Multivectors



Clifford algebra is a type of algebra characterized by the geometric product of scalars, vectors, bivectors, trivectors...etc.

Just as a vector has length so a bivector has area and a trivector has volume.

Just as a vector has direction so a bivector has orientation. In three dimensions a trivector has only one possible orientation and is therefore a pseudoscalar. But in four dimensions a trivector becomes a pseudovector and the quadvector becomes the pseudoscalar.

Multiplication of arbitrary vectors
The dot product of two vectors is:



\begin{split} \mathbf{u} \cdot \mathbf{v} &= {\color{blue}\text{vector}} \cdot {\color{blue}\text{vector}} \\ &= (u_{x} + u_{y})(v_{x} + v_{y}) \\ &= {\color{red}u_{x} v_{x} + u_{y} v_{y}} \end{split} $$

But this is actually quite mysterious. When we multiply $$(a_1 + a_2)$$ and $$(b_1 + b_2)$$ we dont get $$(a_1 b_1 + a_2 b_2)$$ so why is it that when we multiply vectors we only multiply parallel components? Clifford algebra has a surprisingly simple answer. The answer is: We dont! Instead of the dot product or the wedge product Clifford algebra uses the geometric product.



\begin{split} \mathbf{u} \mathbf{v} &= (u_{x} {\color{blue}e_{x}} + u_{y} {\color{blue} e_{y}} ) (v_{x} {\color{blue}e_{x}} + v_{y} {\color{blue} e_{y}} ) \\ &= u_{x} v_{x} {\color{red}e_{x} e_{x}} + u_{x} v_{y} {\color{green}e_{x} e_{y}} + u_{y} v_{x} {\color{green}e_{y} e_{x}} + u_{y} v_{y} {\color{red}e_{y} e_{y}} \\ &= u_{x} v_{x} {\color{red}(1)} + u_{y} v_{y} {\color{red}(1)} + u_{x} v_{y} {\color{green}e_{x} e_{y}} - u_{y} v_{x} {\color{green}e_{x} e_{y}} \\ &= (u_{x} v_{x} + u_{y} v_{y}){\color{red}(1)} + (u_{x} v_{y} - u_{y} v_{x}) {\color{green}e_{xy}} \\ &= {\color{red}\text{scalar}} + {\color{green}\text{bivector}} \end{split} $$

A scalar plus a bivector (or any number of blades of different grade) is called a multivector. The idea of adding a scalar and a bivector might seem wrong but in the real world it just means that what appears to be a single equation is in fact a set of.

For example:



\mathbf{u} \mathbf{v} = 5 $$


 * would just mean that:



(u_{x} v_{x} + u_{y} v_{y}){\color{red}(1)} = 5 \\ \quad \quad \quad \text{and} \\ (u_{x} v_{y} - u_{y} v_{x}) {\color{green}e_{xy}} = 0 $$

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Rules
All the properties of Clifford algebra derive from a few simple rules.

Let $${\color{blue}e_{x}},$$ $${\color{blue}e_{y}},$$ and $${\color{blue}e_{z}}$$ be perpendicular unit vectors.

Multiplying two perpendicular vectors results in a bivector:


 * $$e_x e_y = {\color{green}e_{xy}}$$

Multiplying three perpendicular vectors results in a trivector:


 * $$e_x e_y e_z = {\color{orange}e_{xyz}}$$

Multiplying parallel vectors results in a scalar:


 * $$e_x e_x = {\color{red}e_{xx}} = {\color{red}1}$$

Clifford algebra is associative therefore the fact that multiplying parallel vectors results in a scalar means that:



\begin{split} {\color{green}(e_{x} e_{y})} {\color{blue} (e_{y}) } &= {\color{blue} e_{x}}         {\color{red}(e_{y} e_{y})} \\ &= {\color{blue} e_{x}}         {\color{red}(1)} \\ &= {\color{blue} e_{x}} \end{split} $$


 * and:



\begin{align} {\color{green} (e_{x} e_{y})} {\color{blue} (e_{y} + e_{z}) } &= {\color{blue} e_{xyy}} + {\color{orange}e_{xyz}} \\ &= {\color{blue} e_{x}} + {\color{orange}e_{xyz}} \end{align} $$


 * and:



\begin{align} {\color{green} (e_{x} e_{y}) } {\color{blue} (e_{z}) } &= {\color{orange} e_{xyz}} \end{align} $$

Rotation from x to y is the negative of rotation from y to x:


 * $$e_{xy} = -e_{yx}$$


 * Therefore:



\begin{split} {\color{green}(e_{x} e_{y})} {\color{blue} (e_{x}) }           &= \phantom{-}{\color{blue} e_{x}}        {\color{green}(e_{y} e_{x})} \\ &= {\color{blue} -e_{x}}       {\color{green}(e_{x} e_{y})} \\ &= -{\color{red} (e_{x} e_{x})} {\color{blue} e_{y}} \\ &= -{\color{red} (1)}   {\color{blue} e_{y}} \\ &= -{\color{blue} e_{y}} \end{split} $$

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Multiplication tables

 * In one dimension:



\begin{array}{r|r}

{\color{red} 1   } & {\color{blue} e_{x}}

\\\hline {\color{blue} e_{x}} & {\color{red} 1    }

\end{array} $$


 * In two dimensions:



\begin{array}{r|rr|r}

{\color{red}  1    } & {\color{blue} e_{x}} & {\color{blue} e_{y}} & {\color{green} e_{xy}}

\\\hline {\color{blue} e_{x}} & {\color{red}  1    } & {\color{green} e_{xy}} & {\color{blue} {e_{y}}}

\\ {\color{blue}  e_{y}} & {\color{green} -e_{xy}} & {\color{red}   1     }        & {\color{blue} -e_{x}}

\\\hline {\color{green} e_{xy}} & {\color{blue} -e_{y}} & {\color{blue} {e_{x}}} & {\color{red} {-1}}

\end{array} $$


 * In three dimensions:



\begin{array}{r|rrr|rrr|r}

{\color{red}   1    } & {\color{blue}  e_{x}} & {\color{blue}  e_{y}} & {\color{blue}  e_{z}} & {\color{green} e_{xy}} & {\color{green} e_{xz}} & {\color{green} e_{yz}} & {\color{orange} e_{xyz}}

\\\hline {\color{blue}  e_{x}} & {\color{red}   1    } & {\color{green} e_{xy}} & {\color{green} e_{xz}}  & {\color{blue}  e_{y}} & {\color{blue}  e_{z}} & {\color{orange} e_{xyz}} & {\color{green} e_{yz}}

\\ {\color{blue}   e_{y}} & {\color{green} -e_{xy}} & {\color{red}    1    }        & {\color{green}   e_{yz}}  & {\color{blue}  -e_{x}} & {\color{orange} -e_{xyz}} & {\color{blue}   e_{z}} & {\color{green} -e_{xz}} \\ {\color{blue}  e_{z}}  & {\color{green} -e_{xz}} & {\color{green} -e_{yz}} & {\color{red}   1     } & {\color{orange} e_{xyz}}& {\color{blue} -e_{x}}  & {\color{blue} -e_{y}}  & {\color{green} e_{xy}}

\\\hline {\color{green} e_{xy}} & {\color{blue} -e_{y}} & {\color{blue}  e_{x}} & {\color{orange} e_{xyz}} & {\color{red}  -1} & {\color{green} -e_{yz}} & {\color{green} e_{xz}} & {\color{blue} -e_{z}}

\\ {\color{green}  e_{xz}} & {\color{blue}  -e_{z}} & {\color{orange} -e_{xyz}} & {\color{blue}   e_{x}} & {\color{green}  e_{yz}} & {\color{red}   -1   } & {\color{green} -e_{xy}} & {\color{blue}   e_{y}}

\\ {\color{green} e_{yz}}  & {\color{orange} e_{xyz}} & {\color{blue} -e_{z}}   & {\color{blue}  e_{y}}   & {\color{green} -e_{xz}} & {\color{green} e_{xy}}  & {\color{red}  -1     }  & {\color{blue} -e_{x}}

\\\hline {\color{orange} e_{xyz}} & {\color{green} e_{yz}} & {\color{green} -e_{xz}} & {\color{green} e_{xy}} & {\color{blue} -e_{z}} & {\color{blue}  e_{y}} & {\color{blue} -e_{x}} & {\color{red}  -1    }

\end{array} $$


 * In four dimensions:

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Basis
Every multivector of the Clifford algebra can be expressed as a linear combination of the canonical basis elements. The basis elements of the Clifford algebra Cℓ$3$ are $$\{{\color{red}1}, \, {\color{blue}e_x}, \, {\color{blue}e_y}, \, {\color{blue}e_z}, \, {\color{green}e_{xy}}, \, {\color{green}e_{xz}}, \, {\color{green}e_{yz}}, \, {\color{orange}e_{xyz}} \}$$ and the general element of Cℓ$3$ is given by


 * $$A = a_0{\color{red}(1)} + a_1 {\color{blue}e_x} + a_2 {\color{blue}e_y} + a_3 {\color{blue}e_z} + a_4 {\color{green}e_{xy}} + a_5 {\color{green}e_{xz}} + a_6 {\color{green}e_{yz}} + a_7 {\color{orange}e_{xyz}}.$$

If $$ a_0, a_1, a_2, a_3, a_4, a_5, a_6, a_7 $$ are all real then the Clifford algebra is Cℓ$3$(R). If the coefficients are allowed to be complex then the Clifford algebra is Cℓ$3$(C).

A multivector can be separated into components of different grades:



\langle \mathbf{A} \rangle_0 = a_0{\color{red}(1)}\\

\langle \mathbf{A} \rangle_1 = a_1 {\color{blue}e_x} + a_2 {\color{blue}e_y} + a_3 {\color{blue}e_z}\\

\langle \mathbf{A} \rangle_2 = a_4 {\color{green}e_{xy}} + a_5 {\color{green}e_{xz}} + a_6 {\color{green}e_{yz}}\\

\langle \mathbf{A} \rangle_3 = a_7 {\color{orange}e_{xyz}}\\

$$

The elements of even grade form a subalgebra because the sum or product of even grade elements always results in an element of even grade. The elements of odd grade do not form a subalgebra.

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Relation to other algebras
$$Cℓ_0 (\mathbf{R})$$: Real numbers (scalars). A scalar can (and should) be thought of as zero vectors multiplied together. See Empty product.

$$Cℓ_0 (\mathbf{C})$$: Complex numbers

$$Cℓ_1 (\mathbf{R})$$: Split-complex numbers



\begin{array}{r|r}

{\color{red} 1    } & {\color{blue} e_{x}}

\\\hline {\color{blue} e_{x}} & {\color{red} 1    }

\end{array} $$

$$Cℓ_1 (\mathbf{C})$$: Bicomplex numbers

$$Cℓ_2^0 (\mathbf{R})$$: Complex numbers (The superscript 0 indicates the even subalgebra)



\begin{array}{r|r}

{\color{red} 1}       & {\color{green} e_{xy}}

\\\hline {\color{green} e_{xy}} & {\color{red} {-1}}

\end{array} $$

$$Cℓ_3^0 (\mathbf{R})$$: Quaternions



\begin{array}{r|rrr}

{\color{red} 1} & {\color{green} e_{xy}} & {\color{green} e_{xz}} & {\color{green} e_{yz}}

\\\hline {\color{green} e_{xy}} & {\color{red}  -1     } & {\color{green} -e_{yz}} & {\color{green} e_{xz}}

\\ {\color{green} e_{xz}} & {\color{green} e_{yz}} & {\color{red}  -1     } & {\color{green} -e_{xy}}

\\ {\color{green} e_{yz}} & {\color{green} -e_{xz}} & {\color{green} e_{xy}} & {\color{red} -1      }

\end{array} $$

$$Cℓ_3^0 (\mathbf{C})$$: Biquaternions

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Multivector multiplication using tensors
To find the product



AB = (a_0 {\color{red}  1    } + a_1 {\color{blue}  e_{x}} + a_2 {\color{blue}  e_{y}} + a_3 {\color{green} e_{xy}}) (b_0 {\color{red}  1    } + b_1 {\color{blue}  e_{x}} + b_2 {\color{blue}  e_{y}} + b_3 {\color{green} e_{xy}}) $$

we have to multiply every component of the first multivector with every component of the second multivector.



\begin{split} AB = & \phantom{+}

(a_0 b_0 {\color{red}1}{\color{red}1} + a_0 b_1 {\color{red}1}{\color{blue} e_{x}} + a_0 b_2 {\color{red}1}{\color{blue}  e_{y}} + a_0 b_3 {\color{red}1}{\color{green} e_{xy}}) \\ &+

(a_1 b_0 {\color{blue} e_{x}}{\color{red}1} + a_1 b_1 {\color{blue}  e_{x}}{\color{blue}  e_{x}} + a_1 b_2 {\color{blue}  e_{x}}{\color{blue}  e_{y}} + a_1 b_3 {\color{blue}  e_{x}}{\color{green} e_{xy}}) \\ &+

(a_2 b_0 {\color{blue} e_{y}}{\color{red}1} + a_2 b_1 {\color{blue}  e_{y}}{\color{blue}  e_{x}} + a_2 b_2 {\color{blue}  e_{y}}{\color{blue}  e_{y}} + a_2 b_3 {\color{blue}  e_{y}}{\color{green} e_{xy}}) \\ &+

(a_3 b_0 {\color{green} e_{xy}}{\color{red}1} + a_3 b_1 {\color{green} e_{xy}}{\color{blue} e_{x}} + a_3 b_2 {\color{green} e_{xy}}{\color{blue}  e_{y}} + a_3 b_3 {\color{green} e_{xy}}{\color{green} e_{xy}})

\end{split} $$

Then we reduce each of the 16 resulting terms to its standard form.



\begin{split} AB = & \phantom{+}

(a_0 b_0 {\color{red}1} + a_0 b_1 {\color{blue} e_{x}} + a_0 b_2 {\color{blue}  e_{y}} + a_0 b_3 {\color{green} e_{xy}}) \\ &+

(a_1 b_0 {\color{blue} e_{x}} + a_1 b_1 {\color{red}1} + a_1 b_2 {\color{green}  e_{xy}} + a_1 b_3 {\color{blue} e_{y}}) \\ &+

(a_2 b_0 {\color{blue} e_{y}} - a_2 b_1 {\color{green}  e_{xy}} + a_2 b_2 {\color{red}1} - a_2 b_3 {\color{blue}  e_{x}}) \\ &+

(a_3 b_0 {\color{green} e_{xy}} - a_3 b_1 {\color{blue} e_{y}} + a_3 b_2 {\color{blue} e_{x}} - a_3 b_3 {\color{red}1})

\end{split} $$

Finally we collect like products into the four components of the final multivector.



\begin{split} AB = & \phantom{+} ( a_0 b_0 + a_1 b_1 + a_2 b_2 - a_3 b_3 ) {\color{red}1} \\ &+ ( a_1 b_0 + a_0 b_1 + a_3 b_2 - a_2 b_3 ) {\color{blue} e_{x} } \\ &+ ( a_2 b_0 - a_3 b_1 + a_0 b_2 + a_1 b_3 ) {\color{blue} e_{y} } \\ &+ ( a_3 b_0 - a_2 b_1 + a_1 b_2 + a_0 b_3 ) {\color{green} e_{xy} } \end{split} $$

This is all very tedious and error-prone. It would be nice if there was some way to cut straight to the end. Tensor notation allows us to do just that.

To find the tensor that we need we first need to know which terms end up as scalars, which terms end up as vectors...etc. There is an easy way to do this and it involves the multiplication table.

First lets start with an easy one.

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Complex numbers
The multiplication table for $$Cℓ_2^0 (\mathbf{R})$$ (Which is isomorphic to complex numbers)



\begin{array}{r|r}

{\color{red} 1} & {\color{green} e_{xy}}

\\\hline {\color{green} e_{xy}} & {\color{red}  -1     }

\end{array} $$

We can see then that:

It worked! All the terms in the first row are scalars and all the terms in the second row are bivectors. This is exactly what we are looking for.

Pay special attention to the signs in the final matrix above.

Therefore to find the product



(a_0 {\color{red}  1    } + a_1 {\color{green}  e_{xy}} ) (b_0 {\color{red}  1    } + b_1 {\color{green}  e_{xy}} ) $$

We would multiply:



\left( \begin{array}{rr}

{\color{red}  b_0 } & {\color{green} -b_1 }

\\ {\color{green} b_1 } & {\color{red}  b_0 }

\end{array} \right)

\left( \begin{array}{r}

{\color{red}  a_0 } \\ {\color{green} a_1 }

\end{array} \right) =

\left( \begin{array}{rcr}

{\color{red}  b_0 } {\color{red}  a_0 }& - & {\color{red} b_1 } {\color{red} a_1 }

\\ {\color{green} b_1 } {\color{green}  a_0 }& + & {\color{green}  b_0 } {\color{green} a_1 }

\end{array} \right) $$

Each row of the final matrix has exactly the right terms with exactly the right signs.

The vector above represents a complex number. You should think of the first column of the matrix above as representing another complex number. All the other terms in the matrix are just there to make our lives a little bit easier.

It works. It works so well that complex numbers can be represented as matrices as:



\begin{bmatrix} a & -b \\ b & a \end{bmatrix} $$

Which corresponds perfectly to a multiplication table for complex numbers:



$$ \begin{array}{r|r}

{\color{red} 1} & {\color{green} -i}

\\\hline {\color{green} i} & {\color{red}  1     }

\end{array} $$ $$ \quad \quad \begin{array}{r|r}

{\color{red} 1} & {\color{green} -e_{xy}}

\\\hline {\color{green} e_{xy}} & {\color{red}  1     }

\end{array} $$
 * }

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Quaternions
The multiplication table for $$Cℓ_3^0 (\mathbf{R})$$ (Which is isomorphic to quaternions) is:



\begin{array}{r|rrr}

{\color{red} 1} & {\color{green} e_{xy}} & {\color{green} e_{xz}} & {\color{green} e_{yz}}

\\\hline {\color{green} e_{xy}} & {\color{red}  -1     } & {\color{green} -e_{yz}} & {\color{green} e_{xz}}

\\ {\color{green} e_{xz}} & {\color{green} e_{yz}} & {\color{red}  -1     } & {\color{green} -e_{xy}}

\\ {\color{green} e_{yz}} & {\color{green} -e_{xz}} & {\color{green} e_{xy}} & {\color{red} -1      }

\end{array} \quad \quad \quad \begin{array}{l} i = e_{yz} \\ j = e_{xz} \\ k = e_{xy} \end{array} $$

The entire 2nd row of the multiplication table is just $${\color{green} e_{xy}}$$ multiplied by the entire first row. The entire 3rd row of the multiplication table is just $${\color{green} e_{xz}}$$ multiplied by the entire first row. The entire 4th row of the multiplication table is just $${\color{green} e_{yz}}$$ multiplied by the entire first row.

We can see then that if we multiply each row by the first row again then we get:

This works because we have in effect multiplied each term by a second term twice. In other words we have multiplied every term by the square of another term and the square of every term is either 1 or -1.

Therefore to find the product



(a_0 {\color{red}  1    } + a_1 {\color{green}  e_{xy}} + a_2 {\color{green}  e_{xz}} + a_3 {\color{green} e_{yz}}) (b_0 {\color{red}  1    } + b_1 {\color{green}  e_{xy}} + b_2 {\color{green}  e_{xz}} + b_3 {\color{green} e_{yz}}) $$

We would multiply:



\left( \begin{array}{rrrr}

{\color{red}  b_0 } & {\color{green} -b_1 } & {\color{green} -b_2 } & {\color{green} -b_3 }

\\ {\color{green} b_1 } & {\color{red}  b_0 } & {\color{green} b_3 } & {\color{green} {-b_2 }}

\\ {\color{green}  b_2 } & {\color{green} -b_3 } & {\color{red}   b_0 } & {\color{green} b_1 }

\\ {\color{green} b_3 } & {\color{green} b_2 } & {\color{green} {-b_1 }} & {\color{red}  b_0 }

\end{array} \right)

\left( \begin{array}{rrrr}

{\color{red}  a_0 } \\ {\color{green} a_1 } \\ {\color{green} a_2 } \\ {\color{green} a_3 }

\end{array} \right) =

\left( \begin{array}{rrrr}

{\color{red}  b_0 } {\color{red}  a_0 } & - & {\color{red} b_1 } {\color{red} a_1 }  & - & {\color{red} b_2 } {\color{red} a_2 }  & - & {\color{red} b_3 } {\color{red} a_3 }

\\ {\color{green} b_1 } {\color{green}  a_0 } & + & {\color{green}  b_0 } {\color{green} a_1 } & + & {\color{green} b_3 } {\color{green} a_2 }  & - & {\color{green} {b_2 }} {\color{green} a_3 }

\\ {\color{green}  b_2 } {\color{green}   a_0 } & - & {\color{green} b_3 } {\color{green} a_1 } & + & {\color{green}  b_0 } {\color{green}  a_2 } & + & {\color{green}  b_1 } {\color{green} a_3 }

\\ {\color{green} b_3 } {\color{green}  a_0 } & + & {\color{green} b_2 } {\color{green} a_1 } & - & {\color{green} {b_1 }} {\color{green} a_2 }  & + & {\color{green}  b_0 } {\color{green} a_3 }

\end{array} \right) $$

Just as complex numbers can be represented as matrices, so a quaternion can be represented as:



\left( \begin{array}{rrrr} a & -b & -c & -d \\ b & a & d &-c \\ c & -d & a & b \\ d & c & -b & a \end{array} \right)=

\left( \begin{array}{rr}

\left( \begin{array}{rr} a & -b \\  b & a \end{array} \right) &

- \left( \begin{array}{rr} c & d \\ -d &c \end{array} \right) \\

\left( \begin{array}{rr} c & -d \\ d & c \end{array} \right) &

\left( \begin{array}{rr} a & b \\ -b & a \end{array} \right)

\end{array} \right) $$

Which corresponds to a multiplication table for quaternions:

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CL2
The multiplication table for $$Cℓ_{2} (\mathbf{R})$$ is:



\begin{array}{r|rr|r}

{\color{red}  1    } & {\color{blue} e_{x}} & {\color{blue} e_{y}} & {\color{green} e_{xy}}

\\\hline {\color{blue} e_{x}} & {\color{red}  1    } & {\color{green} e_{xy}} & {\color{blue} {e_{y}}}

\\ {\color{blue}  e_{y}} & {\color{green} -e_{xy}} & {\color{red}   1     }        & {\color{blue} -e_{x}}

\\\hline {\color{green} e_{xy}} & {\color{blue} -e_{y}} & {\color{blue} {e_{x}}} & {\color{red} {-1}}

\end{array} $$

We can see then that:

Therefore to find the product



(a_0 {\color{red}  1    } + a_1 {\color{blue}  e_{x}} + a_2 {\color{blue}  e_{y}} + a_3 {\color{green} e_{xy}}) (b_0 {\color{red}  1    } + b_1 {\color{blue}  e_{x}} + b_2 {\color{blue}  e_{y}} + b_3 {\color{green} e_{xy}}) $$

We would multiply:



\left( \begin{array}{rrrr}

{\color{red}  b_0 } & {\color{blue} b_1 } & \phantom{-} {\color{blue} b_2 } & {\color{green} -b_3 }

\\ {\color{blue} b_1 } & {\color{red}  b_0 } & {\color{green} b_3 } & {\color{blue} {-b_2 }}

\\ {\color{blue}  b_2 } & {\color{green} -b_3 } & {\color{red}   b_0 } & {\color{blue}  b_1 }

\\ {\color{green} b_3 } & {\color{blue} -b_2 } & {\color{blue} {b_1 }} & {\color{red}  b_0 }

\end{array} \right)

\left( \begin{array}{rrrr}

{\color{red}  a_0 } \\ {\color{blue} a_1 } \\ {\color{blue} a_2 } \\ {\color{green} a_3 }

\end{array} \right) =

\left( \begin{array}{rrrr}

{\color{red}  b_0 } {\color{red}  a_0 } & + & {\color{red} b_1 } {\color{red} a_1 }  & + & {\color{red} b_2 } {\color{red} a_2 }  & - & {\color{red} b_3 } {\color{red} a_3 }

\\ {\color{blue} b_1 } {\color{blue}  a_0 } & + & {\color{blue}  b_0 } {\color{blue} a_1 } & + & {\color{blue} b_3 } {\color{blue} a_2 }  & - & {\color{blue} {b_2 }} {\color{blue} a_3 }

\\ {\color{blue}  b_2 } {\color{blue}   a_0 } & - & {\color{blue} b_3 } {\color{blue} a_1 } & + & {\color{blue}  b_0 } {\color{blue}  a_2 } & + & {\color{blue}  b_1 } {\color{blue} a_3 }

\\ {\color{green} b_3 } {\color{green}  a_0 } & - & {\color{green} b_2 } {\color{green} a_1 } & + & {\color{green} {b_1 }} {\color{green} a_2 }  & + & {\color{green}  b_0 } {\color{green} a_3 }

\end{array} \right) $$

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Squares of pseudoscalars are either 1 or -1
In 0 dimensions:


 * $$(1)^2 = 1$$

In 1 dimension:


 * $$(e_{1})^2 = e_{11} = 1$$

In 2 dimensions:


 * $$(e_{12})^2 = e_{1212} = -e_{11 \color{red}{22}} = -1$$

In 3 dimensions:


 * $$(e_{123})^2 = e_{123123} = e_{1212 \color{red}{33}} = -1$$

In 4 dimensions:


 * $$(e_{1234})^2 = e_{12341234} = -e_{123123 \color{red}{44}} = 1$$

In 5 dimensions:


 * $$(e_{12345})^2 = e_{1234512345} = e_{12341234 \color{red}{55}} = 1$$

In 6 dimensions:


 * $$(e_{123456})^2 = e_{123456123456} = -e_{1234512345 \color{red}{66}} = -1$$

In 7 dimensions:


 * $$(e_{1234567})^2 = e_{12345671234567} = e_{123456123456 \color{red}{77}} = -1$$

In 8 dimensions:


 * $$(e_{12345678})^2 = e_{1234567812345678} = -e_{12345671234567 \color{red}{88}} = 1$$

In 9 dimensions:


 * $$(e_{123456789})^2 = e_{123456789123456789} = e_{1234567812345678 \color{red}{99}} = 1$$

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Bivectors in higher dimensions
A simple bivector can be used to represent a single rotation.

In four dimensions a rigid object can rotate in two different ways simultaneously. Such a rotation can only be represented as the sum of two simple bivectors.

In six dimensions a rigid object can rotate in three different ways simultaneously. Such a rotation can only be represented as the sum of three simple bivectors.

From Bivector

The wedge product of two vectors is a bivector, but not all bivectors are wedge products of two vectors. For example, in four dimensions the bivector


 * $$ \mathbf{B} = \mathbf{e}_1 \wedge \mathbf{e}_2 + \mathbf{e}_3 \wedge \mathbf{e}_4 = \mathbf{e}_1 \mathbf{e}_2 + \mathbf{e}_3\mathbf{e}_4 = \mathbf{e}_{12} + \mathbf{e}_{34}$$

cannot be written as the wedge product of two vectors. A bivector that can be written as the wedge product of two vectors is simple. In two and three dimensions all bivectors are simple, but not in four or more dimensions;

A bivector has a real square if and only if it is simple.



\begin{align} (\mathbf{e}_{12})^2 &= e_{1212} \\ &= 1 \end{align} $$


 * But:



\begin{align} (\mathbf{e}_{12} + \mathbf{e}_{34})^2 &= e_{1212} + e_{1234} + e_{3412} + e_{3434} \\ &= -e_{1122} + e_{1234} + e_{1234} - e_{3344} \\ &= -1 + 2 e_{1234} -1 \\ &= 2 e_{1234} - 2 \end{align} $$

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Other quadratic forms
The square of a vector is:



\begin{split} \mathbf{v} \mathbf{v} &= (v_{x} {\color{blue}e_{x}} + v_{y} {\color{blue}e_{y}}) (v_{x} {\color{blue}e_{x}} + v_{y} {\color{blue}e_{y}}) \\ &= v_{x} v_{x} {\color{red}e_{x} e_{x}} + v_{x} v_{y} {\color{green}e_{x} e_{y}} + v_{y} v_{x} {\color{green}e_{y} e_{x}} + v_{y} v_{y} {\color{red}e_{y} e_{y}} \\ &= v_{x} v_{x} {\color{red}(1)} + v_{y} v_{y} {\color{red}(1)} + v_{x} v_{y} {\color{green}e_{x} e_{y}} - v_{y} v_{x} {\color{green}e_{x} e_{y}} \\ &= (v_{x} v_{x} + v_{y} v_{y}){\color{red}(1)} + (v_{x} v_{y} - v_{y} v_{x}) {\color{green}e_{xy}} \\ &= (v_{x}^2 + v_{y}^2){\color{red}(1)} + (0) {\color{green}e_{xy}} \\ &= (v_{x}^2 + v_{y}^2){\color{red}(1)} \\ &= {\color{red}\text{scalar}} \end{split} $$


 * ($$v_{x}^2 + v_{y}^2$$) is called the quadratic form. In this case both terms are positive but some Clifford algebras have quadratic forms with negative terms. Some have both positive and negative terms.

From Clifford algebra:

Every nondegenerate quadratic form on a finite-dimensional real vector space is equivalent to the standard diagonal form:


 * $$Q(v) = v^2 = v_1^2 + \cdots + v_p^2 - v_{p+1}^2 - \cdots - v_{p+q}^2 ,$$

where n = p + q is the dimension of the vector space. The pair of integers (p, q) is called the of the quadratic form. The real vector space with this quadratic form is often denoted Rp,q. The Clifford algebra on R$p,q$ is denoted Cℓ$p,q$(R). The symbol Cℓ$n$(R) means either Cℓ$n,0$(R) or Cℓ$0,n$(R) depending on whether the author prefers positive-definite or negative-definite spaces.

A {ei} for Rp,q consists of n = p + q mutually orthogonal vectors, p of which square to +1 and q of which square to −1. The algebra Cℓ$p,q$(R) will therefore have p vectors that square to +1 and q vectors that square to −1.

From Spacetime algebra:

(STA) is a name for the Cl3,1(R), or equivalently the  G(M4), which can be particularly closely associated with the geometry of  and relativistic. See also.

The spacetime algebra may be built up from an orthogonal basis of one time-like vector $$\gamma_0$$ and three space-like vectors, $$\{\gamma_1, \gamma_2, \gamma_3\}$$, with the multiplication rule
 * $$ \gamma_\mu \gamma_\nu + \gamma_\nu \gamma_\mu = 2 \eta_{\mu \nu} $$

where $$\eta_{\mu \nu}$$ is the with signature (&minus; + + +).

Thus:


 * $$\gamma_0^2 = {-1}$$
 * $$\gamma_1^2 = \gamma_2^2 = \gamma_3^2 = {+1}$$
 * $$\gamma_\mu \gamma_\nu = - \gamma_\nu \gamma_\mu$$

The basis vectors $$\gamma_k$$ share these properties with the, but no explicit matrix representation need be used in STA.

$$Cℓ_{3,0} (\mathbf{R})$$:Algebra of physical space (Time = scalar)

$$Cℓ_{3,1} (\mathbf{R})$$:Spacetime algebra (Time = vector)

$$Cℓ_{0,2} (\mathbf{R})$$:Quaternions (Three quaternions = two vectors that square to -1 and one bivector that squares to -1)

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Pauli matrices and Tensors
Clifford algebra in two and three dimensions has the same multiplication table as Pauli matrices ($$\sigma_x, \sigma_y, \sigma_z$$) so it is tempting to assume that Clifford algebra is derived from Pauli matrices. But in fact, I think it would be more accurate to say that tensors in general owe their properties to, and are ultimately derived from, Clifford algebra rather than the other way around.



\begin{align} \frac{1}{\mathbf{v}} &= \frac{\mathbf{v}}{\mathbf{v}\mathbf{v}}\\ &= \frac{\mathbf{v}}{v_{x}^2 + v_{y}^2} \end{align} $$



\begin{split} \mathbf{u} \otimes \mathbf{v} &=

\begin{bmatrix} u_{x} {\color{blue} e_{x}} \\ u_{y} {\color{blue} e_{y}} \end{bmatrix}

\begin{bmatrix} v_{x} {\color{blue}e_{x}} & v_{y} {\color{blue}e_{y}} \end{bmatrix}           \\ \\ &=

\begin{bmatrix} u_{x} v_{x} {\color{red}  e_{x} e_{x} }  & u_{x} v_{y} {\color{green} e_{x} e_{y} } \\ u_{y} v_{x} {\color{green} e_{y} e_{x} } & u_{y} v_{y} {\color{red}  e_{y} e_{y} } \end{bmatrix}

\end{split}\\ \\ \\ \\ $$



\begin{split} (\mathbf{u} \otimes \mathbf{v}) \cdot (\mathbf{v})^{-1} &=

\begin{bmatrix} u_{x} v_{x} {\color{red}  e_{x} e_{x} }  & u_{x} v_{y} {\color{green} e_{x} e_{y} } \\ u_{y} v_{x} {\color{green} e_{y} e_{x} } & u_{y} v_{y} {\color{red}  e_{y} e_{y} } \end{bmatrix} \cdot

\begin{bmatrix} v_{x} {\color{blue} e_{x}} \\ v_{y} {\color{blue} e_{y}} \end{bmatrix}^{-1} \\ \\ &=

\begin{bmatrix} u_{x} v_{x} {\color{red}  e_{x} e_{x} }  & u_{x} v_{y} {\color{green} e_{x} e_{y} } \\ u_{y} v_{x} {\color{green} e_{y} e_{x} } & u_{y} v_{y} {\color{red}  e_{y} e_{y} } \end{bmatrix} \cdot

\begin{bmatrix} v_{x} {\color{blue} e_{x}} (v_x^2 + v_y^2)^{-1} \\ v_{y} {\color{blue} e_{y}} (v_x^2 + v_y^2)^{-1} \end{bmatrix} \\ \\ &=

\begin{bmatrix} u_{x} v_{x} {\color{red}  e_{x} e_{x} } v_{x} e_{x} (v_x^2 + v_y^2)^{-1}  + u_{x} v_{y} {\color{green} e_{x} e_{y} } v_{y} e_{y} (v_x^2 + v_y^2)^{-1} \\ u_{y} v_{x} {\color{green} e_{y} e_{x} } v_{x} e_{x} (v_x^2 + v_y^2)^{-1} + u_{y} v_{y} {\color{red}  e_{y} e_{y} } v_{y} e_{y} (v_x^2 + v_y^2)^{-1} \end{bmatrix} \\ \\ &=

\begin{bmatrix} (u_{x} v_{x} {\color{red}  e_{x} e_{x} } v_{x} e_{x}  +   u_{x} v_{y} {\color{green} e_{x} e_{y} } v_{y} e_{y}) (v_x^2 + v_y^2)^{-1}  \\ (u_{y} v_{x} {\color{green} e_{y} e_{x} } v_{x} e_{x} +  u_{y} v_{y} {\color{red}   e_{y} e_{y} } v_{y} e_{y}) (v_x^2 + v_y^2)^{-1} \end{bmatrix} \\ \\ &=

\begin{bmatrix} (u_{x} v_{x} v_{x} {\color{red}   e_{x} e_{x} } e_{x}  +   u_{x} v_{y}  v_{y} {\color{green} e_{x} e_{y} } e_{y}) (v_x^2 + v_y^2)^{-1}  \\ (u_{y} v_{x} v_{x} {\color{green} e_{y} e_{x} } e_{x}  +  u_{y} v_{y}  v_{y} {\color{red}   e_{y} e_{y} } e_{y}) (v_x^2 + v_y^2)^{-1} \end{bmatrix} \\ \\ &=

\begin{bmatrix} (u_{x} v_{x} v_{x} e_{x} {\color{red}e_{x} e_{x}} +  u_{x} v_{y} v_{y} e_{x} {\color{red}e_{y} e_{y}}) (v_x^2 + v_y^2)^{-1} \\ (u_{y} v_{x} v_{x} e_{y} {\color{red}e_{x} e_{x}} + u_{y} v_{y} v_{y} e_{y} {\color{red}e_{y} e_{y}}) (v_x^2 + v_y^2)^{-1} \end{bmatrix} \\ \\ &=

\begin{bmatrix} (u_{x} v_{x} v_{x} {\color{blue}e_{x}} +   u_{x} v_{y} v_{y} {\color{blue}e_{x}}) (v_x^2 + v_y^2)^{-1}  \\ (u_{y} v_{x} v_{x} {\color{blue}e_{y}} +  u_{y} v_{y} v_{y} {\color{blue}e_{y}}) (v_x^2 + v_y^2)^{-1} \end{bmatrix} \\ \\ &=

\begin{bmatrix} u_{x} {\color{blue}e_{x}} (v_{x} v_{x} + v_{y} v_{y}) (v_x^2 + v_y^2)^{-1} \\ u_{y} {\color{blue}e_{y}} (v_{x} v_{x} + v_{y} v_{y}) (v_x^2 + v_y^2)^{-1} \end{bmatrix} \\ \\ &=

\begin{bmatrix} u_{x} {\color{blue}e_{x}} \\ u_{y} {\color{blue}e_{y}} \end{bmatrix} \\ \\ \\ &= \mathbf{u}

\end{split} $$

The dot product behaves more like division than multiplication!

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Geometric calculus
In calculus units like feet or seconds behave like constants



\begin{align} \int \! x \, meters \, \cdot d(x \, meters) &= meters \cdot meters \int \! x \, \cdot d(x) \\ &= \frac{x^2}{2} meters^2 \end{align} $$

And:



\begin{align} \frac{d \left( \frac{x^2}{2} meters^2 \right)}{d(x \, meters)} &= \frac{meters^2}{meters} \frac{d \left( \frac{x^2}{2} \right) }{dx} \\ \\ \\ \\ &= x \, meters \end{align} $$

But:


 * $$\frac{1}{e_i} = \frac{e_i}{e_i e_i} = \frac{e_i}{1} = e_i$$

Derivatives in three dimensions
The derivative of a scalar function:



\begin{align} \nabla \mathbf{F} &=

\left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right) \left( f(x,y,z) \right) \\ \\ \\ &= \left( \frac{\partial f}{\partial x} e_x + \frac{\partial f}{\partial y} e_y + \frac{\partial f}{\partial z} e_z \right)

\end{align} $$


 * which is similar to the total differential which is a differential form:



\begin{align} d \mathbf{F} &=

\left( \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz \right)

\end{align} $$

The derivative of a vector function:

The derivative of a bivector function:

The derivative of a pseudoscalar function:



\begin{align} \nabla \mathbf{F} &=

\left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right) \left( f(x,y,z) e_{xyz} \right) \\ \\ \\ &= \left( \frac{\partial f}{\partial x} e_{xyzx} + \frac{\partial f}{\partial y} e_{xyzy} + \frac{\partial f}{\partial z} e_{xyzz} \right) \\ &=

\left( {\color{green}\frac{\partial f}{\partial x} e_{yz}} - {\color{green}\frac{\partial f}{\partial y} e_{xz}} + {\color{green}\frac{\partial f}{\partial z} e_{xy}} \right)

\end{align} $$

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Classical mechanics
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Maxwell's equations
From Mathematical descriptions of the electromagnetic field

E(x, y, z, t) and B(x, y, z, t) are vector fields called the electric field and the magnetic field.

The behaviour of electric and magnetic fields, whether in cases of electrostatics, magnetostatics, or electrodynamics (electromagnetic fields), is governed by Maxwell's equations:


 * {| width="400px" style="white-space: nowrap; background-color:#ECFCF4; padding:6; cellpadding=6;text-align:left;border:2px solid #50C878"


 * text-align="center" colspan="2"|Maxwell's equations (vector fields)
 * $$\nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0}$$ ||   Gauss's law
 * $$\nabla \cdot \mathbf{B} = 0$$ ||   Gauss's law for magnetism
 * $$\nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \mu_0\varepsilon_0 \frac{\partial \mathbf{E}}{\partial t}$$ ||    Ampère–Maxwell law
 * $$\nabla \times \mathbf{E} = -\frac {\partial \mathbf{B}}{\partial t}$$ ||   Faraday's law
 * }
 * $$\nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \mu_0\varepsilon_0 \frac{\partial \mathbf{E}}{\partial t}$$ ||    Ampère–Maxwell law
 * $$\nabla \times \mathbf{E} = -\frac {\partial \mathbf{B}}{\partial t}$$ ||   Faraday's law
 * }
 * $$\nabla \times \mathbf{E} = -\frac {\partial \mathbf{B}}{\partial t}$$ ||   Faraday's law
 * }

where ρ is the charge density, ε0 is the electric constant, μ0 is the magnetic constant, and J is the current per unit area.

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Algebra of physical space
In two  are introduced. One for the field and one for the current.


 * The field multivector, known as the, is


 * $$ \bold{F} = \bold{E} + Ic\bold{B} = E^k\sigma_k + IcB^k\sigma_k = \text{vector + pseudovector}$$


 * The current multivector is


 * $$ c \rho - \bold{J} = c \rho - J^k\sigma_k = \text{scalar + vector}$$


 * where, in the (APS) $$C\ell_{3,0}(\mathbb{R})$$ with the vector basis $$\{\sigma_k\}$$. The unit  is $$I=\sigma_1\sigma_2\sigma_3$$ (assuming an ). Orthonormal basis vectors share the algebra of the, but are usually not equated with them.

After defining the derivative


 * $$ \boldsymbol{\nabla} = \sigma^k \partial_k$$


 * Maxwell's equations are reduced to the single equation


 * $$ \left(\frac{1}{c}\dfrac{\partial }{\partial t} + \boldsymbol{\nabla}\right)\bold{F} = \mu_0 c (c \rho - \bold{J}). $$

In three dimensions, the derivative has a special structure allowing the introduction of a cross product:


 * $$ \boldsymbol{\nabla}\bold{F} = \boldsymbol{\nabla} \cdot \bold{F} + \boldsymbol{\nabla} \wedge \bold{F} = \boldsymbol{\nabla} \cdot \bold{F} + I \boldsymbol{\nabla} \times \bold{F}$$


 * from which it is easily seen that Gauss's law is the scalar part, the Ampère–Maxwell law is the vector part, Faraday's law is the pseudovector part, and Gauss's law for magnetism is the pseudoscalar part of the equation . After expanding and rearranging, this can be written as



\left( \boldsymbol{\nabla} \cdot \mathbf{E} - \frac{\rho}{\epsilon_0} \right)- c \left( \boldsymbol{\nabla} \times \mathbf{B} - \mu_0 \epsilon_0 \frac{\partial {\mathbf{E}}}{\partial {t}} - \mu_0 \mathbf{J} \right)+ I \left( \boldsymbol{\nabla} \times \mathbf{E} + \frac{\partial {\mathbf{B}}}{\partial {t}} \right)+ I c \left( \boldsymbol{\nabla} \cdot \mathbf{B} \right)= 0 $$

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Spacetime algebra
We can identify APS as a subalgebra of the (STA) $$C\ell_{1,3}(\mathbb{R})$$, defining $$\sigma_k=\gamma_k\gamma_0$$ and $$I=\gamma_0\gamma_1\gamma_2\gamma_3$$. The $$\gamma_\mu$$s have the same algebraic properties of the but their matrix representation is not needed. The derivative is now


 * $$\nabla = \gamma^\mu \partial_\mu.$$

The Riemann–Silberstein becomes a bivector


 * $$F = \bold{E} + Ic\bold{B} = E^1\gamma_1\gamma_0 + E^2\gamma_2\gamma_0 + E^3\gamma_3\gamma_0 -c(B^1\gamma_2\gamma_3 + B^2\gamma_3\gamma_1 + B^3\gamma_1\gamma_2),$$

and the charge and current density become a vector


 * $$J = J^\mu \gamma_\mu = c \rho \gamma_0 + J^k \gamma_k = \gamma_0(c \rho - J^k \sigma_k).$$

Owing to the identity


 * $$\gamma_0 \nabla = \gamma_0\gamma^0 \partial_0 + \gamma_0\gamma^k\partial_k = \partial_0 + \sigma^k\partial_k = \frac{1}{c}\dfrac{\partial }{\partial t} + \boldsymbol{\nabla},$$

Maxwell's equations reduce to the single equation


 * $$ \nabla F = \mu_0 c J. $$

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