Hi,

I asked this question already on math.stackexchange but got no answer (link: http://math.stackexchange.com/questions/22155).

Our setting: An Euclidean vector bundle $(E, h, \nabla^E)$ over a Riemannian manifold (M,g) is said to have bounded geometry, if the norms of the curvature tensor $R^E$ and of all its covariant derivatives are bounded. The manifold itself is said to have bounded geometry, if the tangent bundle TM, equipped with the manifold metric and the Levi-Civita connection, has bounded geometry and additionally the metric is complete and the injectivity radius fulfills $\operatorname{inj rad}(x) > \epsilon > 0$ for all x.

The question: We have a Riemannian manifold (M,g) of bounded geometry and some isometric embedding $\iota\colon M \to R^N$. Now we can look at the normal bundle NM over M, equipped with the pull-back metric and pull-back connection. Has this bundle bounded geometry? My intuition says "yes".

I tried it with local computations using the corresponding projection matrices but got nowhere.

I use the fact that a manifold has bounded geometry, if and only if the Christoffel symbols of the Levi-Civita connection and all their derivatives are uniformly bounded functions when computed in Riemannian normal coordinates (where the radii of the coordinate balls are the same for all points p). An analogous statement holds for vector bundles of bounded geometry, where the frames we use for the computation of the Christoffel symbols are acquired by choosing a orthonormal basis for the bundle in the point p and then parallel translate it along the radial geodesics in a normal coordinate ball (also with fixed radius for every point).

So if $\partial_{x_i}$ are the normal coordinates and $\{n_i\}$ is the orthonormal frame for the normal bundle we have the following expression: $\Gamma_{ij}^{k, TM} = g^{kl}\langle \nabla_{\partial_{x_i}} \partial_{x_j}, \partial_{x_l}\rangle$ and analogously $\Gamma_{ij}^{k, NM} = h^{kl}\langle \nabla_{\partial_{x_i}} n_j, n_l\rangle$, where $g^{ij}$ is as usually the inverse matrix of the matrix of the metric g (computed w.r.t. the normal coordinates), $h_{ij}$ the matrix of the metric of the normal bundle and $\langle \cdot, \cdot \rangle$ is the Euclidean metric of $R^N$ (we pushed the coordinates $\partial_{x_i}$ and the frame $\{n_i\}$ forward via the embedding $M \to R^N$). Since the frame we use for the normal bundle is orthonormal, we have $h_{ij}=\delta_{ij}$ and so the formula for the Christoffel symbols of the normal bundle reduces to $\Gamma_{ij}^{k, NM} = \langle \nabla_{\partial_{x_i}} n_j, n_k\rangle$.

For the matrices of the projections $p^{TM}: TR^N \to TM$, resp. $p^{NM}$ we get the following expressions w.r.t. the standard coordinates $\{e_i\}$ of $R^N$: $(p^{TM})_{ij} = g^{kl}\langle e_j, \partial_{x_l}\rangle \langle e_i, \partial_{x_k} \rangle$ and analogously $(p^{NM})_{ij} = h^{kl}\langle e_j, n_l\rangle \langle e_i, n_k \rangle$.

Now I want to deduce that if the Christoffel symbols of TM and all their derivates are uniformly bounded (i.e. the manifold has bounded geometry), then the entries of the projection matrix $p^{TM}$ and all their derivatives are uniformly bounded (which automatically gives the uniform boundedness of the entries of $p^{NM}$ and their derivatives). And from here I want to deduce the uniform boundedness of the Christoffel symbols of NM and all their derivatives. But I do not see how using the equations I got so far.

May we get further equation / information which give the desired results? Or maybe there is some other way to answer the question posed in the third paragraph (not using ugly local computation)? I would be happy with any solution.

Thanks, Alex