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...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.

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    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 $M$.2010-10-10
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    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.2010-10-10
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    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 $TM$.2010-10-10
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    Hmm. Also, regarding your first point, I think conditions (i) and (ii) should deal with this.2010-10-10
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    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.2010-10-11

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