I have the following question. Provided that $E\to M$ is vector bundle and that a Clifford module $Cl(T^*M)$ acts on $\Gamma(E)$ via Clifford multiplication $c$, the Dirac operator of this Clifford bundle is defined to be $$D = c \circ \nabla$$

If $x\in M$ and $e_1,\dots,e_n$ a local frame in $x\in U\subseteq M$, then the local form of the Dirac operator is said to be $$D = \sum_i c(e^i) \nabla_i$$

The question is: This equality holds everywhere in the local area $U$ or only at the origin? And if it holds only at $x$ why do we say "local can be written".

If we take a synchronous frame around $x$ (we extend $e_1,\dots, e_n \in T_pM$ by parallel transport, along radial lines in a small area around $x$) then we can have that

$$\nabla s = \sum d(f_i) e_i + \sum f_i \nabla e_i$$

for $s = \sum f_i e_i$, but the second terms is zero only at $x$. It doesn't seem to hold in all points around $x$.

Anyone who can help?


This all boils down to a very basic principle from linear algebra: if $V$ is a finite-dimensional vector space, then for any basis $\{e_1,\dotsc,e_n\}$ for $V$ with dual basis $\{e^1,\dotsc,e^n\}$ for $V^\ast$, we have $$ \forall \phi \in V^\ast, \quad \phi = \sum_{i=1}^n \phi(e_i)e^i. $$ Now, fix $x \in M$. Since $TM$ is a locally trivial vector bundle, we can find an open neighbourhood $U$ of $x$ such that $TM|_U \cong TU$ is trivial, and as such admits a global frame $\{e_1,\dotsc,e_n\}$ as a vector bundle over $U$; taking dual bases in each fibre as usual, this gives us a global dual frame $\{e^1,\dotsc,e^n\}$ for $T^\ast M|_U \cong T^\ast U$ as a vector bundle over $U$. Applying this basic principle fibrewise, we see that $$ \forall \omega \in \Omega^1(U), \quad \omega = \sum_{i=1}^n \omega(e_i) e^i, $$ and hence that $$ \forall s \in \Gamma(U,E), \quad \nabla s = \sum_{i=1}^n e^i \otimes \nabla_{e_i}s \in \Omega^1(U,E), $$ an equality valid on all of $U$; applying, in turn, the definition of the Dirac operator $D$, this then yields the local expression $$ \forall s \in \Gamma(U,E), \quad D s = c(\nabla s) = c\left(\sum_{i=1}^n e^i \otimes \nabla_{e_i}s\right) = \sum_{i=1}^n c(e^i)\nabla_{e_i}s, $$ which is valid on all of $U$.

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