Weitzenböck formula and comparison of norms Let $M$ be a closed Riemannian manifold with a spin$^\mathbb{C}$ bundle $S$. Now for a spin connection $A,$ and a spinor $\phi,$ it can be shown that $C\lvert\nabla_A\phi\rvert^2\geq \lvert D_A\phi\rvert^2$ for some $C>0$. My question is what's the best value of $C$ one can hope for? Ideally this should depend on the geometry and dimension of the manifold I would think. Then again for a flat manifold and with a trivial line bundle, the usual Weitzenböck formula
\begin{align*}
D_A^2=\nabla_A^*\nabla_A+\frac{s}{4}+\frac{1}{2}F_A
\end{align*}
would imply that $\int_M \lvert D_A\rvert^2=\int_M \lvert\nabla_A\rvert^2$. My hope is: can $C$ be $1$? Is there an easy counter example to see?
I am also interested in a similar question for any unitary connection $B$ on $S$, one can define $D_B:\nabla_B\xrightarrow{\text{Clifford mult.}}D_B$. Again we have $C\lvert\nabla_B\phi\rvert^2\geq \lvert D_B\phi\rvert^2$, does the answer really depend on what connection we use?
 A: I am not an expert in all the delicate points of clifford algebras and spin structures, but I think the following shows the constant can't be 1 in general: let $(M,g)$ be a Riemannian manifold of dimension $n$. Take the usual Dirac operator $d+\delta:\Omega^{*}(M)\rightarrow\Omega^{*}(M)$ and the covariant derivative $\nabla$.
To cook up a concrete example, take $f:M\rightarrow\mathbb{R}$ such that at a point $p\in M$ one has $\nabla df|_{p}=\lambda g$ and take $\omega=df\in\Omega^{1}(M)$. Such a function exists, for example, by taking normal coordinates $(x_i)$ at $p$ and setting $f(x)=\sum_{i\leq j} \lambda\,x^{i}x^{j}$ and mulyiplying by a bump funtion so it is defined on the whole of $M$. Then, since $|g|^{2}=n$, we have $|\nabla \omega|_{p}|^{2}=n\lambda^{2}$ while $|\delta\omega|_{p}|^{2}+|d\omega|_{p}|^{2}=|\mathrm{tr}_{g}\nabla df|_{p}|^{2}=n^{2}\lambda^{2}$. So the constant can't be 1, not even if $(M,g)$ is flat.
Here is a more abstract comparsion: for a general 1-form $\omega\in\Omega^{1}(M)$ we find that $|\nabla\omega|^{2}=\frac{1}{4}|d\omega|^{2}+\frac{1}{4}|\mathcal{L}_{\omega^{\sharp}}g|^{2}$ since $\frac{1}{2}d\omega$ is the projection of $\nabla\omega$ onto the anti-symmetric tensors while half the Lie derivative is the porjection onto the symmetric tensors, and symmetric and anti-symmetric tensors are orthogonal. Now, we can decompose $\mathcal{L}_{\omega^{\sharp}}g$ further into a tracless part, let us denote it as $\mu(\omega)$, and its orthogonal component which is $-\frac{2}{n}\delta\omega\,g$. Since these are orthogonal as well, we find $|\nabla\omega|^{2}=\frac{1}{4}|d\omega|^{2}+\frac{1}{4}|\mu(\omega)|^{2}+\frac{1}{n}|\delta\omega|^{2}$, where we used $|g|^{2}=n$.
This shows that in general $|\nabla\omega|^{2}\geq\min{(\frac{1}{4},\frac{1}{n})}\,|d\omega+\delta\omega|^{2}$, and I think the above example demosntrates that this is the best you can hope for.
