I am reading Weiping Zhang's lecture and I encounter the following:
Suppose $M$ is a $n$-dimensional compact manifold. For any $e\in TM$, let $e^*\in T^*M$ corresponds to $e$ via $g^{TM}$. Let $c(e)$, $\hat{c}(e)$ be the Clifford operators acting on the exterior algebra bundle $\Lambda^*(T^*M)$ defined by $$ c(e)=e^*\wedge{}-i_e,\quad \hat{c}(e)=e^*\wedge{}+i_e $$ where $e^*\wedge$ and $i_e$ are the standard notation for exterior and interior multiplications.
My Question: why we can derive the following identity can be seen in the context of Clifford algebra: $$c(e)c(e')+c(e')c(e)=-2\langle e,e'\rangle$$ can we prove it by direct computation?
Though @ziggurism had answered my question, I typed down the details in consideration of future audience.
Let $\omega \in \bigwedge^{\bullet} (T^{\ast} M)$. Apply it to the operator $c (e) c (e') + c (e') c (e)$. Using the antisymmetric properties of wedge product and interior product, we have $$ \begin{array}{lll} {}[c (e) c (e') + c (e') c (e)] \omega & = & - i_e (e^{\prime \ast} \wedge \omega) - e^{\ast} \wedge (i_{e'} \omega)\\ & & - i_{e'} (e^{\ast} \wedge \omega) - e^{\prime \ast} \wedge (i_e \omega) \end{array} $$ By the product rule of interior product, we have $$ i_e (e^{\prime \ast} \wedge \omega) = i_e^{} (e^{\prime \ast}) \wedge \omega + (- 1) e^{\prime \ast} \wedge i_e \omega $$ and $$ i_{e'} (e^{\ast} \wedge \omega) = i_{e'}^{} (e^{\ast}) \wedge \omega + (- 1) e^{\ast} \wedge i_{e'} \omega . $$ We know that $$ i_e (e^{\prime \ast}) = (e^{\prime \ast}, e) = \langle e', e \rangle $$ where $(e^{\prime \ast}, e)$ is the duality pairing.
Similary, $$i_{e'} (e^{\ast}) = \langle e, e' \rangle.$$ Put in all these results to the first formula, we get $$ [c (e) c (e') + c (e') c (e)] \omega = - 2 \langle e, e' \rangle \omega . $$