So, I've written some notes on the past year regarding Geometric (Clifford) Algebra. I was revisiting these notes, and I've encountered the following assertion:
"The tensor algebra $T(\mathbf V)$ is the vector space $\mathbf V$ together with the tensor product $\otimes:\mathbf V \times \mathbf V \to \mathbf V$. The tensor algebra has the ``free'' algebra flavor, meaning, it's 'largest' algebra one can construct from $\mathbf V$. Thus, all other algebras on $\mathbf V$ are quotients of $T(\mathbf V)$, i.e. we can construct the other bilinear products by introducing equivalence relations... One example is the exterior algebra. To construct it, just impose the following equivalence relation, for every $\mathbf v \in \mathbf V$, $$ \mathbf{v} \otimes \mathbf{v} \cong 0. $$ Note that this condition implies that $\mathbf {v} \otimes \mathbf {u} = - \mathbf{u} \otimes \mathbf{v}$. This follows from
$$ (\mathbf u + \mathbf v) \otimes (\mathbf u + \mathbf v) = \mathbf u \otimes \mathbf u + \mathbf v \otimes \mathbf v + \mathbf u \otimes \mathbf v + \mathbf v \otimes \mathbf u \cong 0. $$
When considering the exterior algebra, we change the product notation from $\otimes$ to $\wedge$.
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Using a similar idea, we arrive at the Geometric (Clifford) Algebra. Instead of the equality to zero as in the exterior algebra, we use:
$$ \mathbf{v} \otimes \mathbf{v} - B(\mathbf v, \mathbf v) \cong 0, $$ where $B$ is a symmetric bilinear form. Note that, this definition actually defines a family of algebras (one for each possible $B$), where the exterior algebra is one of them (just use $B(x,x) = 0$)."
So, this might sound silly, but I have no idea where I found this construction, or if it came from myself. Thus, I was wondering if this is indeed correct. Also, if it is correct, I'd love a reference where perhaps this topic is presented in this fashion.