Given a smooth manifold $M$, we define the differentiable structure on $TM$ in the usual way.

I would like to know examples of two smooth manifolds which are non-diffeomorphic, but their tangent bundles are.

Which is the smallest dimension in which one can find such examples?

What if I ask the same question for $k$ pairwise non-diffeomorphic manifolds?

Can we have $k=\infty$?


4 Answers 4


Here are examples of non-diffeomorphic closed manifolds with diffeomorphic tangent bundles:

  1. 3-dimensional lens spaces have trivial tangent bundles, which are diffeomorphic if and only if the lens spaces are homotopy equivalent, e.g. $L(7,1)$, $L(7,2)$ are not homeomorphic, but their tangent bundles are diffeomorphic. This follows from proofs in [Milnor, John, Two complexes which are homeomorphic but combinatorially distinct. Ann. of Math. (2) 74 1961 575--590].

  2. Tangent bundle to any exotic $n$-sphere is diffeomorphic to $TS^n$ as proved in [De Sapio, Rodolfo, Disc and sphere bundles over homotopy spheres, Math. Z. 107 1968 232--236].

  3. In dimensions $n\ge 5$ one can attack this question via surgery theory. For example, let $f:N\to M$ be a homotopy equivalence of closed $n$-manifolds that has trivial normal invariant (which is a bit more than requiring that $f$ preserves stable tangent bundle). Multiply $f$ by the identity map of $(D^n, S^{n-1})$, where $D^n$ is the closed $n$-disk. Then Wall's $\pi-\pi$ theorem implies that $M\times D^n$ and $N\times D^n$ are diffeomorphic, so if tangent bundles of $M, N$ are trivial, this gives manifolds with diffeomorphic tangent bundles.

  4. To illustrate the method in 3, here is a particular example of infinitely many non-homeomorphic closed manifolds with diffeomorphic tangent bundles. Fix any closed $(4r-1)$-manifold $M$ where $r\ge 2$ such that $TM$ is stably trivial and $\pi_1(M)$ has (nontrivial) elements of finite order. Then results of Chang-Weinberger imply that there are infinitely many closed $n$-manifolds $M_i$ that are simply-homotopy equivalent to $M$ and such that the tangent bundles $TM_i$ are all diffeomorphic (I am not quite sure how to get them be diffeomorphic to $M\times\mathbb R^n$ even though this should be possible). I know how to deduce this from [On invariants of Hirzebruch and Cheeger-Gromov, Geom. Topol. 7 (2003), 311--319].

  5. I have been thinking extensively of related issues, so you might want to look at my papers at arxiv, e.g. this one.

  • $\begingroup$ These examples are great! Do you know any examples of non-homeomorphic closed manifolds whose tangent bundles are nontrivial and diffeomorphic? $\endgroup$
    – Tom Church
    Commented Jul 13, 2010 at 17:16
  • $\begingroup$ @Tom, I think example 4 works even when the tangent bundle to $M$ is not stably trivial. The conclusion will be that there are infinitely many manifolds $M_i$ in the simple homotopy type of $M$ that all $M_i$'s have diffeomorphic tangent bundles. I am not sure how to get them diffeomorphic to $TM$ though (in fact, this may be a hasty claim in my original answer , so I will edit it accordingly). Also in example 2 most bundles are nontrivial I think (as spheres are only parallelizable in dimensions 1, 3, 7). $\endgroup$ Commented Jul 13, 2010 at 18:30

Another good $2$-dimensional example is torus punctured once and sphere punctured three times. These become diffeomorphic when crossed with $\mathbb R$, and they have trivial tangent bundles. More generally, if you puncture a closed genus $g$ surface $n>0$ times then the product with $\mathbb R$ depends only on $2g+n$, so in dimension $2$ you can get arbitrarily large finite $k$.

But maybe you wanted a compact (without boundary) example?

EDIT In the spirit of Igor's (3), suppose we have closed smooth $n$-manifolds $M_1$ and $M_2$ with rank $k$ vector bundles $\xi_1$ and $\xi_2$, and we want to know whether the two $n+k$-manifolds $D(\xi_i)$ (closed disk bundles) are diffeomorphic. There is the necessary condition: there is a simple homotopy equivalence $f:M_1\to M_2$ such that $TM_1\oplus\xi_1$ is isomorphic (as vector bundle) to $f^*(TM_2\oplus\xi_2)$. In some cases this is sufficient. If $k>n$ then $f$ gives an embedding of $M_1$ in $D(\xi_2)$, and the normal bundle is isomorphic to $\xi_1$, so $M_1$ embeds in $D(\xi_2)$ with tubular neighborhood $D(\xi_1)$. The h-cobordism theorem shows that the outside of this is a collar. This works as long as $n+k$ is not too small. You don't need the homotopy equivalence to be simple if you just want a diffeomorphism of open disk bundles. In the case $k=n>2$ I believe this can still be made to work: Whitney trick to make $f$ an embedding and also some care about stable vs unstable normal bundle.

  • $\begingroup$ Thank you, it was a general question, so any answer is actually a good answer. However, it would be interesting to know examples of closed (compact without boundary) manifolds (maybe even orientable) $\endgroup$ Commented Jul 13, 2010 at 14:40
  • 1
    $\begingroup$ Thanks for the edit. I should have included this argument but run out of steam, and since the questioner looks at the case of high codimension the method you described is much more appropriate than the surgery-theoretic approach of (3). Also for the benefit of the questioner I wish to point out that one does not really need $k=n$. It suffices to assume that k is above metastable range so that Haefliger's embedding theorem applies to make $f$ an embedding, and of course one wants to have $k>2$ to run the h-cobordism argument you mention. $\endgroup$ Commented Jul 14, 2010 at 21:15

The smallest dimension is 2 : you may take the open annulus and Möbius band : their tangent bundles are both diffeomorphic to $S^1\times R^3$. I think you can obtain $k=\infty$ in three dimensions, by taking various Whitehead manifolds (contractible 3-manifolds not diffeomorphic to $R^3$ but whose product with $R$ is diffeomorphic to $R^4$, if I'm not mistaking, see this question, answers and comments).


For $k = \infty$ (a continuum, to be precise), the continuum of non-diffeomorphic smooth structures on $\mathbb R^4$ would suffice. The tangent bundle of any $\mathbb R^4$ is trivial (since $\mathbb R^4$ is contractible), therefore homeomorphic to $\mathbb R^8$, but $\mathbb R^8$ has only one smooth structure up to diffeomorphism.

  • 2
    $\begingroup$ You mean "for $k=\infty$, dimension $4$ is sufficient" don't you? $\endgroup$ Commented Jul 13, 2010 at 13:03
  • $\begingroup$ I seem to be having trouble reading today, thanks. $\endgroup$ Commented Jul 13, 2010 at 14:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.