Let $Y$ be a smooth manifold, and let $\pi:E\to Y$ be a smooth vector bundle over $Y$. Suppose that $X$ is an immersed submanifold of $Y$, which lies in $E$ as the zero set of some smooth section $s:Y\to E$. I am trying to show that $E|_X$ is isomorphic to the normal bundle of $X$ in $Y$ (defined as the quotient bundle $TY|_X/TX$). I proceed as follows.

Let $x\in X$ and $z:Y\to E$ be the zero section. The map $\phi_x:E_x\to T_{z(x)}E,v\mapsto\dot\gamma_v(0)$ (where $\gamma_v(t)=(x,tv)$ is injective and $im(\phi_x)=\ker(d\pi_{z(x)})$, so $0\to E_x\to T_{z(x)}E\to T_xY$ is an exact sequence, which splits since $d\pi_{z(x)}\circ ds_x=id_{T_xY}$. Now the map \begin{aligned} \psi_x:T_{z(x)}E&\to E_x\oplus T_xY \\ w&\mapsto(\tilde\phi^{-1}_x(w-dz_x(d\pi_{z(x)}(w))),d\pi_{z(x)}(w)) \end{aligned} is an iso, where $\tilde\phi_x$ is the map $\phi_x$ with codomain restricted to $\ker(d\pi_{z(x)}$. Hence we get map $F:=p_1\circ\psi_x\circ ds_x$, where $p_1:E_x\oplus T_xY\to E_x$ is the projection.

This map should clearly become my isomorphism, and I have two concrete questions about it.

Unraveling the definitions we see that $v\in T_xY$ is in $\ker(p_1\circ\psi_x\circ ds_X)$ if and only if $dz_x(v)=ds_x(v)$, and from here I want to conclude that this implies that $\ker F=T_xX$, but I'm not quite sure how. The inclusion $T_xX\subset\ker F$ is clear since $z|_X=s|_X\implies dz_x|_{T_xX}=ds_X|_{T_xX}$, but I'm not sure how $dz_x(v)=ds_x(v)$ implies that $v\in T_xX$. We do know that $z|_{X\setminus Y}\neq s|_{X\setminus Y}$, but I am not sure that that implies that $dz_x(v)\neq dz_x(v)$ if $v\in T_xY\setminus T_xX$.

Secondly, I need this map to be surjective, but I have no idea how this follows.

Any help or tips or references are greatly appreciated!


I think that the surjectivity question can be rephrased: since $z(x)=s(x)$ it holds that $d\pi_{z(x)}\circ ds_x=id_{T_xY}$, so the surjectivity of $p_1\circ\psi_x\circ ds_x$ is equivalent to the surjectivity of $\tilde\phi^{-1}_x(ds_x-dz_x)$, i.e. if each $v\in E_x$ can be written as $\tilde\phi^{-1}(ds_x(u)-dz_x(u))$ for some $u\in T_xY$.

  • $\begingroup$ Are you interested in a different, perhaps simpler way to approach the problem? If yes, are you familiar with connections on vector bundles? $\endgroup$ – Amitai Yuval Oct 8 '17 at 20:02
  • $\begingroup$ I am and I am. (Although a closing argument on this approach would be optimal.) $\endgroup$ – B. Pasternak Oct 8 '17 at 20:03

The following is a slightly different approach to the problem (although I suspect it may be the same, just offering an alternative reasoning to the process described in the post).

I am assuming the section $s\in\Gamma(Y,E)$ is transverse to the zero section. Let $\nabla$ be any linear connection on $E$. Consider the following morphism of vector bundles over $X$: $$ \varphi:TY|_X\to E|_X,\quad u\mapsto\nabla_us. $$ (It may be noted that $\varphi$ is, in fact, independent of $\nabla$, but this is not crucial for the argument). The morphism $\varphi$ is onto, and its kernel is $TX$ (this is basically equivalent to the transversality assumption). The claim now follows from the first isomorphism theorem.

  • $\begingroup$ What do you precisely by transverse to the zero section? $\endgroup$ – B. Pasternak Oct 8 '17 at 20:20
  • $\begingroup$ @B.Pasternak Both $s$ and the zero section are smooth maps $Y\to E$, so they may be transverse. In fact, they should be, if you want to conclude that $X$ is a smooth submanifold. $\endgroup$ – Amitai Yuval Oct 8 '17 at 20:28
  • $\begingroup$ I simply don't know the meaning of two sections being transverse in this context. $\endgroup$ – B. Pasternak Oct 8 '17 at 20:31
  • $\begingroup$ @B.Pasternak Two smooth maps $f:M\to K$ and $g:N\to K$ are said to be transverse to one another if for every $(m,n)\in M\times N$ with $f(m)=g(n)$ we have $T_{f(m)=g(n)}K=df(T_mM)+dg(T_nN)$. $\endgroup$ – Amitai Yuval Oct 8 '17 at 20:40
  • $\begingroup$ @B.Pasternak And by the way, the claim does not hold in general without the transversality assumption. $\endgroup$ – Amitai Yuval Oct 8 '17 at 21:44

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