# Involution action on $H^3(S^1\times S^2)$

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,

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

and

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

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

• 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. – Pedro Tamaroff Dec 24 '18 at 22:18
• @Michael Albanese Come to think of it, $ds$ seems not $2$ -form. Is it still vaild argument? – Lev Ban Dec 26 '18 at 19:48
• @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). – Michael Albanese Dec 26 '18 at 21:29
• @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? – Lev Ban Dec 26 '18 at 23:38
• 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 $I_2^*\omega = -\omega$. – Michael Albanese Dec 27 '18 at 1:19