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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    $\begingroup$ 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. $\endgroup$ Mar 27, 2012 at 2:13
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    $\begingroup$ Perhaps my ignorance is showing, but I do not see how the Poincare-Hopf index theorem is helpful here. $\endgroup$ Mar 27, 2012 at 3:38
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    $\begingroup$ @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. $\endgroup$ Mar 27, 2012 at 4:54
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    $\begingroup$ 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). $\endgroup$ Mar 27, 2012 at 8:34
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    $\begingroup$ @Georges is a wise man. $\endgroup$
    – Igor Rivin
    Mar 27, 2012 at 13:45

6 Answers 6


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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    $\begingroup$ 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). $\endgroup$ Mar 27, 2012 at 13:25
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    $\begingroup$ @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. $\endgroup$ Mar 27, 2012 at 13:32

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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    $\begingroup$ 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. $\endgroup$ Mar 29, 2012 at 18:35
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    $\begingroup$ @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. $\endgroup$ Mar 30, 2012 at 13:55
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    $\begingroup$ 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. $\endgroup$ Apr 1, 2012 at 23:47
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    $\begingroup$ By the way, I think Juan Carlos' argument actually shows that for a compact $n$-dimensional manifold $M$ with nonzero Euler characteristic, $TM$ and $M \times \mathbb R^n$ cannot be homeomorphic. $\endgroup$ Apr 2, 2012 at 0:52
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    $\begingroup$ @alvarezpaiva: in avoiding cohomology by the price of using the transversality theorem be careful not to overlook larger animals under your carpet. You'd have to use smooth (or at best PL) copies of $S^2$ to compute the intersection numbers. A topological copy of $S^2$, as in your definition of displaceability, might be not approximable by any smooth or PL embedding, so you'd have to approximate it by a smooth map, with self-intersections, as Ryan suggests. $\endgroup$ Apr 2, 2012 at 11:07

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 class 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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    $\begingroup$ 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. $\endgroup$ Mar 27, 2012 at 20:31
  • $\begingroup$ @Tom - thanks for clarifying all that! $\endgroup$
    – zygund
    Mar 27, 2012 at 21:28
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    $\begingroup$ 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. $\endgroup$ Mar 27, 2012 at 22:18

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).


This question recently came up on Math Stack Exchange (here), and I posted an answer that involves only fundamental group and homology. I've copied that answer below.

Let's start by examining the unit tangent bundle $US^2 \subset TS^2$ (i.e. the set of all tangent vectors of unit length) and its analogue $S^1\times S^2 \subset \mathbb{R}^2 \times S^2$. Observe that the action of $SO(3)$ on $S^2$ extends to $US^2$, and this action is transitive and has trivial stabilizer, so it induces a diffeomorphism $SO(3)\to US^2$. It follows that $$ \pi_1(US^2) \;\cong\; \pi_1\bigl(SO(3)\bigr) \;\cong\; \mathbb{Z}_2. $$ Then $US^2$ is not homeomorphic to $S^1\times S^2$, since $\pi_1(S^1\times S^2)\cong \mathbb{Z}$.

Now, to prove that $X = TS^2$ and $Y = \mathbb{R}^2 \times S^2$ are not homeomorphic, let $X\cup\{\infty\}$ and $Y\cup\{\infty\}$ denote their one point compactifications. I claim that $$ H_2(X\cup\{\infty\},X) \;\cong\; H_1(US^2) \;\cong\; \mathbb{Z}_2\tag*{(1)} $$ and $$ H_2(Y\cup\{\infty\},Y) \;\cong\; H_1(S^1\times S^2) \;\cong\; \mathbb{Z}.\tag*{(2)} $$ To see the isomorphism in (1), let $S \subset TS^2$ denote the canonical copy of $S^2$ in $TS^2$, i.e. the set of zero tangent vectors. By excision $$ H_2(X\cup\{\infty\},X) \;\cong\; H_2(X\cup\{\infty\}-S,X-S) $$ But $X\cup\{\infty\}-S$ is clearly contractible, so $$ H_2(X\cup\{\infty\}-S,X-S) \;\cong\; H_1(X-S) $$ by the long exact sequence for the pair $(X\cup\{\infty\}-S,X-S)$. But $X-S$ deformation retracts onto $US^2$, so $H_1(X-S) \cong H_1(US^2)$, as desired. A similar argument establishes the isomorphism in (2).

In general, this argument allows you to show that vector bundles over compact spaces are non-homeomorphic if the associated sphere bundles have different homology.

  • $\begingroup$ Jim, your bard web address seems to be dead. Do you have a new working page? $\endgroup$ Dec 5, 2019 at 1:19
  • $\begingroup$ Jim, the address still seems to be dead. $\endgroup$ Dec 10, 2019 at 0:22

An argument using only basic topology and fundamental group (no homology or higher $\pi_i$).

Consider the condition $\mathcal{P}$, for a topological space: for every compact subset $K\subset X$ there exists a compact subset $K'$ containing $K$, such that $X-K'$ is path-connected and has a finite fundamental group.

Then $T\mathbf{S}^2$ satisfies $\mathcal{P}$ ($K$ can be enlarged to the set $K'$ of vectors of norm $\le R$ for some $R$, and the complement of $K'$ is homeomorphic to the product of $\mathbf{R}$ and the set of vectors of norm $2R$, so is homeomorphic to $\mathbf{R}\times\mathrm{SO}(3)$ and has $\pi_1$ with 2 elements).

While $\mathbf{S}^2\times\mathbf{R}^2$ does not satisfy $\mathcal{P}$: given $K$ containing $\mathbf{S}^2\times\{0\}$, for any $K'$ we can retract the complement of $K'$ onto $\mathbf{S}^2\times \mathbf{S}^1(R)$ for large enough $R$, where $\mathbf{S}^1(R)$ is the circle of radius $R$, so the $\pi_1$ complement of $K'$ has an infinite homomorphism onto the group $\mathbf{Z}$ of integers.

Informally, the fundamental groups at infinity of the 2 spaces differ. (This argument is, anyway, related to the previous ones.)


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