# Bivector Wedge Bivector

I'm struggling with bivector wedge bivector. If i do via the formula $$A\wedge B=\frac12(AB-BA)$$ I get the correct answer but i can't do it directly. For Example, $$A=\hat i\hat j+\hat j\hat k+\hat k\hat i$$ $$B=\hat i\hat j+2\hat j\hat k+3\hat k\hat i$$

$$A\wedge B=(\hat i\hat j+\hat j\hat k+\hat k\hat i)\wedge (\hat i\hat j+2\hat j\hat k+3\hat k\hat i)$$ $$A\wedge B=(\hat i\wedge\hat j+\hat j\wedge\hat k+\hat k\wedge\hat i)\wedge (\hat i\wedge\hat j+2\hat j\wedge\hat k+3\hat k\wedge\hat i)$$

$$A\wedge B=\hat i\wedge\hat j\wedge\hat i\wedge\hat j+2\hat i\wedge\hat j\wedge\hat j\wedge\hat k+3\hat i\wedge\hat j\wedge\hat k\wedge\hat i+$$ $$\hat j\wedge\hat k\wedge\hat i\wedge\hat j+2\hat j\wedge\hat k\wedge\hat j\wedge\hat k+3\hat j\wedge\hat k\wedge\hat k\wedge\hat i+$$ $$\hat k\wedge\hat i\wedge\hat i\wedge\hat j+2\hat k\wedge\hat i\wedge\hat j\wedge\hat k+3\hat k\wedge\hat i\wedge\hat k\wedge\hat i$$

Since $$a\wedge a$$ is $$0$$, $$A\wedge B$$ is also $$0$$

But, $$AB=(\hat i\hat j+\hat j\hat k+\hat k\hat i)(\hat i\hat j+2\hat j\hat k+3\hat k\hat i)$$

$$AB=\hat i\hat j\hat i\hat j+2\hat i\hat j\hat j\hat k+3\hat i\hat j\hat k\hat i+$$ $$\hat j\hat k\hat i\hat j+2\hat j\hat k\hat j\hat k+3\hat j\hat k\hat k\hat i+$$ $$\hat k\hat i\hat i\hat j+2\hat k\hat i\hat j\hat k+3\hat k\hat i\hat k\hat i$$

$$AB=-1-2\hat k\hat i+3\hat j\hat k+$$ $$\hat k\hat i-2-3\hat i\hat j+$$ $$-\hat j\hat k+2\hat i\hat j-3$$

$$AB=-6-\hat i\hat j+2\hat j\hat k-\hat k\hat i$$ Similarly, $$BA=-6+\hat i\hat j-2\hat j\hat k+\hat k\hat i$$ So $$A\wedge B$$ is $$-\hat i\hat j+2\hat j\hat k-\hat k\hat i$$

What did i do wrong in the first example?

Probably the second result is right because bivector is a pseudovector in 3d and $$IA$$ is a vector where $$I=\hat i\hat j\hat k$$. Since vectro wedge vector is a bivector, $$IA\wedge IB=I^2A\wedge B [Since I^2=-1]$$ $$IA\wedge IB=-A\wedge B=B\wedge A$$ which implies bivector wedge bivector is also a bivector

• I am not sure what i,j,k are, but is your wedge product associative? It satisfies the Jacobi identity at most, in general. Aug 22, 2022 at 17:11
• Because you don't give the reasons for each of your steps and you don't number them, we can't really help you. Also because you don't define your notation. Aug 22, 2022 at 17:18
• $\hat i$, $\hat j$ and $\hat k$ are unit vectors along $x$, $y$ and $z-axes$ respectively Aug 23, 2022 at 1:12

$$\newcommand\proj[1]{\langle#1\rangle}$$
Because your formula $$A\wedge B = \frac12(AB - BA)$$ is incorrect if $$A, B$$ are bivectors. This just isn't how the wedge product works. In fact, it can't work like this: even-grade multivectors commute with all other multivectors under the wedge product, precisely because they are made up of an even number of vectors.
The first way you computed the wedge product which resulted in $$A\wedge B = 0$$ is correct, and in fact we must get $$0$$ since $$A$$ and $$B$$ live in the same 3D space which can't have a 4D component (i.e. a 4-vector).
What is true is that if $$v$$ is a vector and $$X$$ is any multivector then $$v\wedge X = \frac12(vX + \hat Xv)$$ where $$\hat X$$ is the grade involution of $$X$$, i.e. odd-grade components are negated. We also get the reversed formula $$X\wedge v = \frac12(Xv + v\hat X).$$ It follows that if $$A_r$$ is homogeneous with odd grade $$r$$ then $$v\wedge A_r = \frac12(vA_r - A_rv).$$ In particular for vectors $$v,w$$ we have the formula you were likely trying to mimic $$v\wedge w = \frac12(vw - wv).$$
What we could do for bivectors, if you were so inclined, is note that $$AB = \proj{AB}_0 + \proj{AB}_2 + \proj{AB}_4,$$ from which it follows that $$\frac12(AB + BA) = \proj{AB}_0 + \proj{AB}_4 = A\cdot B + A\wedge B,$$$$\frac12(AB - BA) = \proj{AB}_2 = A\times B.$$ We see that $$\frac12(AB-BA)$$ isn't even related to the wedge product at all. But we should expect this: even-grade multivectors commute under the wedge product. What we can say for bivectors $$A$$ and $$B$$ clearly reflects this: $$A\wedge B = \frac12(AB + BA) - A\cdot B.$$
• I have a doubt. What is the term $\frac12(AB-BA)=A\times B$? What is the cross product?($\times$) Aug 23, 2022 at 1:09
• @JRBros In a geometric algebra context (which I assume is where you're coming from), we usually define $X\times Y := \frac12(XY - YX)$ for arbitrary multivectors and call it the "commutator product". So when I use it above I'm not saying anything important, just directly applying this definition. The commutator product with a bivector $B$ is particularly important since it's how Lie algebras are represented with bivectors, and since $$B\times\langle X\rangle_r = \langle B\times X\rangle_r,$$ i.e. commutators with bivectors preserve grade. Aug 23, 2022 at 3:31