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?
Best Answer
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 $I$.