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.

The answer to your second question is yes (in some cases). See Abouzaid's paper "Framed bordism and Lagrangian embeddings of exotic spheres", available here.

Let's make an example for Question 1 using classical smoothing theory.

If $K$ is a finite cell complex then for any vector bundle $\xi$ on $K$ you can make a high-dimensional smooth manifold $M$ having a homotopy equivalence $f:M\to K$ such that the tangent bundle is stably $f^\star \xi$. (Embed $K$ in a big $\mathbb R^N$, take a suitable neighborhood, and then take a disk bundle over that.) Furthermore, $M$ is determined by $\xi$ in the sense that if the dimension is big enough and the manifolds are simply connected at $\infty$ then a homotopy equivalence covered by an isomorphism of tangent bundles must be homotopic to a diffeomorphism. (The proof uses the $h$-cobordism theorem.)

Apply this with $K$ being the Moore space $\Sigma^6\mathbb RP^2$. The space of pointed maps $K\to BO$ is the homotopy fiber of the map $\Omega^7BO\to\Omega^7BO$ induced by a degree $2$ map $S^7\to S^7$. Thus, since $\pi_8BO=\mathbb Z$ and $\pi_7BO=0$, there are two vector bundles over $K$ to choose from, stably. This gives, for large $n$, two smooth manifolds of this homotopy type. Their (co)tangent manifolds are diffeomorphic, because they are $2n$-dimensional manifolds of the same homotopy type, both with trivial tangent bundle. (The direct sum of the nontrivial bundle over $K$ with itself is trivial.)

As piecewise linear manifolds $M$ and $M'$ are isomorphic, because the composed maps $K\to BO\to BTop$ are homotopic, because $\pi_8BPL=\mathbb Z$ and $\pi_7BPL=0$ with the map $\pi_8BO\to \pi_8BPL$ taking a generator to an even ($28$) multiple of a generator.