7

Let $M$ be an $n$-dimensional oriented manifold. Let $f:M\to\mathbb{R}$ be a smooth function. Suppose $c$ is a regular value of $f$ with $f^{-1}(c)$ nonempty. Show that $f^{-1}(c)$ is an oriented regular submanifold of $M$.

By constant rank thoerem, I know that $f^{-1}(c)$ is an $(n-1)$-dimensional regular submafold of $M$, but how to show that it is orientable?

Xiang Yu
  • 4,835
  • What definition of orientability are you using? – florence Aug 12 '16 at 10:46
  • 1
    Note that $\forall p\in f^{-1}(c)$, the tangent space to the fibre is $T_{p}(f^{-1}(c)) = \text{Ker } df_p$. If $M$ was $\Bbb R^n$, then this tangent space would be $\langle \nabla f_p\rangle^{\perp}$ so the gradient of $f$ would give an orientation of $f^{-1}(c)$. – paf Aug 12 '16 at 10:49
  • @florence The determinants of the Jacobian of transition maps are positive. – Xiang Yu Aug 12 '16 at 10:51

1 Answers1

3

Suppose $S$ is the manifold $f^{-1}(c)$. Prove that $\triangledown f$ is normal to $S$ in $M$. Now if $S$ is non-orientable then there exists a continuous choice of collection of ordered basis at each point of $S$. Attach $\triangledown f$ with it. This will show that $M$ is non-orientable, a contradiction.

Cusp
  • 218