# Can cotangent bundles see exotic smooth structures?

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.

It's known that exotic $\mathbb{R}^4$'s have diffeomorphic cotangent bundles, as it's known (see here) that the Milnor spheres have isomorphic (topological) cotangent bundles.

1. Can the smooth structure of cotangent bundles distinguish exotic smooth structures on the base?

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?

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...

2. Can the symplectic structure of cotangent bundles see the base?

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 (see this question), but we can impose some further restrictions: what if $M$ and $M'$ are homeomorphic?

UPDATE: on this 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: 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).

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.

• I realise that someone might want me to split the question into two questions. If so, please upvote this comment. – Marco Golla Jun 26 '11 at 9:00
• And since one can't downvote comment (at least, I can't), please upvote this comment if you want to keep the question as it is. – Marco Golla Jun 26 '11 at 9:00
• @Marco Golla: would you clarify what you mean by $T^*M$, $T^*M^\prime$ are isomorphic as vector bundles''? By your assumption the bases are not diffeomorphic so the usual notion of bundle isomorphism does not apply. Do you mean that there is a homeomorphism $M\to M^\prime$ which pulls back $T^*M^\prime$ to $T^*M$? – Igor Belegradek Jun 26 '11 at 12:37
• @Igor Belegradek: what I had written was actually a bit (unwillingly) ambiguous. Let me try and restate it. Thank you for your remark. – Marco Golla Jun 26 '11 at 13:04
• Here is a closely related question mathoverflow.net/questions/31690/… – Igor Belegradek Jun 27 '11 at 1:08

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.

• From the abstract of the paper, it seems it's more related to the second, rather than the first, question: typo? Also, is it obvious that all exotic spheres (or at least the ones considered in the paper) have diffeomorphic cotangent bundles? – Marco Golla Jun 26 '11 at 14:03
• Whoops, that's a typo! And I think that the cotangent bundles are diffeomorphic (or at least homeomorphic), but I haven't had a cup of coffee yet so I'd too tired to think it through. – Andy Putman Jun 26 '11 at 14:13

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.

• Doesn't this only give an example of two non-diffeomorphic manifolds with diffeomorphic cotangent bundles? If true, then there are easier ways to achieve the goal, e.g. all homotopy $n$-spheres have diffeomorphic cotangent bundles. – Igor Belegradek Jun 27 '11 at 11:53
• I suppose so. My example is different in that the tangent bundles of $M$ and $M'$ are stably non-isomorphic. And the hardest facts that it uses are (a) [the main fact of smoothing theory] that smoothing a PL manifold is equivalent to putting a vector bundle structure on its stable tangent microbundle, i.e. making a lift of $M\to BPL$ to $BO$ (b) that the first nontrivial homotopy group of $PL/O$ is $\pi_7$, which has even order, and (c) that $\pi_7(BO)=0$. – Tom Goodwillie Jun 27 '11 at 13:12
• ... and (d) some version of the $h$-cobordism theorem, to show that when dimension is high compared to connectivity a homotopy equivalence covered by a tangent bundle isomorphism guarantees a diffeomorphism, and the $PL$ analogue of this statement. – Tom Goodwillie Jun 27 '11 at 16:31

Edit: I misunderstood the question...

The answer to the first question for closed orientable manifolds in dimension 4 is negative.

The Dold-Whitney theorem states that two oriented 4-plane bundles over the same 4-manifold $M$ are isomorphic if and only if they share the same second Stiefel-Whitney class $w_2$, the same first Pontryagin class $p_1$, and the same Euler class $e$. Each such characteristic class is determined by the homotopy type of $M$ (see below), and hence the cotangent bundle over $M$ is also determined, no matter what differentiable structure one assigns to $M$.

These characteristic classes are indeed easily determined by the homotopy type of $M$: $p_1$ is 3 times the signature $\sigma(M)$ of $M$ by Hirzebruch formula, the Euler class is $\chi(M)$, and $w_2$ is determined by the intersection form thanks to Wu's formula, see for instance this question.

• You mean that the answer is negative, really :) (The rephrasing of the question which follows it is not really a rephrasing...) – Mariano Suárez-Álvarez Jun 26 '11 at 19:43
• ooops... thank you, I have corrected from "positive" to "negative" :-) – Bruno Martelli Jun 26 '11 at 22:34
• @Bruno: Hi! I'm a bit confused.. Correct me if I'm wrong: you're saying that the homeomorphism type determines the isomorphism class of the (topological) vector bundle over the (topological) manifold. If so, what does this tell you about the smooth structure on $T^*M$? – Marco Golla Jun 26 '11 at 23:41
• ooops again.... of course you were talking about smooth structures! I don't know if the smooth structures on $T^*M$ are equivalent for the (sometimes infinitely many) different smooth structures one can put on $M$. – Bruno Martelli Jun 27 '11 at 7:57

The questions at hand are much easier to answer for open manifolds than for closed manifolds. Here is a large family of simple examples: There are uncountably many diffeomorphism types of exotic $$\mathbb{R}^4$$s. As mentioned already in the question, the total space of the cotangent bundle of each exotic $$\mathbb{R}^4$$ is diffeomorphic to $$\mathbb{R}^8$$.

Moreover, the bundle is trivial, because its base is contractible. Hence, for any homeomorphism $$f:M_0\to M_1$$ between exotic $$\mathbb{R}^4$$s, the $$f$$-pullback of the cotangent bundle of $$M_1$$ is base-preservingly isomorphic (as a topological vector bundle) to the cotangent bundle of $$M_0$$. This yields uncountably many (parallelisable) examples where the answer to question 1 is NO.

All these total spaces are also symplectomorphic; this yields uncountably many examples where the answer to question 2 is NO. That the total spaces of the cotangent bundles of any two exotic $$\mathbb{R}^4$$s $$M_0,M_1$$ are symplectomorphic follows from Corollary 1.12.D in Eliashberg/Gromov: Convex symplectic manifolds, because $$M_0,M_1$$ are tangentially equivalent in the Eliashberg/Gromov sense: There is (again by triviality of the bundles) a vector bundle morphism $$T^*M_0\to T^*M_1$$ which is fibrewise bijective and whose underlying map $$M_0\to M_1$$ is a homotopy equivalence.