I am studying about involution action $I^*$ on de Rham cohomology group $H^3(S^1\times S^2)$ induced from an action $I\cdot (z,x)=(\overline{z},-x) $ where $S^1\times S^2\subset \mathbb{C}\times \mathbb{R}^3$. Note that, by Kunneth formula, $$H^3(S^1\times S^2)=\mathbb{R}.$$

I figured out the action of $I^*$ on $H^2(S^1\times S^2)$ in a similar way with my previous question Involution action on $H^1(S^1\times S^2)$.

However, I am stuck at identifying the action on $H^3(S^1\times S^2)$.

I was trying to pick $d\theta \in \Omega^1(S^1)$ and $ds\in \Omega^2(S^2)$ where $\theta(\overline{z})=-\theta(z)$ and $s(-x)=-s(x)$. And it seems leading me to create $\omega=d\theta\wedge ds$ as a generator of $H^3(S^1\times S^2)$. But I am not sure if I am going in right direction since the $\omega$ seems not the one for some anticlimatic reason.

Any help would be appreciated! Thank you in advance!


As you have defined it, $\omega$ doesn't make sense because you can't take the wedge product of a form on $S^1$ with a form on $S^2$. Instead, let $\pi_i : S^1\times S^2 \to S^i$ be the natural projections. Then $\pi_1^*d\theta$ and $\pi_2^*ds$ are both forms on $S^1\times S^2$, so their wedge product $\omega = (\pi_1^*d\theta)\wedge(\pi_2^*ds)$ is defined.

Let $I_1 : S^1\to S^1$ be given by $I_1(z) = \bar{z}$ and $I_2 : S^2\to S^2$ be given by $I_2(x) = -x$, then $I(z, x) = (I_1(z), I_2(x))$. Said another way, $\pi_1\circ I = I_1\circ\pi_1$ and $\pi_2\circ I = I_2\circ \pi_2$. Therefore,

$$I^*\pi_1^*d\theta = (\pi_1\circ I)^*d\theta = (I_1\circ \pi_1)^*d\theta = \pi_1^*(I_1^*d\theta) = \pi_1^*(-d\theta) = -\pi_1^*d\theta$$


$$I^*\pi_2^*ds = (\pi_2\circ I)^*ds = (I_2\circ \pi_2)^*ds = \pi_2^*(I_2^*ds) = \pi_2^*(-ds) = -\pi_2^*ds.$$

Therefore $I^*\omega = (I^*\pi_1^*d\theta)\wedge(I^*\pi_2^*ds) = (-d\theta)\wedge(-ds) = d\theta\wedge ds = \omega$.

  • 1
    $\begingroup$ More generally, there's a Kunneth theorem for deRham cohomology: $H^*(X)\otimes H^*(Y)\longrightarrow H^*(X\times Y)$ given by the formula of this post is an isomorphism of algebras. $\endgroup$ – Pedro Tamaroff Dec 24 '18 at 22:18
  • $\begingroup$ @Michael Albanese Come to think of it, $ds$ seems not $2$ -form. Is it still vaild argument? $\endgroup$ – Lev Ban Dec 26 '18 at 19:48
  • $\begingroup$ @LeB: What do you mean by $ds$? I thought you meant a two-form on $S^2$ (not the exterior derivative of a function, which is not a two-form). $\endgroup$ – Michael Albanese Dec 26 '18 at 21:29
  • $\begingroup$ @MichaelAlbanese I am sorry! I got a bit confused. If it is not too much, could you let me know which kind of 2 form I can use of for this problem? $\endgroup$ – Lev Ban Dec 26 '18 at 23:38
  • 1
    $\begingroup$ The two form $\omega = x\,dy\wedge dz - y\,dx\wedge dz + z\,dx\wedge dy$ on $\mathbb{R}^3$ restricts to a closed form on $S^2$ which generates $H^2_{\text{dR}}(S^2)$, and satisfies $I_2^*\omega = -\omega$. $\endgroup$ – Michael Albanese Dec 27 '18 at 1:19

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