# The double of a smooth manifold with boundary?

$\def\mc#1{\mathcal#1}\def\seq#1{\langle#1\rangle}\def\bbR{\mathbb R}\def\gt{>}\def\dom{{\rm dom\ }}$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 $\seq{0,x_1,\ldots x_{n-1}}$ .

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If you're looking for a reference, there's a detailed write-up in Kosinski's "Differentiable Manifolds" textbook, along the lines of Goodwillie's response. – Ryan Budney Jun 14 '11 at 23:54

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.

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Of course, this is independent of choice not in the sense that two choices give the same smooth structure on the double but in the sense that there is a diffeomorphism between the two smooth doubles. – Tom Goodwillie Jun 14 '11 at 23:27
Indeed, if $f:[0,1[\times\partial M\to M$ is a collar, the charts for the double covering the boundary are obtained by choosing an atlas $\mathcal B$ for the boundary, and then instead taking $\mathcal A_1=\{\check\theta:\theta\in\mathcal B\}$ where the functions $\check\theta$ are given by $m\mapsto f^{-1}(m)=(t,n)\mapsto(t,\theta(n))\in\mathbb R\times\mathbb R^{n-1}\simeq\mathbb R^n$ , and $(m,{\rm w})\mapsto f^{-1}(m)=(t,n)\mapsto(-t,\theta(n))$ with $m\not\in M_1$ , for $\theta\in\mathcal B$ . – TaQ Jun 15 '11 at 0:16

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)

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@TaQ: You're absolutely right. In dim 1, it is canonically $C^1$. In dim 2 it is only $C^0$. – André Henriques Jun 14 '11 at 22:46
@ André Henriques: "The double is only a $C^1$-manifold ..." How it is (is it?) generally even $C^1$ in dimensions at least two, even if one accepts any noncanonical structure without any functoriality? If it is $C^1$ , then one can put there a compatible $C^\infty$ structure. – TaQ Jun 14 '11 at 22:46
I decided to complete my first comment, and for that I first deleted it, and then put the completed comment. Meanwhile, André Henriques commented it, and now the order is a bit "funny". – TaQ Jun 14 '11 at 22:50
Actually the double of $M$ has a canonical $C^\infty$ structure, but in such a way that the resulting $C^\infty$ structures on the two copies of $M$ are not equal to, but merely diffeomorphic to, the original structure. You can map $Diff(\mathbb R_+)$ to $Diff(\mathbb R)$ by sending $\phi$ to $\bar \phi$ where $\bar \phi(x)=xh(x^2)^{1/2}$ with $\phi(x)=xh(x)$. – Tom Goodwillie Jun 14 '11 at 23:51
@Tom: Very nice observation. I see how it works. – André Henriques Jun 15 '11 at 8:55