Given two linear symplectic 4-tori. When are they symplectomorphic?

Question

What are symplectic invariants that can be computed for a linear symplectic 4-torus? The symplectic volume is the simplest one. Is it possible to have something else?

Answer 1

A symplectic form $\omega$ on a compact manifold $M$ determines a non-trivial cohomology class $[\omega] \in H^2_{dR}(M ; \mathbb{R})$. This cohomology class is not quite a symplectic invariant, but it almost is. Any 'scalar' constructed in an 'intrinsic way' from this cohomology class is a symplectic invariant.

Consider two diffeomorphic $2n$-manifolds $M_1$ and $M_2$ and suppose that $(M_1, \omega_1)$ and $(M_2, \omega_2)$ are symplectic manifolds. Are they symplectomorphic? Or more generally: which symplectic invariants do they have equal?

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Consider for instance volume. The volume of $(M_i, \omega_i)$ is defined as $\int_{M_i} \wedge^n \omega_i$, which can be computed in (co)homological terms as $\langle [\wedge^n\omega_i], [M_i] \rangle = \langle \cup^n[\omega_i], [M_i] \rangle$. Here, $[M_i] \in H_{2n}(M_i ; \mathbb{Z})$ is the (positive) generator (with respect to the orientation on $M_i$ induced by $\omega_i$) of the top singular homology group, also called the fundamental class of $M_i$. The volume is a 'scalar' computed in an 'intrinsic way' from $[\omega_i]$ and other 'intrinsic objects' defined on $M_i$. If a symplectomorphism $\phi : (M_1, \omega_1) \to (M_2, \omega_2)$ exists, then

$$ \langle \cup^n[\omega_1], [M_1] \rangle = \langle \cup^n[\phi^*\omega_2], [M_1] \rangle = \langle \cup^n(\phi_{\ast})^*[\omega_2], [M_1] \rangle = \langle (\phi_*)^*\cup^n[\omega_2], [M_1] \rangle = \langle \cup^n[\omega_2], (\phi_*)_*[M_1] \rangle = \langle \cup^n[\omega_2], [\phi(M_1)] \rangle = \langle \cup^n[\omega_2], [M_2] \rangle , $$

so the volume is a symplectic invariant.

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In the same spirit, since a 2-cohomology class naturally acts on a 2-homology class (that is, since a 2-form can be integrated on a 2-surface), we have (so to say) symplectic invariants associated to any 2-homology class.

For example, take $M_1 = M_2 = T^4$ the 4-torus. Its 2-homology group is generated by two 2-tori (whose cartesian product essentially yields $T^4$). Hence, integration of a symplectic form on these two generators gives, in an appropriate sense, two symplectic invariants. I say 'in an appropriate sense' because a symplectomorphism $\phi : (T^4, \omega_1) \to (T^4, \omega_2)$ does not need to fix these two 2-cohomology classes; it might for instance switch them. As such, some care is needed in the identification of the corresponding invariants on the source manifold and on the target manifold.

Anyway, if $(T^4, \omega_1)$ and $(T^4, \omega_2)$ are symplectomorphic, then there exists two bases $\{s_1, s_2\}$ and $\{t_1, t_2\}$ of $H_2(T^4 ; \mathbb{Z})$ such that $\int_{s_i}\omega_1 = \int_{t_i}\omega_2$. However, we also have to verify that ${t_1, t_2}$ can be obtained from ${s_1, s_2}$ from a diffeomorphism. And if that is also the case, it is still not clear a priori that it can be chosen to be a symplectomorphism.

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This kind of symplectic invariants are said to be 'classical' or 'soft' as they can be defined using only notions from algebraic topology and differential topology. However, there are examples of non-symplectomorphic symplectic manifolds which have identical classical invariants. In fact, given a manifold $M$ and two symplectic forms $\omega_1$ and $\omega_2$ such that $[\omega_1] = [\omega_2]$, it sometimes happens that they are not symplectomorphic. Proofs of such statements use 'hard' (i.e. 'elliptic') methods of symplectic topology: $J$-holomorphic curves theory, Gromov-Witten invariants, Floer theories, etc.

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That being said, the case of the 4-torus may be purely 'classical'. Since the group of diffeomorphisms of $T^4$ acts transitively on $H_2(T^4 ; \mathbb{Z})$, the problem can be reduced to the case of two symplectic forms in the same 2-cohomology class. If one can show that these two symplectic forms $\omega_0$ and $\omega_1$ are symplectically isotopic (i.e. that there exists a smooth family of symplectic forms $\omega_t$, $t \in [0,1]$, such that $[\omega_t]$ is constant), then Moser's trick shows the existence of a symplectomorphism. Since the bundle $\Lambda^2T^*T^4$ is trivial, it may be possible to explicitly find such an isotopy. I don't know.