(V)$ is for it to contain all bivector exponentials. Commented May 2 at 5:27