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This question was posed on Math.SE but no one has answered it; it may be suitable for MathOverflow.

Let $D$ be the Dirac operator on $\mathbb{R}^n$ i.e. $D=\sum_{j=1}^nE_j\frac{\partial}{\partial x_j}$ where $E_j$ are $2^r \times 2^r$ matrices (where $n=2r$ or $n=2r+1$) satisfying $E_i^2=I, \ \ E_iE_j+E_jE_i=0$.

Let $spin(n)$ be a spin group (it is a double cover of $SO(n))$. It acts on $\mathbb{R}^n$ via $g \cdot v=gvg^{-1}$ where the multiplication takes place in Clifford algebra. It also acts on $\mathbb{C}^{2^r}$ via $g \cdot \xi:=c(g)(\xi)$ where $c:\mathbb{R}^n \to End(\mathbb{C}^{2^r})$ is defined by $c(v):=\sum_{j=1}^nv_jE_j$ and extended to Clifford algebra (in particular to spin group). It also acts on $End(\mathbb{C}^{2^r})$ via $g \cdot T:=c(g)Tc(g)^{-1}$.

One can show that $c$ is $spin(n)$ equivariant. But $c$ is nothing else than the symbol of $D$ and $D$ is constant coefficient operator.

Why this is enough to claim that the operator $D$ itself is $spin(n)$ equivariant?

Improve this question
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  • $\begingroup$ It is right that given your $spin(n)$ actions the symbol is equivariant as a map $\mathbb R ^n \to End( \mathbb C ^{2^r})$. Dirac is then equivariant by considering the symbol as a map $C^{\infty}(\mathbb R ^n ; \mathbb C ^{2^r}) \to C^\infty(\mathbb R ^n; \mathbb C ^{2^r})$ but with the corresponding $spin(n)$ action : via $spin(n) \to SO(n)$ as you decribed for the first occurence of $\mathbb R^n$ but trivial for the second one. $Spin(n)$ also acts via Clifford on $\mathbb C ^{2^r}$ : $D(g.f\circ g^{-1}) = g.(Df)$ $\endgroup$
    – InfiniteLooper
    Nov 16, 2017 at 0:58
  • $\begingroup$ Could you please rewrite it as an answer? I would also like to see some more details (how these actions are defined? Why the equivariance follows for Dirac once we have it for symbol? I'm asking for a standard answer not only for me but also in order to be able to award the bounty $\endgroup$
    – truebaran
    Nov 22, 2017 at 23:40
  • $\begingroup$ Ok, can you write what the symbol is equivariant means ? It will be pretty much the same for the operator. $\endgroup$
    – InfiniteLooper
    Nov 24, 2017 at 8:39

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