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 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.

  • $\begingroup$ 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 ? $\endgroup$
    – BrianT
    Commented Dec 27, 2018 at 18:27
  • $\begingroup$ 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. $\endgroup$
    – user98602
    Commented Dec 27, 2018 at 20:56
  • $\begingroup$ 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) ? $\endgroup$
    – BrianT
    Commented Dec 27, 2018 at 21:22
  • $\begingroup$ @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 $[\omega] \in H^2(M;\Bbb Z)$. $\endgroup$
    – user98602
    Commented Dec 27, 2018 at 21:23
  • $\begingroup$ Thanks for your help. $\endgroup$
    – BrianT
    Commented Dec 27, 2018 at 21:28

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