A Clifford-algebra combines and generalizes the scalar product and the vector product. When you multiply two elements in a Clifford-algebra, the result can be decomposed into a symmetric scalar product and an antisymmetric vector product as follows:
\(\renewcommand{\vec}[1]{{\bf{#1}}}\)
\begin{align}ab&=a\cdot b+a\wedge b,\\
a\cdot b&=(ab+ba)/2,\\
a\wedge b&=(ab-ba)/2.\end{align}
In one dimension, the Clifford-algebra reduces trivially to the algebra of real numbers, and the vector product is identically zero.
In two dimensions, things start to get interesting. One way to generate a Clifford-algebra is through a slightly unconventional application of complex numbers: instead of ordinary multiplication, we may use multiplication on the left by the complex conjugate of the first multiplicand. The result can be decomposed as follows:
\begin{align}a&=a_1+a_2\vec{i},\\
b&=b_1+b_2\vec{i},\\
ab&=(a_1-a_2\vec{i})(b_1+b_2\vec{i})=(a_1b_1+a_2b_2)+(a_1b_2-a_2b_1)\vec{i}.\end{align}
The scalar and vector products, respectively, are $(a_1b_1+a_2b_2)$ and $(a_1b_2-a_2b_1)\vec{i}$.
There is, however, a more conventional way to represent the two-dimensional Clifford-algebra, a way that makes more structure evident. Instead of starting with complex numbers, we start with two unit vectors, $\vec{e}_1$ and $\vec{e}_2$ ($a=a_1\vec{e}_1+a_2\vec{e}_2$, $b=b_1\vec{e}_1+b_2\vec{e}_2$). The following multiplication rules apply:
\begin{align}\vec{e}_1\vec{e}_1&=\vec{e}_2\vec{e}_2=1,\\
\vec{e}_1\vec{e}_2&=-\vec{e}_2\vec{e}_1=\vec{e}_1\wedge\vec{e}_2.\end{align}
From this we can compute the Clifford product:
\[ab=(a_1b_1+a_2b_2)+(a_1b_2-a_2b_1)(\vec{e}_1\vec{e}_2).\]
Now is a good time to notice an important coincidence:
\[(\vec{e}_1\vec{e}_2)(\vec{e}_1\vec{e}_2)=-(\vec{e}_1\vec{e}_2)(\vec{e}_2\vec{e}_1)=-\vec{e}_1(\vec{e}_2\vec{e}_2)\vec{e}_1=-\vec{e}_1\vec{e}_1=-1.\]
Here we made use without proof or closer scrutiny of the fact that Clifford multiplication is associative. The result is that $(\vec{e}_1\vec{e}_2$) can be identified with the imaginary unit $\vec{i}$, in which case the product of two 2-dimensional "pure" Clifford numbers (no scalar component) is an ordinary complex number.
Generally, an $n$-dimensional Clifford-algebra can be represented using $n$ unit vectors $\vec{e}_1...\vec{e}_n$, the scalar unit 1, and the pseudoscalar $\vec{e}_1\vec{e}_2...\vec{e}_n$. In some cases, the pseudoscalar will be equal to 1; when it isn't, the algebra is called a universal Clifford-algebra.
In three dimensions, some real fun begins. To produce a three-dimensional Clifford-algebra, we need three basis units, $\vec{e}_1$, $\vec{e}_2$ and $\vec{e}_3$.The multiplication rules are derived from anticommutativity:
\begin{align}\vec{e}_1\vec{e}_1&=\vec{e}_2\vec{e}_2=\vec{e}_3\vec{e}_3=1,\\
\vec{e}_1\vec{e}_2&=-\vec{e}_2\vec{e}_1,\end{align}
and similarly,
\begin{align}\vec{e}_2\vec{e}_3&=-\vec{e}_3\vec{e}_2,\\
\vec{e}_3\vec{e}_1&=-\vec{e}_1\vec{e}_3.\end{align}
From this, the multiplication rule follows:
\[ab=(a_1b_1+a_2b_2+a_3b_3)+(a_1b_2-a_2b_1)\vec{e}_1\vec{e}_2+(a_2b_3-a_3b_2)\vec{e}_2\vec{e}_3+(a_3b_1-a_1b_30\vec{e}_3\vec{e}_1.\]
What we recovered here is the dot product and cross product of ordinary three-dimensional geometry. But notice something: in the cross product part, instead of the base vectors $\vec{e}_1$, $\vec{e}_2$ and $\vec{e}_3$, we have the product terms $\vec{e}_1\vec{e}_2$, $\vec{e}_2\vec{e}_3$ and $\vec{e}_3\vec{e}_1$. The meaning of this is very important. Forget what you were taught in high-school: the cross product is not a vector. It is a bivector. It so happens that in three dimensions (and only in three dimensions) a bivector is also three dimensional, so we have a convenient way to identify bivectors with ordinary vectors, but unfortunately when we do so, we hide some real geometric meaning.
Just as a vector represents the direction and the length of a line segment, a bivector represents the orientation and area of a disk. And unlike the 3-dimensional cross product concept, the concept of bivectors naturally generalizes to higher dimensions.
Anyway, back to Clifford algebras. They're generally represented by two numbers: The Clifford algebra $Cl(p,q)$ has $p+q$ base vectors, the square of which will be +1 in $p$ cases and –1 in the remaining $q$ cases. Clifford algebras can always be represented by matrices, but in some cases, simpler representations are possible: for instance, $Cl(0,1)$ can be represented by the complex numbers, and $Cl(0,2)$ by quaternions.
So what makes a Clifford-algebra so special? Take a look at what multiplication by the pseudoscalar does to a unit vector in the two-dimensional case:
\begin{matrix}(\vec{e}_1\vec{e}_2)\vec{e}_1=-(\vec{e}_2\vec{e}_1)\vec{e}_1=-\vec{e}_2(\vec{e}_1\vec{e}_1)=-\vec{e}_2,&\vec{e}_1(\vec{e}_1\vec{e}_2)=(\vec{e}_1\vec{e}_1)\vec{e}_2=\vec{e}_2,\\
(\vec{e}_1\vec{e}_2)\vec{e}_2=\vec{e}_1(\vec{e}_2\vec{e}_2)=\vec{e}_1,&\vec{e}_2(\vec{e}_1\vec{e}_2)=-\vec{e}_2(\vec{e}_2\vec{e}_1)=-(\vec{e}_2\vec{e}_2)\vec{e}_1=-\vec{e}_1.\end{matrix}
This, curiously, has a geometric interpretation. If we take $\vec{e}_1$ and $\vec{e}_2$ to be the unit vectors in the plane pointing in the $x$ and $y$ direction, multiplying on the left rotated $\vec{e}_2$ from the $y$-direction to the $x$-direction, and $\vec{e}_1$ from the $x$-direction to the $-y$-direction, i.e., performed a clockwise rotation by 90º. Similarly, multiplying on the right is equivalent to a counterclockwise rotation by 90º.
A rotation of a vector $\vec{a}$ by an arbitrary angle $\phi$ can be expressed as
\[\vec{a}'=R\vec{a}R^\star,\]
where $R=\cos\phi/2+\vec{e}_1\vec{e}_2\sin\phi/2$ and $R^\star=\cos\phi/2-\vec{e}_1\vec{e}_2\sin\phi/2$. This is very similar to the way rotations in 3D-space were associated with elements of the group SU(2), with one crucial difference: this technique can be generalized to arbitrary dimensions, and doesn't depend on the unique correspondence between SU(2) and SO(3).
Wow.
