The geometric Dirac operator is globally well defined and formally symmetric

On an overlap let $e_i^{\mathrm{\prime }}={\sum }_j{a}_{ij}{e}_j$ be a change of oriented orthonormal frame, so $A=(a_{ij})$ is orthogonal. The dual coframe changes by the same orthogonal coefficient matrix with the corresponding inverse convention, while ${\mathrm{\nabla }}_{e_i^{\mathrm{\prime }}}={\sum }_k{a}_{ik}{\mathrm{\nabla }}_{e_k}$. Therefore ${\sum }_{i}c(e^{\mathrm{\prime }i}){\mathrm{\nabla }}_{e_i^{\mathrm{\prime }}}={\sum }_{i,j,k}{a}_{ij}{a}_{ik}c(e^j){\mathrm{\nabla }}_{e_k}={\sum }_{j}c(e^j){\mathrm{\nabla }}_{e_j}$ by ${\sum }_i{a}_{ij}{a}_{ik}={\delta }_{jk}$. Each chart expression has smooth coefficients, and these overlap equalities glue them to one global first-order operator.

If $\dim M=0$, the defining sum for $D_V$ is empty, so $D_V=0$ is formally symmetric. Now suppose $\dim M\ge 1$. Define the complex vector field $Y={\sum }_i⟨c(e^i)u,v⟩e_i$ for smooth sections $u,v$. At any chosen point take a local orthonormal frame with ${\mathrm{\nabla }}^{LC}{e}_i=0$ there. Using [F2], metric compatibility and $\mathrm{\nabla }{e}_i=0$ at that point gives $\mathrm{div}Y=⟨D_{V}u,v⟩-⟨u,D_{V}v⟩$: the second term has this sign because $c(e^i{)}^{\ast }=-c(e^i)$ and the Hermitian pairing is linear in its first variable. Since this is a pointwise identity independent of the chosen normal frame, it holds globally. Apply [F3] to real and imaginary parts of $Y$ and integrate to get ${\int }_M⟨D_{V}u,v⟩={\int }_M⟨u,D_{V}v⟩$. Thus $D_V$ is formally symmetric. Oddness follows termwise because each $c(e^i)$ is odd and $\mathrm{\nabla }$ preserves grading.