index free proof of dot product of two wedge products I am learning geometric algebra, and meet an identy of (edited according to Andrey's comments below)
$$  (a\wedge b)\cdot(c\wedge d) = (a \cdot d)(b\cdot c) - (a \cdot c)(b \cdot d)$$
as in wiki item of Binet–Cauchy identity.
How can one prove this without using indices?  Like start from geometric algebra context of
 $$a\cdot b=\dfrac {ab+ba}{2}=b\cdot a$$
and $$a\wedge b=\dfrac {ab-ba}{2}$$
 A: Since both sides are multilinear (linear in each variable when the rest are fixed), it is enough to check it for combinations of a basis.
There are not too many distinct cases: e.g. when $a=b$ (and say, $=e_i$), then we get $0$ on both sides. More cases to check: 
$\ (e_i\land e_j)\cdot (e_i\land e_j)$, $\ (e_i\land e_j)\cdot (e_i\land e_k)$, $\ (e_i\land e_j)\cdot (e_k\land e_l)$ (with distinct indices $i,j,k,l$).
A: Sure. You can argue it pretty well just from symmetry. As was said by Berci, it has to be linear in each term, and it has to be antisymmetric under exchange of $ a $ and $ b $ or $c$ and $d$. I'm not sure if you can show that the above is the only such relation. It certainly make me feel more comfortable with the identity at least. But, regardless, you can just work it out by hand.
$$ (a \wedge b) \cdot (c \wedge d) 
= Tr\left[\frac{(a \otimes b - b \otimes a)}{\sqrt{2}}\frac{(c \otimes d - d \otimes c)}{\sqrt{2}}\right]\\
= \frac{Tr[a \otimes d (b\cdot c) - a \otimes c (b\cdot d) + b \otimes c (a\cdot d) - b \otimes d (a \cdot c) ]}{2} \\
= (a \cdot d)(b \cdot c) - (a \cdot c) (b \cdot d)$$
A: Use the associativity of the geometric product and grade projection.
$$\begin{align*}\langle (ab)(cd) \rangle_0 = (a \cdot b) (c \cdot d) + (a \wedge b) \cdot (c \wedge d)\end{align*}$$
Using an $(ab)(cd)$ grouping, but grouping $(bc)$ instead gives
$$\langle a (bc) d \rangle_0 = (a \cdot d)(b \cdot c) + a \cdot [(b \wedge c) \cdot d]$$
Use the BAC-CAB rule on the last term to rewrite it as $(c \cdot d) (b \cdot a) - (b \cdot d)( c \cdot a)$.  The desired identity shortly follows--edit: and Andrey Sokolev is correct.
