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I'm looking for an invariant to distinguish the homeomorphism types of homotopy equivalent spaces. Specifically, how does one show that the total spaces of the tangent bundle to $S^2$ and the trivial bundle $S^2 \times R^2$ are not homeomorphic? (I am not asking for a proof that $TS^2$ is not the trivial bundle.)

Also, is there a way to reduce the question, "Are the total spaces of two vector bundles homeomorphic" to "Are the associated sphere bundles homeomorphic"? In the case of $TS^2$ and $S^2\times R^2$ it's not too difficult to show that the sphere bundles are not homeomorphic, and I'm wondering if there's a way to leverage that.



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Sounds like a homework problem. Have you been learning the Poincare-Hopf index theorem, Euler characteristic and such? That's where to look. Your question will be closed soon as it's off-topic for this forum but math.stackexchange.com is more open to this kind of question. –  Ryan Budney Mar 27 '12 at 2:13
Perhaps my ignorance is showing, but I do not see how the Poincare-Hopf index theorem is helpful here. –  Steven Landsburg Mar 27 '12 at 3:38
@Steven, the one-point compactifications are homeomorphic, and the (co)homologies of these one-point compactifications rel $\infty$ is isomorphic to the (co)homologies of the original manifolds rel the standard boundary. So you have enough structure to define the self-intersection of $H_2$ classes in these manifolds. In particular, you can compute them in the standard smooth structures, so Poincare-Hopf tells you what they are: one gives you a $2$, the other only zeros. –  Ryan Budney Mar 27 '12 at 4:54
Ryan's sophisticated comment proves on the contrary that the question is perfectly legitimate here, and well above math.stackexchange's level . And if my finding it sophisticated proves my ignorance (cf. Steven's comment), so be it. That a professional topologist can answer a question in topology is no reason to close it: read the FAQ! (as closers like to say). –  Georges Elencwajg Mar 27 '12 at 8:34
@Georges is a wise man. –  Igor Rivin Mar 27 '12 at 13:45
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4 Answers

up vote 31 down vote accepted

This is more or less equivalent to Ryan's comment but with more details and a slightly different point of view.

Let $X$ be the total space of the tangent bundle, and put $Y=S^2\times\mathbb{R}^2$. If $X$ and $Y$ were homeomorphic, then their one-point compactifications would also be homeomorphic. We will show that this is impossible by considering their cohomology rings.

Put $X'=\{(p,q)\in S^2\times S^2 : p+q\neq 0\}$. There is a homeomorphism $f:X\to X'$ given by $f(u,v)=((\|v\|^2-1)u+2v)/(\|v\|^2+1)$ (a variant of stereographic projection). It follows that $X_\infty$ can be obtained from $S^2\times S^2$ by collapsing out the antidiagonal. We have $H^*(S^2\times S^2)=\mathbb{Z}[a,b]/(a^2,b^2)$ and it follows that $H^*(X_\infty)$ is the subring generated by $1$, $a+b$ and $ab$. In particular, the squaring map from $H^2$ to $H^4$ is nonzero.

However, $Y$ can be identified with $(S^2\times S^2)\setminus (S^2\times\{point\})$, so $H^*(Y_\infty)$ is isomorphic to the subring generated by $1$, $a$ and $ab$, so the squaring map $H^2\to H^4$ is zero.

Note that the tangent bundle plus a rank-one trivial bundle is trivial, so the suspensions of $X_\infty$ and $Y_\infty$ are homeomorphic.

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Sorry to hijack this beatiful answer with a (presumably) trivial question, but how do you obtain the cohomology ring of a quotient space as a subring of the cohomology ring of the original space (any kind of reference is more than enough... maybe I'm just incapable of searching Hatcher's book thoroughly). –  Tom Bachmann Mar 27 '12 at 13:25
@Tom: Note that for reasonably regular closed subspaces $B\subseteq A$ we have $\tilde{H}^{\ast}(A/B)=H^{\ast}(A,B)$. This is connected to $H^*(A)$ and $H^*(B)$ by a standard long exact sequence. Thus, if the restriction $H^{\ast}(A)\to H^{\ast}(B)$ is surjective (easily verified in the examples above) then $\tilde{H}^{\ast}(A/B)$ is the kernel of the restriction. –  Neil Strickland Mar 27 '12 at 13:32
Thanks for your quick answer! –  Tom Bachmann Mar 27 '12 at 13:45
cool, thank you! –  zygund Mar 27 '12 at 21:26
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These answers look at bit complicated so maybe there is something obviously wrong with the following argument:

Every embedded two-sphere $\Sigma \subset S^2 \times {\mathbb R}^2$ is displaceable: there is a one-parameter group (or family) of homeomorphisms $\varphi_t$ from $S^2 \times {\mathbb R}^2$ to itself such that $\varphi_T (\Sigma)$ is disjoint from $\Sigma$ for some (large) $T$. Indeed, just translate in the second variable far enough.

However, it is impossible to displace the zero section of $TS^2$ because its self-intersection number is $2$.

I read somewhere that to distinguish homeomorphism type of homotopic spaces one could look at the homotopy invariants of configuration spaces. I wonder :

Is the homotopy type of the (two-point) configuration space $C_2(S^2 \times {\mathbb R}^2)$ different from that of $C_2(TS^2)?$.

Edit. It turns out that the answer to the preceeding question is yes as is nicely explained here by Paolo Salvatore. This provides yet another way of proving that $S^2 \times {\mathbb R}^2$ and $TS^2$ are not homeomorphic.

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That is a nice answer, but I think it needs a little more to make it complete (that is, there ARE displaceable spheres in the tangent bundle of $S^2,$ the zero section just does not happen to be one of them, but I think that once one makes things sufficiently precise, your argument works well. As for the configuration space, that is a nice question, and maybe a good separate question?! –  Igor Rivin Mar 27 '12 at 15:02
@Igor: I meant EVERY sphere in $S^2 \times {\mathbb R}^2$ is displaceable and this is a topological property of the space. I'll edit to make this clearer. –  alvarezpaiva Mar 27 '12 at 15:07
This is essentially the point underlying the answer Ryan and Neil give: the self-intersection number of the zero section is what creates the homological invariants. However, it's an excellent point that the answer doesn't need to be couched in that language. –  Greg Friedman Mar 29 '12 at 18:35
@Greg: You're 100% right. The thing is somewhere along the way, we (mathematicians) started speaking like Baldrick here : youtube.com/watch?v=uk37TD_08eA and then everything is hard to understand. –  alvarezpaiva Mar 30 '12 at 13:55
I suppose there's a shortcut that could be used, avoiding cohomology completely. Given two continuous functions $f,g : S^2 \to M$ (an oriented 4-manifold) you could take a smooth approximation and take the transverse intersection number for those maps. A textbook like Hirsch's "Differential Topology" has a proof this is well-defined. That would be a simple tool to give the formal backing for Juan-Carlos's answer. Knowing that the intersection number isn't zero would be a computation, or quoting Poincare-Hopf. –  Ryan Budney Apr 1 '12 at 23:47
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This may be overkill, but to elaborate on Ryan's answer in another way:

Without mentioning either boundaries or any other compactifications, we can define the intersection number of $x\in H_p$ and $y\in H_q$ for homology classes in an oriented $(p+q)$-manifold. First turn them into compactly supported cohomology classes by duality, then cup these to get into $H_c^{p+q}\cong H_0$, etc.

In the smooth case (smooth manifold, and classes represented by smooth compact oriented submanifolds), after putting the submanifolds in general position you can get this same number by counting intersection points with signs.

When $x=y$ this is the same as counting the zeroes of a section of the normal bundle of the submanifold.

This in turn is the same as evaluating the Euler number of the normal bundle of the submanifold on the fundamental class of the submanifold.

Of course, in our examples the ambient manifold is the total space of the normal bundle, so what all of this amounts to is the statement:

The self-intersection number (as defined by algebraic topology) of the zero section of a smooth rank $n$ oriented vector bundle over an oriented $n$-manifold is the result of evaluating the Euler class of the submanifold on the fundamental class.

I don't see that any of this follows from what I call Poincare-Hopf. But, if you combine the last statement with the fact that in the special case of the tangent bundle evaluation of Euler class on the fundamental class gives Euler number, then you get Poincare-Hopf.

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I suppose I tend to think of Poincare-Hopf as the circle of ideas where you argue that the self-intersection number of the 0-section of a tangent bundle is the euler characteristic, or the local index formula for vector fields, and such. –  Ryan Budney Mar 27 '12 at 20:31
@Tom - thanks for clarifying all that! –  zygund Mar 27 '12 at 21:28
Everything related to Euler characteristic and manifolds is such a "circle of ideas" that I sometimes don't know how to teach it without going around in circles. –  Tom Goodwillie Mar 27 '12 at 22:18
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A general form of this question is studied by De Sapio and Walschap in "Diffeomorphism of total spaces and equivalence of bundles" -- a very cool paper (which uses a quite different. at least on the surface, method from that suggested).

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