normal bundle of level set

Let $M$ be a Riemannian manifold and also $S \subset M$ a normal degree set of a smooth function $f:M\rightarrow \mathbb{R}^k$. Just how can I show that the regular package of $S$ is unimportant?

If $k=1$ after that $\text{grad}f$ is an international structure for $NS$ yet I am not exactly sure just how to show it officially for the basic instance. Many thanks!

2022-06-07 14:32:56
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Answers: 2

If $S=f^{-1}(x)$ where $f:M\to N$ and also $x$ is a normal value after that $f_*:T_yM\to T_xN$ is (necessarily of normal value) surjective, $T_yS$ is the bit of $f_*$, therefore $f_*$ is an isomorphism of $(T_yS)^\perp\to T_xN$, i.e. (allowing $y$ differ) it offers an isomorphism $(TS)^\perp\cong S\times T_xN$.

2022-06-07 14:51:50

I would love to safeguard the perspective that this triviality outcome for the regular package has in fact really little to do with the riemannian framework of $M$.

As customer 8268 mention, $f_{\ast y}: T_{\ast y}M\to T_{\ast f(y)}N$ is surjective for each and every $y\in M$ and also this is converted right into the specific series of vector packages on $S$ $$0\to T(S) \to TM_{|S}\to f^\ast T\mathbb (R^k)_{|S} \to 0$$ Now the critical point is that the real regular package of $S$ in $M$ is necessarily the quotient package $N_{S/M}=T(M)_{|S}/TS$. This real regular package is below isomorphic to $ f^\ast T\mathbb (R^k)_{|S}$, therefore unimportant as asked for.

And also what has this to do with the riemannian framework? It is simply that we can after that, if we so dream, change the real regular package, which is a ratio of $T(M)_{|S}$ by the orthogonal enhance $T(N)^\perp$ 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 regular package in the riemannian instance is that everyone favors subspaces to quotient spaces, yet we need to remember that what I have actually called the real regular package is entirely approved whereas the extra concrete isomorphic riemannian regular package differs with the riemannian framework we trouble $M$.

2022-06-07 14:51:32