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!
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.