where $a\in Cl^2(V)$, and $\tau(a)(v)=[a,v]$ for $v\in V$. I was wondering is the first equality correct as written? Jost earlier defined the clifford algebra as the quotient of the tensor algebra by ideal generated by the relations $v\otimes v+\|v\|^2=0$, which yields $$-2\langle v,w\rangle=vw+wv$$ upon polarizing. Back to the computation, I'm computing $$\langle \tau(a)v,w\rangle=-\frac{1}{2}(\tau(a)vw+w\tau(a)v)$$ while $$-\frac{1}{2}[[a,v],w]=-\frac{1}{2}[\tau(a)v,w]=-\frac{1}{2}(\tau(a)vw-w\tau(a)v)$$
These differ by a sign. Did I make an error somewhere?
This is a follow up on the comment sequence above.
The following are taken from Heat Kernels and Dirac Operators.
On page 39,
(2) is "the Jacobi identity." To verify it (in the case of a the bracket being the [super]commutator of multiplication), use that the parity of a product is the sum $\pmod 2$ of that of the individual terms:
$$ |a\cdot b| = |a| + |b| $$ (assuming that parity makes sense - i.e., $a$ and $b$ are 'pure').
Next, on page 103,
Finally, on page 105,
The reason that $[a,[v,w]] = 0$ is that $[v,w] = vw +wv = -2 Q(v,w)\in {\mathbb R}$, and so in the center.