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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?