# actual dimension of concrete moduli space of holomorphic curves vs its virtual dimension

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 $2E$.

## 2 Answers

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.

Ok. I have managed to convince myself that there are no other curves. Basically because the cohomology ring of $M$ is given by $$H^*(M;\mathbb{Z})=\mathbb{Z}[\alpha,\beta]/(\alpha^3,\beta^3,\alpha \beta, \alpha^2 + \beta^2),$$ where $\alpha$ and $\beta$ are the poincare duals of the corresponding $P^1$'s, and so if $u$ is a k-covered sphere in the class $2E$, with $k\geq 2$, and $A$ is the homology class of the underlying simple sphere, we can write $A=a PD\alpha + b PD\beta$ for some integers $a,b$ and so $$kA=2E=2PD \alpha$$ implies $b=0$, $a=1$ and $k=2$, so that $A=E$ and hence the underlying simple sphere is $E$ itself, by the uniqueness of spheres mentioned above. I.e $u$ is a degree 2 cover of $E$.

Now, I am still trying to convince myself of the much more basic fact that a degree 2 map is determined by its two multiplicity 2 branch points, up to reparametrizaton. Can you ellaborate? Maybe this is obvious and someone should slap me in the face...

I cant comment below you, for some reason..

• Fair enough. But say, why is it true that if you have two maps with different branch points, then they do not differ by an automorphism preserving these points? I want to make sure I am not counting more things...That is what I dont see. – Agustín Moreno May 19 '15 at 9:54
• The branch points are in the image; it's the critical values that you're really counting. – Chris Gerig May 19 '15 at 16:06
• Oh, ok. That makes more sense... terminology failed me. Sure, if you fix $0$ and $\infty$ to be your branch points, and look at the maps that fix these two points and ramify at them to order 2, then they are of the form $z \mapsto az^2$, which up to reparametrization (i.e dilation) is just the map $z^2$. If you now have two other different ramifying points, take an automorphism mapping them to $0$ and $\infty$, and so, precomposing with it (and a dilation, if necessary) you get the map $z^2$ again. I.e the branch points determine your map up to parametrization, and now I am happy. – Agustín Moreno May 19 '15 at 16:53