Is Clifford product only for scalars already defined?

Question

hi i'm a high school student interested in Clifford algebra. Since it is hard for me to understand several things thus i hope to know as follows:

Is Clifford product only for scalars already defined?

i mean there are some equations on Internet which said $(e_1e_2)^2=-1$

To avoid any mistakes, i will not do any abbreviation.

the symbol $*$ means clifford product (aka Geometric product)

definition:

$e_1*e_1:=1$ and $e_1*e_2:=-e_2*e_1$ vice versa

equation:

$(e_1*e_2)*(e_1*e_2)= (e_1*(e_2*e_1)*e_2)$

$=(e_1*((-e_1)*e_2)*e_2)$ $=-(e_1*e_1)*(e_2*e_2)$ $=-1*1$

and then i got stuck in the last sequence

how could $-1*1$ be $-1$?

is there any a defined notion in Clifford algebra?

Answer 1

A Clifford algebra is a particular kind of associative algebra with identity. "Associative" means $a*(b*c) = (a*b)*c$ for any elements $a, b, c$. The "identity", which we'll call $\mathbb I$, is the unique element such that $a*\mathbb I = \mathbb I*a = a$ for all elements $a$. A Clifford algebra is also a vector space, meaning among other things that there is a notion of scalar multiplication: for any scalar $\alpha$ and element $b$ of the Clifford algebra, we can form another Clifford algebra element which I will write as $\alpha\bullet b$. One rule this scalar multiplication is required to obey is $$ \alpha\bullet(\beta\bullet c) = (\alpha\beta)\bullet c $$ where $\alpha,\beta$ are scalars and $c$ is a Clifford algebra element.

What we mean when talking about scalars as elements of the Clifford algebra is actually $\alpha\bullet\mathbb I$. That is to say that your relation $e_1*e_1 = 1$ is more properly written $e_1*e_1 = 1\bullet\mathbb I$ (and one of the rules of a vector space says that $1\bullet a = a$ for any $a$ so we can actually write $e_1*e_1 = \mathbb I$).

The word algebra in "Clifford algebra" has a very particular meaning in this case: it means that the Clifford product $*$ is "compatible" with the vector space operations. In particular, for scalar multiplication it means that $$ (\alpha\bullet b)*c = b*(\alpha\bullet c) = \alpha\bullet(a*b). $$

With all these rules layed out we can now compute the Clifford product of two scalars $\alpha, \beta$ (and what we actually mean by that is the Clifford product of $\alpha\bullet\mathbb I$ and $\beta\bullet\mathbb I$): $$ (\alpha\bullet\mathbb I)*(\beta\bullet\mathbb I) = \mathbb I*(\alpha\bullet(\beta\bullet\mathbb I)) = \mathbb I*((\alpha\beta)\bullet I) = (\alpha\beta)\bullet(\mathbb I*\mathbb I) = (\alpha\beta)\bullet\mathbb I. $$ $(\alpha\beta)\bullet\mathbb I$ is how we represent the scalar $\alpha\beta$ in the Clifford algebra, so another way to state the above is that the Clifford algebra of two scalars is the same as "normal" scalar multiplication.

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The above is not specific to Clifford algebras in the slightest; it applies to any algebra with identity $\mathbb I$, where again "algebra" is meant in the very particular sense where a product is compatible with a vector space structure.