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Are there any examples of non-orientable closed 4-(or 3-)manifolds $M$ whose cohomology ring $H^*(M,\mathbb{Z}_n)$ and/or $H^*(M,\mathbb{Z})$ are known.

Here we assume that $M$ is not a product of lower dimensional manifolds, and is not a connected sum.

The following paper give a class of examples for 3-manifold based on torus bundle: https://arxiv.org/abs/1307.0518 . I wonder if there are examples for 4-manifolds and other examples for 3-manifolds.

This question is motivated from a physics problem. Manifolds with known cohomology ring are like probes that can detect topological orders in quantum matter. We like to have as much probes as possible in order to distinguish all topological orders.

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    $\begingroup$ $\mathbb{R}P^4$? $\endgroup$
    – Tyrone
    Commented Dec 17, 2016 at 15:01
  • $\begingroup$ Yes $\mathbb{R}P^4$ is certainly the simplest example. I am not so clear in my question. I am looking for a sequence of examples, with different torsions in $H^*(M,\mathbb{Z})$. $\mathbb{R}P^4$ has only 2-torsion. Are there examples with other torsion? Also I wonder if cohomology ring $H^*(\mathbb{R}P^4; \mathbb{Z}_n)$ is known? $\endgroup$ Commented Dec 17, 2016 at 15:17
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    $\begingroup$ @Xiao-GangWen it should be possible to compute $H^*(\mathbb{RP}^4, \mathbb{Z}/n)$ in the same way as for $\mathbb Z$ coefficients, which Hatcher's book discusses following Thm. 3.19 (pp. 220-22). $\endgroup$ Commented Dec 17, 2016 at 16:20
  • $\begingroup$ As for torsion in $\mathbb Z$ coefficients, lens spaces (Hatcher, Exm. 2.43, p. 144) are a simple example, though they're orientable. $\endgroup$ Commented Dec 17, 2016 at 16:24

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The Grassmannian of unoriented 2-planes in $\mathbb{R}^4$ is 4-dimensional. A product of an unorientable surface and an orientable one is 4-dimensional.

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  • $\begingroup$ The cohomology is computed, for example, in McCleary's A User's Guide to Spectral Sequences. $\endgroup$
    – Ben McKay
    Commented Dec 17, 2016 at 16:22
  • $\begingroup$ Here, we are looking for examples which are not a product of lower dimensional manifolds, and is not a connected sum. $\endgroup$ Commented Dec 17, 2016 at 17:20
  • $\begingroup$ Is the Grassmannian of unoriented 2-planes $\mathbb{R} P^4$? $\endgroup$ Commented Dec 17, 2016 at 17:23
  • $\begingroup$ No. For computations of the cohomology rings of oriented and unoriented real Grassmannians, see for instance arxiv.org/abs/1610.07968. $\endgroup$
    – Alex Suciu
    Commented Dec 17, 2016 at 18:17
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    $\begingroup$ Why isn't this Grassmannian orientable? $\endgroup$ Commented Jan 2, 2017 at 8:10

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