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Let $f:M\to N$ be a smooth map between smooth manifolds and assume $M$ is oriented. Suppose $y\in N$ is a regular value of $f$ with $S:=f^{-1}(y)$ nonempty. According to this question: normal bundle of level set, the normal bundle of $S$ in $M$ is trivial (explicitly, $NS\to S\times T_yN$, $(p,v)\mapsto (p,df(v))$ is a trivialization) and in particular orientable. This implies that $S$ is orientable. My question is: Is there a "canonical" orientation on $S$? This is of course equivalent to choosing a canonical orientation of the normal bundle on $S$.

In some speical situations, this seems possible:

If $\dim (N)=1$, then we can regard $f$ as a map $f:M\to \Bbb R$ (locally) and in this case we can choose a canonical orientation on $S$ using $\text{grad}(f)$ .

If $N$ is oriented, then we have an orientation on $T_yN$, and we can choose orientation on $NS$ by declaring the above trivialization $NS\to S\times T_yN$ to be orientation-preserving.

But I can't handle the general case.

blancket
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No, you can't. Intuitively, the mapping from non-empty regular levels to $N$ is a bijection, so a canonical orientation on the normal bundle to an arbitrary regular level induces an orientation on $N$.

The simplest example is probably to take $M$ to be a solid torus, $N$ a Möbius strip embedded in $M$ as a rotating segment in the disks perpendicular to the central circle, and $f:M \to N$ the orthogonal projection, which is clearly a submersion. Each level of $f$ is a line segment $S$ in $M$, whose normal bundle is (isomorphic to) $S \times \mathbf{R}^{2}$. But if we fix an orientation on the normal bundle of $S$ and try to extend by continuity, we run into trouble when we walk around the central circle.

Andrew D. Hwang
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