I am looking at exercise 6.3.3 in Mcduff's and Salamon's book J-holomorphic curves and Symplectic topology, which basically gives an example of a moduli space whose actually dimension is greater than its virtual dimension. I am wondering how to compute the actual one.


The example is the following: 
Take your symplectic manifold to be the blowup of $\mathbb{C}P^2$ at a point, that is, $M=\mathbb{C}P^2 \# \overline{\mathbb{C}P^2}$, and let $A=2E$ be two times the class $E$ of the exceptional divisor, also denoted $E$. I want to see that the moduli space $\mathcal{M}\_{0,0}(M,A,J)$ of (not necessarily simple) unmarked $J$-holomorphic spheres in the homology class $A$ has dimension 4 for generic compatible $J$.


Its virtual dimension is actually 2, since we have that $c\_1(\mathbb{C}P^2)= 3 \alpha$, where $\alpha=PD[\mathbb{C}P^1]$ is the generator of $H^2(\mathbb{C}P^2;\mathbb{Z})$, and using the fact that the normal bundle to $E$ in $M$ is given by $\nu\_E=\mathcal{O}(-1)$, we get $TM|\_E= TE \oplus \mathcal{O}(-1)$, and hence $$c\_1(M)(E)=c\_1(E) + c\_1(\mathcal{O}(-1))=\chi(S^2)-1=2-1=1$$ Uisng the index formula with $g=k=0$, $n =2$, we then have that $$\mbox{virt-dim}\; \mathcal{M}\_{0,0}(M,A,J) = 2n+2c\_1(A)+2k-6 = 4+4c\_1(M)(E)-6 = 2$$


Thing is, I have no idea how to compute the dimension of a moduli space which does not achieve transversality. My guess is that I would have to explicitly compute this moduli space by hand and see that I get a 4-dimensional family of spheres. For instance, I know that there cannot be any simple curves in this space, because its intersection number with $E$ would have to be $E$.$2E= -2$, which is impossible by positivity of intersections. So I have only multiple covers. Similarly, the moduli space $\mathcal{M}\_{0,0}(M,E,J)$ is a one-point space (only has $E$), since any two curves there have intersection number $E.E=-1$, but it must be positive, again by positivity of intersections.


So I guess I would have to look at, say, degree 2 covers of $E$ itself, and the space of such covers. But it is not obvious to me how to count these, and if they are actually the only ones achieving the class $2E$. 



actual dimension of concrete moduli space of holomorphic curves vs its virtual dimension
**Moral:** This is giving a glimpse at the troubles with *branched* multiply-covered $J$-holomorphic curves. The virtual-dimension is a Fredholm index, involving $\chi(C)$ of your curve. If $C$ is multiply-covered with $b$ branch points then the index scales as $b$. But the actual dimension scales as $2b$ (the branch points can move around with 2 degrees of freedom), so we're off by $b$. In your exercise, $b=2$.


That space $\mathcal{M}$ of degree-2 self-maps $S^2\to S^2$ is 4-dimensional (topologically $\mathbb{C}P^2$), because they are determined by their 2 branch points (critical values) which move on the 2-dimensional sphere in the range. To elaborate, two maps in $\mathcal{M}$ are equivalent if they differ by (precomposition with) a holomorphic automorphism $h:S^2\to S^2$, and we now invoke Riemann's Existence Theorem, i.e. what $h$ can do once you know about the branch points. *Note:* There is another stratum where the two branch points coincide. When two branch points run into each other, i.e. when you enter $\overline{\mathcal{M}}-\mathcal{M}$, you get the map $S^2\vee S^2\to S^2$ which sends the intersection point to the single branch point.


To see that this gives all curves representing $A$, identify a tubular neighborhood of $E$ in $M$ with a neighborhood of the zero-section of $\mathcal{O}(-1)$ to see that there can't be other curves near the double covers.


