Definition of a Manifold from Guillemin Pollack

Question

I have been studying differential topology from Guillemin and Pollack (GP). Unlike many other books that define differentiable manifolds using maximal atlases GP starts by saying $ X \subset R^{N}$ for some ambient space $R^{N}$ and then goes on to define a $k$ dimensional manifold. But I know that this containment comes due to a weak version of Whitney's theorem.

Later on when they prove Whitney's theorem it is done so by induction on $N >= 2k+1$. But how to I justify that $ X \subset R^{N}$ in the first place? How can it just be assumed in definition like that?

I need help getting from the general definition of manifolds using atlases to the weak version of Whitney.

Thanks

Answer 1

There are two parts to the (weak) Whitney embedding theorem:

1) Any abstract manifold can be embedded in $\mathbb{R}^N$ for some $N$.

2) Any $k$-dimensional submanifold of $\mathbb{R}^N$ can in fact be embedded in $\mathbb{R}^{2k+1}$.

They prove part (2) of this. For a proof of (1), there's a nice exposition in Lee, Smooth Manifolds.

Here's the idea for (1) in the special case where the $k$-dimensional abstract manifold is compact: Let $U_i, \phi_i$ for $1 \leq i \leq m$ be a finite chart (possible by compactness), with $\phi_i: U_i \rightarrow \mathbb{R}^k$, and let $\rho_i: U_i \rightarrow \mathbb{R}$ be a subordinate partition of unity.

Then embed by $(\rho_1, \ldots \rho_m, \rho_1 \phi_1, \ldots, \rho_m \phi_m)$ to $\mathbb{R}^N$ for $N = m (k + 1)$.

Now prove this is injective and with injective derivative at every point. There's a hint for injective in the comments.

Answer 2

Compare with:

One can define a vector space of dimension $k$ as certain subsets of $\mathbb R^N$ for some large $N$ with certain algebraic properties. Later I show, that it is infact isomorphic to $\mathbb R^k$. Now, I can 're-axiomatize' the vector space without reference to $\mathbb R^N$ and show every basis vector can be represented as some $[0,0,\ldots ,1, \ldots, 0]^t$ with respect to itself.

The important point is that manifolds are objects made by patching together euclidean spaces of the same dimension.

Answer 3

It seems that GP mentioned the first part of the (weak) Whitney's theorem, i.e., any $k$-manifold $\mathcal{M}$ ($\subset \mathbb{R}^{N}$) can be one-to-one immersed in any larger $\mathbb{R}^{M}$ with $M>N$ at the bottom of page 48.

In fact, $\mathcal{M}$ is naturally one-to-one immersed into $\mathbb{R}^{N}$ by inclusion $i_0$ (resulting $\mathrm{d}i_0$ which is also an inclusion and obviously injective). Furthermore, any $\mathbb{R}^{N}$ can be canonically immersed into larger $\mathbb{R}^{M}$, where $M>N$, by catenating $M-N$ zeros, i.e., $(x_1,\dots,x_N)\rightarrow(x_1,\dots,x_N,0,\dots,0)$, denoted as $i_1$. Thus, the composite $i_1\circ i_0$ ($\mathcal{M}\rightarrow \mathbb{R}^N \rightarrow \mathbb{R}^M$) is the one-to-one immersion we wanted.

Using the above operations, after injectively immersing $\mathcal{M}$ into any large euclidean space $\mathbb{R}^M$ with $M>2k+1$, we can perform the procedure in the proof of the Whitney's theorem in GP to further collapse the ambient space to dimension $2k+1$.