# Symplectic Embedding of Torus Cotangent bundle into Cylinder

Consider the cotangent bundle of the $$2$$-torus, $$T^*\mathbb{T}^2$$. The question is whether there is a symplectic embedding from it into $$B^2(1) \times \mathbb{R}^2$$, both with their standard symplectic forms.

This was given as an exercise some time after we discussed Gromov's non-squeezing theorem, but so far I don't really have a starting point. One idea I had was that if there was an embedding $$\varphi: B^4(r) \hookrightarrow T^*\mathbb{T}^2,$$ for which $$r>1$$, then Gromov's theorem would give that there can be no symplectic embedding into $$B^2(1) \times \mathbb{R}^2$$.

Endowing $$\mathbb{T}^2 = S^1 \times S^1$$ with canonical angular coordinates $$(q^1, q^2)$$, we'd have that the cotangent bundle looks like $$T^*\mathbb{T}^2 = \{((q^1, q^2), p^1dq^1 + p^2 dq^2) \mid q^1, q^2 \in [0, 2\pi), \; p^1, p^2 \in \mathbb{R} \} \cong \mathbb{T}^2 \times \mathbb{R}^2$$. However, I couldn't think of any constraints on $$\varphi$$ that would let me deduce that $$r>1$$.

Any hints or other approaches? Thanks in advance!

## 1 Answer

I will show that there isn't any symplectic embedding into $$B^2(r) \times \mathbb R^2$$ for any $$r>0$$. (Not a symplectic geometer, but I cannot find any mistake)

The cotangent bundle $$T^* \mathbb T^2$$ can be identified as

$$T^* \mathbb T^2 = \mathbb R^4/ \sim$$

where $$\sim$$ is given by the translation

$$(q^1, q^2, p^1, p^2 )\sim (q^1 + nR_1, q^2 + mR_2, p^1, p^2), \ \ \ \ \forall m, n\in \mathbb Z$$

and the symplectic form is $$\omega_0 = dq^1 \wedge dp^1 + dq^2 \wedge dp^2$$, where $$R_1, R_2 >0$$ are arbitrary: to see this, note for any $$R_1, R_2$$,

$$f :\mathbb R^2/\mathbb Z^2 \times \mathbb R^2 \to \mathbb R^4 /\sim, \ \ \ f(q^1, q^2, p^1, p^2) = (R_1 q^1, R_2 q^2, R_1^{-1} p^1, R_2^{-1} p^2)$$

satisfies $$f^*\omega_0 = \omega_0$$.

Note that the projection $$\pi: \mathbb R^4 \to \mathbb R^4 /\sim$$ also have $$\pi^*\omega_0 = \omega_0$$. Thus it suffices to find $$r' >r$$ so that $$\pi|_{B^4(r')} :B^4(r') \to \mathbb R^4 /\sim$$ is a diffeomorphism. But this is easy since $$R_1, R_2$$ can be as large as we want.