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 $C^1$.

• 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)

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 $C^1$.

• 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)

The double is only a $C^1$-manifold, unless you put more structure on your manifold with boundary.
This can already be seen in one dimension.

But 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.

• 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)

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 $C^1$.

• 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)