# Why is the manifold structure on the tangent bundle unique?

...subject to the conditions that (i) the projection be smooth and that (ii) smooth sections correspond to smooth vector fields.

This homework problem is really bugging the hell out of me. Of course it can be checked locally, so we'll look at an open neighborhood $U\in \mathbb{R}^n$ and check that $TU\cong U\times \mathbb{R}^n$ (where the isomorphism is as topological spaces with sheaves of functions that we declare to be smooth). On the one hand, it's easy to see that $U \times \mathbb{R}^n$ satisfies these two conditions, since the projection is literally just a projection (in coordinates) and the correspondence is

• $X$ a vector field --> $s = (x_1,\ldots, x_n, X(\partial/\partial x_1), \ldots, X(\partial/\partial x_n))$
• $s=(s_1,\ldots, s_n)$ a smooth section --> $X = \sum_i s_i\cdot\partial/\partial x_i$.

On the other hand, I'm having a lot of trouble showing that the obvious sheaf (which I'll denote $O_{TU}$) is the unique one with properties (i) and (ii).

For example, suppose $\mathcal{F}$ is another sheaf of functions on $U\times \mathbb{R}^n$. Suppose that $f \in \mathcal{F}(V) \backslash \mathcal{O}_{TU}$ for some open set $V\subseteq U$. There really doesn't seem to be any way to show that this violates property (i), because what it really means is that for any $g\in O_U$, $g \circ \pi \in O_{TU}$, and the hypothesis of this statement begins with a function on $U$, not $TU$. On the other hand, I don't see any way to make a vector field on $U$ out of $f$, and (according to this question) it feels likely that I can't necessarily get a section that detects the failure of $f$ to be $O_{TU}$-smooth. Nor am I having much luck deriving a contradiction from $f \in O_{TU}(V) \backslash \mathcal{F}(V)$.

So, I'd welcome any suggestions as to how I should proceed.

• Note that you can't solve this problem by thinking locally. I.e., if $(M,\mathcal{E})$ and $(M,\mathcal{F})$ are two distinct differentiable structures on $M$, then the restrictions to a small enough neighbourhood will always be isomorphic. I don't have an actual answer to your question, but one approach to think about is to show that any choice of an appropriate differentable structure on $TM$ defines a unique differentiable structure on $M$. – yasmar Oct 10 '10 at 18:40
• Really?? This doesn't seem right to me. E.g. my impression is that for exotic $\mathbb{R}^4$, you consider $\mathbb{R}\times \mathbb{R}^3$ and glue your $\mathbb{R}^3$'s together as time passes in crazy ways. This seems distinct from $(\R^4,O_{std})$ at every point. In any case, I'm hoping to prove this without reference to the word "differentiable", i.e. I'd like it to work equally well for "analytic", "holomorphic", or whatever other word you want to throw in there. – Aaron Mazel-Gee Oct 10 '10 at 19:54
• Also, as for your last statement, presumably you mean that $M$ should inherit a sheaf of functions from its inclusion as the 0-section. I'd imagine that two distinct sheaves of functions on $TM$ can still agree there, which would mean that that's not a fine enough 'invariant' of the differentiable structure on $TM$. – Aaron Mazel-Gee Oct 10 '10 at 19:57
• Hmm. Also, regarding your first point, I think conditions (i) and (ii) should deal with this. – Aaron Mazel-Gee Oct 10 '10 at 20:28
• Okay, maybe I'm finally cluing in. My comment (and its elaboration below) is ignoring the special structure that we were explicitly asked to pay attention to. – yasmar Oct 11 '10 at 2:37

This is actually easier if you use charts. Suppose your smooth structure is given by a maximal atlas C. Near any point $p$ in $TU$ you have a neighborhood $V=\pi^{-1} W$ for $W$ - a chart neighborhood of $\pi(p)$, so that on $V$ which there are functions $x_i \cdot\pi$ which are smooth by property (1) and functions $\frac{\partial}{\partial x_i}$, which are smooth by property (2). But they also define a chart - a map $V$ to $R^{2n}$. So this chart has to be smooth. So we have a cover by charts smooth in both the standard atlas and the maximal atlas C. Hence the two smooth structures coincide.

Trying to do it in sheave seems a bit strange. If your structure sheaf is F, you get $2n$ functions in $F(V)$ and the fact that they define a chart should mean that you can pull back (restriction of) any function f in F to a function on part of $R^{2n}$, so that restriction is in the standard structure sheaf, and hence, (by sheaf axiom) f is too. This seems slightly fishy, even though can probably be fixed, in which case it will be the sheaf translation of the above argument.

• This seems really close, but here's what I'm nervous about. Consider the real line with coordinate $t$, and suppose we happen to put on a "coordinate" $x=t^3$. Certainly $x$ is a smooth function from the line to itself, but it's somehow not fundamental enough, because $t$ is supposed to be a smooth function too, but $t=x^{1/3}$ and this is not smooth at 0. – Aaron Mazel-Gee Oct 20 '10 at 17:18
• It's worse. Since there are no 'vertical' sections, I can't actually prove the vertical coordinates to be continuous in the smooth structure of C. – Max Oct 21 '10 at 20:44
• I've talked more with the professor, and it sounds like the problem itself might be wrong. When I get a chance, I'll try to see if what I've said actually makes an explicit counterexample... – Aaron Mazel-Gee Oct 27 '10 at 8:57

First, I am pretty rusty on this stuff, so hopefully somebody with more confidence will chime in. I started writing this as another comment, but it didn't fit.

Anyway, I still think what I said makes sense. If $\mathcal{E}_{\mathbb{R}^n}$ is the standard differentiable structure on $\mathbb{R}^n$, and $\mathcal{C}_M$ is the sheaf of continuous functions on a topological manifold $M$, then a subsheaf $\mathcal{F}$ of $\mathcal{C}_M$ is a differentiable structure on $M$ if (and only if)

1) at each point $p\in M$ there exists a neighbourhood $V$, and $x_1,...x_n \in \mathcal{F}(V)$ such that $q \mapsto (x_1, ..., x_n)$ is a homeomorphism onto an open $U \subset \mathbb{R}^n$, and

2) $f \in \mathcal{F}(V)$ iff there exists $F \in \mathcal{E}_{\mathbb{R}^n}(U)$, such that $f = F(x_1,...,x_n)$.

I think this definition is derived from Warner ("Foundations on differentiable manifolds and Lie groups"), but it has paraphrased by me and possibly been distorted through the lens of an absurdly long time. In particular, that second condition should probably be reworked to be more obviously a purely sheaf theoretic statement.

In any event, a proper definition of a differentiable structure is going to resemble something like this (likewise for holomorphic structure, or analytic structure or whatever). And the point is, that the \emph{definition} will require that your sheaf of differentiable functions on $M$ will be locally isomorphic to the \emph{standard} sheaf of differentiable functions on $\mathbb{R}^n$.

This should apply even if the topological manifold $M$ is $\mathbb{R}^n$. This is why I say that you can't solve the problem locally: Locally all differentiable structures are isomorphic to the standard one on $\mathbb{R}^n$ by definition.

As to my second statement, about an "appropriate" differentiable structure on $TM$ defining a unique differentiable structure on $M$, this is mostly a guess. However, the qualifier "appropriate" is important here. In particular, whatever differentiable structure on $TM$ you choose, it has to restrict to the standard structure on $\mathbb{R}^n$ on the fibres. In this way, I don't think there is any choice left once you've defined the restriction on the zero section.