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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 $k<m-1$.

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    $\begingroup$ You don't mean $k$-frames that are orthogonal to $N$. You mean adapted orthonormal frames, where the first $n$ are tangent to $N$ and the remaining $k-n$ are orthogonal to $N$. Is that right? Or do you mean $k$-frames for the normal bundle of $N$ in $M$ with $1\le k<m-n$? $\endgroup$ – Ted Shifrin Dec 2 '13 at 20:20
  • $\begingroup$ I really mean orthonormal k-frames, which are orthogonal to N, like I said. You're right. $k< m-n$ follows. $\endgroup$ – Tina Dec 2 '13 at 20:27
  • $\begingroup$ And now the fiber is $V_k(\Bbb R^{m-n})$. $\endgroup$ – Ted Shifrin Dec 2 '13 at 20:41
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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.

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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!)

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  • $\begingroup$ More simply, the Klein bottle contains an embedded circle whose normal bundle is a Moebius band. $\endgroup$ – lavinia Dec 4 '13 at 3:10
  • $\begingroup$ @lavinia: I originally thought the same as well, but the question specifically asks for $N$ to have codimension at least 2. Hence, in this sense, John's answer is optimal. (And I upvoted it, as should others.) $\endgroup$ – Jason DeVito Dec 4 '13 at 16:10
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    $\begingroup$ Yeah but just cross the Klein bottle with a circle. $\endgroup$ – lavinia Dec 12 '13 at 15:15

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