Clarification on the Definition of a Submanifold of $\mathbb{R}^N$ (and the differentiability of it’s charts)

differential-geometrymanifoldssmooth-manifoldssubmanifold

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Question

My question is rather simpler and is only related to a special case: on submanifolds of $\mathbb{R}^N$. Specifically, looking at the definition of a submanifold in Janich's book at page 8, where the submanifold charts are simply restrictions of certain charts of the parent manifold (which do preserve the topological structure), let's then proceed to the definition of the tangent space of submanifolds $M$ of $\mathbb{R}^N$ on page 26. There, as far as I understand, the differentiability of the charts $(U,h)$ of the $M$ themselves is assumed.

Janich page 26

Naturally, in the general case, the charts of differentiable manifolds are not differentiable themselves, rather only their transition maps. Further I do understand that the actual definitions of the tangent space (i.e. linear space of curve equivalence classes or linear space of derivations on germs) do not have such assumptions. Going back to submanifolds of $\mathbb{R}^N$ and looking at Munkres' text, where only this special case is treated, the charts on a manifold are defined as differentiable from the start (see page 196),

Munkres page 196

I am trying to understand how (where) it is implied in Janich's book that manifold charts, when the manifold is a submanifold of $\mathbb{R}^N$, are differentiable. I do have the following interpretation which I submit for your review. Guided by the following excerpt from Spivak's volume 1, page 30

Normally, of course, we will suppress mention of the atlas for a differentiable manifold, and speak elliptically of the "differentiable manifold $M$"… It will always be understood that $\mathbb{R}^N$ refers to the pair $(\mathbb{R}^N,{\cal U})$

where in said text, $\cal U$ refers to the maximal atlas containing the identity chart $I(x) = x,\ x\in\mathbb{R}^N$, will it not then be the case that since $(U,h)$ of $M$ (when not restricted) is a chart of $(\mathbb{R}^N,{\cal U})$ and thus is differentiably compatible with the chart $(\mathbb{R}^N,I)$? It then follows immediately that $h$ is differentiable on $\mathbb{R}^N$ since $h\circ I = h$ will be differentiable

Edit 1

Please see below the Janich's definition of a submanifold

enter image description here

Best Answer

Jänich uses the same convention as Spivak. At the end of chapter 1.1 on p.3 he writes

We also agree that, without an explicit statement to the contrary, a chart $(U,h)$ on the manifold $M$ always means a chart in the differentiable structure.

On $\mathbb R^N$ one has a canonical atlas $\mathfrak A(\mathbb R^N)$ consisting of the single chart $(\mathbb R^N,id_{\mathbb R^N})$. This atlas is trivially differentiable because all its charts (there is only one!) are differentiably related.

The atlas $\mathfrak A(\mathbb R^N)$ generates the standard differentiable structure $\mathfrak D(\mathbb R^N)$ on $\mathbb R^N$. The charts in $\mathfrak D(\mathbb R^N)$ are precisely the $N$-dimensional charts $(U,h)$ on $\mathbb R^N$ which are differentiably related to $id_{\mathbb R^N}$. The map $h$ is a homeomorphism $h : U \to U'$ between open subsets of $\mathbb R^N$. The two transition functions between $(U,h)$ and $(\mathbb R^N,id_{\mathbb R^N})$ are $$h \circ id_{\mathbb R^N}^{-1} : id_{\mathbb R^N}(U \cap \mathbb R^N) \to h(U \cap \mathbb R^N) \tag{1}$$ and $$id_{\mathbb R^N} \circ h^{-1} : h(U \cap \mathbb R^N) \to id_{\mathbb R^N} (U \cap \mathbb R^N) \tag{2} .$$ But $U \cap \mathbb R^N = U$, thus the transition functions are nothing else than $h : U \to U'$ and $h^{-1} : U' \to U$. By the definition of "differentiably related" both must be $C^\infty$. In other words, $h$ must be a diffeomorphism in the sense of classic multilinear calculus.