Rescaling a symplectic form and integral cohomology

Question

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.

Answer 1

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.