normal bundle on a submanifold

Can you give me an example of a nontrivial normal bundle of a submanifold (of any manifold)? There is standard example of the core circle of a mobius band, but can you give an example of a submanifold of an orientable manifold?

• Do you also want the submanifold to be orientable? Nov 21 '14 at 6:33

Proposition: Let $M$ be a submanifold of $N$, where $N$ is orientable and $M$ is not. Then the corresponding normal bundle is non-trivial.

Proof: Suppose the normal bundle is trivial, then it has a global frame. This frame, together with an orientation of $N$, induces an orientation of $M$.

In view of the above proposition, you can take any non-orientable submanifold of $\mathbb{R}^n$, for example an open Moebius band inside $\mathbb{R}^3$ or a Klein bottle inside $\mathbb{R}^4$.

I assume you don't want to consider points. Take any manifold $X$ which isn't stably parallelizable, e.g. because one of its Stiefel-Whitney classes or Pontryagin classes doesn't vanish, and embed it into a manifold which is stably parallelizable, e.g. $\mathbb{R}^n$ or $S^n$. Then the normal bundle cannot be stably trivial, and in particular cannot be trivial. Manifolds which aren't stably parallelizable are plentiful, and they can all be embedded into $\mathbb{R}^n$ or $S^n$ for sufficiently large $n$, so that gives a large source of examples.

The interesting question, I guess, is how low the dimension of $X$ can be in this construction. If we require $X$ to be closed and orientable then $\dim X \ge 4$. This is because for $\dim X \le 3$ any such manifold is in fact stably parallelizable (which implies that the oriented bordism groups $MSO_1, MSO_2, MSO_3$ all vanish). When $\dim X = 4$ we can take $X$ to be any closed orientable $4$-manifold with nonzero signature, since that implies that $p_1 \neq 0$. Arguably the simplest example of such a manifold is $\mathbb{CP}^2$ (which generates $MSO_4 \cong \mathbb{Z}$), which embeds into $\mathbb{R}^8$ or $S^8$ by Whitney.

Edit #2: In the above construction, if we don't require $X$ to be orientable then $X$ can be chosen to be any non-orientable manifold (since then $w_1$ won't vanish, so $X$ cannot be stably parallelizable). An example of minimal dimension is $X = \mathbb{RP}^2$, which embeds into $\mathbb{R}^4$ or $S^4$ by Whitney.

Edit: I don't know much about the case of not-necessarily-closed manifolds, but if you want to allow the target of the embedding to vary, so an embedding $X \to Y$ of a closed orientable manifold into another closed orientable manifold, then the comments above show that the normal bundle is stably trivial if $\dim Y \le 3$. In fact the normal bundle must be trivial: if $\dim X = 1$ then a vector bundle on $X$ is trivial iff it is stably trivial iff $w_1$ vanishes, and if $\dim Y = 3$ and $\dim X = 2$ then a line bundle on $X$ is trivial iff it is stably trivial iff $w_1$ vanishes.

So the minimum possible dimension of $X$ is $2$ and the minimum possible dimension of $Y$ is $4$ (again assuming that both $X$ and $Y$ are closed and orientable). This minimum is achievable: take $X$ to have nontrivial tangent bundle and take the embedding to be the diagonal embedding $X \to X \times X$. The normal bundle is just the tangent bundle again, so it is also nontrivial. In particular we can take $X$ to be a closed orientable surface with nonzero Euler characteristic, say $S^2$ (since then the Euler class does not vanish).

• If the OP cares about non-orientable closed things, the smallest example of an embedding of closed manifolds $Y \to X$ for which $Y$ has nontrivial normal bundle is $\Bbb{RP}^1 \hookrightarrow \Bbb{RP}^2$ (consider the homeomorphism type of a tubular neighborhood); the smallest example for which $X$ is orientable but $Y$ needn't be is $\Bbb{RP}^2 \hookrightarrow \Bbb{RP}^3$ (consider the Stiefel-Whitney classes).
– user98602
Nov 21 '14 at 6:52

If you consider the the manifold $X=\mathbb P^2(\mathbb R)$ with coordinates $x:y:z$, its submanifold $Y$ given by $z=0$ is isomorphic to $X=\mathbb P^1(\mathbb R)$ and the normal bundle $N(Y/X)$ corresponds in that isomorphism to the non-trivial tautological line bundle $\xi$ of $\mathbb P^1(\mathbb R)$.
Thus the normal bundle $N(Y/X)$ is non-trivial.