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I'm reading An Introduction to Manifolds (Tu) and got confused on p.123 Theorem 11.13. Let me briefly explain what was done before that.

The author defines an embedding between two manifolds $f: N\rightarrow M$ as an injective immersion (differential is injective everywhere) with the additional property that $f(N)$ as a subspace of $M$ is homeomorphic via $f$ to $N$. Then a canonical form theorem is proven, namely, that for any point $p \in N$, there are charts $(U,\phi = x^1, \ldots, x^n) \ni p$ and $(V,\psi = y^1, \ldots, y^m) \ni f(p)$, such that the transition map takes the following form:

$$\psi \circ f \circ \phi^{-1} : (r^1,\ldots,r^n) \mapsto (r^1,\ldots,r^n,0,\ldots,0).$$

(I guess it is implicitly assumed that $f(U) \subset V$).

Now in the process of proving that $f(N) \subset M$ is an embedded submanifold, he states that $f(U) \subset V$ is defined by the vanishing of the coordinates $y^{n+1},\ldots,y^m$. I can prove that $$ f(U) \subset \{q \in V: y^{n+1}(q) = \cdots = y^m(q)\} $$ but not the reverse inclusion. Can anyone explain why this should be true?

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The reverse inclusion doesn't necessarily follow from the canonical form theorem you've stated, since $V$ can just happen to be too large (or, if you like, $U$ can be too small): given $U$ and $V$ as you have one can always replace $U$ by a much smaller neighborhood $U'$ of $0$, and still guarantee that $U'$ and $V$ have charts satisfying your theorem -- and also be sure that the inclusion $\supset$ can't hold.

The solution, of course, is just to prove a slightly better canonical form theorem, by reducing the possible size of $V$:

Theorem. For any $p \in N$, there exist charts $(U, \phi = x^1, \dots, x^n)$ and $(V, \psi = y^1, \dots, y^m)$ satisfying $$ \psi f \phi^{-1} : (r^1, \dots, r^n) \mapsto (r^1, \dots, r^n, 0, \dots, 0) $$ and such that $$ f(U) = \lbrace q \in V : y^{n+1}(q) = \cdots = y^m(q) = 0 \rbrace. $$

Proof. The proof is simple once you have your canonical form theorem; if $(U, \phi)$ and $(V', \psi)$ are the charts guaranteed by your theorem, then you can take $$ V = \psi^{-1} \Big( \psi(V') \cap (\phi(U) \times \mathbb{R}^{m-n}) \Big). $$ Then $(U, \phi)$ and $(V, \psi|_V)$ satisfy the conditions of this new theorem.

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  • $\begingroup$ Thank you so much! this does it (had to change U to \phi(U) in the last equation) but that was an obvious fix. $\endgroup$ – user25959 Jan 16 '15 at 23:49

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