Seeking intuitive explanation of Clifford Algebra Is there a simple intuitive graphical explanation of Clifford Algebra for the layman? Since Clifford Algebra is a Geometric Algebra, surely the best way to present those concepts is with graphical figures.
 A: I'm not sure I agree with the premise of the question; I would say that the point of introducing Clifford algebras is to work with certain geometric data that cannot be easily visualized.
But Clifford algebras do make contact with conventional geometry via the twisted adjoint representation.  Given a unit vector $x$ in a Euclidean space $V$, $x$ acts on $V$ via:
$$\rho_x(v) := -x v x^{-1}$$
where the product on the right-hand side is given by multiplication in the Clifford algebra $Cl(V)$.  It is not immediately obvious that $\rho_x(v)$ is an element of $V$, but a simple calculation shows that $\rho_x(v)$ is in fact the reflection of $v$ across the hyperplane perpendicular to $x$.  Every rotation is the product of an even number of reflections, so we have a surjective group homomorphism
$$\rho \colon Spin(V) \to SO(V)$$
where $Spin(V)$ is by definition the multiplicative subgroup of $Cl(V)$ generated by products of an even number of unit vectors in $V$.  To really understand $Cl(V)$, you have to understand why this map isn't an isomorphism.
Let's work in dimension $3$.  Every rotation has an axis, and there are two choices of unit vector along that axis.  Choose the unit vector which has the property that the rotation occurs counter-clockwise if the vector is pointing at your eye.  Now multiply the vector by the angle of the rotation (a number from $0$ to $\pi$), yielding a point in the ball of radius $\pi$ in $\mathbb{R}^3$.  This space is nearly a model of $SO(3)$, but we need to account for the fact that a clockwise rotation by $\pi$ is the same as a counter-clockwise rotation by $\pi$.  Thus $SO(3)$ is really the ball of radius $\pi$ with antipodal points on the boundary identified; topologically this is $\mathbb{R}P^3$.
You may recall that $\mathbb{R}P^3$ is double covered by the sphere $S^3$, and indeed topologically the map $Spin(3) \to SO(3)$ is just the double cover $S^3 \to \mathbb{R}P^3$.  For higher dimensional $V$ it is no longer true that $SO(n) = \mathbb{R}P^n$, but $Spin(n)$ is still a simply connected double cover of $SO(n)$.  So in a sense $Cl(V)$ keeps track of an extra bit of orientation data that you can't really see in the symmetries of $V$.
A: To get some geometrical meaning you can look at some special Clifford algebras:
$\mathcal{Cl}_{0,1}$ is isomorphic to the complex plane.
$\mathcal{Cl}_{2,0}$ is isomorphic to the Euclidean plane.
$\mathcal{Cl}_{3,0}$ is isomorphic to the 3D Euclidean space.
$\mathcal{Cl}_{3,1}$ is isomorphic to the 4D Minkowski spac-time.
