Lichnerowicz formula for a geometric Dirac operator

At $p$, ${\mathrm{\nabla }}^{LC}{e}_i=0$ and $[e_i,e_j]=0$. Compatibility [F2] gives ${\mathrm{\nabla }}_i^V(c(e^j))=0$ there, so expanding $D_V^2={\sum }_{i,j}{c}_i{c}_j{\mathrm{\nabla }}_i{\mathrm{\nabla }}_j$ yields diagonal terms $-{\sum }_i{\mathrm{\nabla }}_i^2$ and off-diagonal terms ${\sum }_{i<j}{c}_i{c}_j[{\mathrm{\nabla }}_i,{\mathrm{\nabla }}_j]$. The first term is $({\mathrm{\nabla }}^V{)}^{\ast }{\mathrm{\nabla }}^V$ at $p$; the commutators are $F_{ij}^V$. Since $p$ was arbitrary, $D_V^2=({\mathrm{\nabla }}^V{)}^{\ast }{\mathrm{\nabla }}^V+{\sum }_{i<j}{c}_i{c}_j{F}_{ij}^V$ globally.

Applying two covariant derivatives to ${\mathrm{\nabla }}_X^V(c(\xi )v)=c({\mathrm{\nabla }}_X^{LC}\xi )v+c(\xi ){\mathrm{\nabla }}_X^{V}v$ and antisymmetrizing in $X,Y$ gives $[F_{ij}^V,c(\xi )]=c(R_{ij}^{T^{\ast }M}\xi )$. Direct Clifford expansion gives $[c_a{c}_b,c_k]=2({\delta }_{ak}{c}_b-{\delta }_{bk}{c}_a)$; substituting the displayed coefficient $F_{ij}^S$ shows $[F_{ij}^S,c(\xi )]=c(R_{ij}^{T^{\ast }M}\xi )$ with the cotangent action induced by the fixed curvature convention. Subtracting proves $[F_{ij}^{V\mathrm{/}S},c(\xi )]=0$ for every $\xi$.

Substituting $F_{ij}^S$ gives ${\sum }_{i<j}{c}_i{c}_j{F}_{ij}^S=\frac{1}{8}{\sum }_{i,j,a,b}{\mathrm{Rm}}_{ijab}{c}_i{c}_j{c}_a{c}_b$. The Clifford relation expands $c_i{c}_j{c}_a{c}_b$ into exterior degrees $4,2,0$ as $c_i\wedge c_j\wedge c_a\wedge c_b-{\delta }_{ij}{c}_a\wedge c_b+{\delta }_{ia}{c}_j\wedge c_b-{\delta }_{ib}{c}_j\wedge c_a-{\delta }_{ja}{c}_i\wedge c_b+{\delta }_{jb}{c}_i\wedge c_a+{\delta }_{ij}{\delta }_{ab}-{\delta }_{ia}{\delta }_{jb}+{\delta }_{ib}{\delta }_{ja}$. The degree-four coefficient is the total alternation of ${\mathrm{Rm}}_{ijab}$ and is zero by the cyclic first-Bianchi identity [F3]. For the degree-two part, the ${\delta }_{ij}$ term vanishes by skewness in the first curvature pair; the other four contractions give $4{\sum }_i{\mathrm{Rm}}_{ipiq}$ as the coefficient of $c_p\wedge c_q$, which is symmetric in $p,q$ by pair interchange and therefore contracts to zero. The scalar coefficient is ${\sum }_{i,j,a,b}{\mathrm{Rm}}_{ijab}({\delta }_{ij}{\delta }_{ab}-{\delta }_{ia}{\delta }_{jb}+{\delta }_{ib}{\delta }_{ja})=2{\sum }_{i,j}{\mathrm{Rm}}_{ijji}=2{\mathrm{scal}}_g$: the first contraction vanishes and each of the last two equals the scalar contraction. Multiplication by $1\mathrm{/}8$ gives ${\mathrm{scal}}_g\mathrm{/}4$.

Substituting $F_{ij}^V=F_{ij}^S+F_{ij}^{V\mathrm{/}S}$ in step 1.1 and using step 3.1 gives the displayed formula. The principal symbol of $D_V$ at covector $\xi$ is $ic(\xi )$, so ${\sigma }_2(D_V^2)(\xi )=\mathrm{\mid }\xi {\mathrm{\mid }}^{2}I$ by [F1]. If $M$ is compact and boundaryless with $n\ge 2$, The geometric Dirac operator is globally well defined and formally symmetric supplies formal symmetry of $D_V$; therefore $D_V^2$ is formally symmetric and [F4] applies. In dimension zero, $T^{\ast }M=0$, so $D_V=0$ and the formula is direct. The empty and zero-rank cases have the zero operator and the same formula with all endomorphisms zero.