Cross Product
Cross product of two vectors
\( \vec{\mathbf{u}} = \langle u_1, u_2, u_3 \rangle \)
\( \vec{\mathbf{v}} = \langle v_1, v_2, v_3 \rangle \)
is originaly defined in three dimensions as
\(
\vec{\mathbf{u}} \times \vec{\mathbf{v}} = \begin{vmatrix}
\vec{\mathbf{i}} & \vec{\mathbf{j}} & \vec{\mathbf{k}} \\
u_1 & u_2 & u_3 \\
v_1 & v_2 & v_3
\end{vmatrix}
=
(u_2 v_3 - u_3 v_2) \vec{\mathbf{i}} +
(u_3 v_1 - u_1 v_3) \vec{\mathbf{j}} +
(u_1 v_2 - u_2 v_1) \vec{\mathbf{k}}
\)
This is a vector.
\( \vec{\mathbf{u}} = \langle u_1, u_2 \rangle \)
\( \vec{\mathbf{v}} = \langle v_1, v_2 \rangle \)
To find a two-dimensional definition, we first consider them as 3-vectors,
\(
\vec{\mathbf{u}} \times \vec{\mathbf{v}} = \begin{vmatrix}
\vec{\mathbf{i}} & \vec{\mathbf{j}} & \vec{\mathbf{k}} \\
u_1 & u_2 & 0 \\
v_1 & v_2 & 0
\end{vmatrix}
=
(u_1 v_2 - u_2 v_1) \vec{\mathbf{k}}
\)
Then we return to two dimensions
\(
\vec{\mathbf{u}} \times \vec{\mathbf{v}} = u_1 v_2 - u_2 v_1
\)
This is a scalar.
\( \vec{\mathbf{u}} = \langle u_1, u_2, u_3, u_4 \rangle \)
\( \vec{\mathbf{v}} = \langle v_1, v_2, v_3, v_4 \rangle \)
\(
\vec{\mathbf{u}} \times \vec{\mathbf{v}} =
\begin{vmatrix}
\vec{\mathbf{e_1}} & \vec{\mathbf{e_2}} & \vec{\mathbf{e_3}} & \vec{\mathbf{e_4}} \\
\vec{\mathbf{e_1}} & \vec{\mathbf{e_2}} & \vec{\mathbf{e_3}} & \vec{\mathbf{e_4}} \\
u_1 & u_2 & u_3 & u_4 \\
v_1 & v_2 & v_3 & v_4
\end{vmatrix}
\)
Since the tensor product is not commutative, this pseudo-determinant is not zero.
\(
\begin{matrix}
+(u_3 v_4-u_4 v_3) (\vec{\mathbf{e_1}} \otimes \vec{\mathbf{e_2}})
+(u_4 v_2-u_2 v_4) (\vec{\mathbf{e_1}} \otimes \vec{\mathbf{e_3}})
+(u_2 v_3-u_3 v_2) (\vec{\mathbf{e_1}} \otimes \vec{\mathbf{e_4}})
+(u_1 v_4-u_4 v_1) (\vec{\mathbf{e_2}} \otimes \vec{\mathbf{e_3}})
+(u_3 v_1-u_1 v_3) (\vec{\mathbf{e_2}} \otimes \vec{\mathbf{e_4}})
+(u_1 v_2-u_2 v_1) (\vec{\mathbf{e_3}} \otimes \vec{\mathbf{e_4}})
\\
-(u_3 v_4-u_4 v_3) (\vec{\mathbf{e_2}} \otimes \vec{\mathbf{e_1}})
-(u_4 v_2-u_2 v_4) (\vec{\mathbf{e_3}} \otimes \vec{\mathbf{e_1}})
-(u_2 v_3-u_3 v_2) (\vec{\mathbf{e_4}} \otimes \vec{\mathbf{e_1}})
-(u_1 v_4-u_4 v_1) (\vec{\mathbf{e_3}} \otimes \vec{\mathbf{e_2}})
-(u_3 v_1-u_1 v_3) (\vec{\mathbf{e_4}} \otimes \vec{\mathbf{e_2}})
-(u_1 v_2-u_2 v_1) (\vec{\mathbf{e_4}} \otimes \vec{\mathbf{e_3}})
\end{matrix}
\)
Actually, we can use the wedge product
\(\vec{\mathbf{u}} \wedge \vec{\mathbf{v}} = \vec{\mathbf{u}} \otimes \vec{\mathbf{v}}-\vec{\mathbf{v}} \otimes \vec{\mathbf{u}} \)
to write down the final result as:
\(
(u_3 v_4-u_4 v_3) (\vec{\mathbf{e_1}} \wedge \vec{\mathbf{e_2}})
+(u_4 v_2-u_2 v_4) (\vec{\mathbf{e_1}} \wedge \vec{\mathbf{e_3}})
+(u_2 v_3-u_3 v_2) (\vec{\mathbf{e_1}} \wedge \vec{\mathbf{e_4}})
+(u_1 v_4-u_4 v_1) (\vec{\mathbf{e_2}} \wedge \vec{\mathbf{e_3}})
+(u_3 v_1-u_1 v_3) (\vec{\mathbf{e_2}} \wedge \vec{\mathbf{e_4}})
+(u_1 v_2-u_2 v_1) (\vec{\mathbf{e_3}} \wedge \vec{\mathbf{e_4}})
\)
This is a 2-tensor.
\( \vec{\mathbf{u}} = \langle u_1, u_2, u_3, u_4, u_5 \rangle \)
\( \vec{\mathbf{v}} = \langle v_1, v_2, v_3, v_4, v_5 \rangle \)
In five dimensions where triple wedge product is
\(
\vec{\mathbf{u}} \wedge \vec{\mathbf{v}} \wedge \vec{\mathbf{w}} =
\vec{\mathbf{u}} \otimes \vec{\mathbf{v}} \otimes \vec{\mathbf{w}}
-\vec{\mathbf{u}} \otimes \vec{\mathbf{w}} \otimes \vec{\mathbf{v}}
+\vec{\mathbf{v}} \otimes \vec{\mathbf{w}} \otimes \vec{\mathbf{u}}
-\vec{\mathbf{v}} \otimes \vec{\mathbf{u}} \otimes \vec{\mathbf{w}}
+\vec{\mathbf{w}} \otimes \vec{\mathbf{u}} \otimes \vec{\mathbf{v}}
-\vec{\mathbf{w}} \otimes \vec{\mathbf{v}} \otimes \vec{\mathbf{u}}
\)
we have
\(
\vec{\mathbf{u}} \times \vec{\mathbf{v}} =
\begin{vmatrix}
\vec{\mathbf{e_1}} & \vec{\mathbf{e_2}} & \vec{\mathbf{e_3}} & \vec{\mathbf{e_4}} & \vec{\mathbf{e_5}} \\
\vec{\mathbf{e_1}} & \vec{\mathbf{e_2}} & \vec{\mathbf{e_3}} & \vec{\mathbf{e_4}} & \vec{\mathbf{e_5}} \\
\vec{\mathbf{e_1}} & \vec{\mathbf{e_2}} & \vec{\mathbf{e_3}} & \vec{\mathbf{e_4}} & \vec{\mathbf{e_5}} \\
u_1 & u_2 & u_3 & u_4 & u_5 \\
v_1 & v_2 & v_3 & v_4 & v_5
\end{vmatrix} =
\begin{matrix}
+(u_1 v_2-u_2 v_1) (\vec{\mathbf{e_3}} \wedge \vec{\mathbf{e_4}} \wedge \vec{\mathbf{e_5}})
-(u_3 v_1-u_1 v_3) (\vec{\mathbf{e_2}} \wedge \vec{\mathbf{e_4}} \wedge \vec{\mathbf{e_5}}) \\
+(u_1 v_4-u_4 v_1) (\vec{\mathbf{e_2}} \wedge \vec{\mathbf{e_3}} \wedge \vec{\mathbf{e_5}})
-(u_1 v_5-u_5 v_1) (\vec{\mathbf{e_2}} \wedge \vec{\mathbf{e_3}} \wedge \vec{\mathbf{e_4}}) \\
+(u_2 v_3-u_3 v_2) (\vec{\mathbf{e_1}} \wedge \vec{\mathbf{e_4}} \wedge \vec{\mathbf{e_5}})
-(u_2 v_4-u_4 v_2) (\vec{\mathbf{e_1}} \wedge \vec{\mathbf{e_3}} \wedge \vec{\mathbf{e_5}}) \\
+(u_2 v_5-u_5 v_2) (\vec{\mathbf{e_1}} \wedge \vec{\mathbf{e_3}} \wedge \vec{\mathbf{e_4}})
+(u_3 v_4-u_4 v_3) (\vec{\mathbf{e_1}} \wedge \vec{\mathbf{e_2}} \wedge \vec{\mathbf{e_5}}) \\
-(u_3 v_5-u_5 v_3) (\vec{\mathbf{e_1}} \wedge \vec{\mathbf{e_2}} \wedge \vec{\mathbf{e_4}})
+(u_4 v_5-u_5 v_4) (\vec{\mathbf{e_1}} \wedge \vec{\mathbf{e_2}} \wedge \vec{\mathbf{e_3}})
\end{matrix}
\)
This is a 3-tensor.
1773 Lagrange algebraic formula
1843 Hamilton quaternion product, scalar, vector
1844 Grassmann bi-vector, wedge product
1881 Gibbs cross product