8
$\begingroup$

This is exercise 1.4.4 on Guillemin and Pollack's Differential Topology

Suppose that $Z \subset X \subset Y$ are manifolds, and $z \in Z$. Then there exist independent functions $g_1, \dots, g_l$, on a neighborhood $W$ of $z$ in $Y$ such that $$Z \cap W = \{y \in W : g_1(y) = 0, \dots, g_l(y) = 0\},$$ $$X \cap W = \{y \in W : g_i(y) = 0, \dots , g_m(y) = 0\},$$ where $l-m$ is the codimension of $Z$ in $X$.

I tried to set up the proof as following:

Suppose that $Z \subset X \subset Y$ are manifolds, and $z \in Z$. Let $Z$ and $X$ have codimensions $l$ and $m$ in $Y$, $Z$ has codimension $l-m$ in $X$. From the partial converse to the preimage theorem, there exist independent functions $f_1, \dots f_m$ on a neighborhood $U$ of $z$ in $Y$ such that $X \cap U$ is the common vanishing set of the $f_i$.

We also know that there exist independent functions $h_{m+1}, \dots, h_l$ on a neighborhood $V$ of $z$ in $X$ such that $Z \cap V$ is the common vanishing set of the $h_i$.

And then I don't know why $h_i$s are smooth, and how should I continue.

Any ideas? Thank you.

$\endgroup$
  • $\begingroup$ why you assumed that $Z$&$X$ have codimensions $l$ and $m$ in Y? $\endgroup$ – Idonotknow Oct 9 '18 at 14:23
  • $\begingroup$ In the question in the second intersection it is $g_{1}$ not $g_{i}$ $\endgroup$ – hopefully Oct 17 '18 at 13:05
3
$\begingroup$

tl;dr: straighten the neighborhood in $Y$ so that $Z$ looks flat, straighten the neighborhood in $Z$ so that $X$ looks flat, and extend the latter straightening some way into $Y$.

I don't know the theorem names off the top of my head, but here's how I'd approach this:

WLOG $z=0$ and $Y = \mathbb{R}^n$ for some $n$. Since $Z$ is a smooth submanifold of codimension $l$, there is some smooth $f\times g\colon U\cong V\times W$ where $U\subseteq\mathbb{R}^n$, $V\subseteq\mathbb{R}^l$, $W\subseteq\mathbb{R}^{n-l}$, and for $x\in U$, $x\in Z$ iff $f(x)=0$. (I want to say this is the domain-straightening theorem?) The neighborhood of $x$ in $Z$ looks like $W$, so we'll identify the two for convenience.

Now $X$ is a smooth submanifold of $Z$ (codimension $l$), so by the same theorem, after possibly shrinking $W$ (and $U$ to match), there is some smooth $h\times k\colon W\cong S\times T$ where $S\subseteq\mathbb{R}^{l-m}$, $T\subseteq\mathbb{R}^{n-m}$, and for $x\in W$, $x\in X$ iff $h(x)=0$.

Of the $g_i$, the first $l$ are the components of $f$, and the remaining $m-l$ are the components of $h\circ g$. Apologies for the notational weirdness.

$\endgroup$
  • $\begingroup$ Could you provide more details please? $\endgroup$ – Idonotknow Oct 9 '18 at 17:40
  • $\begingroup$ what do you mean by "so we'll identify the two for convenience."? sorry my mother tongue is not English. $\endgroup$ – hopefully Oct 17 '18 at 10:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.