Gromov's symplectic nonsqueezing theorem asserts that in the symplectic space ${\bf R}^{2n}$ with canonical coordinates $p_1,\dots,p_n,q_1,\dots,q_n$, and two radii $0 < r < R$, it is not possible to symplectomorphically map the ball $B(0,R)$ into the cylinder $\{ p_1^2+q_1^2 \leq r^2 \}$. It is often known as the "symplectic camel theorem", with several authors interpreting this theorem as an assertion that a symplectic camel (in the sense of a ball $B(0,R)$ that can only evolve by a continuous one-parameter family of symplectomorphisms cannot pass "through" the eye of a needle of radius $r$ less than $R$.

In the past I have accepted this interpretation uncritically, but on closer inspection it seems that the standard formulation of the nonsqueezing theorem does not actually imply a statement of this form. To make precise what an "eye of a needle" is, one needs a codimension one obstacle to serve as the "needle". I will somewhat arbitrarily choose the hyperplane $\{q_n=0\}$, minus the above-mentioned cylinder, as the needle, in which case one has a precise mathematical question.

Question. Let $n \geq 2$ and $0 < r < R$, and let $N \subset {\bf R}^{2n}$ denote the "needle" $N := \{ q_n = 0; p_1^2+q_1^2 > r^2 \}$. Does there exist a family $S(t): B(0,R) \to {\bf R}^{2n} \backslash N$, $t \in [0,1]$ of symplectomorphisms varying continuously in $t$ (say in the uniform topology), such that $S(0)$ takes values in the left half-space $\{ q_n<0\}$ and $S(1)$ takes values in the right half-space $\{q_n>0\}$?

Certainly a counterexample to Gromov's non-squeezing theorem (using a symplectomorphism that is connected to the identity) would allow one to construct a positive answer to this question, by first moving the ball far away from the needle, transforming it into a subset of the cylinder, sliding that subset through the needle and then far on the other side, then undoing the transformation. However it is not clear to me that the non-squeezing theorem in its standard formulation prevents some more exotic way to slide this "camel" through the "needle" (for instance, if it is possible to symplectically deform the ball into an "L-shape" object that resembles the union of two half-cylinders, which could then be maneuvred through the needle).


Eliashberg & Gromov sketched a proof in their paper "Convex symplectic manifolds" (Section 3.4). Written in the 4-dimensional case it says:

For $r>0$ define the subspace $X(r)\subset\mathbb{R}^4$ to be the union of the half-space $\lbrace q_2<0\rbrace$ and the half-space $\lbrace q_2>0\rbrace$ and the 3-ball $\lbrace(p_1,p_2,q_1,0)\;|\;p_1^2+p_2^2+q_1^2<r^2\rbrace$. For $R>r$ there is no 1-parameter family of symplectic embeddings $S_t:(B(0,R),\omega_\text{std})\to (X(r),\omega_\text{std})$ with the image of $S_{t\le0}$ in $\lbrace q_2<0\rbrace$ and the image of $S_{t\ge1}$ in $\lbrace q_2>0\rbrace$.

McDuff & Traynor ("The 4-dimensional symplectic camel and related results") go on to show that $X(r_1)$ and $X(r_2)$ are symplectomorphic if and only if $r_1=r_2$. Oh they also give Eliashberg--Gromov's proof of the camel theorem (Theorem 5.2), reducing it to the monotonicity lemma as in the usual Gromov nonsqueezing theorem.

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    $\begingroup$ Chris' answer is correct, but it should be noted that the camel problem is also resolved in Viterbo's "Symplectic topology as the geometry of generating functions," specifically in Section 5. The proof is truly different, in that there is no invocation of any pseudo-holomorphic curve techniques, so you never need to invoke the monotonicity lemma, for example. $\endgroup$ – KSackel Mar 27 at 2:47

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