Differential forms as a sheaf on the site of all manifolds vs. sheaf on an individual manifold

Question

The de Rham complex can be viewed as a sheaf $\Omega$ on the entire site $\mathsf{Man}$ of smooth manifolds via the usual pullback of differential forms $(f: M \to N) \mapsto (f^*: \Omega(N) \to \Omega(M))$. So, the de Rham complex is a smooth space. Diffeological spaces, for example, are concrete sheaves on the site of smooth manifolds, but I don't know enough about sheaves to say whether or not the de Rham complex is a diffeological space.

We can also consider the category of open sets of $M$ as a site $O(M)$, where the arrows are inclusions. Then we have the sheaf $\Omega_M \in \mathsf{Sh}(M)$ of differential forms on $M$. Likewise, we have the sheaf $\Omega_N \in \mathsf{Sh}(N)$ of differential forms on $N$.

Now, as I understand it, there are morphisms $i_M: \mathsf{Man} \to M$ and $i_N: \mathsf{Man} \to N$ of sites, given by the respective inclusion functors $i_M^t: O(M) \to \mathsf{Man}$ and $i_N^t: O(N) \to \mathsf{Man}$.

Can we describe the sheaves of differential forms $\Omega_M$ and $\Omega_N$ on $M$ and $N$ in terms of these morphisms of sites? I figure they should somehow be induced by the sheaf $\Omega$ on all of $\mathsf{Man}$, but I am not very experienced with sheaves on sites yet. Moreover, given a smooth map $f: M \to N$, there should a relationship between the inverse image functor and the pullback of forms.

Answer 1

The sheaf $ \Omega^k $ ($k>0$) on the site $ \mathsf{Man} $ is not a diffeological space, because it is not a concrete sheaf: Obviousely, $ \Omega^k(\mathbb{R}^k)\ni\alpha\mapsto \underline{\alpha}:\mathrm{hom}(\mathbb{R}^0,\mathbb{R}^k)\rightarrow\Omega^k(\mathbb{R}^0)=0 $ is not one-to-one.

Sheaves $ \Omega_M $ and $ \Omega_N $ are the restriction of the sheaf $ \Omega $ to the sites $ O(M) $ and $ O(N) $, respectively. For a smooth map $f:M \rightarrow N $, the induced inverse image functor commutates with the restrictions. The pullback induced by $ f $ is just the global section map.

Answer 2

The short answer is that you have a morphism of sheaves $f^\ast \Omega_N \to \Omega_M$; equivalently, by adjunction, a morphism $\Omega_N \to f_\ast\Omega_M$. This map of sheaves gives the "global" pullback map on forms as $$\Omega(N) = \Gamma(N,\Omega_N) \to \Gamma(N,f_\ast \Omega_M) = \Omega(M).$$

Two examples:

• Suppose $M \hookrightarrow N$ is a submanifold. Then $f^\ast \Omega^1_N \to \Omega^1_M$ is a surjection between vector bundles (where we identify a vector bundle with its sheaf of sections). The kernel is the conormal bundle.

• Suppose $M \to N$ is a submersion. Then $f^\ast \Omega^1_N$ is a subbundle of $\Omega^1_M$. The quotient bundle is the vertical cotangent bundle.