Low-Dimensional Exterior Algebras | M.E. Irizarry-Gelpí

Mon 12 June 2017
Maths
#exterior-algebra, #parabolic

The Cayley-Dickson construction can be generalized to yield other kinds of algebras other than quaternions and octonions. The parabolic constructs involve adjoining elements that square to zero. At first I thought that these constructs were the dual numbers (2-dimensional) and the hyper numbers (4-dimensional). It turns out that the parabolic Cayley-Dickson constructs are closer to exterior algebras.

Recall that an exterior algebra $\Lambda(V)$ over an $n$-dimensional vector space $V$ is a direct sum of spaces:

$$ \Lambda(V) = \Lambda_{0}(V) \oplus \Lambda_{1}(V) \oplus \Lambda_{2}(V) \oplus \ldots \oplus \Lambda_{n}(V). $$

Here, $\Lambda_{0}$ is the space spanned by the 0-blade, which in our case, are scalar real numbers; $\Lambda_{1}$ is the space spanned by the 1-blades; $\Lambda_{2}$ is the space spanend by the 2-blades, and in general, $\Lambda_{k}$ is the space spanned by the $k$-blades (with $0 \leq k \leq n$). A member of $\Lambda_{1}$ is a vector. A 2-blade is obtained by multiplying two 1-blades with the wedge product. Given two 1-blades $X$ and $Y$, the wedge product

$$ XY \equiv X \wedge Y. $$

The wedge product of two 1-blades is anti-symmetric:

$$ YX = - XY. $$

In general, the dimension of $\Lambda_{k}$ is the binomial coefficient:

$$ \dim{(\Lambda_{k})} = {n \choose k}. $$

Thus,

$$ \dim{(\Lambda)} = \sum_{k = 0}^{n} {n \choose k} = 2^{n}. $$

We are going to explore the cases $n = 1$, $n = 2$, $n = 3$, and $n = 4$.

One-Dimensional

In the 1-dimensional case, the dimension of the exterior algebra is 2. This exterior algebra coincides with the dual numbers:

$$ \Lambda(\mathbb{R}) = \left\lbrace a + b W \mid (a, b) \in \mathbb{R}^2 \right\rbrace. $$

If $x$ and $y$ are elements of $\Lambda(\mathbb{R})$, then their wedge product is given by

$$ xy = (a_{1} + b_{1}W)(a_{2} + b_{2}W) = a_{1}a_{2} + (a_{1}b_{2} + b_{1}a_{2})W + b_{1}b_{2} (WW). $$

The third term is zero because of the anti-symmetry of the wedge product of two 1-blades. However, note that in this 1-dimensional case the multiplication is commutative:

$$ yx = (a_{2} + b_{2}W)(a_{1} + b_{1}W) = a_{2}a_{1} + (a_{2}b_{1} + b_{2} a_{1})W. $$

This is the only case when the multiplication is commutative.

Hodge Star Conjugate

The Hodge star conjugate relates a $k$-blade to an $(n - k)$-blade. In this case, it provides a relation between 0-blades and 1-blades. If

$$x = a + bW, $$

then the Hodge star conjugate on $x$ is

$$ x^{\star} = b + aW. $$

That is, the star conjugate swaps the real and unreal parts of $x$. Note that

$$ (x^{\star})^{\star} = x.$$

That is, the 1-dimensional star conjugate is an involution.

Note that

$$x (x^{\star}) = ab + (a^{2} + b^{2})W.$$

Two-Dimensional

In the 2-dimensional case, the dimension of the exterior algebra is 4. This algebra can be obtained by performing a parabolic Cayley-Dickson construct with the dual numbers. At first I thought that this lead to the hyper numbers. However, the nilpotent units in the hyper numbers commute. Because of the anti-symmetry of the wedge product, the nilpotent units in the exterior algebra anti-commute. The full algebra is:

$$ \Lambda(\mathbb{R}^{2}) = \left\lbrace a + b W + c X + d WX \mid (a, b,c,d) \in \mathbb{R}^4 \right\rbrace. $$

You can make the Cayley-Dickson construct obvious by writing $\Lambda(\mathbb{R}^{2})$ like

$$ \Lambda(\mathbb{R}^{2}) = \left\lbrace A + BX \mid (A, B) \in \Lambda(\mathbb{R}) \times \Lambda(\mathbb{R}) \right\rbrace. $$

If $x$ and $y$ are elements of $\Lambda$, then their wedge product is given by

$$ xy = \left[a_{1} + b_{1}W + c_{1}X + d_{1}WX\right]\left[a_{2} + b_{2}W + c_{2}X + d_{2}WX\right]. $$

More explicitly:

$$ xy = a_{1}a_{2} + (a_{1}b_{2} + b_{1}a_{2})W + (a_{1}c_{2} + c_{1}a_{2})X + (a_{1}d_{2} + d_{1}a_{2} + b_{1}c_{2} - c_{1}b_{2})WX. $$

Because of the anti-symmetry of $WX$, the product of two elements of $\Lambda(\mathbb{R}^{2})$ is non-commutative. However, it is associative.

Hodge Star

The Hodge star operation relates a $k$-blade to an $(n - k)$-blade. In the 2-dimensional case, it provides a relation between 0-blades and 2-blades; and a relation between 1-blades:

$$ 1^{\star} = WX, \qquad W^{\star} = X, \qquad X^{\star} = - W, \qquad (WX)^{\star} = 1.$$

Note that

$$ (1^{\star})^{\star} = 1, \qquad ((WX)^{\star})^{\star} = WX; $$

but

$$ (W^{\star})^{\star} = -W, \qquad (X^{\star})^{\star} = -X. $$

To be specific, if

$$x = a + b W + c X + d WX; $$

then the Hodge star operation on $x$ is

$$ x^{\star} = d - c W + b X + a WX. $$

Thinking of $x$ (an element of $\Lambda(\mathbb{R}^{2})$) as a pair construct with $A$ and $B$ (elements of $\Lambda(\mathbb{R})$) gives

$$x = A + BX, \qquad x^{\star} = (\operatorname{Conj}(B))^{\star} + A^{\star}X.$$

This is a way to express the 2-dimensional Hodge star operation in terms of the one-dimensional Hodge star operation and the one-dimensional conjugation operation.

Acting a second time with the Hodge star gives

$$ (x^{\star})^{\star} = a - b W - c X + d WX. $$

Acting a third time gives

$$((x^{\star})^{\star})^{\star} = d + c W - bY + aWX. $$

Finally, acting a fourth time gives the identity operation

$$(((x^{\star})^{\star})^{\star})^{\star} = x. $$

Note that

$$ x(x^{\star}) = ad - (ac - bd)W + (ab + cd)X + (a^{2} + b^{2} + c^{2} + d^{2})(WX). $$

Three-Dimensional

In the 3-dimensional case, the dimension of the exterior algebra is 8. This algebra can be obtained by performing a parabolic Cayley-Dickson construct with $\Lambda(\mathbb{R}^{2})$. The full algebra is

$$ \Lambda(\mathbb{R}^{3}) = \lbrace a + bW + cX + d WX + eY + f WY + g XY + h (WX)Z \mid (a,b,c,d,e,f,g,h) \in \mathbb{R}^{8} \rbrace. $$

Note that the wedge product of two elements of $\Lambda(\mathbb{R}^{3})$ is non-commutative and non-associative.

Hodge Star

If you write the 3-dimensional Hodge star operation in terms of the 2-dimensional Hodge star operation and the 2-dimensional conjugation operation as before,

$$ x^{*} = (\operatorname{Conj}{(B)})^{\star} + A^{\star}Y; $$

for

$$x = a + bW + cX + d WX + eY + f WY + g XY + h (WX)Y;$$

you get

$$ x^{*} = h + gW - fX - e WX + dY - c WY + b XY + a (WX)Y.$$

However, this is not the Hodge star operation. Instead, it should be

$$ x^{\star} = (B^{\dagger})^{\star} + A^{\star} Y; $$

where

$$ (a + bW + cX + d WX)^{\dagger} = a - bW - cX + d WX. $$

This gives the star conjugate:

$$ x^{\star} = h + gW - fX + e WX + dY - c WY + b XY + a (WX)Y.$$

That is:

$$ 1^{\star} = (WX)Y, \qquad ((WX)Y)^{\star} = 1; $$

$$ W^{\star} = XY, \qquad X^{\star} = -WY, \qquad Y^{\star} = WX; $$

$$ (WX)^{\star} = Y, \qquad (WY)^{\star} = -X, \qquad (XY)^{\star} = W. $$

Note that

$$ (1^{\star})^{\star} = 1, \qquad (((WX)Y)^{\star})^{\star} = (WX)Y; $$

$$ (W^{\star})^{\star} = W, \qquad (X^{\star})^{\star} = X, \qquad (Y^{\star})^{\star} = Y; $$

$$ ((WX)^{\star})^{\star} = WX, \qquad ((WY)^{\star})^{\star} = WY, \qquad ((XY)^{\star})^{\star} = XY. $$

That is, the 3-dimensional Hodge star operation is an involution.

Four-Dimensional

In the 4-dimensional case, the dimension of the exterior algebra is 16. This algebra can be obtained by performing a parabolic Cayley-Dickson construct with $\Lambda(\mathbb{R}^{3})$. An element has the form

$$ x = x_{0} + x_{1} + x_{2} + x_{3} + x_{4}; $$

where

$$ x_{0} = a_{0}; $$

$$ x_{1} = a_{W} W + a_{X} X + a_{Y} Y + a_{Z} Z; $$

$$ x_{2} = a_{WX} WX + a_{WY} WY + a_{WZ} WZ + a_{XY} XY + a_{XZ} XZ + a_{YZ} YZ; $$

$$ x_{3} = a_{WXY} (WX)Y + a_{WXZ} (WX)Z + a_{WYZ} (WY)Z + a_{XYZ} (XY)Z; $$

$$ x_{4} = a_{WXYZ} ((WX)Y)Z. $$

Hodge Star

In the 4-dimensional case, the Hodge star operation provides a relation between 0-blades and 4-blades, a relation between 1-blades and 3-blades, and a relation between 2-blades. Explicitly, you have

$$ 1^{\star} = ((WX)Y)Z; $$

$$ W^{\star} = (XY)Z, \qquad X^{\star} = -(WY)Z; $$

$$ Y^{\star} = (WX)Z, \qquad Z^{\star} = -(WX)Y; $$

$$ (WX)^{\star} = YZ, \qquad (WY)^{\star} = -XZ, \qquad (WZ)^{\star} = XY; $$

$$ (XY)^{\star} = WZ, \qquad (XZ)^{\star} = -WY, \qquad (YZ)^{\star} = WX; $$

$$ ((WX)Y)^{\star} = Z, \qquad ((WX)Z)^{\star} = -Y; $$

$$ ((WY)Z)^{\star} = X, \qquad ((XY)Z)^{\star} = -W; $$

$$ (((WX)Y)Z)^{\star} = 1. $$

Note that, like in 2-dimensions, in 4-dimensions the Hodge star operation is not an involution.