Homeomorphism type of tangent bundles

Question

If two closed smooth manifolds are homeomorphic, are the total spaces of their tangent bundle also homeomorphic? I bet this is not true but I do not know any counterexample.

Answer 1

Yes, if two smooth manifolds are (PL) homeomorphic, then their tangent bundles are (PL) homeomorphic. And if two PL manifolds are homeomorphic, then their “tangent bundles” are homeomorphic. Something stronger is also true. Smooth manifolds don’t just have a tangent bundle with fiber $\mathbb R^n$; they also have a tangent disk bundle with fiber $D^n$, which you can think of as vectors of length $\leqslant 1$ or as a projective compactification of $\mathbb R^n$. If two smooth manifolds are (PL) homeomorphic, then the total spaces of their disk bundles are (PL) homeomorphic.

First I will give an elementary argument to answer your narrow question. Then I will discuss the strengthening and the difficulties of the obvious/systematic approach, discussed in the other answer and its comments.

First, reduce the question from tangent bundles to normal bundles. The normal bundle to the diagonal $\Delta\subset M^2$ is the tangent bundle. Thus the question reduces to showing that if a smooth embedding $M\subset N$ is (PL) homeomorphic to another smooth embedding $M’\subset N’$, that the total spaces of the normal bundles are (PL) homeomorphic. The advantage of the setting of normal bundles is the tubular neighborhood theorem: a neighborhood of $M$ can be identified with the total space of the normal bundle. Applying the (PL) homeomorphism, this gives an open neighborhood of $M’$. Being open is a purely topological property, so this space can be thought of as a subset of either $N$ or $N’$. So we have made the normal bundle of $M$ live inside $N’$, which is the key step to comparing them.

A better version of the tubular neighborhood theorem is that one can choose a properly embedded tubular neighborhood and identify it with the space of normal vectors of length $<1$ and its closure with the space of vectors of length $\leqslant1$. Thus the boundary of the tubular neighborhood is a smooth manifold diffeomorphic to the tangent sphere bundle. Its image under the (PL) homeomorphism is not smooth in the other structure, but it still is a (PL) manifold. From here on, (tubular) neighborhoods are closed. Take a tubular neighborhood $A_0\supset M$. Then think of its interior as an open subset of $N’$ and take another tubular neighborhood $A_1\supset M’$. Switch back and take another just by scaling: $A_2=\varepsilon A_0\subset A_1$. Finally, scale the other: $A_3=\delta A_1\subset A_2$. So we have $M=M’\subset A_3\subset A_2\subset A_1\subset A_0$. A word of warning: $A_0$ and $A_1$ have different smooth structures, so the two scalings are very different.

The space between two tubular neighborhoods $A_0-\mathring A_1$ is a PL manifold with two boundaries, one the sphere bundle of the one normal bundle of $M$ and the other of $M’$. A manifold with divided boundary is called a cobordism. This one looks like a cylinder from the point of view of homotopy, so it is called an $h$-cobordism. Moreover, it is an invertible cobodism: if we glue together $A_0-\mathring A_1$ with $A_1-\mathring A_2$, we get $A_0-\mathring A_2=A_0-\varepsilon \mathring A_0$, which is just the sphere bundle times the interval. The middle $h$-cobordism $A_1-\mathring A_2$ has inverses both outside $A_0-\mathring A_1$ and inside $A_2-\mathring A_3$. By the usual argument that the left and right inverse are equal, the two inverses are (PL) homeomorphic.

Let us abstract this a little. Let $W$ and $W’$ be cobordisms between the same manifolds: $\partial W=W_0\cup W_1=\partial W’$. Further, assume that they are inverse cobodisms: $W\cup_{W_1} W’=W_0\times I$ and $W’\cup_{W_0}W=W_1\times I$. Then I claim that the interiors are cylinders: $W-\partial=W’-\partial=W_0\times \mathbb R=W_1\times \mathbb R$. Furthermore, if you just remove one end, they are still cylinders: $W-W_1=W_0\times [0,\infty)=W’-W_1$. This is an Eilenberg-Mazur swindle, just like $0=1-1+1-1\ldots=1$. Geometrically: $$W_0\times[0,\infty)=W_0\times I\cup W_0\times I\cup \ldots=(W\cup W’)\cup (W\cup W’)\cup\ldots$$ $$W\cup(W’\cup W)\cup (W’\cup W)\cup\ldots =W\cup W_1\times I \cup W_1\times I\ldots =W\cup W_1\times [0,\infty)$$ where all $\cup$ unions mean gluing on boundaries. Finally, boundaries are collared, so adding a half-open collar is the same as removing the boundary: $W_0\times[0,\infty)=W\cup W_1\times [0,\infty)$. Thus, $W_0\times[0,\infty)=W-W_1$.

Apply this to $W=A_0-\mathring A_1$, $W_1=\partial A_1$. Then $A_0=A_1\cup W$. Removing boundaries to get tangent bundles, the one tangent bundle is the other plus a half-open invertible cobordism: $$\mathring A_0=A_1\cup(W-W_0)=A_1\cup W_1\times[0,\infty)=\mathring A_1$$

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The $h$- or $s$-cobordism theorem says that in high dimensions $h$-cobordisms, as identified by homotopical properties, are invertible; and it parameterizes them by a specific group. We did not need this because we had direct access to the inverse cobordism.

But the $s$-cobordism theorem says that there aren’t very many invertible cobordisms, so we have to wonder if the cobordisms were not just invertible, but already cylinders. Indeed, this is true: this is the strengthening I mentioned at the beginning, that it is not just the total spaces of the tangent bundles that are (PL) homeomorphic, but also that of the disk bundles.

The obvious technique is to suppose that PL or topological manifolds have tangent bundles, fiber bundles with fiber $\mathbb R^n$ and structure group (PL) homeomorphisms of $\mathbb R^n$ or $D^n$. Then the smooth tangent bundle would, presumably, be the topological tangent bundle, and so the two smooth tangent bundles would be the same topological tangent bundle. While the Kister-Mazur theorem says that topological manifolds have tangent fiber bundles, well-defined up to a contractible choice, I don’t believe that they have disk bundles. In any event, the microbundle is inadequate for this purpose because it is not a fiber bundle; it does not have a total space. It is only defined up to equivalence. It is true that the tangent bundle is a representative of the PL or topological tangent microbundle, but if we have two smooth structures, equivalence between their microbundles doesn’t immediately tell us anything about the total spaces. It just tells us that they embed in each other, as with the tubular neighborhoods.

I claim that PL and topological manifolds don’t have disk fiber bundles, but they do have something that functions as the total space of a putative disk bundle. This substitute is the regular neighborhood. The regular neighborhood of a polyhedron is the closure of the union of all simplices touching the subset, in a sufficiently subdivided triangulation. It is a basic theorem of PL topology that the regular neighborhood is unique up to PL isomorphism, even up to isotopy. The regular neighborhood of the diagonal is thus a PL invariant of a PL manifold. The disk bundle of a smooth manifold is a regular neighborhood, so if two smooth manifolds are PL isomorphic, so are the total spaces of their disk bundles. Topological manifolds also have regular neighborhoods, but this is a more difficult subject. Since the $s$-cobordism theorem has reared its head, one should expect this to be on par with proving that topological manifolds have Whitehead torsion.

Answer 2

I am not sure if this is exactly what you are asking for, but this: https://math.stackexchange.com/a/477495/195021 answer and this: Examples of non-diffeomorphic smooth manifolds with diffeomorphic tangent bundle question could be of some interest.

Answer 3

Ben gave a wonderful answer, but this is actually (much) easier in the smooth category. See page 50 of Guillemin and Pollack.