I want to prove following statement:

If $TM$ and $TN$ are diffeomorphic, then $M$, and $N$ are diffeomorphic.

First my trial was using a natural projection between tangent bundle and manifold \begin{align} &\pi_M : TM \rightarrow M \\ &\pi_N : TN \rightarrow N \end{align} what I know is this natural projection is smooth and surjective.

And from assumption there is a diffeomorphism $\phi : TM \rightarrow TN$. Using this I can construct a map from $M$ to $N$ as $\pi_N \circ \phi \circ \pi_M^{-1} : M \rightarrow N$, and my guess is this map is diffeomorphism.

But I am having trouble that showing this is indeed diffeomorphism. To see this first I need to show its bijectivity and smoothness. At this moment I am not sure whether $\pi^{-1}$ is smooth or not.

Is there any idea to show this composition map is diffeomorphism?

By the way is my approach right? If I am right, how can I prove $\pi_N \circ \phi \circ \pi_{M}^{-1}$ is a diffeomorphism?

  • 1
    $\begingroup$ If $v, w \in T_pM$, then $\pi_M(v) = \pi_M(w) = p$, so $\pi_M$ is not injective. In particular, $\pi_M$ is not invertible. $\endgroup$ Commented Apr 12, 2017 at 12:32

2 Answers 2


I don't think this is true, but I am sure that there are easier ways of showing it.

The tangent bundle of $S^7$ is trivial with respect to any of its smooth structures (see this MathOverflow question). Let $M$ be an homotopy $7$-sphere (i.e. a manifold homeomorphic to $S^7$ but equipped with a non-standard smooth structure) which bounds a parallelizable manifold, and let $N$ be $S^7$ with its standard smooth structure. Then $TM$ is diffeomorphic to $M\times\mathbb{R}^7$ (with the product smooth structure) and $TN$ is diffeomorphic to $N\times\mathbb{R}^7$ (with the product smooth structure).

There is a theorem of Hirsch (see Theorem 5.4 of Smale's On the structure of manifolds for example), which states that if $M$ is a $n$-dimensional homotopy sphere which bounds a parallelizable manifold, then $M\times D^3$ is diffeomorphic to $S^n\times D^3$. In particular, $M\times D^3 \cong N\times D^3$. As $\mathbb{R}^3$ is diffeomorphic to $D^3$, we have

$$TM \cong M\times\mathbb{R}^7 \cong M\times\mathbb{R}^3\times\mathbb{R}^4 \cong M\times D^3\times\mathbb{R}^4 \cong N\times D^3\times\mathbb{R}^4 \cong N\times\mathbb{R}^3\times\mathbb{R}^4 \cong N\times\mathbb{R}^7 \cong TN$$

where $\cong$ is being used to denote 'diffeomorphic to', and throughout, the smooth structure on products is the product smooth structure.

  • $\begingroup$ I wanted to write such an answer, but I was lacking the argument provided by Hirsch's theorem... I wonder if Milnor's examples $M_1 = L(7,1) \times S^4$ and $M_2 = L(7,2) \times S^4$, which are nonhomeomorphic but $h$-cobordant (so $M_1 \times \mathbb R$ and $M_2 \times \mathbb R$ are diffeomorphic by Mazur's trick) could provide another example... $\endgroup$
    – PseudoNeo
    Commented Apr 12, 2017 at 12:56

Consider (1) the trivial line bundle on the circle, i.e. the cylinder, and (2) the Möbius strip. They are not diffeomorphic (see Moebius band not homeomorphic to Cylinder.), but their tangent bundles are both diffeomorphic to $\mathbb{S}^1\times \mathbb{R}^3$ (see Tangent bundle of open annulus is diffeomorphic to $\mathbb{S}^1 \times \mathbb{R}^3$ and Tangent bundle of mobius strip is diffeomorphic to $\mathbb{S}^1 \times \mathbb{R}^3$).


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