# Constructing diffeomorphisms of moduli spaces of $J$-holomorphic curves

Let $$M^{2n}$$ be a smooth manifold admitting two almost complex structures $$J_0$$ and $$J_1$$.

Suppose that $$J_0$$ and $$J_1$$ are both regular in the sense that the moduli space $$\mathcal{M}_i:=\mathcal{M}(A; J_i)/PSL(2,\mathbb{C})$$ of $$J_i$$-holomorphic spheres representing the homology class $$A\in H_2(M;\mathbb{Z})$$ is a smooth, compact manifold for both $$i=0,1$$. Suppose further that there exists a path $$(J_t)_{t\in [0,1]}$$ of regular almost complex structures on $$M$$ connecting $$J_0$$ and $$J_1$$.

In McDuff and Salamon's book they show that the manifolds $$\mathcal{M}_t:=\mathcal{M}(A;J_t)/PSL(2,\mathbb{C})$$ are oriented cobordant for all $$t\in [0,1]$$.

QUESTION: Are the moduli spaces $$\mathcal{M}_t$$ diffeomorphic for all $$t\in [0,1]$$? If not, how does the diffeomorphism fail?

The issue here is precisely what you mean when you say "regular path $$J_t$$ of almost complex structures".

I am not particularly expert in the theory of holomorphic curves, so I won't comment on that; I expect the situation is precisely analagous to the finite-dimensional setting I will describe below (I'm sure there are more technical issues involving the corner points).

Let $$M$$ be a smooth manifold, and $$S \subset N$$ a closed submanifold. Given a map $$f: M \to N$$ transverse to $$S$$, we have that $$\mathcal M_f = f^{-1}(S)$$ is a smooth manifold (with corners etc if $$S$$ and $$N$$ have them).

The condition "$$f \pitchfork S$$" is the precise analogy to "$$J_t$$ is regular" in this finite-dimensional setting.

Suppose you have a path of functions $$f$$, written as a map $$f_t: I \times M \to N$$. Then I would say that this is a regular path of functions if the map $$I \times M \to N$$ is transverse to $$S$$, and the same for $$\{0,1\} \times M \to N$$. Then the "parameterized moduli space" $$\mathcal M_I$$ is a smooth manifold with boundary equal to $$\mathcal M_0 \sqcup \mathcal M_1$$. This space is equipped with a smooth map $$\mathcal M_I \to I$$ which is a submersion near the boundary.

The point is that each individual $$f_t$$ is not assumed to be regular; spelling it out, it is entirely possible that for $$(t,x)$$, we have $$f_t(x) \in S$$, and $$\text{Im}(df_t) \subset N_{f_t(x)} S$$ is not the entirety of the normal space, but rather of codimension 1; and so that together with the additional vector $$\frac{d}{dt} f_t(x) \in N_{f_t(x)} S$$, we span the whole of the normal space.

The second point is that an individual $$\mathcal M_t$$ is regular if and only if $$t \in I$$ is a regular value of the projection $$\mathcal M_I \to I$$.

So if, in addition, $$f_t$$ is regular for each $$t$$, then by definition the projection $$\mathcal M_I \to I$$ is a submersion; as you know well, a proper smooth submersion is a fiber bundle, and this provides a global diffeomorphism $$\mathcal M_I \cong I \times \mathcal M_0$$.

In almost every interesting application, given any two regular functions / almost complex structures / whatever, it will be essentially impossible to find a path through regular (blahs) between them; but it will be possible to find a "regular path" of (blahs) between them in the sense given above. A simple case to think about the discussion above with is $$S$$ a hypersurface and $$M$$ a point! If $$S$$ disconnects the codomain, then to get from one side to another you will need to cross $$S$$. In this situation, $$\mathcal M_I$$ will be a finite set of points which project to the interior of $$I$$.