Dtjytyk

The math book i'm using states that the cross product for two vectors is defined over $R^3$:

$$u = (a,b,c)$$
$$v = (d,e,f)$$

is:

$$u times v = begin{vmatrix}
hat{i} & hat{j} & hat{k} \
a & b & c \
d & e & f \
end{vmatrix}
$$

and the direction of the resultant is determined by curling fingers from vector v to u with thumb pointing in direction of the cross product of u x v.

Out of curiosity, what's the cross product if u and v are defined over $R^2$ instead of $R^3$ instead:

$$u = (a,b)$$
$$v = (d,e)$$

Is there a "degenerate" case for the cross product of $R^2$ instead $R^3$? like this is some type of 2x2 determinant instead?

for instance if had a parameterization:

$$Phi(u, v) = ( f(u), g(v) )$$

and needed to calculate in $R^2$:

$$
D = Bigg| frac{partial{Phi}}{partial{u}} times frac{partial{Phi}}{partial{v}} Bigg|
$$

There are plenty of examples in the book for calculating the determinate D in $R^3$ but none at all for $R^2$ case.

As in:

$$
iint_{V} f(x,y) dx dy = iint_{Q} f(Phi(u,v) Bigg| frac{partial{Phi}}{partial{u}} times frac{partial{Phi}}{partial{v}} Bigg|
$$

$$
Phi(u,v)=(2u cos v, u sin v)
$$

In $n$ dimensions, the Levi-Civita symbol has $n$ indices, two of which contract with those of two vectors whose wedge product is sought in the geometry's exterior algebra. This obtains an object with $n-2$ indices, so whereas we get a familiar vector if $n=3$, we get a scalar if $n=2$. In particular, $binom{a}{b}landbinom{c}{d}=ad-bc$.

The cross product in 2 dimensions is a scalar give my a 2x2 determinant:

$$
(a, b) times (c, d) = begin{vmatrix}
a & b \
c & d \
end{vmatrix} = ad - cb
$$

The cross product in 3 dimensions is a vector given by the 3x3 determinant:

$$
(a, b, c) times (d, e, f) = begin{vmatrix}
e_x & e_y & e_z \
a & b & c \
d & e & f \
end{vmatrix}
$$

E. A. Abbott describes a 2D cross product nicely in his mathematical fantasy book "Flatland":

Flatland describes life and customs of people in a 2-D world: in this universe vectors can be summed together and projected, areas are calculated, rotations are clock-wise or counter clock-wise, reflection is possible... but cross product does not exist; otherwise, 2-D inhabitants should have great fantasy to imagine a 3rd dimension to contain a vector orthogonal to their plane.

By the way, in 2-D a single scalar number is sufficient to describe a force’s moment.

$$
M(vec{r}, vec{F}) = r_1 F_2 - r_2 F_2
$$

With such a definition, this operation respects all algebraic properties of cross product, but the result is a scalar.