# Is the sheaf associated to a differential structure of a specific type?

On a set $X$, let us define a set $\mathcal{D}$ of functions from $X$ to $\mathbb{R}$. Consider first the initial topology $\tau_\mathcal{D}$ on $X$ with respect to $\mathcal{D}$, i.e. the coarsest topology that makes all functions in $\mathcal{D}$ continuous (with the usual topology on $\mathbb{R}$). Given a $X$ and a set of functions $\mathcal{D}$, for $W \subseteq X$ we say $f:W \to \mathbb{R}$ verifies the property $P_W$ if

$$\forall x \in W, \ \exists (V,g) \in \tau_\mathcal{D} \times \mathcal{D} \text{ such that } V \ni x, \text{ and } f|_V = g|_V.$$

Now, a differential structure on a set $X$ is a set $\mathcal{D}$ of functions from $X$ to $\mathbb{R}$ that verifies the following axioms

1. $\mathcal{D}$ is a $\mathcal{C}^\infty$-ring: if $u_1,\ldots,u_N \in \mathcal{D}$ and $g \in \mathcal{C}^\infty(\mathbb{R}^N,\mathbb{R})$, then $g \circ (u_1,\ldots,u_N) \in \mathcal{D}$
2. It is locally determined: If $f:X \to \mathbb{R}$ verifies $P_X$ then $f \in \mathcal{D}$.

We can consider the functor that associates to an open set $U \in \tau_\mathcal{D}$ the set $$\mathbf{C}_\mathcal{D}(U) := \{ f:U \to \mathbb{R} \mid f \text{ verifies the property } P_U \}.$$ The two axioms ensure this is a sheaf of $\mathcal{C}^\infty$-ring.

The set of smooth functions of a paracompact manifold defines a sheaf of $\mathcal{C}^\infty$-ring which is fine, and it is for sure an important feature. Differential structure are meant to mimick these functions, when one does not have a manifold structure. On the other hand, we have by definition $\mathcal{D} = \Gamma(X,\mathbf{C}_\mathcal{D})$, so it seems the associated sheaf is somehow "determined" by its global sections.

I understand there is a whole typology of sheaves related to the local-VS-global properties: injective, flabby, soft, acyclic, fine. Is $\mathbf{C}_\mathcal{D}$ of one of these types? Can I actually define a differential structure as a sheaf of $\mathcal{C}^\infty$-ring of some well-known type ? Note that it is ok if you need some topological assumption, like $(X, \tau_\mathcal{D})$ is paracompact.

• What's the topology on $X$? $x \in U \in \tau_D$, but $\tau_D$ was defined as a topology on $D$ and not on $X$. In any case, the property of being fine is only defined, as far as I know, for paracompact Hausdorff spaces. – user40276 May 30 '18 at 1:12
• Given the usual topology on $\mathbb{R}$, the topology $\tau_\mathcal{D}$ is defined as the initial topology with respect to $D$, i.e. the thinnest topology on $X$ that makes all maps in $\mathcal{D}$ continuous, so it is a topology on $X$. As for your second comment, you're right, but that's why I mentionned the answer might require extra topological assumptions: if there is a well-known typology for $\mathbb{C}_\mathcal{D}$, but only if $(X,\tau_\mathcal{D})$ is paracompact, then it is ok for me. – Christophe Wacheux May 31 '18 at 7:21
• Ok. I can understand most of your notation after your edit. However that $P_X$ on the definition of the sheaf should probably be $P_U$, otherwise your question is trivial. In any case, I'm almost sure that none of the properties that you want will be satisfied even when you restrict the conditions on the topology of $X$. I'm less sure about the acyclicity under some conditions. – user40276 May 31 '18 at 23:17
• I'm too lazy to write down everything in the general case and I also don't have a scanner now. So let me just sketch the picture that I have in mind: pick $X = S^1$ and $3$ connected open sets $U, V$ and $W$, each one intersecting exactly two open sets. Think about these sets as open sets of $X$. Pick two functions $f$ and $g$ that agree only on $U \cap V$ and no other function agrees with $f$ or $g$ outside of $U \cup V$. Define a new function $h$ on $U \cup V$ that agrees with $f$ on $U$ and $g$ on $V$, then you can't extend $h$ (you get a kind of holonomy). – user40276 May 31 '18 at 23:29
• To get an example with other kind of topologies just keep adding open sets by adding other smooth functions on the circle and also choose $f$ and $g$ very non-smooth so that no function agrees with $f$ and $g$ outside $U \cup V$. – user40276 May 31 '18 at 23:38