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Last semester I studied manifold follwing the Lee's book 'Introduction to Smooth Manifolds', 2ed, and I have a doubt about the orientation of hypersurface. I'll write down the Proposition 15.21 (here, $N\newcommand{\into}{\mathbin{\lrcorner}}\into\ w$ denotes interior multiplication):

Suppose M is an oriented smooth n-manifold with or without boundary, S is an immersed hypersurface with or without boundary in M, and N is a vector field along S that is nowhere tangente to S. Then S has a unique orientation such that for each $p \in S$, $(E_1,\ldots,E_{n-1})$ is an oriented basis for $T_pS$ if and only if $(N_p,E_1,\ldots,E_{n-1})$ is an oriented basis for $T_pM$. If $w$ is an orientarion form for $M$, then $i_S^*(N\into w)$ is an orientation form for $S$ with respect to this orientation, where $i_S: S\longrightarrow M$ is inclusion.

The demonstration starts defining $i_S^*(N\into w)$ as an $n-1$-form on $S$. The problem is: we have the definition of interior multiplication only for $X \in \mathfrak{X}$ and $w \in \Omega^k(M)$ (page 362). So, since $N$ is defined only at points of $S$, we don't have the definition of $N\into w$ as an $n-1$ form in an open subset of $M$. If $S$ is an embedded submanifold, then we can work with an extension $\tilde N$ of $N$, but here $S$ is just an immersed manifold.

We can give a pointwise definition of $N\into w$ (at a point of $S$), but then I think that the pullback is unnecessary. Another solution is to extend $N$ locally and then we consider the pullback (locally). The final definition will not depend on the extensions, but in this case we are not doing exactly the pullback by the inclusion, so we have to prove the properties of this 'new pullback' again, doesn't we?

So, what is the precise definition of $i_S^*(N\into w)$?

Jack Lee
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  • I edited your post to use $\lrcorner$ to denote interior multiplication; it's much clearer than $>$. – Jack Lee May 29 '15 at 22:59

1 Answers1

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The pullback is indeed defined pointwise: For any $p\in S$, $$ (\iota_S^*(N\newcommand{\into}{\mathbin{\lrcorner}}\into\omega))_p= d(\iota_S)_p((N\into\omega)_p) = d(\iota_S)_p ( N_p \into \omega_p). $$ The (pointwise) pullback is definitely needed here, because $(N\into\omega)_p$ is an $(n-1)$-tensor on the $n$-dimensional tangent space $T_pM$, but its pullback is an $(n-1)$-tensor on $T_pS$. Of course, since $\iota_S$ is just an inclusion map, the pointwise pullback is just restriction to $T_pS$; but it can make a difference because a nonzero tensor on $T_pM$ can become zero when it's pulled back to $T_pS$.

Jack Lee
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