Geometric Algebra

• \(\langle A\rangle_r\)

Take the grade \(r\) part.

\[ a\wedge b := \frac{1}{2}(ab-ba) \]

It is extended to exactly match the structure of the exterior algebra. It is defined to satisfy for each graded algebra:

\begin{align*} 1\wedge a_i &= a_i\wedge 1 = a_i\\ a_1\wedge a_2\wedge \cdots \wedge a_r &= \frac{1}{r!}\sum_{\sigma\in S_r}\mathrm{sgn}(\sigma) a_{\sigma(1)}a_{\sigma(2)}\cdots a_{a\sigma(r)} \end{align*}

for all vectors \(a_i\), and extended by liearity to the linear combination of them.

Equivalently, \[ C\wedge D := \sum_{r,s}\langle \langle C\rangle_r\langle D\rangle_s\rangle_{r+s}. \]

Commutator Product \[ C\times D := \frac{1}{2}(CD-DC) \]

Vector Inner product \(a\cdot b := g(a,b)\) satisfies \[ \frac{1}{2}((a+b)^2 - a^2 - b^2) = a\cdot b, \] the .

\[ C\mathbin{\rfloor} D := \sum_{r,s}\langle \langle C\rangle_r\langle D\rangle_s\rangle_{s-r} \] Identity for any vector \(a\) and multivector \(B\): \[ aB = a\rfloor B + a\wedge B. \]

\[ C\mathbin{\lfloor} D := \sum_{r,s}\langle \langle C\rangle_r\langle D\rangle_s\rangle_{r-s} \]

\[ C\ast D := \sum_{r,s}\langle\langle C\rangle_r\langle D\rangle_s\rangle_0 \] See Clifford scalar product.

\[ C\mathbin{\bullet} D := \sum_{r,s}\langle \langle C\rangle_r\langle D\rangle_s\rangle_{|s-r|} \]

The basis vectors can be represented as

\begin{align*} \hat{\mathbf{r}} &= \mathbf{e}_1e^{i\theta} \\ \hat{\vb*{\theta}} &= \mathbf{e}_2e^{i\theta} \end{align*}

where \(i\) is the pseudoscalar \(\mathbf{e_1}\mathbf{e}_2\).

Note that \(\hat{\mathbf{r}}i = \hat{\vb*{\theta}}\).

• Circular velocity and acceleration with geometric algebra - YouTube

\begin{align*} a\cdot (b\wedge c) &= a\cdot (ib\times c) = \frac{1}{2}(aib\times c - ib\times ca)\\ &=\frac{i}{2}(ab\times c - b\times c a) = ia\wedge(b\times c)\\ &= -a\times (b\times c) \end{align*}

Note \(ai = ia\).

\begin{align*} (a\wedge b)\cdot c &= \frac{1}{2}(ia\times b c - cia\times b)\\ &= i (a\times b)\wedge c = -(a\times b)\times c \end{align*}

A vector \[ a_0e_0+\sum_{k=1}^{n}a_ke_k \] corresponds to a n-dimensional hyperplane: \[ a_0+\sum_{k=1}^{n}a_kx_k=0. \] Notice that \(e_0\) is the basis vector for the zeroth homogeneous coordinate.

It can also be interpreted as the kernel of , or the plane with such normal vector. \(e_0\) is the hyperplane at infinity.

Hyperplane here is general term for n-dimensional linear space. For 2-dimensional PGA, it is a line, and 3-dimensional PGA it is plane, and so on.

\(n\)-vector \[ e_{\check{0}}+\sum_{k=1}^n x_ke_{\check{k}} \] is a point \((x_k)\). Here, \(e_{\check{\imath}}\) means the geometric product of basis vectors except the \(i\)th one, and in the correct rotating order.

In 2-dimensional PGA, \(e_{\check{0}}\) is the bivector in the direction zeroth homogeneous coordinate, representing the point at infinity.

The addition of two lines result in a line that passes through the intersection of two lines. This is similar to the covector addition.

Addition of a scalar multiple of \(e_0\) shifts the line, in the direction of the covector. In a sense, it shows different slice in \(z\) direction.

Scalar multiple of the line is the same line. See real projective space.

The inner product in a Euclidean space is given by: \[ a\cdot b = \frac{1}{2}(Q(a+b) - Q(a) - Q(b)). \]

• Meet of two objects \(A\land B\) is an intersection.

It does not directly give the coordinates. One need to normalize it to \(a_0 =1\). Hence, projective.

There are few remarks:

• Intersection point of two lines are given by the Cramer's rule.
• The normalized cross product of the normal vectors, in three dimensional space.
  • It is the bivector in the direction of the cross product.

• Regressive Product

Join of two objects \(A\lor B\) is the intersecting subspace. In 2D PGA, it is the line that passes through both points.

It is the dual of meet. In fact, projective dual as well. The product is a vector that is normal to both vectors to each points, then the orthogonal projective plane is then going to contain both points.

A line perpendicular to \(a\) and passes through a point \(B\) is given by: \[ a\cdot B \] where \(\cdot\) being the fat dot product, or left contraction . By taking the dot product, the line is guaranteed to pass through the point , becuase it is the projection onto the plane perpendicular to the direction of the point.

The perpendicular part is hard to see…

• The projection of point \(B\) onto the line \(a\) is then, \((a\cdot B)\wedge a = (a\cdot B)a\).
• Parallel transport of \(a\) onto \(B\) is similarly, \((a\cdot B)\cdot B = (a\cdot B)B\).
• Projection of \(A\) to \(B\) is done by \((A\cdot B)B\).

• Rigid Transformation
• Reflection is done by sandwiching a vector.
• Rotation and translation is done using Motor \(R\), \(R^\dagger A R\).
  • Motor is a geometric product of two vectors \(R = uv\).
• When rotated around the point at infinity, the line simply translates.

• \[ \mathbf{a} = p_x \mathbf{e}_1 + p_y\mathbf{e}_2 + p_z \mathbf{e}_3 + \mathbf{e}_4 + \frac{p^2+r^2}{2}\mathbf{e}_5 \]