(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 $C^{\infty}$.

  • 1
    $\begingroup$ 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... $\endgroup$ – 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 $k<r$), 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 $k<r$.

  • $\begingroup$ 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: $\endgroup$ – peek-a-boo Jul 17 '19 at 4:56
  • $\begingroup$ 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. $\endgroup$ – peek-a-boo Jul 17 '19 at 4:56
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    $\begingroup$ That's exactly right. $\endgroup$ – Eric Wofsey Jul 17 '19 at 4:58
  • $\begingroup$ thank you very much! this question just emphasized to me once again the importance of having a crystal clear definition and understanding it, and reminds me of a quote in Spivak's preface to Calculus on Manifolds: "There are good reasons why the theorems should all be easy and definitions hard." $\endgroup$ – peek-a-boo Jul 17 '19 at 5:08

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