# Calculation bivector (not based on my intuition)

I am trying to understand more about bivectors (2-blade). While a vector (1-blade) represents a line with length and a direction, a bivector represents a plane with an orientation. I have read that a bivector is produced with the wedge product.

$$a \wedge b = a_1 e_1 \wedge b_1 e_1 \\ + a_1 e_1 \wedge b_2 e_2 \\ + a_2 e_2 \wedge b_1 e_1 \\ + a_2 e_2 \wedge \ b_2 e_2$$

The commutative law give me: $$a \wedge b = a_1 b_1e_1 \wedge e_1 \\ + a_1 b_2 e_1 \wedge e_2 \\ + a_2 b_1 e_2 \wedge e_1 \\ + a_2 e_2 b_2 \wedge \ e_2 \\$$

But then I got stuck.

When I draw a picture it is clear to me that it is not possible to use the wedge product on the two basis vectors $e_1 \wedge e_1$. But is there it possible to understand this by calculation? It's a little awkward to do a calculation based on my intuition.

• $a\wedge a=0$... geometrically, the degenerate parallelogram is merely a line segment and thus has zero area. Commented Apr 22, 2013 at 22:09
• The original title of the question reminded me of this. Commented Apr 23, 2013 at 0:23
• I need to go to school now. When I am back I will study your answers. Commented Apr 23, 2013 at 6:26

Well $e_1\wedge e_1=0$ since they are collinear.

You can also view it this way:

$$\|e_1\wedge e_1\|=\|e_1\|\|e_1\||\sin(\theta)|$$ where $\theta$ is the angle between $e_1$ and $e_1$. In other words $\theta=0$ and thus $\sin(\theta)=0$ and $\|e_1\wedge e_1\|=0$.

Hence $e_1\wedge e_1=0$

Edit: The parallelogram explanation given by @oldrinb is good too !

It's slightly unclear what you mean, but I can give you the algebraic rules which the wedge product of two vectors obeys:

$$a \wedge b = -b \wedge a,\qquad (\lambda a + \lambda' a') \wedge (\mu b + \mu' b') = \lambda\mu (a \wedge b) + \lambda\mu' (a \wedge b') + \cdots$$

That is, it is alternating and bilinear.

For example, the first property tells you that

$$a \wedge a = -a \wedge a \quad \implies \quad a \wedge a = 0$$

So in your two-dimensional calculation, you get

$$a \wedge b = 0 + a_1 b_2 (e_1 \wedge e_2) + a_2 b_1 (e_2 \wedge e_1) + 0 = (a_1 b_2 - a_2 b_1) (e_1 \wedge e_2)$$

You can also look at bivectors in a real clifford algebra, called a "geometric" algebra. This is supported by a geometric product of vectors.

Geometric product rules: \begin{align*} e_1 e_1 &= e_2 e_2 = 1 \\ e_1 e_2 &= -e_2 e_1 \\ (a + b)c &= ac + bc \\ abc &= (ab)c = a(bc)\end{align*}

This makes it possible to evaluate $ab$ as

\begin{align*}ab = (a_1 e_1 + a_2 e_2)(b_1 e_1 + b_2 e_2) &= a_1 b_1 e_1 e_1 + a_2 b_2 e_2 e_2 + a_1 b_2 e_1 e_2 + a_2 b_1 e_2 e_1 \\ &= a_1 b_1 (1) + a_2 b_2 (1) + (a_1 b_2 - a_2 b_1) e_1 e_2 \\ &= a \cdot b + a \wedge b\end{align*}

That $e_1 \wedge e_1$ does follow from antisymmetry of the wedge product, sure, but with the power of the geometric product, you can think in a more generalized fashion. Parallel vectors can't form higher-dimensional objects like areas, volumes and so on. Instead, they combine and annihilate each other, forming scalars, telling us how parallel vectors are (or how much two bivectors are oriented in the same direction, etc.).

All answers are quite good and pointed me in a direction that anwsered my qustion. So thanks for the help.

I have read that the length of a bivector equals.

$$\ |B| = |a| |b| \sin \theta$$

I took a pen and paper and drawed two lines perpendicular to eachother and calculated the sine of it. What I concluded was that the result was 1. When the angle between two lines equals zero degrees than the result of the sine equals zero. So this implies that the area of the wedge product on the same vector equals zero. Therefore it is not possible to produce a bivector, ofcourse that is the goal of the wedge product/ exterior product. That is why $$\ a \wedge a =0$$.

http://en.wikipedia.org/wiki/Bivector#Geometric_interpretation