Frame bundle of orthonormal frames orthogonal to a submanifold.

Question

Suppose we have a smooth manifold $M$ of dimension $m$ with a Riemannian metric and a connected submanifold $N$ of dimension $n$ in $M$ with $n<m-1$.

Let $n\le k<m-1$ and consider the bundle over $N$, which consists of all orthonormal $k$-frames of $M$, which are orthogonal to $N$.

Is this bundle always trivial?

The fiber of this bundle is the Stiefel manifold $V_k(\mathbb{R}^{m-1})$, which is connected, since $k<m-1$.

Answer 1

Let $N$ be any closed manifold with nontrivial Stiefel-Whitney classes. This includes any non-orientable manifold or any manifold with odd Euler characteristic.

By Whitney's Embedding Theorem, $N$ embeds into $\mathbb{R}^n$ for $n$ large enough.

I claim that the orthonormal frame bundle of $N$ is not trivial. The point is that it on the vector bundle level, we have $T\mathbb{R}^n|_N = TN\oplus \nu $ where $\nu$ is the normal vector bundle. Since $\mathbb{R}^n$ has trivial tangent bundle, we have $w_k(\mathbb{R}^n) = 0$ for all $k$. Hence, $$0 = w_k(T\mathbb{R}^n|_N) = \sum_{i+j = k} w_i(TN)\cup w_j(\nu)\in H^k(N,\mathbb{Z}/2\mathbb{Z}).$$

Since $w(TN)$ is nontrivial by assumption, this impiles $w(\nu)$ is nontrivial as well, which implies $\nu$ is nontrivial as a bundle. Now one just notes that $\nu$ is the associated bundle to the principal bundle of orthonormal frames in $M$ orthogonal to $N$. Hence, the fact that $\nu$ is nontrivial implies that this principal bundle is non-trivial as well.

Answer 2

Consider $M = (I \times D^2)/\sim$, where $\sim$ is the relation $(0, (r, \theta)) \sim (1, (r, -\theta))$, i.e., a kind of "filled in" Klein bottle. The subset $N = I \times {(0, 0)} / \sim$ is a circle embedded in $M$. The bundle of $1$-frames in $M$ orthogonal to $N$ is... well... at each point $(x, (r, \theta))$ of $N$, you've got a circle of $1$-frames, i.e., unit vectors perpendicular to the core circle $N$, i.e., pointing outwards in the disk $\{x\} \times D^2$. And the endpoints of those vectors are exactly the points $I \times S^1 / \sim$, i.e., the Klein bottle. And the Klein bottle's not a trivial circle-bundle over $S^1$. So I believe the answer is "no". (Hi, Ted!)