The double of a smooth manifold with boundary?

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.


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


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 $C^1$.


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)