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!


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

  • $\begingroup$ Thank you for the answer! But could I ask why $\theta(\overline{z})=-\theta(z)$? $\endgroup$ – Lev Ban Dec 24 '18 at 5:06
  • $\begingroup$ 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 $\theta(\bar{z}) = -\alpha = -\theta(z)$. $\endgroup$ – Michael Albanese Dec 24 '18 at 5:08
  • $\begingroup$ 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? $\endgroup$ – Lev Ban Dec 24 '18 at 5:11
  • $\begingroup$ @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. $\endgroup$ – Michael Albanese Dec 24 '18 at 5:16
  • $\begingroup$ Ah.. I got it.. Thank you! $\endgroup$ – Lev Ban Dec 24 '18 at 5:20

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