# tangent bundle projection is $C^{\infty}$ even though the manifold is only $C^{k+1}$?

(All the definitions I'm using are from Abraham Marsden Ratiu's Manifolds, Tensor Analysis and Applications).

Here's the setup of my problem. Let $$M$$ be a $$C^{k+1}$$ manifold modelled on a Banach space $$E$$, and let $$TM$$ be its tangent bundle, with $$\pi: TM \to M$$ being the natural projection. I've already shown that the differentiable structure on $$M$$ allows us to construct one for $$TM$$, which makes $$TM$$ into a $$C^k$$ Banach manifold, modelled on $$E \times E$$. With this, I would expect that the projection map $$\pi$$ is at most $$C^k$$ smooth, but I have shown that it is $$C^{\infty}$$, which I think is quite absurd.

So, my question is, whether this is reasonable (which I doubt), and if it isn't then what is the fault in my reasoning below?

The definition of tangent space used in the book is that of equivalence classes of curves through a point. If $$\mathcal{A}$$ is an atlas for $$M$$, then we constructed the atlas $$T\mathcal{A} := \{(TU, T\alpha): (U,\alpha) \in \mathcal{A}\}$$ for the tangent bundle, where $$T \alpha : TU \to E \times E$$ is defined by \begin{align} [c] \mapsto \left( (\alpha \circ \pi)([c]), (\alpha \circ c)'(0) \right). \end{align} In words, it takes an equivalence class of curves $$[c]$$, and maps it to a tuple in $$E \times E$$, where the first coordinate is the chart-representative of the "base point", and the second coordinate is the "velocity vector" of the curve $$c$$ under the chart map $$\alpha$$.

Now, in the book, we defined $$f: M \to N$$ to be $$C^r$$ if for every $$x \in M$$, and every chart $$(U,\alpha)$$ of $$N$$ which contains $$f(x)$$, there is a chart $$(W, \beta)$$ of $$M$$ such that $$f(W) \subset U$$, and $$\alpha \circ f \circ \beta^{-1}$$ is $$C^r$$ as a map between open subsets of Banach spaces.

Hence, applying this definition in our context, for any $$[c] \in TM$$, and for any chart $$(U, \alpha)$$ of $$M$$ containing $$\pi([c])$$, the the chart $$(TU, T\alpha)$$ is such that $$\pi(TU) = U$$, and the "chart-representative" of $$\pi$$ is the map $$\alpha \circ \pi \circ (T\alpha)^{-1}$$, which after unravelling the definitions is \begin{align} (x,v) \mapsto x \end{align} (from a certain open subset of $$E \times E$$ into $$E$$). However, this is just projection onto the first factor which is linear and continuous and hence $$C^{\infty}$$. Hence, I concluded that $$\pi : TM \to M$$ is $$C^{\infty}$$.

So... what's the error in this reasoning? In the book, they also state that the chart representative of $$\pi$$ is $$(x,v) \mapsto x$$, but then they only conclude that $$\pi$$ is $$C^k$$ rather than $$C^{\infty}$$.

• The trouble is that the notion of a $C^\infty$-map to a $C^k$-smooth manifold $M$ is not well-defined. Depending on which charts in $M$ you take, starting from the same map $N\to M$, you get maps (between domains in Euclidean spaces) of different degree of smoothness (at least $C^k$, but sometimes not $C^{k+1}$). You have chosen charts which make $\pi$ to be $C^\infty$, I can choose a different set of charts, where it is not... – Moishe Kohan Jul 17 '19 at 3:48

According to the definition of $$C^r$$ which you have quoted, $$\pi$$ is indeed $$C^\infty$$. However, normally the notion of "$$C^r$$ map" is only defined between manifolds that are $$C^r$$ (or $$C^k$$ for $$k\geq r$$), and the definition in your book is nonstandard if it applies to manifolds that are not $$C^r$$.
Why shouldn't we define $$C^r$$ maps between manifolds that are not $$C^r$$? The reason is that we don't want $$C^r$$-ness of a map to depend on what charts we use. In other words, it is crucial that a map is $$C^r$$ iff it looks $$C^r$$ in every pair of charts on the domain and codomain. Your book's definition is a priori weaker, since it only requires that for each chart on the codomain, there exists a chart on the domain for which the map is $$C^r$$. However, on a $$C^r$$ manifold, this implies that actually every chart on the domain works, since any two charts on the domain differ by a $$C^r$$ diffeomorphism.
On the other hand, on a manifold that is not $$C^r$$ (say, it is $$C^k$$ for some $$k), a map which is $$C^r$$ by your definition can fail to be $$C^r$$ in some choices of charts. That's because if you change the chart used on the domain, you will change $$\alpha\circ f\circ\beta^{-1}$$ by composing with a $$C^k$$ diffeomorphism, and this composition need not be $$C^r$$ since $$k.
• It definitely makes sense that we want $C^r$-ness to be a chart-independent notion, and I guess in the book's definition (and the theorems which followed) I guess it was implicitly agreed (and I guess logically so) that the manifold's smoothness is $\geq$ the map's smoothness. But just to ensure I understand all the details regarding quantifiers and smoothness etc, could you verify the following definition for smoothness: – peek-a-boo Jul 17 '19 at 4:56
• Let $M$ and $N$ be $C^m$ and $C^n$ manifolds $(m,n \geq 1)$, and $f:M \to N$ a map. Then $f$ is said to be $C^r$ ($1 \leq r \leq \min\{m,n\}$) if for every pair of charts $(U,\alpha)$ of $M$ and $(W,\beta)$ of $N$ belonging to the maximal atlas, the chart representative map $\beta \circ f \circ \alpha^{-1}$ is $C^r$ in the Banach space sense. – peek-a-boo Jul 17 '19 at 4:56