# Why does squaring the pseudoscalar of Clifford algebra $Cl_{2, 0}(\mathbb{R})$ equal 1?

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 $$\mathbb{R}$$?

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.

$$e_ie_j$$ 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.

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