The double of a smooth manifold with boundary?

Question

$\def\mc#1{\mathcal#1}\def\seq#1{\langle#1\rangle}\def\bbR{\mathbb R}\def\gt{>}\def\dom{{\rm dom\ }}$In some instances, I have seen an appeal to the concept of "the double" of a smooth manifold with non-empty boundary. Wikipedia gives a pure nonsense for this: "Precisely, the double is $M \times \{0,1\} / \sim $ where $ (x,0) \sim (x,1) $ for all $ x \in \partial M $ ." The essential problem here is how to construct pairwise smoothly compatible charts covering the boundary. I ask whether anyone reading this knows how to do this.

My natural idea of constructing "the" double would be the following. Let $\mc A$ be an atlas for the given $n-$dimensional smooth compact manifold with boundary. Let $M_0$ be the "interior" and $M_1$ the "boundary". Let $\mc A_0$ contain the charts $\phi:\dom\phi\to\bbR^n$ belonging to $\mc A$ with $\dom\phi\cap M_1=\emptyset$ , and let $\mc A_3$ contain the rest. So the functions belonging to $\mc A_3$ are bijections from some subset of $M_0\cup M_1$ onto some set of points $x=\seq{x_0,x_1,\ldots x_{n-1}}\in\bbR^n$ where $x_0\ge 0$ , and mapping points $m\in M_1$ to $x$ with $x_0=0$ .

Then one would take as a generating atlas of a "doubled manifold" the set $\bar{\mc A}=\mc A_0\cup\mc A_1\cup\mc A_2$ where $\mc A_2$ contains the functions $\tilde\phi:(m,{\rm w})\mapsto\phi(m)$ where $\phi\in\mc A_0$ and the fixed ${\rm w}$ is chosen so that $(m,{\rm w})\not\in M_0\cup M_1$ for $m\in M_0$ . As the elements of $\mc A_1$ one would take the functions $\bar\phi$ for $\phi\in\mc A_3$ constructed as follows. Let $P:\bbR^n\to\bbR^n$ be the bijection $\seq{x_0,x_1,\ldots x_{n-1}}\mapsto\seq{-x_0,x_1,\ldots x_{n-1}}$ . Then define $\bar\phi$ so that $\dom\phi\owns m\mapsto\phi(m)$ and $(M_0 \cap \dom \phi ) \times \{ {\rm w} \} \owns(m,{\rm w})\mapsto P(\phi(m))$ .

Then $\bar{\mc A}$ defines a "continuous atlas for the double", i.e. the chart changes $\psi\circ\phi^{-1}$ are homeomorphisms between some open subsets of $\bbR^n$ for $\phi,\psi\in\bar{\mc A}$ , but they need not be differentiable at points $\seq{0,x_1,\ldots x_{n-1}}$ .

Answer 1

I believe that the usual remedy is a collar. That is, for any smooth manifold there is a suitable diffeomorphism from a neighborhood of $\partial M$ to $[0,1)\times \partial M$, or in other words a smooth embedding $[0,1)\times \partial M\to M$ that is "the identity" on the boundary. This allows you to glue along the boundary and get a smooth manifold. To see that the result is independent of the choice of an embedding you use the fact that any two such embeddings are smoothly isotopic.

Answer 2

The doubled manifold is only a (piecewise smooth) $C^0$-manifold, unless you put more structure on the initial manifold with boundary.

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In dimension one, then you get a little bit more: you get a $C^1$-structure on the double.
But still, you do not get a $C^2$-structure.

Here's how it goes:
Take $\mathbb R_+$ with its standard smooth structure. Its double is $\mathbb R$.

Now let's analyze this further:
If you want that construction to be functorial (w.r.t diffeomorphisms), then you would like the diffeomorphism group of $\mathbb R_+$ to act on $\mathbb R$. In other words, you want a group homomorphisms $$ Di\!f\!f(\mathbb R_+) \longrightarrow Di\!f\!f(\mathbb R),\qquad \varphi\mapsto\bar\varphi, $$ where $\bar\varphi$ is defined by $\bar\varphi(x):=\varphi(x)$ for positive $x$, and $\bar\varphi(x):=-\varphi(-x)$ for negative $x$.

Now take $\varphi(x):=x+x^2$. One easily checks that $\bar\varphi$ is not $C^2$!

$\qquad$ Conclusion:
$\qquad$ the double of $\mathbb R_+$ is only equipped with a canonical $C^1$ structure.
$\qquad$ It does NOT have a canonical $C^2$ structure.

Note: The same argument as above with $\mathbb R\times \mathbb R_+$, and the map $\varphi(x,y):=(x+y,y)$ shows that the double of $\mathbb R\times \mathbb R_+$ is not $C^1$.

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On the positive side, here are two situations where it is possible to equip the double with a canonical smooth strucutre:

• if your manifold is Riemannian structure, and the boundary totally is geodesic.

• in two dimensions, a complex structure induces a smooth structure on the double.
  (no compatibility required between the cx structure and the boundary)