# Examples of the varying strengths of topological invariants

In my first algebraic topology class, I remember being told that the simplest reason for homology was to distinguish spaces. For example, if is X=circle and a Y= wedge of a circle and a 2-sphere then X and Y have the same fundamental group, so the fundamental group isn't strong enough to distinguish them. We need to look at the other homotopy groups or homology to tell them apart. I'm looking for a variety of other examples of this nature. The examples I'm wondering about are

1. Same homology groups
2. Same cohomology groups, but different cohomology rings
3. Same cohomology rings (but maybe different Steenrod operations?)

If I put more thought into it, I could come up with others questions like these. Any other examples/thoughts along these lines would be very welcome! (I have examples for the first one, but I'm wondering what others will say.)

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Upon thinking, this is big-list, and should be community wiki. –  Charles Siegel Jan 2 '10 at 18:46
I've converted it to wiki. Thanks for flagging it. –  Anton Geraschenko Jan 2 '10 at 18:47

## 4 Answers

To change up the nature of the responses some, IMO a good theorem to think about is the Kan-Thurston theorem. It states that given any space $X$ you can find a $K(\pi, 1)$ space $Y$ and a map $f : Y \to X$ inducing isomorphisms $f_* : H_i Y \to H_i X$, $f^* : H^i X \to H^i Y$ for all coefficients (it can be souped-up to allow local coefficients) and all $i$. The map $\pi_1 Y \to \pi_1 X$ is onto.

So from the point of view of cohomology algebras with Steenrod operations, these spaces are the same. One way to "spin" this would be to say the fundamental group is a far stronger invariant than anything (co)homological.

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Here's a fun one: two spaces with ALL homotopy groups identical, but that aren't homotopy equivalent: $\mathbb{RP}^2\times S^3$ and $S^2\times\mathbb{RP}^3$. They both have $\pi_1=\mathbb{Z}/2\mathbb{Z}$, and they have the same universal cover, so all the higher homotopy groups are the same. However, their homology distinguishes them, as I recall.

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Homology does distinguish them as the first is nonorientable while the second is orientable. Hence, H_5(RP^2 x S^3) = 0 while H_5(S^2 x RP^3) = Z –  Jason DeVito Jan 2 '10 at 18:52
Ah, good. I knew it was something like that, but I'm in-and-out today, so I put up the example and figured that the proof that homology distinguishes them would come later. Thanks Jason. –  Charles Siegel Jan 2 '10 at 19:02

For 1 and 2, consider the $S^2$ bundles over $S^4$ (with structure group SO(3)). Using clutching functions, one can see that there is a $\mathbb{Z}$s worth of such bundles indexed by, say, k.

Using the Gysin sequence, one finds that the cohomology groups (and homology groups) are the same as those of $S^2\times S^4$, mainly, a $\mathbb{Z}$ in dimension 0,2,4, and 6, and 0 elsewhere.

However, for $k\neq \pm k'$ the bundles corresponding to $k$ and $k'$ have nonisomorphic ring structures.

By Poincare duality, the only question about the ring structure is the following: what is the square of the degree 2 generator? Turns out, the square of the degree two generator is equal to $\pm k$ times the square of the degree 4 generator. Incidentally, the case $k=1$, one gets the cohomology ring structure of $\mathbb{C}P^3$. In fact, the total space of the bundle is diffeomorphic to $\mathbb{C}P^3$.

For your third question, I'd inspect $S^5$ bundles over $S^2$. Again, by clutching function analysis, there must be precisely two such bundles - the trivial bundle and one other. By the Gysin sequence, these must have the same cohomology groups and by Poincare duality, the ring structures must in fact agree.

However, the second Stieffel Whitney class of the trivial bundle is trivial, while the second Stieffel Whitney class of the nontrivial bundle is nontrivial, and hence the two total spaces are not homotopy equivalent. (Stieffel Whitney classes are closely related to Steenrod operations).

And just to anticipate, the spaces $S^3\times \mathbb{R}P^2$ and $S^2\times \mathbb{R}P^3$ have all the same homotopy groups, but are not homotopy equivalent (as homology will tell you).

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I just wanted to add that everything I wrote about $S^5$ bundles over $S^2$ applies equally well to $S^k$ bundles over $S^2$ for any $k\geq 3$, though for $k=4$ one must do a bit more work to make sure the ring structures on the two bundles agree. For $k=2$, the two total spaces are distinguished by their cohomology RING structure (but not groups). In fact, the unique nontrivial $S^2$ bundle over $S^2$ is diffeomorphic to $\mathbb{C}P^2$#$\overline{\mathbb{C}P^2}$, which Anton talks about below. –  Jason DeVito Jan 2 '10 at 22:16

Here's one for number 2 that's pretty standard. Let $X$ be $S^2$ with two loops attached to it at points. Then the cohomology groups are $\mathbb{Z},\mathbb{Z}^2,\mathbb{Z}$, and we've seen that somewhere else: the torus $S^1\times S^1$. However, the cup product is different, because it's trivial on the sphere with loops but nontrivial on the torus.

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