I'm reading McDuff and Salamon's book on $J$-holomorphic curves and am curious about some examples.

They say that if $(M^{2n},J)$ is an almost complex manifold, $(\mathbb{CP}^1,j)$ the Riemann sphere, $A\in H_2(M;\mathbb{Z})$ a homology class, and $G=PSL(2,\mathbb{C})$ the group of Moebius transformations of $\mathbb{CP}^1$, then the space $\mathcal{M}(A,J)/G$ of $(j,J)$-holomorphic curves $u:(\Sigma,j)\to (M,J)$ in the homology class $A$, modulo the action of $G$, is a manifold of dimension $$\dim \mathcal{M}(A,J)/G=2n+2c_1(A)-6.$$

As an example, suppose $(M,J)$ is $\mathbb{CP}^n$ with the standard Kaehler structure and $A$ is the homology class of linear embeddings of $\mathbb{CP}^1$ in $\mathbb{CP}^n$. Note that in this case $c_1(A)=n+1$, so $\mathcal{M}(A,J)/G$ has real dimension $4(n-1)$. Is this moduli space compact? What are its other topological properties, like CW structure, homology, homotopy, etc. Is it a "familiar" compact manifold?

I would also appreciate other examples where $\mathcal{M}(A,J)/G$ has been "determined", and references to where they have been computed.


To the best of my knowledge, the main strategy to compute a moduli space $M(A, J)$ consists in:

  • finding a symplectic form $\omega$ on $M$ which (at least) tames $J$ (ideally, $J$ and $\omega$ are compatible) and for which $J$ is regular i.e. a regular value for the projection map $\mathcal{M}(A, \mathcal{J}_{\omega}) := \bigcup_{J' \in \mathcal{J}_{\omega}} \mathcal{M}(A, J') \to \mathcal{J}_{\omega}$;

  • finding a path $\{J_t\}_{t \in [0,1]} \subset \mathcal{J}_{\omega}$ of regular almost complex structure such that $J_0 = J$ and $J_1$ is integrable i.e. a genuine complex structure on $M$ (if $M$ supports such a structure), so that the different $\mathcal{M}(A, J_t)$ are all diffeomorphic;

  • computing $\mathcal{M}(A, J_1)$ using techniques from algebraic geometry.

An important aspect of Gromov's use of almost complex structure and (pseudo)holomorphic curves was to explain the first two bullets.

In your example, $J$ is already integrable (being the standard complex structure on $M = \mathbb{C}P^3$), so the first two bullets above are irrelevant. I shall describe the moduli space $\mathcal{M}(A, J)$ for $M = \mathbb{C}P^n$ equipped with its standard (integrable) complex structure $J$ and $A$ the homology class of linear embeddings of $\mathbb{C}P^1$ into $\mathbb{C}P^n$.

Claim 1: The image of $u \in \mathcal{M}(A, J)$ is an algebraic subvariety of $\mathbb{C}P^n$.

Proof: We know that $u : \mathbb{C}P^1 \to \mathbb{C}^n$ is a smooth $(j, J)$-holomorphic map. Though we don't know if it is embedded at this point, one could argue (along the lines of this argument for instance) that the image is a (complex) analytic subvariety. But then Chow's theorem implies that the image of $u$ is an algebraic subvariety. We are therefore really allowed to use algebraic geometric tools. $\square$

Claim 2: $u \in \mathcal{M}(A, J)$ is a diffemorphic parametization of a projective line in $\mathbb{C}P^n$ i.e. the vanishing locus of $\mathrm{dim}_{\mathbb{C}}(\mathbb{C}P^n) - \mathrm{dim}_{\mathbb{C}}(\mathbb{C}P^1) = n-1$ linearly independent complex-linear functions.

Proof: It is a fact that given two distincts points in $\mathbb{C}P^n$, there is a unique projective line containing these two points. It is also a fact that the intersection number of a projective line with a generic hyperplane (i.e. the zero-locus of a single complex-linear map) is $+1$. Consider $u \in \mathcal{M}(A, J)$. Since $[u(\mathbb{C}P^1)] = A$, its intersection number (which is a homological invariant) with a generic hyperplane $H$ is $+1$. Pick two distincts points $p,q$ in the image $Im(u)$ of $u$, consider the unique projective line $L$ passing through these two points and consider any hyperplane $H$ which contains $L$. By Bertini's theorem, either $H$ contains $Im(u)$ or it intersects it in finitely many isolated points. If the second possibility were possible, then $H$ and $Im(L)$ would intersect at least in $p,q$; by positivity of intersection, each intersection would contribute at least $+1$, which contradicts $H \cdot Im(u) = 1$. Hence $H \supset Im(u)$. Since this is true for all $H$ which contains $L$ and since $L$ is the intersection of all such $H$, we deduce $Im(u) \subset L$. Since $[Im(u)] = [L]$, the map $u : \mathbb{C}P^1 \to L$ not only has to be surjective (for otherwise $Im(u)$ would be contractible), but it also has to be injective and a submersion (for otherwise, by the open mapping theorem, it would be a multi-covering). $\square$

So the problem is reduced to computing the moduli space of projective lines. Each projective line is the quotient to $\mathbb{C}P^n$ of a unique complex subspace $K \subset \mathbb{C}^{n+1}$ which is the common zero-locus of $n-1$ linearly independent complex-linear maps to $\mathbb{C}$, that is the kernel of a surjective complex-linear map $l : \mathbb{C}^{n+1} \to \mathbb{C}^{n-1}$. We can interpret $l$ as an complex $(n-1)$-frame in $\mathbb{C}^{n+1}$. Two such maps $l, l' : \mathbb{C}^{n+1} \to \mathbb{C}^{n-1}$ have the same kernel $K$ if and only if there is an element $M \in Gl(n-1, \mathbb{C})$ such that $l' = M \circ l$.

We deduce that $\mathcal{M}(A, J)/G$ is the quotient by $Gl(n-1, \mathbb{C})$ of the manifold of complex $(n-1)$-frames in $\mathbb{C}^{n+1}$, namely the Grassmannian $Gr_{\mathbb{C}}(n-1, n+1)$. This is a compact manifold of real dimension $$2(n-1)[(n+1)-(n - 1)] = 4(n-1) = 2n + 2(n+1) - 6 = 2n + c_1(A) - 6 \; .$$

Remark: the compacity of $\mathcal{M}(A,J)/G$ could have been deduced directly from Gromov's compactness theorem, since the energy $\omega(A)$ is constant (hence bounded) and $A$ cannot be decomposed as a positive sum of other homology classes (since it generates $H_2(\mathbb{C}P^n, \mathbb{Z})$).

One can expect that there are a lot of triplets $(M, J, A)$ for which $\mathcal{M}(A, J)$ is known, since algebraic geometry is such an old, developped and active subject. It is thus difficult to refer to any specific place in the literature for any specific computations, as there are a lot of 'folklore' scattered here and there. (However this might be only a personal lack of expertise on my part.) I would thus mention only one other moduli $\mathcal{M}(A, J)$ which was considered by Gromov in order to prove (among other things) his symplectic nonsqueezing theorem. If $(V, \omega)$ is a compact symplectic manifold which is symplectically aspherical (i.e. the symplectic form vanishes on $\pi_2(V)$), we can consider $(M = V \times S^2, \omega \oplus \omega_0)$ where $(S^2, \omega_0)$ is the standard symplectic sphere (of some area). Considering $A = [\{pt\} \times S^2]$, then for a generic choice $J$ of compatible almost complex structure on $M$, the moduli space $\mathcal{M}(A, J)/G \cong V$; in fact, through each point of $M$, there is one and only one $J$-holomorphic curve in the class $A$ (there is no bubbling phenomenon thanks to the asphericity condition).

  • $\begingroup$ In the second bullet point above, why are all of the $\mathcal{M}(A,J_t)$ for $t\in [0,1]$ diffeomorphic? I know that they are oriented cobordant, but how do we know they are also diffeomorphic? $\endgroup$ – smnas Jun 14 '18 at 23:30
  • 1
    $\begingroup$ @smnas This is a strategy to compute a moduli space, but it might be difficult or impossible to implement in practice (e.g. the third bullet to work out). However, when considering a path $J_t$ among the regular values of the map $\mathcal{M}(A, \mathcal{J}_{\omega}) \to \mathcal{J}_{\omega}$, the preimage of this path (if compact) is a fiber bundle. It's like in Morse theory: all level sets over an interval of regular values are diffeomorphic. It's the crossing of an irregular value which spoils the diffeomorphicity of the level sets, yet preserve their cobordance. $\endgroup$ – Jordan Payette Jun 15 '18 at 10:44

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