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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

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Do you have an example of two isometric embedding such that the induced curvatures on the normal bundle are different? My intuition says: if you CAN then answer is "NO" and if you CAN NOT then obviously "YES". –  Anton Petrunin Mar 2 '11 at 23:58
    
Thanks everyone for the helpful answers and comments! –  AlexE Mar 4 '11 at 12:41

2 Answers 2

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Interesting question. The answer is no: surfaces with bounded geometry can have normal bundles with unbounded curvature.

To set the stage, it's worth first noting that you can have a surface with extreme geometry isometrically embedded in $\mathbb E^3$, where the normal bundle, being one-dimensional, has a trivial connection; or include this into $\mathbb E^4$ (Euclidean 4-space) where the normal bundle is 2-dimensional, but the curvature is still 0. This at least illustrates that bounded geometry of the normal bundle and tangent bundle are decoupled.

I'll now describe an isometric embedding of the $\mathbb E^2$ into $\mathbb E^6$ where the connection on the normal bundle has unbounded curvature. The embedding have local 1-parameter groups of symmetry, which makes it easier to keep track of curvature without needing to write down equations.

Start by visualizing a helical curve in $\mathbb E^3$. The tangent vector to a helix goes repeatedly around a circle in its spherical image. The connection on the tangent bundle is induced by this Gauss map from the connection on $S^2$, so the parallel translation of the normal bundle rotates the plane by an angle equal to the area enclosed inside this circle once every coil of the helix.

Now consider a similar curve in $\mathbb E^5$, thought of as $\mathbb E \times \mathbb E^2 \times \mathbb E^2$. In the $\mathbb E$-direction, the curve makes uniform progress, while going around circles of possibly different radii at possibly different rates in the $\mathbb E^2$ directions. If the term weren't otherwise engaged, one could call this a double helix. It is invariant by a 1-parameter group of isometries of $\mathbb E^5$ that translates in the $\mathbb E$ direction while spinning the two perpendicular planes at their own rates.

The normal bundle splits into two $\mathbb E^2$ subbundles, its intersection with the two 3-dimensional $\mathbb E \times \mathbb E^2$'s. The connection preserves this splitting, rotating the two $\mathbb E^2$'s indendently.

Now add an extra "parameter" dimension, making the ambient space $\mathbb E^6$. Modify the curve in $\mathbb E^5$ by increasing the radius of one helix while decreasing the radius of the other, balancing the changes so the curve remains invariant by the same 1-parameter group, and its arc length remains constant (as measured by the time parameter of the 1-parameter group). It's easy to see, since it's locally isometric to a surface of revolution because of the symmetry, that the resulting surface is isometric to $\mathbb E^2$. We can make the circle in on $\mathbb E^2$ go all the way to 0. After making sure it has $C^\infty$ contact to a straight line in this projection, we can then start making this projection helical again, but with a steeper and tighter helix, adjusting by letting the other helical projection shrink to a line. The local symmetry group in $\mathbb E^6$ has changed, but the induced symmetry on the surface remains the same. We can go back and forth, alternating between helical effects in the two factors, inexorably tightening the screws without distoring the surface.

The curvature of the induced connection on the normal bundle becomes arbitrarily high, as you can see by following the connection around a small rectangle, with two edges in the "parameter" direction, one edge where say the first helix has become straight and the fourth edge where the first helix is wound in very small tight coils.

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Thanks a lot! This makes rethink my intuition. ^^ –  AlexE Mar 4 '11 at 12:40

First, the general advice: Using co-ordinates, even normal co-ordinates, to do a computation like this is generally not a good idea. It is much better to use orthonormal frames and Cartan's moving frame approach using 1-forms.

So I did this and to my surprise found the curvature to be identically zero. Then I realized that this is obvious (if you use the "induced" connection on the normal bundle).

ADDED: The above discussion applies to the normal bundle viewed as a Riemannian manifold itself. That's how I originally interpreted the question.

However, Bill Thurston, as well as Anton Petrunin below, point out the other possible interpretation: View the normal bundle as a vector bundle with the connection induced by the Euclidean connection on $R^n$. Bill gives a pretty thorough geometric answer. You can also easily compute the curvature of the normal connection and see that it is a quadratic formula involving the second fundamental form (but not equal to the Riemann curvature of the original submanifold). It follows that bounded intrinsic geometry of the submanifold is not enough to imply bounded curvature of the normal bundle but bounded extrinsic geometry is.

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Do you want to say that normal bundle has zero curvature? (I do not think so --- say normal bundle of $\mathbb C\mathrm P^2$...) –  Anton Petrunin Mar 2 '11 at 17:31
    
Anton, if the Riemannian metric is the one induced by the embedding of, say, a small tubular neighborhood of the zero section in Euclidean space, then it is almost tautological to say that the metric is flat. But maybe I'm using the wrong metric on the normal bundle? –  Deane Yang Mar 2 '11 at 18:52
    
@Deane Yang: The normal bundle $NM \subset TR^N$ is the orthogonal complement of $TM$ w.r.t. the standard Euclidean metric on R^N. Here the embedding $TM \subset TR^N$ comes from the isometric embedding $M \to R^N$. Now we use on $NM \subset TR^N$ the induced metric and the induced connection (which is given by the standard Euclidean connection on R^N composed with the orthogonal projection onto NM). Perhaps my words "pull-back metric/connection" were not well chosen - "induced" is better. –  AlexE Mar 2 '11 at 19:12
    
Alex, that appears to be exactly the Riemannian metric on the normal bundle that I used. In other words, there is a natural immersion of the normal bundle into $R^N$, and the metric on the normal bundle is just restriction of the metric on $R^N$ to $T_*N_*M$. –  Deane Yang Mar 2 '11 at 20:41

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