Does a vector space need a quadratic form in order to define a wedge product?

Question

Geometric (Clifford) algebras require the vector space to be endowed with a quadratic form in order to define the geometric product.

Meanwhile, an exterior algebra has the wedge product as its defining product. It is not clear to me if a quadratic form is needed to specify a wedge product. So: does a wedge product need the vector space to be equipped with a quadratic form in order to be defined?

It seems like maybe not. For example, in 2D, the determinant, which can be calculated from a basis independent formula, is related to the definition of the wedge product.

Now, suppose we had a wedge product which was defined without relying on a quadratic form. Could the wedge product be then used to motivate a “dot product” or quadratic form?

For example, lets say that the wedge product was defined without a quadratic form, and we defined the lengths and angle of two vectors via: $$|a \wedge b|=|a||b|\sin \theta$$ Could this definition of length and angle then be used to define a dot product?

Answer 1

The wedge product in geometric algebra is typically defined in one of two ways, the first as the completely antisymmetrized product $$\mathbf{a} \wedge \mathbf{b} = \frac{1}{{2}} \left( { \mathbf{a} \mathbf{b} - \mathbf{b} \mathbf{a} } \right),$$ and the second, as the grade-2 selection of a product of two vectors $$\mathbf{a} \wedge \mathbf{b} = {\left\langle{{ \mathbf{a} \mathbf{b} }}\right\rangle}_{2}.$$ In the first case, a metric for the vector space is required for this definition to expand in the usual fashion. In the second case, the definition is really of no use whatsoever without the metric.

Let's look at the antisymmetrized product and see where the metric dependencies are implied. Suppose that we have a two dimensional vector space with basis $\left\{ {\mathbf{e}_1, \mathbf{e}_2} \right\}$, where $$\begin{aligned} \mathbf{a} &= a^1 \mathbf{e}_1 + a^2 \mathbf{e}_2 \\ \mathbf{b} &= b^1 \mathbf{e}_1 + b^2 \mathbf{e}_2.\end{aligned}$$ If we imposed a metric $ \mathbf{e}_i \cdot \mathbf{e}_j = g_{ij} $, then it gives us a notion of length and orthonormality, but do we need that for the wedge product, if defined as an antisymmetric sum? That antisymmetric sum expands as $$\begin{aligned}\mathbf{a} \wedge \mathbf{b} &= \frac{1}{{2}} \left( { \mathbf{a} \mathbf{b} - \mathbf{b} \mathbf{a} } \right) \\ &= \frac{1}{{2}} \sum_{i,j = 1}^2 \left( { a^i b^j \mathbf{e}_i \mathbf{e}_j - b^i a^j \mathbf{e}_i \mathbf{e}_j } \right) \\ &= \frac{1}{{2}} \sum_{i \ne j} \left( { a^i b^j \mathbf{e}_i \mathbf{e}_j - b^i a^j \mathbf{e}_i \mathbf{e}_j } \right) + \frac{1}{{2}} \sum_{i = 1}^2 \left( { a^i b^i \mathbf{e}_i \mathbf{e}_i - b^i a^i \mathbf{e}_i \mathbf{e}_i } \right) \\ &= \sum_{i \ne j} a^i b^j \frac{1}{{2}} \left( { \mathbf{e}_i \mathbf{e}_j - \mathbf{e}_j \mathbf{e}_i } \right) \end{aligned}$$

This wedge product definition, even without a metric, allows us to conclude that $$\begin{aligned} \mathbf{a} \wedge \mathbf{a} &= 0 \\ \mathbf{a} \wedge \mathbf{b} &= -\mathbf{b} \wedge \mathbf{a},\end{aligned}$$ but we don't know how to reduce an expression like $$\frac{1}{{2}} \left( { \mathbf{e}_i \mathbf{e}_j - \mathbf{e}_j \mathbf{e}_i } \right),$$ nor can we give any meaning to any of $ \left\lVert {\mathbf{a}} \right\rVert $, $\left\lVert {\mathbf{b}} \right\rVert$, $ \left\lVert { \mathbf{a} \wedge \mathbf{b} } \right\rVert $.

In geometric algebra, the metric is usually introduced by way of the contraction axiom $$ \mathbf{a}^2 = \mathbf{a} \cdot \mathbf{a}.$$ When expanded in coordinates, this brings the metric into the mix explicitly $$\begin{aligned} \mathbf{a}^2 &= \sum_{i,j = 1}^N \left( { a^i \mathbf{e}_i } \right) \cdot \left( { a^j \mathbf{e}_j } \right) \\ &= \sum_{i,j = 1}^N a^i a^j \mathbf{e}_i \cdot \mathbf{e}_j \\ &= \sum_{i,j = 1}^N a^i a^j g_{ij}.\end{aligned}$$ Given this axiom, we have $$ \left( { \mathbf{a} + \mathbf{b} } \right)^2 = \mathbf{a}^2 + \mathbf{b}^2 + \mathbf{a} \mathbf{b} + \mathbf{b} \mathbf{a},$$ but also $$ \left( { \mathbf{a} + \mathbf{b} } \right)^2 = \left( { \mathbf{a} + \mathbf{b} } \right) \cdot \left( { \mathbf{a} + \mathbf{b} } \right) = \mathbf{a}^2 + \mathbf{b}^2 + 2 \mathbf{a} \cdot \mathbf{b},$$ so we are able to see that the symmetric sum is the dot product, that is $$ \mathbf{a} \cdot \mathbf{b} = \frac{1}{{2}} \left( { \mathbf{a} \mathbf{b} + \mathbf{b} \mathbf{a} } \right),$$ and $$\mathbf{b} \mathbf{a} = -\mathbf{a} \mathbf{b} + 2 \mathbf{a} \cdot \mathbf{b}.$$ In particular, for $i \ne j$ $$\begin{aligned} \mathbf{e}_i \wedge \mathbf{e}_j &=\frac{1}{{2}} \left( { \mathbf{e}_i \mathbf{e}_j - \mathbf{e}_j \mathbf{e}_i } \right) \\ &=\frac{1}{{2}} \left( { \mathbf{e}_i \mathbf{e}_j - \left( { -\mathbf{e}_i \mathbf{e}_j + 2 \mathbf{e}_i \cdot \mathbf{e}_j } \right) } \right) \\ &=\mathbf{e}_i \mathbf{e}_j - \mathbf{e}_i \cdot \mathbf{e}_j.\end{aligned}$$ Only by virtue of having a metric in play, do we see the notion of grade fall out of the mix in a natural fashion (here, the wedge product of our basis vectors, is observed to be the portion of the vector product that has the scalar part of the product subtracted off.)

Answer 2

$ \newcommand\Ext{{\textstyle\bigwedge}} \newcommand\Cl{\mathrm{Cl}} \newcommand\tr{\mathrm{tr}} $Let $V$ be a finite-dimensional real vector space.

Does the exterior algebra require a quadratic form to be defined?

No (not really, but sort of). I think this becomes clear if we adopt the following sort-of-informal definitions of the exterior algebra and Clifford algebras:

• The exterior algebra $\Ext V$ of $V$ is the associative algebra generated by $V$ subject only to the relations $v\wedge v = 0$ for all $v \in V$. (Note that if we use typical algebra notation then this would be written $v^2 = 0$.)

• The Clifford algebra $\Cl(Q)$ of a quadratic form $Q$ over $V$ is the associative algebra generated by $V$ subject only to the relations $v^2 = Q(v)$ for all $v \in V$.

These definitions are formalized by talking about universal properties or quotients of the tensor algebra. It should be immediately obvious from these definitions that $\Ext V$ is just $\Cl(0)$, where by "$0$" I mean the trivial quadratic form $Q(v) = 0$ for all $v$. In this sense, $\Ext V$ does require a quadratic form: the one where you don't "actually" have one!

There is, however, a very real sense in which $\Ext V$ is what you get when you "take away" the quadratic form of $\Cl(Q)$. We can show that there is a canonical choice of vector space isomorphism $\Cl(Q) \cong \Ext V$ for any and every $Q$, or in other words there is a specific, preferred way of viewing $\Cl(Q)$ as $\Ext V$ endowed with another product. This is essentially covered between the question and answers here. Once we have this fact, we of course also have that $\Cl(Q) \cong \Cl(Q')$ as vector spaces for any $Q, Q'$. It is in this way that $\Cl(Q)$ is what you get when you "deform" the product of $\Ext V$ with $Q$, $\Ext V$ is what you get when you you "remove" $Q$ from the product of $\Cl(Q)$, and all Clifford algebras are just different "deformations" of the same $\Ext V$.

Does the exterior algebra $\Ext V$ induce a quadratic form on $V$?

No, other than the trivial $Q(v) = v\wedge v = 0$. I don't have much to say here since as far as I know there isn't anything to say. The issue with your question here is that (1) you haven't defined $|a\wedge b|$ and (2) we have no way of breaking up the product of three terms $|a||b|\cos\theta$ without further information. Once you try specifying that information, you're probably just going to end up specifying a quadratic form on $V$ anyway.

To go off in a different direction, the one "trick" I know that we could try is to define the trace $\tr(Y)$ of $Y \in \Ext V$ as the trace of $X \mapsto Y\wedge X$, and then define a quadratic form by $Y \mapsto \frac1{2^n}\tr(Y\wedge Y)$ with $2^n$ being the dimension of $\Ext V$ when $n = \dim V$. But you will find that this is just the square of the scalar part of $Y$, and in particular gives us nothing on vectors.