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The product of two spheres admits a diagonal $\mathbb{Z}_2$-action, $(x,y)\mapsto (-x,-y)$. I'm trying to compute the integral singular cohomology ring of the orbit space $X$ of this action. $X$ is not just $\mathbb{R}P^2\times\mathbb{R}P^2=S^2\times S^2/\mathbb{Z}_2\times \mathbb{Z}_2$. In particular it has embedded spheres like $S^2\times \{*\}$.

I think that it would be possible to work this out via the Serre spectral sequence, but it was suggested as practice in a course that did not cover spectral sequences-so is there something elementary to be done? The two other strategies I've thought of are trying to write down a CW structure, as one does for $\mathbb{R}P^n$-but it seems harder to give an explicit description of the points of $X$ in terms of, say, pairs of lines-and to use the Gysin sequence-but I don't think the line bundle of which $S^2\times S^2\to X$ is the sphere bundle is orientable, so this looks unhelpful for $\mathbb{Z}$ coefficients.

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  • $\begingroup$ Is this quotient homeomorphic to the quotient $S^2\times S^2/\mathbb Z_2$ where $\mathbb Z_2$ is generated by $(x,y)\mapsto (y,x)$? if so, then this quotient is the symmetric product $SP^2(S^2)$ which is homeomorphic to $\mathbb CP^2$ $\endgroup$ May 10 '14 at 6:53
  • $\begingroup$ Thanks for the comment-the symmetric action has fixed points, whereas this diagonal action does not. Seen another way, our $ X$ has elements of order 2 in its $\pi_1$ for the same reason the real protective plane does. $\endgroup$ May 10 '14 at 11:38
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Here are some tricks you can pull. Let $X$ denote this space (which, incidentally, is the Grassmannian $\text{Gr}_2(\mathbb{R}^4)$ of $2$-planes in $\mathbb{R}^4$, although we won't use this fact). Below homology and cohomology are with integer coefficients.

To start, since $X$ is the quotient of a simply connected space by a free action of $\mathbb{Z}_2$, we have $\pi_1(X) \cong H_1(X) \cong \mathbb{Z}_2$. This gives $H^1(X) \cong 0$ by universal coefficients. Since the $\mathbb{Z}_2$ action on $S^2 \times S^2$ is orientation-preserving, $X$ is orientable, so $H^4(X) \cong H_4(X) \cong \mathbb{Z}$. By Poincaré duality we conclude that $H_3(X) \cong H^1(X) \cong 0$ and $H^3(X) \cong H_1(X) \cong \mathbb{Z}_2$.

It remains to work out $H^2(X) \cong H_2(X)$ and then to work out the ring structure on cohomology. Since $\chi(S^2 \times S^2) = 4$, we have $\chi(X) = 2$, and since the first and third Betti numbers of $X$ vanish we conclude that $H_2(X)$ and $H^2(X)$ must both be torsion. By universal coefficients

$$H^2(X) \cong \text{Ext}^1(H_1(X), \mathbb{Z}) \cong \mathbb{Z}_2$$

(since $H_2(X)$ is torsion the other term vanishes). Since $H^1(X)$ vanishes, the only possible interesting cup product is $H^2(X) \times H^2(X) \to H^4(X)$, but since $H^4(X)$ has no torsion this cup product vanishes too. In summary, the cohomology groups are

$$\mathbb{Z}, 0, \mathbb{Z}_2, \mathbb{Z}_2, \mathbb{Z}$$

and there aren't any interesting cup products.

In general, suppose $X$ is a compact oriented $4$-manifold whose fundamental group and Euler characteristic you know. Then $H_1(X) \cong \pi_1(X)/[\pi_1(X), \pi_1(X)], H^1(X) \cong \text{Hom}(H_1(X), \mathbb{Z})$, and by Poincaré duality this determines $H^3(X)$ and $H_3(X)$. Universal coefficients identifies the torsion subgroup of $H^2(X) \cong H_2(X)$ with the torsion subgroup of $H_1(X)$, and the torsion-free part is determined by the Euler characteristic. This determines all the homology and cohomology groups of $X$.

To get information about the cup product, if $H^1(X)$ vanishes (that is, if $H_1(X)$ is torsion), then the only interesting cup product is $H^2(X) \times H^2(X) \to H^4(X)$, which you can deduce things about if you can compute the signature of $X$; see this blog post, where this is done for simply connected compact oriented $4$-manifolds.


Edit: Here are some details for the identification with $\text{Gr}_2(\mathbb{R}^4)$. First, $\text{Gr}_2(\mathbb{R}^4)$ has a natural double cover, the Grassmannian $\text{Gr}_2^{+}(\mathbb{R}^4)$ of oriented $2$-planes in $\mathbb{R}^4$. The claim is that this is $S^2 \times S^2$ and moreover that the action of $\mathbb{Z}_2$ given by reversing orientations is given by the diagonal action in the OP.

The oriented $2$-plane spanned by $v, w \in \mathbb{R}^4$ can be represented (this is a variation on the Plücker embedding) by the bivector $v \wedge w \in \Lambda^2(\mathbb{R}^4)$ up to multiplication by a positive scalar. Now, $\Lambda^2(\mathbb{R}^4)$ is known to decompose under the action of $\text{SO}(4)$ as the direct sum of two $3$-dimensional representations. Explicitly, the Hodge star operation

$$\star : \Lambda^2(\mathbb{R}^4) \to \Lambda^2(\mathbb{R}^4)$$

defines a map squaring to $1$ and commuting with the action of $\text{SO}(4)$, and the two $3$-dimensional representations are its eigenspaces. Very explicitly, if $e_1, e_2, e_3, e_4$ is an oriented orthonormal basis for $\mathbb{R}^4$, then the defining relation of $\star$ is that

$$\alpha \wedge \star \beta = e_1 \wedge e_2 \wedge e_3 \wedge e_4.$$

We compute that

$$\star (e_1 \wedge e_2) = e_3 \wedge e_4, \star (e_3 \wedge e_4) = e_1 \wedge e_2$$ $$\star (e_1 \wedge e_3) = -e_2 \wedge e_4, \star (e_2 \wedge e_4) = -e_1 \wedge e_3$$ $$\star (e_1 \wedge e_4) = e_2 \wedge e_3, \star (e_2 \wedge e_3) = e_1 \wedge e_4$$

and hence that the bivectors $e_1 \wedge e_2 + e_3 \wedge e_4, e_1 \wedge e_3 - e_2 \wedge e_4, e_1 \wedge e_4 + e_2 \wedge e_3$ span the $3$-dimensional $+1$-eigenspace of $\star$ (the self-dual bivectors) while the corresponding bivectors with the opposite signs span the $3$-dimensional $-1$-eigenspace of $\star$ (the anti-self-dual bivectors).

In any case, a bivector $v \wedge w \in \Lambda^2(\mathbb{R}^4)$ up to positive scaling now induces a pair of vectors in $\mathbb{R}^3$ up to positive scaling. Since $\star (v \wedge w)$ necessarily describe a $2$-plane orthonormal to the $2$-plane spanned by $v$ and $w$, these bivectors can neither be self-dual nor anti-self-dual, so neither of these vectors in $\mathbb{R}^3$ vanish. Thus we get a pair of elements in $\mathbb{R}^3$ minus the origin up to positive scale; this is naturally identified with $S^2$. Altogether we've described a natural map

$$\text{Gr}_2^{+}(\mathbb{R}^4) \to S^2 \times S^2$$

but at this point I've probably said too much already. There should be a shorter argument starting from that $\text{Spin}(4)$ observation in the comments but maybe it's less geometrically enlightening.

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    $\begingroup$ To at least be convinced that it's plausible, note that the universal cover of $\text{SO}(4)$ is $\text{Spin}(4) \cong \text{SU}(2) \times \text{SU}(2)$. I'll edit in something more rigorous in a bit. $\endgroup$ Nov 3 '14 at 5:08
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    $\begingroup$ @Kevin: the decomposition of $\Lambda^2(\mathbb{R}^4)$ I described above gives a natural map $\text{SO}(4) \to \text{SO}(3) \times \text{SO}(3)$ which turns out to be a double cover. Taking universal covers from here gives the result. $\endgroup$ Nov 3 '14 at 5:42
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    $\begingroup$ Okay, so everything works out. The point is that $\text{Gr}_2^{+}(\mathbb{R}^4)$ is $\frac{\text{SO}(4)}{\text{SO}(2) \times \text{SO}(2)}$, so it's also $\text{Spin}(4)$ modulo the preimage of $\text{SO}(2) \times \text{SO}(2)$ in $\text{Spin}(4)$, which turns out to be the image of $\text{Spin}(2) \times \text{Spin}(2)$ in $\text{Spin}(3) \times \text{Spin}(3)$. And the quotient $\text{Spin}(3)/\text{Spin}(2)$ is just the quotient $\text{SO}(3)/\text{SO}(2) \cong S^2$ as expected. $\endgroup$ Nov 3 '14 at 5:54
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    $\begingroup$ (In particular your first description of $\text{Gr}_2(\mathbb{R}^4)$, the nonoriented Grassmannian, is off by a factor of $2$: it should be either $\frac{\text{O}(4)}{\text{O}(2) \times \text{O}(2)}$ or $\frac{\text{SO}(4)}{\text{S}(\text{O}(2) \times \text{O}(2))}$.) $\endgroup$ Nov 3 '14 at 5:55
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    $\begingroup$ Hmm, no, I think I'm off by a factor of $2$ as well. I'm not totally confident that the preimage of $\text{SO}(2) \times \text{SO}(2)$ in $\text{Spin}(4)$ is $\text{Spin}(2) \times \text{Spin}(2)$. But I'm not sure what I'm doing wrong. $\endgroup$ Nov 3 '14 at 6:25

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