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Inspired by this question I attempted to pull back the 2-form $dt\wedge ds \in \bigwedge^2(\Bbb R^2)$ using the product of the stereographic projections. I'd like to have my calculations checked, for I have the faintest idea of what I'm doing here.

I'll just report the pull back by $\phi:S^1\times S^1 \setminus \{(0,1,0,1)\} \rightarrow \Bbb R^2$ with $$\phi(x,y,z,w) = \left(\frac{x}{1-y},\frac{z}{1-w}\right)$$ Now, knowing that $\phi^*$ commutes with the wedge product and $d$ I got: $$\phi^*(dt\wedge ds) = d\left(\frac{x}{1-y}\right)\wedge d\left(\frac{z}{1-w}\right) = \left(\frac{dx}{1-y} + \frac{xdy}{(1-y)^2}\right)\wedge \left(\frac{dz}{1-w} + \frac{zdw}{(1-w)^2}\right)$$ then expanding everything by linearity of $\wedge$. This form is non vanishing on $S^1\times S^1$ because $dt\wedge ds$ is non vanishing on $\Bbb R^2$ right?

Bernard
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barmanthewise
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  • Do you want just any non-vanishing differential form at all? Why are you not happy with $\operatorname{d}!\theta_1\wedge \operatorname{d}!\theta_2$ (the natural volume form)? – Arnaud Mortier Feb 17 '18 at 23:52
  • @ArnaudMortier I'm perfectly happy with it, I'd just like to have my reslut checked. Is it true that $\phi^*(dt\wedge ds)$ is non vanishing becuse $dt\wedge ds$ is? Is this a legitimate way of pulling back differential forms using charts? – barmanthewise Feb 17 '18 at 23:56
  • Actually your map $\phi$ is not well-defined on a larger set: the domain should be $\mathbb S^1\times \mathbb S^1\setminus{ y=1 \text{ or } w=1}$. –  Feb 18 '18 at 00:07

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