# 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 $\seq{0,x_1,\ldots x_{n-1}}$ .

• 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. 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.

• 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. 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
• @TomGoodwillie: If one had instead glued two copies of $M\setminus\partial M$ along collars via the map $(0, 1)\times \partial M \to (0, 1)\times \partial M$, $(t, x) \mapsto (1-t, x)$, would this result in a diffeomorphic manifold? This approach is more akin to the description of connected sum introduced by Kervaire and Milnor (as opposed to cutting out open disks and gluing along boundaries). Jul 19 at 15:04
• @Michael Albanese: Yes. Jul 19 at 18:41

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)

• @TaQ: You're absolutely right. In dim 1, it is canonically $C^1$. In dim 2 it is only $C^0$. 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 $\phi(x)=xh(x)$. Jun 14 '11 at 23:51
• @Tom: Very nice observation. I see how it works. Jun 15 '11 at 8:55