# 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?

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.