I have two questions that are inspired by a couple of questions here on MO (referenced below), as well as by a conversation with some other grad students at a summer school.


*Caveat*: I'm not a symplectic geometer, nor a differential topologist in the 'classical' sense, so my questions might have a well-known answer, or be open.




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It's known that exotic $\mathbb{R}^4$'s have *diffeomorphic* cotangent bundles, as it's known (see [here](https://mathoverflow.net/questions/58131/parallelizability-of-the-milnors-exotic-spheres-in-dimension-7)) that the Milnor spheres have *isomorphic* (topological) cotangent bundles.



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> **1**. Can the *smooth* structure of cotangent bundles distinguish exotic smooth structures on the base?
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Phrased differently, does it exist a pair of smooth manifolds $M,M'$ that are homeomorphic, such that $T^\*M$ and $T^\*M'$ are isomorphic (see footnote) as vector bundles, but not diffeomorphic as smooth manifolds?




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As said before, exotic smooth structures on $\mathbb{R}^4$ are not detected by the cotangent bundle as a smooth manifold. But the cotangent bundle admits a canonical symplectic structure, so...



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> **2**. Can the *symplectic* structure of cotangent bundles see the base?
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*I.e.* Does it exist a pair $M,M'$ of smooth manifolds that have diffeomorphic tangent bundles, but such that their symplectic cotangent bundles are/are not symplectomorphic? In this phrasing, I'm including also pairs like $\mathbb{R}^3$ and the [Whitehead manifold](http://en.wikipedia.org/wiki/Whitehead_manifold) (see [this](https://mathoverflow.net/questions/24970/exotic-differentiable-structures-on-r4) question), but we can impose some further restrictions: what if $M$ and $M'$ are homeomorphic?


**UPDATE**: on [this](https://mathoverflow.net/questions/31690/examples-of-non-diffeomorphic-smooth-manifolds-with-diffeomorphic-tangent-bundle/) page, Igor Belegradek gives a reference to a paper where it's proved that homeomorphic $n$-spheres have diffeomorphic cotangent bundles. As Andy Putman points out in his answer, Mohammed Abouzaid found examples of spheres with non-symplectomorphic cotangent bundles, so the answer is **YES**: cotangent bundles can (at least sometimes) see the base.


**UPDATE** (16/07/2012): regarding question 2, Tobias Ekholm and Ivan Smith have an interesting [preprint](http://arxiv.org/abs/1111.5932): Corollary 1.4 says that the symplectic topology on $T^\*(S^1\times S^{8k-1})$ detects the smooth topology of the underlying manifold. Their paper is about double points of Lagrangian immersions (mentioned in Tim Perutz's answer, below).




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As Igor Belegradek remarked, the word "isomorphic" might be a bit confusing, in this context. What I mean is that there is a homeomorphism $M\to M'$ that pulls back $T^\*M'$ to $T^\*M$. This can be strengthened/weakened by asking that this homeomorphism is topologically isotopic to the identity (since $M$ and $M'$ are homeomorphic, this makes sense). For example, I might want $M$ and $M'$ to be both parallelisable.



Can cotangent bundles see exotic smooth structures?
I wrote a little expository piece about this and related matters in the Newsletter of the European Mathematical Society:


<http://www.ems-ph.org/journals/newsletter/pdf/2010-03-75.pdf>


The classical topology of $X:=T^\ast L$ can be taken to include a little more than its diffeomorphism type: there's also an almost complex structure $J$ on $X$, canonical up to homotopy. The pair $(X,J)$ knows the Pontryagin classes of $L$, because $$c\_{2k}(TX,J)|\_{L}=c\_{2k}(TL\otimes\mathbb{C})=(-1)^k p\_k(TL),$$
so $(-1)^k p\_k(TL)$ pulls back to $c\_{2k}(L)$ under projection $X\to L$. However, even with this embellishment, the smooth topology of $X$ doesn't determine $L$.


**Faithfulness conjecture:** the exact symplectomorphism type of the cotangent bundle $(X,\omega=d\lambda\_L)$ of a compact manifold $L$ determines $L$. 


An exact symplectomorphism $T^\ast L \to T^\ast L'$ is a diffeomorphism $f$ such that $f^\*\lambda\_{L'}-\lambda\_L= dh$ for $h$ a compactly supported function. The conjecture (but not the name) is standard.


Attempts to use symplectic *invariants* of $X$ to distinguish smooth structures on $L$ have so far been a complete failure. For example, the symplectic cohomology ring $SH^\*(X)$ is isomorphic to loopspace homology $H\_{-\*}(\mathcal{L}L; w)$ (the coefficients are the local system $w$ of $\mathbb{Z}$-modules determined by $w\_2$), with the string product. This invariant is determined by the homotopy type of $L$.


**"Arnol'd's conjecture"** (scare quotes because Arnol'd really made a much more circumspect conjecture). Any exact Lagrangian embedding $\Lambda \to X$ (with $\Lambda$ compact) is exact-isotopic to the embedding of the zero-section. 


This would immediately imply the faithfulness conjecture. 


There has been progress towards Arnol'd's conjecture of three kinds: 


(1) It's true for $L=S^2$ (Hind, <http://arxiv.org/abs/math/0311092>). 


(2) The work of several authors (Fukaya-Seidel-Smith <http://arxiv.org/abs/0705.3450>, Nadler <http://arxiv.org/abs/math/0612399>, Abouzaid <http://arxiv.org/abs/1005.0358>, and Kragh's work in progress) cumulatively shows that the projection from an exact Lagrangian to the zero-section is a homotopy equivalence. This is good evidence for the truth of the conjecture, but for the application to faithfulness one might as well make homotopy-equivalence an assumption. 


(3) As Andy mentioned, Abouzaid <http://arxiv.org/abs/0812.4781> has shown that a homotopy $(4n+1)$-sphere $S$, such that $T^\*S$ contains an exact embedded Lagrangian $S^{4n+1}$, bounds a parallelizable manifold. This is proved by a stunning analysis of the geometry of a space of pseudo-holomorphic discs.


The existence of exact Lagrangian *immersions* is governed by homotopy theory (there is an h-principle which finds such an immersion given suitable homotopical data). Just as the subtleties of 4-manifold topology can be located at the impossibility of removing double points of immersed surfaces (the failure of the Whitney trick), so the subtlety of the symplectic structure of cotangent bundles comes down to the question of removability of double points of Lagrangian immersions.


