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

I am studying about an action $$I^*$$ on a de Rham cohomology group $$H^1(S^1\times S^2)$$ 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, $$H^1(S^1\times S^2)=\mathbb{R}.$$

Thus, I want to find a nonzero element in $$H^1(S^1\times S^2)$$ and want to see how $$I^*$$ acts to the element.

And my teacher taught me as below.

Let $$d\theta \in \Omega^1(S^1)$$ be a generator of $$H^1(S^1)=\mathbb{R}.$$ And let $$\pi:S^1\times S^2\rightarrow S^1$$ and let $$\omega=\pi^*(d\theta)$$. Then clearly, $$d\omega=0$$ so $$[\omega]$$ is nonzero element in $$H^1(S^1\times S^2)$$. If $$\iota : S^1\times \{\text{north pole}\}\hookrightarrow S^1\times S^2$$ is an embedding, observe that $$\pi\circ\iota =Id \implies \iota^*\pi^*=Id \implies \iota^*(\omega)=d\theta \implies \iota^*[\omega]=[d\theta]$$

Thus, $$I^*\omega=I^*\pi^*(d\theta)=\pi^*(I_1^*d\theta)=\pi^*(-d\theta)=-\omega.$$

So we get $$I^*=-Id$$.

But I got stuck at why $$I_1^*d\theta=-d\theta$$.

If we see carefully, $$I_1^*d\theta(z)=d\theta(\overline{z})$$. I don't know where is wrong and where I am missing.

I would very appreciate for any help and solution for this issue! Thank you!

## 1 Answer

Note that $$I_1^*d\theta = d(\theta\circ I_1)$$. As $$(\theta\circ I_1)(z) = \theta(I_1(z)) = \theta(\bar{z}) = -\theta(z)$$, so $$\theta\circ I_1 = -\theta$$ and hence $$I_1^*d\theta = d(\theta\circ I_1) = d(-\theta) = -d\theta$$.

• Thank you for the answer! But could I ask why $\theta(\overline{z})=-\theta(z)$? – Lev Ban Dec 24 '18 at 5:06
• Recall that $\theta(z)$ is the argument of $z$. So if $z = e^{i\alpha}$, then $\theta(z) = \alpha$. On the other hand $\bar{z} = e^{-i\alpha}$, so $\theta(\bar{z}) = -\alpha = -\theta(z)$. – Michael Albanese Dec 24 '18 at 5:08
• I guess that is that. But the reason why I got confused it that my teacher has not specify what was $\theta$. He just said $d\theta$ is the generator. Should $\theta$ be such a function in order for $d\theta$ be a generator? – Lev Ban Dec 24 '18 at 5:11
• @LeB: Yes. It is common to denote the generator of $H^1_{\text{dR}}(S^1)$ by $d\theta$ where $\theta$ is the argument function. – Michael Albanese Dec 24 '18 at 5:16
• Ah.. I got it.. Thank you! – Lev Ban Dec 24 '18 at 5:20