I've been self studying Doran and Lasenby's Geometric Algebra for Physicists, and I'm getting stuck on a small derivation in section 4.4.3 about the adjoint of linear functions. It's not big enough to keep me from proceeding through the rest of the book, but it's annoying the heck out of me that I can't figure it out. Especially because it's probably something exceedingly simple.

The derivation is supposed to show that the defining property of the adjoint ($a \cdot \bar{F}(b)=F(a) \cdot b$) also applies when a linear function acts on bivectors:

$$ (a_1 \wedge a_2) \cdot F(b_1 \wedge b_2)=a_1 \cdot F(b_2)a_2 \cdot F(b_1)-a_1 \cdot F(b_1)a_2 \cdot F(b_2)$$ $$ =\bar{F}(a_1) \cdot b_2\bar{F}(a_2) \cdot b_1-\bar{F}(a_1) \cdot b_1\bar{F}(a_2) \cdot b_2=\bar{F}(a_1 \wedge a_2) \cdot (b_1 \wedge b_2). $$

Now, I understand how the function F acting on the bivector got split up (since $F(a \wedge b)=F(a) \wedge F(b)$), but I don't understand anything else about the first step. What relationship is being applied here? I think it might be similar to something derived way back in section 4.1.2 about the inner product of a vector with an arbitrary r-blade:

$$ a \cdot (a_1 \wedge a_2 \wedge \dots \wedge a_r)=\sum_{k=1}^r (-1)^{k+1} a \cdot a_k a_1 \wedge \dots \wedge \check{a}_k \wedge \dots \wedge a_r, $$

but of course, in this case I'm dealing with the inner product of two different blades.


1 Answer 1


The requisite identity can be derived through grade projection. For brevity, I will refer to $f_1 = F(b_1)$ and $f_2 = F(b_2)$. See that

$$\langle (a_1 a_2)( f_1 f_2) \rangle_0 = (a_1 \wedge a_2) \cdot (f_1 \wedge f_2) + (a_1 \cdot a_2)(f_1 \cdot f_2)$$

But using a different associative grouping, we get

$$\langle [(a_1 a_2)f_1]f_2 \rangle_0 = [(a_1 \wedge a_2) \cdot f_1] \cdot f_2 + (a_1 \cdot a_2)(f_1 \cdot f_2)$$

Thus we conclude the first terms in both expressions are equivalent. You can then apply the formula you have for a 1-blade and an $r$-blade dotted together, but for this case, it's really just the GA version of the BAC-CAB rule:

$$(a_1 \wedge a_2) \cdot f_1 = a_1 (a_2 \cdot f_1) - a_2 (a_1 \cdot f_1)$$

Edit: the derived identity $(a \wedge b) \cdot (c \wedge d) = ([a \wedge b] \cdot c) \cdot d$ is itself quite handy to know for many kinds of problems.

  • $\begingroup$ Ah, that's perfect. I suppose I need to develop a more intuitive sense for grade projection. I understand how it works, but I never think of it as a useful tool. Thanks! $\endgroup$
    – Itserpol
    Jan 9, 2014 at 0:21
  • $\begingroup$ It is an immensely useful tool, and this method I used here is my preferred method for deriving useful algebraic identities, as well as geometric calculus identities. It's really the associativity that makes it all work; grade projection is just a fancy way of saying, "I want to look at the grade-$k$ component," the same way when two vectors are equal, each of the components are equal, and you often use that in vector algebra. - Ironically, though, the BAC-CAB identity is actually very tedious to derive this way, as you have to do a cycle of 3 permutations. $\endgroup$
    – Muphrid
    Jan 9, 2014 at 2:42

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