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Let $E$ be a Clifford module with Clifford multiplication $c$. On page 117 of Heat Kernels and Dirac Operators it is claimed that "any operator satisfying \begin{equation}\tag{1} \forall f \in C^\infty (M):[D, f] = c(df) \end{equation} is a Dirac operator" (i.e. such a $D$ is odd and $D^2$ is a generalized Laplacian). The key observation is the following identity: \begin{equation}\tag{2} \forall f \in C^\infty (M):[[D^2,f],f]=2[D,f]^2 \end{equation} I tried to simply write out the commutators and repeatedly use $(1)$, but I soon realized that I would be facing a crazy amount of terms and according to the authors there is a smarter way to prove $(2)$:

They say that to derive $(2)$ they "have used the fact that that $[[D, f], f] = 0$ and that $[D, [D, A]] = [D^2, A]$ for any odd operator $D$ and even operator $A$".

I do not see how the last two inline equations help us to derive $(2)$, can someone perhaps help me put the pieces together?

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Note that for an odd operator $B$ we have $[B, B] = BB - (-1)^{|B||B|}BB = 2B^2$. Therefore, we can deduce $(2)$ as follows:

\begin{align*} [[D^2, f], f] &= [[D, [D, f]], f]\\ &=-[f, [D, [D, f]]]\\ &= -[[f, D], [D, f]] - [D, [f, [D, f]]]\\ &= [[D, f], [D, f]] + [D, \underbrace{[[D, f], f]}_{0}]\\ &= 2[D, f]^2. \end{align*}

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  • $\begingroup$ Another way to interpret this calculation is as an application of the super-Jacobi identity for super-commutators. $\endgroup$
    – Branimir Ćaćić
    Commented Nov 21, 2023 at 11:42
  • $\begingroup$ @BranimirĆaćić: I used the super-Jacobi identity in the third equality. Do you mean the other version of the super-Jacobi identity (i.e. the first one here)? $\endgroup$
    – Michael Albanese
    Commented Nov 21, 2023 at 12:31
  • $\begingroup$ Since $D$ is odd and $f$ is even we can replace all super-commutators with commutators and anti-commutators and we obtain a proof that also works if $D$ is not odd (or if no grading is given). This is relevant because $(1)$ alone does not imply that $D$ is odd. $\endgroup$
    – Filippo
    Commented Nov 22, 2023 at 1:38

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