Geometric product of 2 r-blades

Question

In some lecture notes the geometric product of a $r$-blade $A_r$ and an $s$-blade $B_s$ is given as:

$A_r B_s = \langle A_r B_s\rangle_{|r-s|} + \langle A_r B_s\rangle_{|r-s|+2} +... + \langle A_r B_s\rangle_{r+s}$

where dot means the lowest and wedge means the highest terms in this expansion:

$A_r \cdot B_s = \langle A_r B_s\rangle_{|r-s|}\\ A_r \wedge B_s = \langle A_r B_s\rangle_{r+s}$

So if I have $A_2 = e_1 \wedge e_2$ and $B_2 = b e_1 \wedge e_2$ where $b$ is a scalar and $I_2$ means the pseudoscalar $(I_2^2=-1)$ I get:

$ \overbrace{(e_1 \wedge e_2)(e_1 \wedge e_2)}^{I_2^2=-1}=\overbrace{\Big\langle(e_1 \wedge e_2)(e_1 \wedge e_2)\Big\rangle_0}^{=1} +\Big\langle(e_1 \wedge e_2)(e_1 \wedge e_2)\Big\rangle_{2}+\overbrace{\Big\langle(e_1 \wedge e_2)(e_1 \wedge e_2)\Big\rangle_{4}}^{e_1\wedge e_2 \wedge e_1\wedge e_2=0}$

My question is now what is this middle term in the expansion and is this expansion even true when $r=s$, because then this middle term would need to give $-2$, which seems very weird. Or is the problem that $A$ and $B$ are basically the same blade up to a scalar? Where are my mistakes here?

Answer 1

I'm assuming $e_1$ and $e_2$ are orthonormal, so $e_1\wedge e_2=e_1e_2$.

Let $X=(e_1\wedge e_2)(e_1\wedge e_2)$. You've noted on the left side of your last equation that this is

$$X=(e_1e_2)(e_1e_2)=e_1e_2e_1e_2=e_1(e_2e_1)e_2=e_1(-e_1e_2)e_2=-(e_1e_1)(e_2e_2)=-(1)(1)$$ $$=-1.$$

This is a scalar, i.e. a $0$-vector, so $\langle X\rangle_0=X=-1$. It has no higher-grade parts, so $\langle X\rangle_2=\langle X\rangle_4=0$.

But then on the right side of your last equation you claim that $\langle X\rangle_0=+1$. Where did this come from? Have you been given several definitions of the dot product of two blades? In fact we have $(e_1\wedge e_2)\cdot(e_1\wedge e_2)=-1$, not $+1$.

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The expansion in your first equation is correct, for any $r$ and $s$, for any $A=\langle A\rangle_r$ and $B=\langle B\rangle_s$. (But when $s=1$ there are only two terms, grades $r-1$ and $r+1$. And when $s=0$ there is only one term, grade $r$.) That expansion is not a definition, though. The geometric product is often defined implicitly, by the properties it's supposed to have.

See this answer of mine for an alternative, explicit definition, though it takes some work to show that it's associative, and that it doesn't really depend on the basis, etc.