## Eqivalency of two norms on the symmetric two tensor-fields on a compact Riemannian manifold.

Let $(M,g)$ be a closed Riemannian manifold and $D$ denote the Riemannian connection corresponding to $g$. Let $S^2(T^*M)$ denote the space of $C^2$ symmetric 2-tensor filds on $M$. Let $h\in S^2(T^*M)$. Then $$D^2_{x,y}h(z,w)= D_xD_yh(z,w)-D_{D_xy}h(z,w)$$ where $x,y,z,w$ are vector-fields on $M$. Let $D^*$ denote the formal adjoint of $D$.\ Define, $$\|h\|^2=\int_M|h|dv_g$$ where $|h|$ denote the point-wise norm on $h$ and $dv_g$ denote the volume form defined by $g$.\ Define, $$\|h\|^2_1=\|D^2h\|^2+\|Dh\|^2+\|h\|^2$$ and $$\|h\|^2_2=\|D^*Dh\|^2+\|Dh\|^2+\|h\|^2$$ Are $\|.\|_1$ and $\|.\|_2$ equivalent?

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We have $\|D^2h-DD^\ast h\| \le \|D-D^*\|_{L^\infty(M,T^\ast M\otimes End(T^\ast M\otimes TM))} \|Dh\|,$ proving the uniform equivalence of the metrics. Notice that the difference of two connections is an ordinary tensor field, whose $L^\infty$-norm is bounded due to the compactness of $M$.

Another way to see this is that both metrics are locally equivalent to the metric $\xi\mapsto \|\nabla^2\xi\|+\|\xi\|,$ where now $\nabla$ refers to the "flat" connection and the $L^2$-norm refers to the local Lebesgue measure. Since $M$ is compact it can be covered by a finite number of such neighborhoods and in particular both norms are equivalent.

EDIT: I will keep the stuff I wrote at first (I thought of $D^*$ as a connection as well, which is not true, as Brian pointed out), but the correct answer involves elliptic regularity. I will consider the following simpler version (which also implies the general case, but once this simpler case is understood the more complicated case becomes easy.). Let $\nabla$ be a connection on $M.$ Then the following two norms are equivalent ($f\in C^\infty(M)$):
1. $\|f\|_{W^{2,2}},$
2. $\|\Delta f\|_{L^2}+\|f\|_{L^2}.$
Clearly the second norm is dominated by the first. The other inequality follows from the elliptic estimate
$\|D^2 f\|_{L^2}\leq \|\Delta f\|_{L^2},$ valid for $f\in C^\infty_0(R^n),$ which is actually trivial to prove.

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 thank you very much. – Soma Nov 10 2011 at 9:39 I may be missing the point, but I don't think you can subtract $D^2 h$ and $D D^* h$ (or $D$ and $D^*$), since the former is a (0,4)-tensor and the latter is a (0, 2)-tensor. – Brian Clarke Nov 10 2011 at 19:41 @Brian: The latter expression is a (0,4)-tensor. h is a (0,2) tensor and any covariant derivative adds a (0,1). – Orbicular Nov 10 2011 at 22:44 The adjoint of a covariant derivative is not a covariant derivative. If $h$ and $k$ are of type (r,s-1) and $(r,s)$, respectively, then for the $L^2$ inner product we have $(Dh, k) = (h, D^* k)$. For the product on the right to make sense, $D^* k$ has to be of type (r,s-1), not (r,s+1). – Brian Clarke Nov 10 2011 at 23:40 The math does not display properly. I don't know why and would appreciate help. – Orbicular Nov 11 2011 at 11:34