1
$\begingroup$

The Clifford algebra $Cl_{2, 0}(\mathbb{R})$ is a 4-dimensional algebra with vector space $G^2$ spanned by the basis vectors $\{1, e_1, e_2, e_1e_2\}$ where \begin{align*} e_i^2 &= +1 &&1 \leq i \leq 2 \\ e_ie_j &= -e_je_i &&i \neq j. \end{align*} Why does $e_1e_2$ square to $+1$ (according to page 3 of this paper)?

We have $(e_ie_j)^2=e_ie_je_ie_j=e_i(e_je_i)e_j=e_i(-e_ie_j)e_j=-e_i^2e_j^2=-1$?

While I am here: The grade of a Clifford algebra basis element is the dimension of the subspace it represents. Why does the basis element $1$ have a grade 0? Doesn't it represent $\mathbb{R}$?

$\endgroup$

2 Answers 2

2
$\begingroup$

The authors are wrong. $(e_1e_2)^2 = -1$ for both $\mathrm{Cl}_{2,0}(\mathbb R)$ and $\mathrm{Cl}_{0,2}(\mathbb R)$.

For your grade question: note how non-zero vectors represent lines, but projectively. Two vectors represent the same line iff they are non-zero multiplies of each other (so that they then have the same span). The same is true of a blade $A$: another blade $B$ represents the same subspace iff $B = \alpha A$ for some nonzero $\alpha \in \mathbb R$. We can more concretely define the subspace $[A]$ that the blade $A$ represents via $$ [A] = \{v \in \mathbb R^n \;:\; v\wedge A = 0\}. $$ To see that this definition makes sense, recall that the wedge product of vectors is zero iff those vectors are linearly dependent. Now if $A \not= 0$ is grade $0$, i.e. a scalar, for any vector $v$ we have $v\wedge A = vA$; this is non-zero except when $v = 0$, so $[A] = \{0\}$.

Thus any non-zero grade zero element (i.e. non-zero scalar) represents the the unique $0$-dimensional subspace otherwise known as the origin, just like how any $k$-blade represents a $k$-dimensional subspace.

$\endgroup$
1
$\begingroup$

$e_ie_j$ squares to $-1$: $$ (e_ie_j)^2=e_ie_je_ie_j=-e_ie_ie_je_j=-(e_i)^2(e_j)^2=-1. $$ Actually, the even sub algebra of this algebra is the algebra of complex numbers, identifying $e_ie_j$ with $i$.

The scalars are sort of exceptional. Grade $1$ is reserved for vectors, as they correspond to $1$-dimensional spaces. It is convenient to set the grade of scalars to be zero, so one can make statements such as: "the inner product of two homogeneous multi vectors of grades $s$ and $r$ is $|s-r|$". If $s=r$ you get a scalar, so the formula also works if you set the grade of a scalar equal to zero. And other statements of the same sort are more compact with this definition.

$\endgroup$
4
  • $\begingroup$ So the paper is wrong? $\endgroup$
    – user572780
    Feb 9, 2023 at 16:53
  • $\begingroup$ Scalars are not exceptional in the slightest. See my answer. $\endgroup$ Feb 9, 2023 at 17:11
  • $\begingroup$ Agree. Projectively speaking, they are not exceptions. $\endgroup$
    – GReyes
    Feb 9, 2023 at 17:56
  • 1
    $\begingroup$ The paper has a typo. $\endgroup$
    – GReyes
    Feb 9, 2023 at 22:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .