# Open subsets of the space of smooth structures

Let $M$ be a closed manifold admitting an almost complex structure and let $\mathcal{S}$ be the set of smooth structures on $M$ with the $C^{\infty}$ topology.

When constructing a moduli space of pseudoholomorphic curves on $M$, the almost complex structure on $M$ is allowed to vary in the space $\mathcal{J}_{\tau}(M,\omega)$ of $\omega$-tame almost complex structures, for some symplectic form on $M$. This forms an open subset of $\mathcal{S}$.

McDuff and Salamon (section 3.1) say that in fact any open (in the $C^{\infty}$ topology) subset $\mathcal{J}\subset \mathcal{S}$ of almost complex structures works just as well as $\mathcal{J}_{\tau}(M,\omega)$. In particular, a symplectic form on $M$ is not necessary.

My question is about replacing $\mathcal{J}_{\tau}(M,\omega)$ with a larger set when constructing a moduli space.

Question: Let $\mathcal{J}(c_1)\subset \mathcal{S}$ be the subset consisting of almost complex structures on $M$ with first Chern class $c_1$. Is $\mathcal{J}(c_1)$ an open subset of $\mathcal{S}$?

Note: Since $\mathcal{J}_{\tau}(M,\omega)$ is contractible, the first Chern class $c_1$ is constant in $\mathcal{J}_{\tau}(M,\omega)$, as shown here. Therefore, for any symplectic form $\omega$ on $M$ (if one exists), $\mathcal{J}_{\tau}(M,\omega)$ will always be contained in $\mathcal{J}(c_1)$ for some $c_1$.

• It is implicit in the argument you linked to that $c_1(TM, J)$ depends continuously on $J$. Hence the map $c_1(TM, -) : \mathcal{J} \to H^2_{dR}(M; \mathbb{R})$ is continuous. But this map factors through $H^2(M; \mathbb{Z})$ which is discrete and has the discrete topology, hence its singletons are open subset, so the preimages of $c_1(TM, -)$ are open subsets of $\mathcal{J}$. – Jordan Payette Jun 5 '18 at 19:35
• As to whether the set $\mathcal{J}$ of (smooth) almost complex structure on $M$ is open inside the set $\mathcal{S}$ of (smooth) endomorphisms of $TM$ (that is what I figure to be your 'set of smooth structures'): observe that the map $q : \mathcal{S} \to \mathcal{S} : S \mapsto -S^2$ is continuous, hence $\mathcal{J} = q^{-1}(\{Id\})$ is closed but not open. – Jordan Payette Jun 7 '18 at 14:51