# Closed manifolds with isomorphic cohomology rings, but different cohomology modules over the Steenrod algebra

For any $n > 2$, $\mathbb{CP}^n/\mathbb{CP}^{n-2}$ and $S^{2n}\vee S^{2n-2}$ have the same cohomology groups: for any ring $R$, we have

$$H^k(\mathbb{CP}^n/\mathbb{CP}^{n-2}; R) \cong H^k(S^{2n}\vee S^{2n-2}; R) \cong \begin{cases} R & k = 0, 2n-2, 2n\\ 0 & \text{otherwise}. \end{cases}$$

Moreover, the two spaces have the same cohomology ring structure (the product of any two elements of positive degree is necessarily zero).

However, for $n$ even, the two spaces are not homotopy equivalent. As is discussed here and here, this can be shown by demonstrating that $\operatorname{Sq}^2 : H^{2n-2}(\mathbb{CP}^n/\mathbb{CP}^{n-2}; \mathbb{Z}_2) \to H^{2n}(\mathbb{CP}^n/\mathbb{CP}^{n-2}; \mathbb{Z}_2)$ is an isomorphism, while $\operatorname{Sq}^2 : H^{2n-2}(S^{2n}\vee S^{2n-2}; \mathbb{Z}_2) \to H^{2n}(S^{2n}\vee S^{2n-2}; \mathbb{Z}_2)$ is the zero map. In particular, we see that the cohomology of the two spaces are not isomorphic as modules over the Steenrod algbera.

I am looking for a similar example where the spaces are closed manifolds.

Are there examples of closed manifolds with isomorphic cohomology rings, but different cohomology modules over the Steenrod algebra?

• Do you want the cohomology rings to be isomorphic over every ring?
– user98602
Dec 31, 2016 at 17:07
• @MikeMiller: Preferably, but I'd be interested in seeing other examples as well. Dec 31, 2016 at 18:18
• It seems to me that the construction here of a spin and non-spin 5-manifold with second homology $\Bbb Z_2^2$ should furnish examples. There is no third or fourth homology, so over every ring in which 2 is invertible the cohomology rings are clearly the same. Over $\Bbb Z_2$ I think they automatically need to be isomorphic by the nondegeneracy of the cup product. Then the key point is that $w_2$ is zero in one and nonzero in another, and is equal to $v_2$, so that $\text{Sq}^2$ is zero in only one.
– user98602
Dec 31, 2016 at 18:56
• $S^3\times \Bbb CP^n$ and $S^1\times \Bbb CP^n/S^1\times\{x_0\}$ have isomorphic cohomology rings but can be distinguished by their module structure over the Steenrod algebra. Jan 1, 2017 at 13:09

Here is an example of two simply connected, closed manifolds $$M_0$$ and $$M_1$$ such that $$H^*(M_0; R)$$ and $$H^*(M_1; R)$$ are isomorphic as rings for any commutative ring with identity $$R$$, but $$H^*(M_0; \mathbb{Z}_2)$$ and $$H^*(M_1; \mathbb{Z}_2)$$ are not isomorphic as modules over the Steenrod algebra. Moreover, $$M_0$$ and $$M_1$$ have the same homotopy groups.

First note that $$\operatorname{Vect}^4(S^2) = [S^2, BSO(4)] = \pi_1(SO(4)) \cong \mathbb{Z}_2$$. So, up to isomorphism, there are two real rank four vector bundles over $$S^2$$: the trivial one $$E_0 \cong \varepsilon^4$$ and a non-trivial one $$E_1$$. Every rank four bundle over $$S^2$$ splits as $$F\oplus\varepsilon^2$$ where $$F$$ is a real rank two bundle. As $$S^2$$ is simply connected, $$F$$ is orientable and can therefore be viewed as a complex line bundle. Identifying $$S^2$$ with $$\mathbb{CP}^1$$, we see that $$F \cong \mathcal{O}(k)$$ for some integer $$k$$. If $$k = 2m + 1$$, then $$w_2(\mathcal{O}(2m+1)\oplus\varepsilon^2) \neq 0$$ so $$\mathcal{O}(2m+1)\oplus\varepsilon^2$$ is not the trivial bundle and is therefore isomorphic to $$E_1$$; in particular, we see that $$w_2(E_1) \neq 0$$. If $$k = 2m$$, $$w_2(\mathcal{O}(2m)\oplus\varepsilon^2) = 0$$ so $$\mathcal{O}(2m)\oplus\varepsilon^2$$ is not isomorphic to $$E_1$$ and therefore must be trivial.

Equip $$E_0$$ and $$E_1$$ with Riemannian metrics, and let $$M_0$$ and $$M_1$$ be their corresponding sphere bundles.

There are fibre bundles $$S^3 \to M_i \to S^2$$. From the long exact sequence in homotopy associated to a fibration, we have an exact sequence

$$\dots \to \pi_2(S^3) \to \pi_2(M_i) \to \pi_2(S^2) \to \pi_1(S^3) \to \pi_1(M_i) \to \pi_1(S^2) \to \dots$$

As $$\pi_1(S^2) = \pi_1(S^3) = \pi_2(S^3) = 0$$ and $$\pi_2(S^2) \cong \mathbb{Z}$$, we see that $$M_i$$ is simply connected and $$\pi_2(M_i) \cong \mathbb{Z}$$. By the Hurewicz theorem, $$H_2(M_i; \mathbb{Z}) \cong \mathbb{Z}$$. By the universal coefficient theorem for cohomology, $$H^1(M_i; \mathbb{Z}) = 0$$ and $$H^2(M_i;\mathbb{Z}) \cong \mathbb{Z}$$. Since $$M_i$$ is simply connected, $$M_i$$ is orientable, so by Poincaré duality $$H_3(M_i; \mathbb{Z}) \cong H^2(M_i; \mathbb{Z}) \cong \mathbb{Z}$$ and $$H_4(M_i; \mathbb{Z}) \cong H^1(M_i; \mathbb{Z}) = 0$$. Now by the universal coefficient theorem again, for any commutative ring with identity $$R$$, we have

$$H^k(M_i; R) \cong \begin{cases} R & k = 0, 2, 3, 5\\ 0 & \text{otherwise}. \end{cases}$$

Alternatively, we could have used the cohomological Serre spectral sequence to arrive at this result. Now Poincaré duality with $$R$$ coefficients uniquely determines the ring structure on $$H^*(M_i; R)$$, up to isomorphism, so $$H^*(M_0; R) \cong H^*(M_1; R)$$ as graded rings.

Let $$\tilde{\pi} : E \to B$$ be a smooth vector bundle, then $$TE \cong \tilde{\pi}^*TB\oplus\tilde{\pi}^*E$$. If $$i : S(E) \to E$$ is the natural inclusion, then $$i^*TE = TS(E)\oplus\nu$$ where $$\nu$$ is the normal line bundle. So

$$TS(E)\oplus\nu = i^*TE = i^*\tilde{\pi}^*(TB\oplus E) = (\tilde{\pi}\circ i)^*(TB\oplus E) = \pi^*(TB\oplus E)$$

where $$\pi : S(E) \to B$$ is the projection of the sphere bundle.

As $$M_i$$ is simply connected, $$\nu$$ is trivial; more generally, the normal bundle of a sphere bundle is always trivial, see this note. So

$$w(TM_i) = w(TS(E_i)) = w(\pi^*(TS^2\oplus E_i)) = \pi^*(w(TS^2\oplus E_i)) = \pi^*(w(TS^2)w(E_i)) = \pi^*w(E_i) = \pi^*(1 + w_2(E_i)) = 1 + \pi^*w_2(E_i).$$

It follows from the Gysin sequence that $$\pi^* : H^2(S^2; \mathbb{Z}_2) \to H^2(M_i; \mathbb{Z}_2)$$ is an isomorphism. Alternatively, we can view this map as an edge homomorphism in the cohomological Serre spectral sequence (see Theorem $$5.9$$ of A User's Guide to Spectral Sequences by McCleary), from which it is clear the map is an isomorphism. Therefore, $$w(M_0) = 1$$ while $$w(M_1) = 1 + a$$ where $$a$$ is the unique non-zero element of $$H^2(M_2; \mathbb{Z}_2)$$. So $$M_0$$ and $$M_1$$ are not homotopy equivalent. In particular, $$\operatorname{Sq}^2 : H^3(M_i; \mathbb{Z}_2) \to H^5(M_i; \mathbb{Z}_2)$$ is zero for $$i = 0$$ and an isomorphism for $$i = 1$$; that is, $$H^*(M_0; \mathbb{Z}_2)$$ and $$H^*(M_1; \mathbb{Z}_2)$$ are not isomorphic as modules over the Steenrod algebra.

Finally, both $$E_0$$ and $$E_1$$ admit a nowhere section, so the associated sphere bundles $$M_i \to S^2$$ admit a section. This implies that the long exact sequence in homotopy of the fibration splits into short exact sequences

$$0 \to \pi_n(S^3) \to \pi_n(M_i) \to \pi_n(S^2) \to 0.$$

Moreover, for $$n \geq 2$$, $$\pi_n(M_i)$$ is abelian, so the existence of a right splitting $$\sigma_* : \pi_n(S^2) \to \pi_n(M_i)$$ implies that the sequence splits, i.e. $$\pi_n(M_i) = \pi_n(S^3)\oplus\pi_n(S^2)$$. As for $$n = 1$$, we saw above that $$M_0$$ and $$M_1$$ are simply connected.

It should be noted that as $$E_0$$ is trivial, $$E_0 = S^2\times \mathbb{R}^4$$ so $$M_0 = S^2\times S^3$$. With this description, we could have calculated the cohomology ring, Stiefel-Whitney classes, and homotopy groups of $$M_0$$ using simpler techniques.

Note, the same computation shows that for $$k \geq 4$$, the sphere bundles of $$\varepsilon^k$$ and $$\mathcal{O}(1)\oplus\varepsilon^{k-2}$$ are simply connected, closed $$(k + 1)$$-dimensional manifolds with isomorphic cohomology rings (for any commutative ring with identity), isomorphic homotopy groups, but are not homotopy equivalent; the first is spin, and the second isn't (and therefore their $$\mathbb{Z}_2$$ cohomologies are not isomorphic as modules over the Steenrod algebra).

• Is it easy to see that the non-trivial bundle admits a section? (My obstruction theory is very weak). An alternative computation of the homotopy groups is that any linear $S^k$ bundle over $S^2$ for $k\geq 2$ is the quotient of $S^k\times S^3$ by a circle action, which acts via the Hopf action on the $S^3$ factor and a linear action on the $S^k$ factor. Then the LES assoiated to $S^1\rightarrow S^k\times S^3\rightarrow (S^k\times S^3)/S^1$ gives the result on homotopy groups. Dec 12, 2018 at 17:44
• @JasonDeVito: The non-trivial $S^3$ bundle over $S^2$ is the sphere bundle of $\mathcal{O}(2k+1)\oplus\varepsilon^2$; the former admits a section because the latter admits a nowhere-zero section. Dec 12, 2018 at 18:59
• Oh, duh. Thanks! Dec 12, 2018 at 19:12