# Is Clifford product only for scalars already defined?

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?

• Are you familiar with vector spaces, linear algebra, or any abstract algebra? If not then it's hard to give any answer that will be understandable to you other than "you can just assume the Clifford product of scalars is the same as scalar multiplication". Commented Sep 12, 2023 at 17:26
• @NicholasTodoroff i don't know linear algebra and abstract algebra as well. thus i just hope to know if clifford product of scalars is the same with our normal multiplication. ive seen your comment and it seems it is possible though. thanks a lot! i hope to reach the level that i am fluent at clifford algebra someday! Commented Sep 12, 2023 at 17:33
• It's not just "possible", it is true. But there isn't an explanation that will be understandable to you unless you study more abstract algebra, not just Clifford algebras. I wrote this explanation out though in the hope that it will help you focus on what to study. Commented Sep 12, 2023 at 18:15

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.

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.

• ive read all you wrote. little hard to me yet, but this answer is really a gem! thanks a lot! Commented Sep 12, 2023 at 19:27
• now i understand all you wrote!! since im tired yesterday i felt it hard. but today it is really easy to understand!! thank you, the key point to investigate how clifford product of scalars works is using Idendity! Commented Sep 13, 2023 at 6:30
• i saw what you wrote (math.stackexchange.com/a/4516892/1219610) and you save my life..really thank you my teacher... there are some books that state AB = A•B + A^B where A,B are bivectors. it seemed weird and i visited your answer one more time... really thank you Commented Oct 31, 2023 at 16:43