$L^p$ norm of tensor Let $T_{ab}$ be a $(0,2)$ tensor on a Riemannian manifold $M$ with component $T_{ab} = \nabla_a S_b - \nabla_b S_a - 
g_{ab} \text{ div $S$}$. Here, $S$ is a 1-form and $g_{ab}$ is the metric tensor. By definition,  the tensor norm of $T_{ab}$ is given by $\lvert T \rvert^2 = T_{ab}T^{ab}$. However, I have some trouble computing the $L^p$ norm of $T$. In particular, I'm wondering if $\lVert T \rVert_p$ is equivalent to $\lVert \nabla S \rVert_p + \lVert \text{ div $S$} \rVert_p$. I tried to express the norm in components but it gets too tedious at some point. Any insight would be appreciated.
 A: By definition,
$$ \|A\|_{L^p} = \left( \int_M|A|^p d\mu\right)^{1/p}$$
and $|T|^2 = T^{ab}T_{ab}$. Define
$$\langle A, B\rangle = A^{ab}B_{ab}, $$
then $|A|^2 = \langle A, A\rangle$ and $\langle \cdot, \cdot \rangle$ is an inner product.
Write $B_{ab} = \nabla_a S_b - \nabla_b S_a$. The calculation in your case is easier: we have
$$\langle B_{ab} ,
g_{ab}\rangle = g^{ab} B_{ab} = 0$$
since $g^{ab}$ is symmetric and $B$ is antisymmetric. Thus
$$|T|^2 = |B_{ab} - g_{ab} \operatorname{div}(S)|^2 = |B_{ab}|^2 + |g_{ab} \operatorname{div}(S)|^2= |B_{ab}|^2 + n|\operatorname{div}(S)|^2$$
This implies
$$|B|^2 + |\operatorname{div}(S)|^2 \le |T|^2 \le n(|B|^2 + |\operatorname{div}(S)|^2). $$
To see a bit more, note that $B= dS$ and $\operatorname{div} S = d^*S$.From here we can see that $|T|$ is not equivalent to $|\nabla S|^2 + |\operatorname{div}S|^2$: Take $S$ to be a non-zero harmonic one form on a compact manifold with negative Ricci curvature (e.g. a surface of genus $g\ge 2$ with hyperbolic metric). Then $dS =d^*S = 0$, but $|\nabla S| \neq 0$. To see why $|\nabla S|^2 \neq 0$, note from the Ricci identity,
\begin{align}
\int_M (|B|^2 + | \operatorname{div}(S)|^2)d\mu &= \int_M (\langle dS, dS\rangle + (d^*S)^2)d\mu\\
&= \int_M ( \langle S , d^*dS \rangle + \langle S, dd^*S\rangle) \\
&= \int_M \langle S , \Delta_d S\rangle d\mu \\
&= \int_M \langle \langle S , -\nabla^* \nabla S\rangle + \langle S, Rc(S)\rangle d\mu \\
&= \int_M |\nabla S|^2 d\mu + \int_M Rc(S, S) d\mu. 
\end{align}
If $S$ is harmonic, we have
$$\int_M |\nabla S|^2 d\mu = - \int_M Rc(S, S) d\mu.$$
So if $M$ has negative Ricci curvature then $\|\nabla S\|_{L^2} \neq 0$.
