Multivectors

Clifford algebra (Euclidean algebra) is a type of algebra characterized by the geometric productw of scalarsw, vectorsw, bivectorsw, trivectorsw...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 pseudoscalarw.

In three dimensions
 Bivector
 Psuedoscalar

But in four dimensions a trivector becomes a pseudovectorw and the quadvector becomes the
pseudoscalar.

## Clifford algebra

See also: Lp space
Consider the square (quadratic form) of a single vector:

Q
(
c
# )

c
2

=
⟨
c
,
c
# ⟩
(
a

e
1

b

e
2

)
2

{\displaystyle Q(c) = c^2 = \langle c , c \rangle = (a\mathbf{e\_1} + b\mathbf{e\_2})^2}

Q
(
c
# )

c
2

=
a
a

e
1

e
1

b
b

e
2

e
2

a
b

e
1

e
2

b
a

e
2

e
1

{\displaystyle Q(c) = c^2 = aa\mathbf{e\_1}\mathbf{e\_1} + bb\mathbf{e\_2}\mathbf{e\_2} + ab\mathbf{e\_1}\mathbf{e\_2} + ba\mathbf{e\_2}\mathbf{e\_1}}

Q
(
c
# )

c
2

=

a
2

(

e
1

e
1

)

b
2

(

e
2

e
2

)
a
b
(

e
1

e
2

e
2

e
1

)

{\displaystyle Q(c) = c^2 = a^2(\mathbf{e\_1}\mathbf{e\_1}) + b^2(\mathbf{e\_2}\mathbf{e\_2}) + ab(\mathbf{e\_1}\mathbf{e\_2} + \mathbf{e\_2}\mathbf{e\_1})}

From the Pythagorean theorem we know that:

c
2

=

a
2

(
1
)

b
2

(
1
)
a
b
(
0
# )
S
c
a
l
a
r

{\displaystyle c^2 = a^2(1) + b^2(1) + ab(0) = Scalar}

So we know that

e
1

e
1

=

e
2

e
2

=
1

{\displaystyle e\_1 e\_1 = e\_2 e\_2 = 1}

e
1

e
2

=
−

e
2

e
1

{\displaystyle e\_1 e\_2 = -e\_2 e\_1 }

This particular Clifford algebra is known as Cl2,0. The subscript 2 indicates that the 2 basis vectors are square roots of +1. See Metric signature. If we had used

c
2

=
−

a
2

−

b
2

{\displaystyle c^2 = -a^2 -b^2}

 then the result would have been Cl0,2.

### Multiplication of arbitrary vectors

Instead of the dot product or the wedge productw Clifford algebra uses the geometric productw.

u

v

=
(

u

x

e

x

u

y

e

y

)
(

v

x

e

x

v

y

e

y

)

=

u

x

v

x

e

x

e

x

u

x

v

y

e

x

e

y

u

y

v

x

e

y

e

x

u

y

v

y

e

y

e

y

=

u

x

v

x

(
1
)

u

y

v

y

(
1
)

u

x

v

y

e

x

e

y

−

u

y

v

x

e

x

e

y

=
(

u

x

v

x

u

y

v

y

)

(
1
)

(

u

x

v

y

−

u

y

v

x

)

e

x
y

=

u

⋅

v

u

∧

v

=

scalar

bivector

{\displaystyle
\begin{align}
\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}} \\ &=
\mathbf{u} \cdot \mathbf{v} + \mathbf{u} \wedge \mathbf{v} \\ &=
{\color{red}\text{scalar}} +
{\color{green}\text{bivector}}
\end{align}
}

A scalar plus a bivector (or any number of bladesw of different grade) is called a multivectorw. 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 simultaneous equations.

For example:

u

v

=
5

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

would just mean that:

(

u

x

v

x

u

y

v

y

)

(
1
)

=
5

and

(

u

x

v

y

−

u

y

v

x

)

e

x
y

=
0

{\displaystyle
\begin{align}
(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
\end{align}
}

Back to top

### Rules

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

Let

e

x

,

{\displaystyle {\color{blue}e\_{x}},}

e

y

,

{\displaystyle {\color{blue}e\_{y}},}

 and

e

z

{\displaystyle {\color{blue}e\_{z}}}

 be perpendicular unit vectorsw.

Multiplying two perpendicular vectors results in a bivector:

e
x

e
y

=

e

x
y

{\displaystyle e\_x e\_y = {\color{green}e\_{xy}}}

Multiplying three perpendicular vectors results in a trivector:

e
x

e
y

e
z

=

e

x
y
z

{\displaystyle e\_x e\_y e\_z = {\color{orange}e\_{xyz}}}

Multiplying parallel vectors results in a scalar:

e
x

e
x

=

e

x
x

=

1

{\displaystyle 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:

(

e

x

e

y

)

(

e

y

)

=

e

x

(

e

y

e

y

)

=

e

x

(
1
)

=

e

x

{\displaystyle
\begin{align}
{\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{align}
}

and:

(

e

x

e

y

)

(

e

y

e

z

)

=

e

x
y
y

e

x
y
z

=

e

x

e

x
y
z

{\displaystyle
\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:

(

e

x

e

y

)

(

e

z

)

=

e

x
y
z

{\displaystyle
\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

x
y

=
−

e

y
x

{\displaystyle e\_{xy} = -e\_{yx}}

Therefore:

(

e

x

e

y

)

(

e

x

)

=

e

x

(

e

y

e

x

)

=

−

e

x

(

e

x

e

y

)

=
−

(

e

x

e

x

)

e

y

=
−

(
1
)

e

y

=
−

e

y

{\displaystyle
\begin{align}
 {\color{green}(e\_{x} e\_{y})} {\color{blue} (e\_{x}) } &=
 {\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{align}
}

Back to top

### Multiplication tables

In one dimension:

1

e

x

e

x

1

{\displaystyle
\begin{array}{r|r}

{\color{red} 1 } &
{\color{blue} e\_{x}}

\\\hline
{\color{blue} e\_{x}} &
{\color{red} 1 }

\end{array}
}

In two dimensions:

1

e

x

e

y

e

x
y

e

x

1

e

x
y

e

y

e

y

−

e

x
y

1

−

e

x

e

x
y

−

e

y

e

x

−
1

{\displaystyle
\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:

1

e

x

e

y

e

z

e

x
y

e

x
z

e

y
z

e

x
y
z

e

x

1

e

x
y

e

x
z

e

y

e

z

e

x
y
z

e

y
z

e

y

−

e

x
y

1

e

y
z

−

e

x

−

e

x
y
z

e

z

−

e

x
z

e

z

−

e

x
z

−

e

y
z

1

e

x
y
z

−

e

x

−

e

y

e

x
y

e

x
y

−

e

y

e

x

e

x
y
z

−
1

−

e

y
z

e

x
z

−

e

z

e

x
z

−

e

z

−

e

x
y
z

e

x

e

y
z

−
1

−

e

x
y

e

y

e

y
z

e

x
y
z

−

e

z

e

y

−

e

x
z

e

x
y

−
1

−

e

x

e

x
y
z

e

y
z

−

e

x
z

e

x
y

−

e

z

e

y

−

e

x

−
1

{\displaystyle
\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:

1

e

1

e

2

e

3

e

4

e

12

e

13

e

14

e

23

e

24

e

34

e

123

e

124

e

134

e

234

e

1234

e

1

1

e

12

e

13

e

14

e

2

e

3

e

4

e

123

e

124

e

134

e

23

e

24

e

34

e

1234

e

234

e

2

−

e

12

1

e

23

e

24

−

e

1

−

e

123

−

e

124

e

3

e

4

e

234

−

e

13

−

e

14

−

e

1234

e

34

−

e

134

e

3

−

e

13

−

e

23

1

e

34

e

123

−

e

1

−

e

134

−

e

2

−

e

234

e

4

e

12

e

1234

−

e

14

−

e

24

e

124

e

4

−

e

14

−

e

24

−

e

34

1

e

124

e

123

−

e

1

e

234

−

e

2

−

e

3

−

e

1234

e

12

e

13

e

23

−

e

123

e

12

−

e

2

e

1

e

123

e

124

−
1

−

e

23

−

e

24

e

13

e

14

e

1234

−

e

3

−

e

4

−

e

234

e

134

−

e

34

e

13

−

e

3

−

e

123

e

1

e

134

e

23

−
1

−

e

34

−

e

12

−

e

1234

e

14

e

2

e

234

−

e

4

−

e

124

e

24

e

14

−

e

4

−

e

124

−

e

123

e

1

e

24

e

34

−
1

e

1234

−

e

12

−

e

13

−

e

234

e

2

e

3

e

123

−

e

23

e

23

e

123

−

e

3

e

2

e

234

−

e

13

e

12

e

1234

−
1

−

e

34

e

24

−

e

1

−

e

134

e

124

−

e

4

−

e

14

e

24

e

124

−

e

4

−

e

234

e

2

−

e

14

−

e

1234

e

12

e

34

−
1

−

e

23

e

123

−

e

144

−

e

123

e

3

e

13

e

34

e

134

e

234

−

e

4

e

3

e

1234

−

e

14

e

13

−

e

24

e

23

−
1

−

e

124

e

123

−

e

1

−

e

2

−

e

12

e

123

e

23

−

e

13

e

12

e

1234

−

e

3

e

2

e

234

−

e

1

−

e

134

e

124

−
1

−

e

34

e

24

−

e

14

−

e

4

e

124

e

24

−

e

14

−

e

1234

e

12

−

e

4

−

e

234

e

2

e

123

−

e

1

−

e

123

e

34

−
1

−

e

23

e

13

e

3

e

134

e

34

e

1234

−

e

14

e

13

e

234

−

e

4

e

3

−

e

124

e

123

−

e

1

−

e

24

e

23

−
1

−

e

12

−

e

2

e

234

−

e

1234

e

34

−

e

24

e

23

−

e

134

e

124

−

e

123

−

e

4

e

3

−

e

2

e

14

−

e

13

e

12

−
1

e

1

e

1234

−

e

234

e

134

−

e

124

e

123

−

e

34

e

24

−

e

23

−

e

14

e

13

−

e

12

e

4

−

e

3

e

2

−

e

1

1

{\displaystyle

{\color{red} 1 } &
{\color{black} e\_{1}} &
{\color{black} e\_{2}} &
{\color{black} e\_{3}} &
{\color{black} e\_{4}} &
{\color{black} e\_{12}} &
{\color{black} e\_{13}} &
{\color{black} e\_{14}} &
{\color{black} e\_{23}} &
{\color{black} e\_{24}} &
{\color{black} e\_{34}} &
{\color{black} e\_{123}} &
{\color{black} e\_{124}} &
{\color{black} e\_{134}} &
{\color{black} e\_{234}} &
{\color{red} e\_{1234}}

\\\hline

{\color{black} e\_{1}} &
{\color{red} 1 } &
{\color{black} e\_{12}} &
{\color{black} e\_{13}} &
{\color{black} e\_{14}} &
{\color{black} e\_{2}} &
{\color{black} e\_{3}} &
{\color{black} e\_{4}} &
{\color{black} e\_{123}} &
{\color{black} e\_{124}} &
{\color{black} e\_{134}} &
{\color{black} e\_{23}} &
{\color{black} e\_{24}} &
{\color{black} e\_{34}} &
{\color{red} e\_{1234}} &
{\color{black} e\_{234}}

\\

{\color{black} e\_{2}} &
{\color{black} -e\_{12}} &
{\color{red} 1 } &
{\color{black} e\_{23}} &
{\color{black} e\_{24}} &
{\color{black} -e\_{1}} &
{\color{black} -e\_{123}} &
{\color{black} -e\_{124}} &
{\color{black} e\_{3}} &
{\color{black} e\_{4}} &
{\color{black} e\_{234}} &
{\color{black} -e\_{13}} &
{\color{black} -e\_{14}} &
{\color{red} -e\_{1234}} &
{\color{black} e\_{34}} &
{\color{black} -e\_{134}}

\\

{\color{black} e\_{3}} &
{\color{black} -e\_{13}} &
{\color{black} -e\_{23}} &
{\color{red} 1 } &
{\color{black} e\_{34}} &
{\color{black} e\_{123}} &
{\color{black} -e\_{1}} &
{\color{black} -e\_{134}} &
{\color{black} -e\_{2}} &
{\color{black} -e\_{234}} &
{\color{black} e\_{4}} &
{\color{black} e\_{12}} &
{\color{red} e\_{1234}} &
{\color{black} -e\_{14}} &
{\color{black} -e\_{24}} &
{\color{black} e\_{124}}

\\

{\color{black} e\_{4}} &
{\color{black} -e\_{14}} &
{\color{black} -e\_{24}} &
{\color{black} -e\_{34}} &
{\color{red} 1 } &
{\color{black} e\_{124}} &
{\color{black} e\_{123}} &
{\color{black} -e\_{1}} &
{\color{black} e\_{234}} &
{\color{black} -e\_{2}} &
{\color{black} -e\_{3}} &
{\color{red} -e\_{1234}} &
{\color{black} e\_{12}} &
{\color{black} e\_{13}} &
{\color{black} e\_{23}} &
{\color{black} -e\_{123}}

\\\hline

{\color{black} e\_{12}} &
{\color{black} -e\_{2}} &
{\color{black} e\_{1}} &
{\color{black} e\_{123}} &
{\color{black} e\_{124}} &
{\color{red} -1 } &
{\color{black} -e\_{23}} &
{\color{black} -e\_{24}} &
{\color{black} e\_{13}} &
{\color{black} e\_{14}} &
{\color{red} e\_{1234}} &
{\color{black} -e\_{3}} &
{\color{black} -e\_{4}} &
{\color{black} -e\_{234}} &
{\color{black} e\_{134}} &
{\color{black} -e\_{34}}

\\

{\color{black} e\_{13}} &
{\color{black} -e\_{3}} &
{\color{black} -e\_{123}} &
{\color{black} e\_{1}} &
{\color{black} e\_{134}} &
{\color{black} e\_{23}} &
{\color{red} -1 } &
{\color{black} -e\_{34}} &
{\color{black} -e\_{12}} &
{\color{red} -e\_{1234}} &
{\color{black} e\_{14}} &
{\color{black} e\_{2}} &
{\color{black} e\_{234}} &
{\color{black} -e\_{4}} &
{\color{black} -e\_{124}} &
{\color{black} e\_{24}}

\\

{\color{black} e\_{14}} &
{\color{black} -e\_{4}} &
{\color{black} -e\_{124}} &
{\color{black} -e\_{123}} &
{\color{black} e\_{1}} &
{\color{black} e\_{24}} &
{\color{black} e\_{34}} &
{\color{red} -1 } &
{\color{red} e\_{1234}} &
{\color{black} -e\_{12}} &
{\color{black} -e\_{13}} &
{\color{black} -e\_{234}} &
{\color{black} e\_{2}} &
{\color{black} e\_{3}} &
{\color{black} e\_{123}} &
{\color{black} -e\_{23}}

\\

{\color{black} e\_{23}} &
{\color{black} e\_{123}} &
{\color{black} -e\_{3}} &
{\color{black} e\_{2}} &
{\color{black} e\_{234}} &
{\color{black} -e\_{13}} &
{\color{black} e\_{12}} &
{\color{red} e\_{1234}} &
{\color{red} -1 } &
{\color{black} -e\_{34}} &
{\color{black} e\_{24}} &
{\color{black} -e\_{1}} &
{\color{black} -e\_{134}} &
{\color{black} e\_{124}} &
{\color{black} -e\_{4}} &
{\color{black} -e\_{14}}

\\

{\color{black} e\_{24}} &
{\color{black} e\_{124}} &
{\color{black} -e\_{4}} &
{\color{black} -e\_{234}} &
{\color{black} e\_{2}} &
{\color{black} -e\_{14}} &
{\color{red} -e\_{1234}} &
{\color{black} e\_{12}} &
{\color{black} e\_{34}} &
{\color{red} -1 } &
{\color{black} -e\_{23}} &
{\color{black} e\_{123}} &
{\color{black} -e\_{144}} &
{\color{black} -e\_{123}} &
{\color{black} e\_{3}} &
{\color{black} e\_{13}}

\\

{\color{black} e\_{34}} &
{\color{black} e\_{134}} &
{\color{black} e\_{234}} &
{\color{black} -e\_{4}} &
{\color{black} e\_{3}} &
{\color{red} e\_{1234}} &
{\color{black} -e\_{14}} &
{\color{black} e\_{13}} &
{\color{black} -e\_{24}} &
{\color{black} e\_{23}} &
{\color{red} -1 } &
{\color{black} -e\_{124}} &
{\color{black} e\_{123}} &
{\color{black} -e\_{1}} &
{\color{black} -e\_{2}} &
{\color{black} -e\_{12}}

\\\hline

{\color{black} e\_{123}} &
{\color{black} e\_{23}} &
{\color{black} -e\_{13}} &
{\color{black} e\_{12}} &
{\color{red} e\_{1234}} &
{\color{black} -e\_{3}} &
{\color{black} e\_{2}} &
{\color{black} e\_{234}} &
{\color{black} -e\_{1}} &
{\color{black} -e\_{134}} &
{\color{black} e\_{124}} &
{\color{red} -1 } &
{\color{black} -e\_{34}} &
{\color{black} e\_{24}} &
{\color{black} -e\_{14}} &
{\color{black} -e\_{4}}

\\

{\color{black} e\_{124}} &
{\color{black} e\_{24}} &
{\color{black} -e\_{14}} &
{\color{red} -e\_{1234}} &
{\color{black} e\_{12}} &
{\color{black} -e\_{4}} &
{\color{black} -e\_{234}} &
{\color{black} e\_{2}} &
{\color{black} e\_{123}} &
{\color{black} -e\_{1}} &
{\color{black} -e\_{123}} &
{\color{black} e\_{34}} &
{\color{red} -1 } &
{\color{black} -e\_{23}} &
{\color{black} e\_{13}} &
{\color{black} e\_{3}}

\\

{\color{black} e\_{134}} &
{\color{black} e\_{34}} &
{\color{red} e\_{1234}} &
{\color{black} -e\_{14}} &
{\color{black} e\_{13}} &
{\color{black} e\_{234}} &
{\color{black} -e\_{4}} &
{\color{black} e\_{3}} &
{\color{black} -e\_{124}} &
{\color{black} e\_{123}} &
{\color{black} -e\_{1}} &
{\color{black} -e\_{24}} &
{\color{black} e\_{23}} &
{\color{red} -1 } &
{\color{black} -e\_{12}} &
{\color{black} -e\_{2}}

\\

{\color{black} e\_{234}} &
{\color{red} -e\_{1234}} &
{\color{black} e\_{34}} &
{\color{black} -e\_{24}} &
{\color{black} e\_{23}} &
{\color{black} -e\_{134}} &
{\color{black} e\_{124}} &
{\color{black} -e\_{123}} &
{\color{black} -e\_{4}} &
{\color{black} e\_{3}} &
{\color{black} -e\_{2}} &
{\color{black} e\_{14}} &
{\color{black} -e\_{13}} &
{\color{black} e\_{12}} &
{\color{red} -1 } &
{\color{black} e\_{1}}

\\\hline

{\color{red} e\_{1234}} &
{\color{black} -e\_{234}} &
{\color{black} e\_{134}} &
{\color{black} -e\_{124}} &
{\color{black} e\_{123}} &
{\color{black} -e\_{34}} &
{\color{black} e\_{24}} &
{\color{black} -e\_{23}} &
{\color{black} -e\_{14}} &
{\color{black} e\_{13}} &
{\color{black} -e\_{12}} &
{\color{black} e\_{4}} &
{\color{black} -e\_{3}} &
{\color{black} e\_{2}} &
{\color{black} -e\_{1}} &
{\color{red} 1 }

\end{array}
}

Back to top

### 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 Cℓ3 are

{

1

,

e
x

,

e
y

,

e
z

,

e

x
y

,

e

x
z

,

e

y
z

,

e

x
y
z

}

{\displaystyle \{{\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 Cℓ3 is given by

# A

a
0

(
1
)

a
1

e
x

a
2

e
y

a
3

e
z

a
4

e

x
y

a
5

e

x
z

a
6

e

y
z

a
7

e

x
y
z

.

{\displaystyle 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

{\displaystyle a\_0, a\_1, a\_2, a\_3, a\_4, a\_5, a\_6, a\_7 }

 are all real then the Clifford algebra is Cℓ3(**R**). If the coefficients are allowed to be complexw then the Clifford algebra is Cℓ3(**C**).

A multivectorw can be separated into components of different grades:

⟨

A

⟩
0

=

a
0

(
1
)

⟨

A

⟩
1

=

a
1

e
x

a
2

e
y

a
3

e
z

⟨

A

⟩
2

=

a
4

e

x
y

a
5

e

x
z

a
6

e

y
z

⟨

A

⟩
3

=

a
7

e

x
y
z

{\displaystyle
\begin{align}

\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}}

\end{align}

}

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.

Back to top

### Hodge dual

Multiplying (using geometric product) any multivector by the pseudoscalar converts the multivector to its dual.

The **Hodge star operator** is a linear operator on the exterior algebra of *V*, mapping *k*-vectors to (*n* – *k*)-vectors, for

0
≤
k
≤
n

{\displaystyle 0\leq k \leq n}

. It has the following property, which defines it completely:

α
∧
(

⋆

β
# )
⟨
α
,
β
⟩

ω

{\displaystyle \alpha \wedge ({\star} \beta) = \langle \alpha,\beta \rangle \,\omega }

 for every pair of *k*-vectors

α
,
β
∈

⋀

k

V
.

{\displaystyle \alpha,\beta\in {\textstyle\bigwedge}^{\!k}V .}

The Hodge star relates the exterior and cross product in three dimensions:

⋆

(

u

∧

v

# )

u

×

v

⋆

(

u

×

v

# )

u

∧

v

.

{\displaystyle {\star} (\mathbf{u} \wedge \mathbf{v}) = \mathbf{u} \times \mathbf{v} \qquad {\star} (\mathbf{u} \times \mathbf {v}) = \mathbf{u} \wedge \mathbf{v} .}

And

⋆
(

w
1

∧
⋯
∧

w
k

)

=

u
1

∧
⋯
∧

u

n
−
k

,

{\displaystyle \star(w\_1\wedge\cdots\wedge w\_k) \,=\, u\_1\wedge\cdots\wedge u\_{n-k},}

Back to top

### Relation to other algebras

C

ℓ
0

(

R

)

{\displaystyle C\ell\_0 (\mathbf{R})}

: Real numbersw (scalarsw). A scalar can (and should) be thought of as zero vectors multiplied together. See Empty productw.

C

ℓ
0

(

C

)

{\displaystyle C\ell\_0 (\mathbf{C})}

: Complex numbersw

C

ℓ
1

(

R

)

{\displaystyle C\ell\_1 (\mathbf{R})}

: Split-complex numbersw

1

e

x

e

x

1

{\displaystyle
\begin{array}{r|r}

{\color{red} 1 } &
{\color{blue} e\_{x}}

\\\hline
{\color{blue} e\_{x}} &
{\color{red} 1 }

\end{array}
}

C

ℓ
1

(

C

)

{\displaystyle C\ell\_1 (\mathbf{C})}

: Bicomplex numbersw

C

ℓ
2
0

(

R

)

{\displaystyle C\ell\_2^0 (\mathbf{R})}

: Complex numbersw (The superscript 0 indicates the even subalgebra)

1

e

x
y

e

x
y

−
1

{\displaystyle
\begin{array}{r|r}

{\color{red} 1} &
{\color{green} e\_{xy}}

\\\hline
{\color{green} e\_{xy}} &
{\color{red} {-1}}

\end{array}
}

C

ℓ
3
0

(

R

)

{\displaystyle C\ell\_3^0 (\mathbf{R})}

: Quaternionsw

1

e

x
y

e

x
z

e

y
z

e

x
y

−
1

−

e

y
z

e

x
z

e

x
z

e

y
z

−
1

−

e

x
y

e

y
z

−

e

x
z

e

x
y

−
1

{\displaystyle
\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}
}

C

ℓ
3
0

(

C

)

{\displaystyle C\ell\_3^0 (\mathbf{C})}

: Biquaternionsw

Back to top

### Multivector multiplication using tensors

To find the product

A
# B
(

a
0

1

a
1

e

x

a
2

e

y

a
3

e

x
y

)
(

b
0

1

b
1

e

x

b
2

e

y

b
3

e

x
y

)

{\displaystyle
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.

A
# B

(

a
0

b
0

1

1

a
0

b
1

1

e

x

a
0

b
2

1

e

y

a
0

b
3

1

e

x
y

)

(

a
1

b
0

e

x

1

a
1

b
1

e

x

e

x

a
1

b
2

e

x

e

y

a
1

b
3

e

x

e

x
y

)

(

a
2

b
0

e

y

1

a
2

b
1

e

y

e

x

a
2

b
2

e

y

e

y

a
2

b
3

e

y

e

x
y

)

(

a
3

b
0

e

x
y

1

a
3

b
1

e

x
y

e

x

a
3

b
2

e

x
y

e

y

a
3

b
3

e

x
y

e

x
y

)

{\displaystyle
\begin{align}
AB = &

(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{align}
}

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

A
# B

(

a
0

b
0

1

a
0

b
1

e

x

a
0

b
2

e

y

a
0

b
3

e

x
y

)

(

a
1

b
0

e

x

a
1

b
1

1

a
1

b
2

e

x
y

a
1

b
3

e

y

)

(

a
2

b
0

e

y

−

a
2

b
1

e

x
y

a
2

b
2

1

−

a
2

b
3

e

x

)

(

a
3

b
0

e

x
y

−

a
3

b
1

e

y

a
3

b
2

e

x

−

a
3

b
3

1

)

{\displaystyle
\begin{align}
AB = &

(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{align}
}

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

A
# B

(

a
0

b
0

a
1

b
1

a
2

b
2

−

a
3

b
3

)

1

(

a
1

b
0

a
0

b
1

a
3

b
2

−

a
2

b
3

)

e

x

(

a
2

b
0

−

a
3

b
1

a
0

b
2

a
1

b
3

)

e

y

(

a
3

b
0

−

a
2

b
1

a
1

b
2

a
0

b
3

)

e

x
y

{\displaystyle
\begin{align}
AB = &
( 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{align}
}

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.

Back to top

#### Complex numbers

The multiplication table for

C

ℓ
2
0

(

R

)

{\displaystyle C\ell\_2^0 (\mathbf{R})}

 (Which is isomorphic to complex numbers)

1

e

x
y

e

x
y

−
1

{\displaystyle
\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:

(

1

e

x
y

e

x
y

1

)

(

1

e

x
y

)

=

(

1

⋅

1

e

x
y

e

x
y

e

x
y

1

1

e

x
y

)

=

(

1

−

1

e

x
y

e

x
y

)

{\displaystyle
\begin{align}

\left(
\begin{array}{rrrr}

{\color{red} 1} &
{\color{green} e\_{xy}}

\\
{\color{green} e\_{xy}} &
{\color{red} 1 }

\end{array}
\right)

\left(
\begin{array}{rrrr}

{\color{red} 1} \\
{\color{green} e\_{xy}}

\end{array}
\right) &=

\left(
\begin{array}{lcl}

{\color{red} 1} \cdot
{\color{red} 1} &
+ &
{\color{green} e\_{xy}}
{\color{green} e\_{xy}}

\\
{\color{green} e\_{xy}}
{\color{red} 1} &
+ &
{\color{red} 1 }
{\color{green} e\_{xy}}

\end{array}
\right) \\ &=

\left(
\begin{array}{lcl}

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

\\
{\color{green} e\_{xy}} &
+ &
{\color{green} e\_{xy}}

\end{array}
\right)

\end{align}
}

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.

(

b
0

⋅

1

b
1

⋅

e

x
y

b
1

⋅

e

x
y

b
0

⋅

1

)

(

a
0

⋅

1

a
1

⋅

e

x
y

)

=

(

b
0

a
0

⋅

1

⋅

1

b
1

a
1

⋅

e

x
y

e

x
y

b
1

a
0

⋅

e

x
y

1

b
0

a
1

⋅

1

e

x
y

)

=

(

b
0

a
0

⋅

1

−

b
1

a
1

⋅

1

b
1

a
0

⋅

e

x
y

b
0

a
1

⋅

e

x
y

)

{\displaystyle
\begin{align}

\left(
\begin{array}{rrrr}

b\_0 & \cdot &
{\color{red} 1} \quad &
b\_1 & \cdot &
{\color{green} e\_{xy}}

\\
b\_1 & \cdot &
{\color{green} e\_{xy}} \quad &
b\_0 & \cdot &
{\color{red} 1 }

\end{array}
\right)

\left(
\begin{array}{rrrr}

a\_0 & \cdot &
{\color{red} 1} \\
a\_1 & \cdot &
{\color{green} e\_{xy}}

\end{array}
\right) &=

\left(
\begin{array}{lcl}

b\_0 a\_0 & \cdot &
{\color{red} 1} \cdot
{\color{red} 1} &
+ &
b\_1 a\_1 & \cdot &
{\color{green} e\_{xy}}
{\color{green} e\_{xy}}

\\
b\_1 a\_0 & \cdot &
{\color{green} e\_{xy}}
{\color{red} 1} &
+ &
b\_0 a\_1 & \cdot &
{\color{red} 1 }
{\color{green} e\_{xy}}

\end{array}
\right) \\ &=

\left(
\begin{array}{lcl}

b\_0 a\_0 & \cdot &
{\color{red} 1} &
- &
b\_1 a\_1 & \cdot &
{\color{red} 1 }

\\
b\_1 a\_0 & \cdot &
{\color{green} e\_{xy}} &
+ &
b\_0 a\_1 & \cdot &
{\color{green} e\_{xy}}

\end{array}
\right)

\end{align}
}

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

Therefore to find the product

(

a
0

1

a
1

e

x
y

)
(

b
0

1

b
1

e

x
y

)

{\displaystyle
(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:

(

b
0

−

b
1

b
1

b
0

)

(

a
0

a
1

)

=

(

b
0

a
0

−

b
1

a
1

b
1

a
0

b
0

a
1

)

{\displaystyle

\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. See Advanced\_mathematics#Column\_vectors.

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

[

a

−
b

b

a

]

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

Which corresponds perfectly to a multiplication table for complex numbers:

1

−
i

i

1

{\displaystyle
\begin{array}{r|r}

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

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

\end{array}
}

1

−

e

x
y

e

x
y

1

{\displaystyle \quad \quad
\begin{array}{r|r}

{\color{red} 1} &
{\color{green} -e\_{xy}}

\\\hline
{\color{green} e\_{xy}} &
{\color{red} 1 }

\end{array}
}

Back to top

#### Quaternions

The multiplication table for

C

ℓ
3
0

(

R

)

{\displaystyle C\ell\_3^0 (\mathbf{R})}

 (Which is isomorphic to quaternions) is:

1

e

x
y

e

x
z

e

y
z

e

x
y

−
1

−

e

y
z

e

x
z

e

x
z

e

y
z

−
1

−

e

x
y

e

y
z

−

e

x
z

e

x
y

−
1

# i

e

y
z

# j

e

x
z

# k

e

x
y

{\displaystyle
\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

e

x
y

{\displaystyle {\color{green} e\_{xy}}}

 multiplied by the entire first row.

The entire 3rd row of the multiplication table is just

e

x
z

{\displaystyle {\color{green} e\_{xz}}}

 multiplied by the entire first row.

The entire 4th row of the multiplication table is just

e

y
z

{\displaystyle {\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:

(

1

e

x
y

e

x
z

e

y
z

e

x
y

1

e

y
z

e

x
z

e

x
z

e

y
z

1

e

x
y

e

y
z

e

x
z

e

x
y

1

)

(

1

e

x
y

e

x
z

e

y
z

)

=

(

1

⋅

1

e

x
y

e

x
y

e

x
z

e

x
z

e

y
z

e

y
z

e

x
y

1

1

e

x
y

e

y
z

e

x
z

e

x
z

e

y
z

e

x
z

1

e

y
z

e

x
y

1

e

x
z

e

x
y

e

y
z

e

y
z

1

e

x
z

e

x
y

e

x
y

e

x
z

1

e

y
z

)

=

(

1

−

1

−

1

−

1

e

x
y

e

x
y

e

x
y

−

e

x
y

e

x
z

−

e

x
z

e

x
z

e

x
z

e

y
z

e

y
z

−

e

y
z

e

y
z

)

{\displaystyle
\begin{align}

\left(
\begin{array}{rrrr}

{\color{red} 1} &
{\color{green} e\_{xy}} &
{\color{green} e\_{xz}} &
{\color{green} e\_{yz}}

\\
{\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}
\right)

\left(
\begin{array}{rrrr}

{\color{red} 1} \\
{\color{green} e\_{xy}} \\
{\color{green} e\_{xz}} \\
{\color{green} e\_{yz}}

\end{array}
\right) &=

\left(
\begin{array}{rrrr}

{\color{red} 1} \cdot
{\color{red} 1} &
+ &
{\color{green} e\_{xy}}
{\color{green} e\_{xy}} &
+ &
{\color{green} e\_{xz}}
{\color{green} e\_{xz}} &
+ &
{\color{green} e\_{yz}}
{\color{green} e\_{yz}}

\\
{\color{green} e\_{xy}}
{\color{red} 1} &
+ &
{\color{red} 1 }
{\color{green} e\_{xy}} &
+ &
{\color{green} e\_{yz}}
{\color{green} e\_{xz}} &
+ &
{\color{green} e\_{xz}}
{\color{green} e\_{yz}}

\\
{\color{green} e\_{xz}}
{\color{red} 1} &
+ &
{\color{green} e\_{yz}}
{\color{green} e\_{xy}} &
+ &
{\color{red} 1 }
{\color{green} e\_{xz}} &
+ &
{\color{green} e\_{xy}}
{\color{green} e\_{yz}}

\\
{\color{green} e\_{yz}}
{\color{red} 1} &
+ &
{\color{green} e\_{xz}}
{\color{green} e\_{xy}} &
+ &
{\color{green} e\_{xy}}
{\color{green} e\_{xz}} &
+ &
{\color{red} 1 }
{\color{green} e\_{yz}}

\end{array}
\right) \\ &=

\left(
\begin{array}{llll}

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

\\
{\color{green} e\_{xy}} &
+ &
{\color{green} e\_{xy}}&
+ &
{\color{green} e\_{xy}} &
- &
{\color{green} e\_{xy}}

\\
{\color{green} e\_{xz}} &
- &
{\color{green} e\_{xz}} &
+ &
{\color{green} e\_{xz}} &
+ &
{\color{green} e\_{xz}}

\\
{\color{green} e\_{yz}} &
+ &
{\color{green} e\_{yz}} &
- &
{\color{green} e\_{yz}} &
+ &
{\color{green} e\_{yz}}

\end{array}
\right)

\end{align}
}

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

1

a
1

e

x
y

a
2

e

x
z

a
3

e

y
z

)
(

b
0

1

b
1

e

x
y

b
2

e

x
z

b
3

e

y
z

)

{\displaystyle
(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:

(

b
0

−

b
1

−

b
2

−

b
3

b
1

b
0

b
3

−

b
2

b
2

−

b
3

b
0

b
1

b
3

b
2

−

b
1

b
0

)

(

a
0

a
1

a
2

a
3

)

=

(

b
0

a
0

−

b
1

a
1

−

b
2

a
2

−

b
3

a
3

b
1

a
0

b
0

a
1

b
3

a
2

−

b
2

a
3

b
2

a
0

−

b
3

a
1

b
0

a
2

b
1

a
3

b
3

a
0

b
2

a
1

−

b
1

a
2

b
0

a
3

)

{\displaystyle

\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 matricesw, so a quaternion can be represented as:

(

a

−
b

−
c

−
d

b

a

d

−
c

c

−
d

a

b

d

c

−
b

a

)

=

(

(

a

−
b

b

a

)

−

(

c

d

−
d

c

)

(

c

−
d

d

c

)

(

a

b

−
b

a

)

)

{\displaystyle
\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:

1

−

e

x
y

−

e

x
z

−

e

y
z

e

x
y

1

e

y
z

−

e

x
z

e

x
z

−

e

y
z

1

e

x
y

e

y
z

e

x
z

−

e

x
y

1

{\displaystyle
\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}
}

Back to top

#### CL2

The multiplication table for

C

ℓ

2

(

R

)

{\displaystyle C\ell\_{2} (\mathbf{R})}

 is:

1

e

x

e

y

e

x
y

e

x

1

e

x
y

e

y

e

y

−

e

x
y

1

−

e

x

e

x
y

−

e

y

e

x

−
1

{\displaystyle
\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:

(

1

e

x

e

y

e

x
y

e

x

1

e

x
y

e

y

e

y

e

x
y

1

e

x

e

x
y

e

y

e

x

1

)

(

1

e

x

e

y

e

x
y

)

=

(

1

⋅

1

e

x

e

x

e

y

e

y

e

x
y

e

x
y

e

x

1

1

e

x

e

x
y

e

y

e

y

e

x
y

e

y

1

e

x
y

e

x

1

e

y

e

x

e

x
y

e

x
y

1

e

y

e

x

e

x

e

y

1

e

x
y

)

=

(

1

1

1

−

1

e

x

e

x

e

x

−

e

x

e

y

−

e

y

e

y

e

y

e

x
y

−

e

x
y

e

x
y

e

x
y

)

{\displaystyle
\begin{align}

\left(
\begin{array}{rrrr}

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

\\
{\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}}

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

\end{array}
\right)

\left(
\begin{array}{rrrr}

{\color{red} 1 } \\
{\color{blue} e\_{x}} \\
{\color{blue} e\_{y}} \\
{\color{green} e\_{xy}}

\end{array}
\right) &=

\left(
\begin{array}{rrrr}

{\color{red} 1 } \cdot
{\color{red} 1 } &
+ &

{\color{blue} e\_{x}}
{\color{blue} e\_{x}} &
+ &
{\color{blue} e\_{y}}
{\color{blue} e\_{y}} &
+ &

{\color{green} e\_{xy}}
{\color{green} e\_{xy}}

\\
{\color{blue} e\_{x}}
{\color{red} 1 } &
+ &

{\color{red} 1 }
{\color{blue} e\_{x}} &
+ &
{\color{green} e\_{xy}}
{\color{blue} e\_{y}} &
+ &

{\color{blue} {e\_{y}}}
{\color{green} e\_{xy}}

\\
{\color{blue} e\_{y}}
{\color{red} 1 } &
+ &
{\color{green} e\_{xy}}
{\color{blue} e\_{x}} &
+ &
{\color{red} 1 }
{\color{blue} e\_{y}} &
+ &
{\color{blue} e\_{x}}
{\color{green} e\_{xy}}

\\
{\color{green} e\_{xy}}
{\color{red} 1 } &
+ &
{\color{blue} e\_{y}}
{\color{blue} e\_{x}} &
+ &
{\color{blue} {e\_{x}}}
{\color{blue} e\_{y}} &
+ &
{\color{red} {1}}
{\color{green} e\_{xy}}

\end{array}
\right) \\ &=

\left(
\begin{array}{llll}

 {\color{red}1} & + & {\color{red}1} & + & {\color{red}1} & - & {\color{red}1} \\
 {\color{blue}e\_{x}} & + & {\color{blue}e\_{x}} & + & {\color{blue}e\_{x}} & - & {\color{blue}e\_{x}} \\
 {\color{blue}e\_{y}} & - & {\color{blue}e\_{y}} & + & {\color{blue}e\_{y}} & + & {\color{blue}e\_{y}} \\
 {\color{green}e\_{xy}} & - & {\color{green}e\_{xy}} & + & {\color{green}e\_{xy}} & + & {\color{green}e\_{xy}}

\end{array}
\right)

\end{align}
}

Therefore to find the product

(

a
0

1

a
1

e

x

a
2

e

y

a
3

e

x
y

)
(

b
0

1

b
1

e

x

b
2

e

y

b
3

e

x
y

)

{\displaystyle
(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:

(

b
0

b
1

b
2

−

b
3

b
1

b
0

b
3

−

b
2

b
2

−

b
3

b
0

b
1

b
3

−

b
2

b
1

b
0

)

(

a
0

a
1

a
2

a
3

)

=

(

b
0

a
0

b
1

a
1

b
2

a
2

−

b
3

a
3

b
1

a
0

b
0

a
1

b
3

a
2

−

b
2

a
3

b
2

a
0

−

b
3

a
1

b
0

a
2

b
1

a
3

b
3

a
0

−

b
2

a
1

b
1

a
2

b
0

a
3

)

{\displaystyle

\left(
\begin{array}{rrrr}

{\color{red} b\_0 } &
{\color{blue} b\_1 } &
{\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)
}

Back to top

### Squares of pseudoscalars are either 1 or -1

In 0 dimensions:

(
1

)
2

=
1

{\displaystyle (1)^2 = 1}

In 1 dimension:

(

e

1

)
2

=

e

11

=
1

{\displaystyle (e\_{1})^2 = e\_{11} = 1}

In 2 dimensions:

(

e

12

)
2

=

e

1212

=
−

e

11

22

=
−
1

{\displaystyle (e\_{12})^2 = e\_{1212} = -e\_{11 \color{red}{22}} = -1}

In 3 dimensions:

(

e

123

)
2

=

e

123123

=

e

1212

33

=
−
1

{\displaystyle (e\_{123})^2 = e\_{123123} = e\_{1212 \color{red}{33}} = -1}

In 4 dimensions:

(

e

1234

)
2

=

e

12341234

=
−

e

123123

44

=
1

{\displaystyle (e\_{1234})^2 = e\_{12341234} = -e\_{123123 \color{red}{44}} = 1}

In 5 dimensions:

(

e

12345

)
2

=

e

1234512345

=

e

12341234

55

=
1

{\displaystyle (e\_{12345})^2 = e\_{1234512345} = e\_{12341234 \color{red}{55}} = 1}

In 6 dimensions:

(

e

123456

)
2

=

e

123456123456

=
−

e

1234512345

66

=
−
1

{\displaystyle (e\_{123456})^2 = e\_{123456123456} = -e\_{1234512345 \color{red}{66}} = -1}

In 7 dimensions:

(

e

1234567

)
2

=

e

12345671234567

=

e

123456123456

77

=
−
1

{\displaystyle (e\_{1234567})^2 = e\_{12345671234567} = e\_{123456123456 \color{red}{77}} = -1}

In 8 dimensions:

(

e

12345678

)
2

=

e

1234567812345678

=
−

e

12345671234567

88

=
1

{\displaystyle (e\_{12345678})^2 = e\_{1234567812345678} = -e\_{12345671234567 \color{red}{88}} = 1}

In 9 dimensions:

(

e

123456789

)
2

=

e

123456789123456789

=

e

1234567812345678

99

=
1

{\displaystyle (e\_{123456789})^2 = e\_{123456789123456789} = e\_{1234567812345678 \color{red}{99}} = 1}

Back to top

### Bivectors in higher dimensions

Multiplying a complex number by an imaginary number rotates the complex number 90° in the complex plane and shrinks or enlarges it.

To perform a general rotation without enlarging it requires

e

θ
i

=
cos
⁡

θ

i
sin
⁡

θ

,

{\displaystyle e^{\theta i} = \cos{\theta} + i \sin{\theta},}

Multipying a vector by a simple bivector rotates the vector 90° in the plane of the bivector and shrinks or enlarges it.

To perform a general rotation without enlarging it requires

e

θ
i

=

e

θ

e

12

=
cos
⁡

θ

e

12

sin
⁡

θ

,

{\displaystyle e^{\theta i} = e^{\theta\mathbf{e}\_{12}} = \cos{\theta} + \mathbf{e}\_{12}\sin{\theta},}

which when multiplied by any vector rotates it through an angle *θ* about the origin:

(

x
′

e

1

y
′

e

2

# )
(
x

e

1

y

e

2

)

e

i
θ

.

{\displaystyle (x'\mathbf{e}\_1 + y'\mathbf{e}\_2) = (x\mathbf{e}\_1 + y\mathbf{e}\_2)e^{i\theta}.}

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 Wikipedia: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

B

=

e

1

∧

e

2

e

3

∧

e

4

=

e

1

e

2

e

3

e

4

=

e

12

e

34

{\displaystyle \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.

(

e

12

)
2

=

e

1212

=
−
1

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

But:

(

e

12

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

{\displaystyle
\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}
}

Back to top

### Other quadratic forms

The square of a vector is:

v

v

=
(

v

x

e

x

v

y

e

y

)
(

v

x

e

x

v

y

e

y

)

=

v

x

v

x

e

x

e

x

v

x

v

y

e

x

e

y

v

y

v

x

e

y

e

x

v

y

v

y

e

y

e

y

=

v

x

v

x

(
1
)

v

y

v

y

(
1
)

v

x

v

y

e

x

e

y

−

v

y

v

x

e

x

e

y

=
(

v

x

v

x

v

y

v

y

)

(
1
)

(

v

x

v

y

−

v

y

v

x

)

e

x
y

=
(

v

x

2

v

y

2

)

(
1
)

(
0
)

e

x
y

=
(

v

x

2

v

y

2

)

(
1
)

=

scalar

{\displaystyle
\begin{align}
\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{align}
}

(

v

x

2

v

y

2

{\displaystyle v\_{x}^2 + v\_{y}^2}

) is called the quadratic formw. 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 Wikipedia: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

⋯

v
p
2

−

v

p
1

2

−
⋯
−

v

p
q

2

,

{\displaystyle 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 signature of the quadratic form. The real vector space with this quadratic form is often denoted **R***p*,*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 Cℓ0,*n*(**R**) depending on whether the author prefers positive-definite or negative-definite spaces.

A standard basisw {*e**i*} for **R***p*,*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* basis vectors that square to +1 and *q* basis vectors that square to −1.

From Wikipedia:Spacetime algebra:

Spacetime algebra (STA) is a name for the Clifford algebraw Cl1,3(**R**), or equivalently the geometric algebraw G(M4), which can be particularly closely associated with the geometry of special relativityw and relativistic spacetimew. See also Algebra of physical space.

The spacetime algebra may be built up from an orthogonal basis of one time-like vector

γ
0

{\displaystyle \gamma\_0}

 and three space-like vectors,

{

γ
1

,

γ
2

,

γ
3

}

{\displaystyle \{\gamma\_1, \gamma\_2, \gamma\_3\}}

, with the multiplication rule

γ
μ

γ
ν

γ
ν

γ
μ

=
2

η

μ
ν

{\displaystyle \gamma\_\mu \gamma\_\nu + \gamma\_\nu \gamma\_\mu = 2 \eta\_{\mu \nu} }

where

η

μ
ν

{\displaystyle \eta\_{\mu \nu}}

 is the Minkowski metricw with signature ( + − − − ).

Thus:

γ
0
2

=

1

{\displaystyle \gamma\_0^2 = {1}}

γ
1
2

=

γ
2
2

=

γ
3
2

=

−
1

{\displaystyle \gamma\_1^2 = \gamma\_2^2 = \gamma\_3^2 = {-1}}

γ
μ

γ
ν

=
−

γ
ν

γ
μ

{\displaystyle \gamma\_\mu \gamma\_\nu = - \gamma\_\nu \gamma\_\mu}

The basis vectors

γ
k

{\displaystyle \gamma\_k}

 share these properties with the Gamma matrices, but no explicit matrix representation need be used in STA.

C

ℓ

3
,
0

(

R

)

{\displaystyle C\ell\_{3,0} (\mathbf{R})}

:Algebra of physical spacew (Time = scalar)

C

ℓ

3
,
1

(

R

)

{\displaystyle C\ell\_{3,1} (\mathbf{R})}

:Spacetime algebraw (Time = vector)

Back to top

### Regressive product

The **regressive product** (usually referred to as the "meet") is the dual of the exterior product (or "join" in this context)

The dual specification of elements permits, for blades

A

{\displaystyle A}

 and

B

{\displaystyle B}

, the intersection (or meet) where the duality is to be taken relative to the smallest grade blade containing both

A

{\displaystyle A}

 and

B

{\displaystyle B}

 (the join).

C
∨
D
:=
(
(
C

I

−
1

)
∧
(
D

I

−
1

)
)
I

{\displaystyle
C \vee D := ((CI^{-1}) \wedge (DI^{-1}))I
}

Which in 2 or 3 dimensions is

C
∨
D
:=
(
(
−
C
I
)
∧
(
−
D
I
)
)
I

{\displaystyle
C \vee D := ((-CI) \wedge (-DI))I
}

with

I

{\displaystyle I}

 the unit pseudoscalar of the algebra. The regressive product, like the exterior product, is associative.

For example:

(
(

e
x

∧

e
y

)
∨
(

e
y

∧

e
z

)
)
I

=
(
(
−

e
x

∧

e
y

∧
I
)
∧
(
−

e
y

∧

e
z

∧
I
)
)
I

=
(
(
−

e
x

∧

e
y

∧

e
x

∧

e
y

∧

e
z

)
∧
(
−

e
y

∧

e
z

∧

e
x

∧

e
y

∧

e
z

)
)
I

=
(
(

e
z

)
∧
(

e
x

)
)
I

=

e
z

∧

e
x

∧
(

e
x

∧

e
y

∧

e
z

)

=
−

e
y

{\displaystyle
\begin{align}
((e\_x \wedge e\_y) \vee (e\_y \wedge e\_z)) I
&= ((-e\_x \wedge e\_y \wedge I) \wedge (-e\_y \wedge e\_z \wedge I)) I \\
&= ((-e\_x \wedge e\_y \wedge e\_x \wedge e\_y \wedge e\_z) \wedge (-e\_y \wedge e\_z \wedge e\_x \wedge e\_y \wedge e\_z)) I \\
&= ((e\_z) \wedge (e\_x))I \\
&= e\_z \wedge e\_x \wedge (e\_x \wedge e\_y \wedge e\_z) \\
&= -e\_y
\end{align}
}

Back to top