I saw this Identity in "QFT for the gifted amateur" when explaining non-Abelian gauge theory and I just don't get any intuition on why it's true, perhaps some geometric algebra intuition would clear it up for me:
$$(\vec{\tau} \cdot \mathbf{a})(\vec{\tau} \cdot \mathbf{b})=(\mathbf{a} \cdot \mathbf{b})+i \vec{\tau} \cdot \mathbf{a} \times \mathbf{b}\tag{46.19}$$
where $\vec{\tau} = (\sigma_1, \sigma_2, \sigma_3)$ is a vector whose three components are the Pauli matrices.
This is a confusing form of the Clifford-algebraic identity $ab = a\cdot b + a\wedge b$, where $a$ and $b$ are vectors. It's confusing because it mixes Clifford-algebraic and 3D-vector-algebraic operations.
$τ\cdot a$ is the injection of the vector $a$ into the Clifford algebra (sometimes written $/\!\!\!a$), so the left-hand side of the identity is just the Clifford product of the vectors $a$ and $b$.
In three dimensions, $a\wedge b$ is the Hodge dual of the cross product $a\times b$ (and vice versa). In the Clifford algebra, you can get the Hodge dual by multiplying by the pseudoscalar of the algebra. In odd-dimensional Clifford algebras, the pseudoscalar is central (commutes with everything). It squares to $+1$ or $-1$, depending on the metric signature. In the Pauli algebra, with signature $+{+}+$, it squares to $-1$. This means that you can identify it with the complex unit $i$, at least if you didn't introduce another $i$ by complexifying the algebra.
Putting that all together, $iτ\cdot(a\times b)$ means to take the cross product, convert it to a Clifford-algebra vector, then take its Hodge dual, which gets you the wedge product as a Clifford bivector.
If you want to know why $ab = a\cdot b + a\wedge b$, see, e.g., this answer.
