# Rescaling a symplectic form and integral cohomology

Let $$(M,\omega)$$ be a symplectic manifold. I am trying to understand a procedure which seems so obvious that its implications are omitted in any article I could read. I encountered the following: assume that the cohomology class $$[\omega]$$ of the symplectic form lies in the image of the natural homomorphism $$\rho : H^2(M,\mathbb{Z}) \to H^2(M,\mathbb{R}).$$ Then up to rescaling $$\omega$$, we can assume that it is integral, that is $$[\omega] \in H^2(M,\mathbb{Z})$$.

## Here are my questions:

1. first of all, I don't understand why $$\rho$$ could not be an embedding. Does this come from the Universal Coefficient Theorem ?
2. If it is indeed not an embedding, then all I can say is that there exists a closed two form $$\tau$$ such that $$\rho([\rho]) = [\omega]$$. Then how could I obtain $$[\omega] \in H^2(M,\mathbb{Z})$$, even by rescaling the form ?
3. What does $$[\omega] \in H^2(M,\mathbb{Z})$$ mean in terms of values $$\omega$$ can take on $$2$$-dimensional submanifolds of $$M$$ ?
4. a more general question: rescaling a symplectic form seems to change the symplectic structure. Why does this procedure seem so meaningless ?

Any help will be appreciated, thanks a lot.

1) Sure, if you like, you can phrase this in terms of the universal coefficient theorem. We have the natural sequences $$0 \to \text{Ext}^1_{\Bbb Z}(H_{i-1}(X;\Bbb Z), R) \to H^2(X;R) \to \text{Hom}(H_2(X;\Bbb Z), R) \to 0.$$

In the case $$R = \Bbb R$$, the first term is zero. When $$M$$ is a compact triangulable space (such as a manifold) and $$R = \Bbb Z$$, the first term is $$H_{i-1}(X;\Bbb Z)_{\text{tors}}$$, the subgroup of torsion classes. (See eg here, Corollary 21.)

The naturality of this sequence implies that the map $$H^2(X;\Bbb Z) \to H^2(X;\Bbb R)$$ has kernel equal to $$H_{i-1}(X;\Bbb Z)_{\text{tors}} \subset H^2(X;\Bbb Z)$$; this is the torsion subgroup of $$H^2(X;\Bbb Z)$$. (Because $$\Bbb R$$ is divisible, these kernel elements should not surprise you.) The map $$H^2(X;\Bbb Z)/\text{Tors} \to H^2(X;\Bbb R)$$ is therefore injective, with domain identified with $$\text{Hom}(H_2(X;\Bbb Z), \Bbb Z) \cong \Bbb Z^{b_2}$$, and codomain identified with $$\text{Hom}(H_2(X;\Bbb Z), \Bbb R) \cong \Bbb R^{b_2}$$; the inclusion picks out a lattice in $$\Bbb R^{b_2}$$, and the induced map $$\left(H^2(X;\Bbb Z)/\text{Tors}\right) \otimes_{\Bbb Z} \Bbb R \to H^2(X;\Bbb R)$$ is an isomorphism.

1b) Notice, however, that this does not say that every symplectic form may be rescaled to be integral: we're asking that a line in $$\Bbb R^{b_2}$$ intersect some element other than $$0$$ in $$\Bbb Z^{b_2}$$. This is a very non-generic situation; it is true for a dense, measure 0 set of lines. That last isomorphism means that I can write every element of $$H^2(X;\Bbb R)$$ as a finite sum $$\sum c_i \omega_i$$, where $$c_i \in \Bbb R$$ and $$\omega_i \in H^2(X;\Bbb Z)/\text{Tors}$$.

2) A symplectic form lives in $$H^2(X;\Bbb R)$$. To say it is integral means it is in the lattice defined above, the image of $$H^2(X;\Bbb Z) \to H^2(X;\Bbb R)$$. This is equivalent to saying that $$\int_{\Sigma} \omega \in \Bbb Z$$ for every closed surface $$\Sigma \subset X$$. This is the content of saying we live in $$\text{Hom}(H_2 X, \Bbb Z) \subset \text{Hom}(H_2 X, \Bbb R).$$ You don't end up picking out a particular class in $$H^2(X;\Bbb Z)$$, but that's okay, that wouldn't be a differential form.

3) See above.

4) It changes the volume of submanifolds but not really any of the other symplectic geometry of $$M$$, other than by scaling factors. For instance, for fixed almost complex structure $$J$$, then (maybe after a scaling of $$J$$, I haven't checked) you expect the same moduli spaces of holomorphic discs, hence the same Gromov-Witten type invariants. It's the silliest thing you could do to a symplectic structure that technically gives you something new.

• Thanks a lot. Allow me to ask you some questions about this. Regarding 1): what does it mean for a homology class to lie in a cohomology group (namely you wrote $H_1(X,\mathbb{Z})_{\text{tor}} \subset H^2(X,\mathbb{Z})$) ? Regarding 1b): if I understand well you identify the codomain $H^2(X,\mathbb{R})$ of the injection $H^2(X,\mathbb{Z}) / Tor \to H^2(X,\mathbb{R})$ with $Hom(H_2(X,\mathbb{Z}),\mathbb{R}) = \mathbb{R}^{b_2}$ ? In this case, a line would represent a generator of the real second cohomology of $X$, and we ask the line representing $[\omega]$ to intersect the lattice ? Commented Dec 27, 2018 at 18:27
• 1) It's sloppy notation. I'm using the universal coefficient theorem, which identifies the space of torsion elements as a certain Ext-space, which here is that subgroup of $H_1$. 1b) Exactly. In fact, instead of measure 0, I should have said only countably many lines are admissible.
– user98602
Commented Dec 27, 2018 at 20:56
• 1b) By countably many, you mean as many as there are rational lines right ? Also, regarding the notation $[\omega] \in H^2(M,\mathbb{Z})$, consider the case of the existence of a prequantization bundle over a symplectic manifold $(M,\omega)$. It this existence ensured by the condition $[\omega] \in H^2(M,\mathbb{Z})$, or by $[\omega] \in H^2(M,\mathbb{Z}) / Tor$ (which is equivalent to saying that it is in the image of the homomorphism) ? Commented Dec 27, 2018 at 21:22
• @BrianT Yes, there are countably many rational lines. I don't know what a prequantization bundle means, but the only place $[\omega]$ lives when $\omega$ is a symplectic form is in de Rham cohomology. So it will never make sense to write $[\omega] \in H^2(M;\Bbb Z)$.
– user98602
Commented Dec 27, 2018 at 21:23
• Thanks for your help. Commented Dec 27, 2018 at 21:28