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Let $M$ be a Riemannian manifold and $S \subset M$ a regular level set of a smooth function $f:M\rightarrow \mathbb{R}^k$. How can I show that the normal bundle of $S$ is trivial?

If $k=1$ then $\text{grad}f$ is a global frame for $NS$ but I am not sure how to show it formally for the general case. Thanks!

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up vote 6 down vote accepted

If $S=f^{-1}(x)$ where $f:M\to N$ and $x$ is a regular value then $f_*:T_yM\to T_xN$ is (by definition of regular value) surjective, $T_yS$ is the kernel of $f_*$, hence $f_*$ is an isomorphism of $(T_yS)^\perp\to T_xN$, i.e. (letting $y$ vary) it gives an isomorphism $(TS)^\perp\cong S\times T_xN$.

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I'd like to defend the point of view that this triviality result for the normal bundle has actually very little to do with the riemannian structure of $M$.

As user 8268 points out, $f_{\ast y}: T_{\ast y}M\to T_{\ast f(y)}N$ is surjective for each $y\in M$ and this is translated into the exact sequence of vector bundles on $S$ $$0\to T(S) \to TM_{|S}\to f^\ast T\mathbb (R^k)_{|S} \to 0$$ Now the crucial point is that the genuine normal bundle of $S$ in $M$ is by definition the quotient bundle $N_{S/M}=T(M)_{|S}/TS$. This genuine normal bundle is here isomorphic to $ f^\ast T\mathbb (R^k)_{|S}$, hence trivial as requested.

And what has this to do with the riemannian structure? It is just that we can then, if we so wish, replace the genuine normal bundle, which is a quotient of $T(M)_{|S}$ by the orthogonal complement $T(N)^\perp$ of $T(N)$ inside $T(M)_{|S}$, which is a subbundle of $T(M)_{|S}$, isomorphic to $N_{S/M}$. The success of this vision of the normal bundle in the riemannian case is that everybody prefers subspaces to quotient spaces, but we have to keep in mind that what I have called the genuine normal bundle is completely canonical whereas the more concrete isomorphic riemannian normal bundle varies with the riemannian structure we impose on $M$.

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