Show that $\overline{z_1}dz_1+\overline{z_2}dz_2$ is invariant under right translations.

Question

I want to show that $(R_{e^{i\theta}})^*(\overline{z_1}dz_1+\overline{z_2}dz_2)=\overline{z_1}dz_1+\overline{z_2}dz_2$.

We are working with the principal $U(1)$ bundle $p:S^{2n+1}\rightarrow \mathbb CP^n$. And using $x^k,y^k$ as coordinates. I am trying to keep things simple and work with $n=1$ for the start.

I have found the following in the book Analysis and algebra on Manifolds. But I do not understand:

1. how they go from $dx$ and $dy$ to $d(\arctan(y/x))$

2. how they go from $R_{e^{i\theta}}^*d(\arctan(y/x))$ to $d(\arctan(y/x)+\theta)$. As far as I understand $R_{e^{i\theta}}^*d(\arctan(y/x))=d(dR_{e^{i\theta}}\arctan(y/x))$. Then $dR_{e^{i\theta}}=R_{e^{i\theta}}$ so we get $d(e^{i\theta}\arctan(y/x))$ not $d(\arctan(y/x)+\theta)$

   

Answer 1

I’ll be changing the proof because working with arctangents lets one wonder about vanishing points for the coordinates, branches, and so on.

So instead, notice that the $1$-form $\omega:=\overline{z_1}dz_1+\overline{z_2}dz_2$ is well-defined over $\mathbb{R}^4$, and that $R=R_{e^{i\theta}}$ is as-well defined over $\mathbb{R}^4$.

It is thus enough to find the relation between $R^*\omega$ and $\omega$ in $\mathbb{R}^4$. If we find that they are equal at each point of $S^3$ (on the full four-dimensional tangent space to $\mathbb{R}^4$), we can then pull-back by the inclusion $i: S^3 \rightarrow \mathbb{R}^4$ and we are good.

So: in $\mathbb{R}^4$, $$R^*\omega=R^*\overline{z_1}d(R^*z_1)+R^*\overline{z_2}d(R^*z_2)=\overline{z_1} \circ R d(z_1 \circ R) + \overline{z_2} \circ R d(z_2 \circ R)=\overline{e^{i\theta}z_1}d(e^{i\theta}z_1)+\overline{e^{i\theta}z_2}d(e^{i\theta}z_2)=\omega$$ as $e^{i\theta}$ is a constant.

The derivation actually answers your question 2: for any $p$-form $\omega$ on $N$ and any smooth $f:M \rightarrow N$, $f^*\omega$ is a $p$-form on $M$ and $d(f^*\omega)=f^*(d\omega)$.

If $\omega$ is a $0$-form, corresponding to a smooth function $g: N \rightarrow \mathbb{R}$ then $f^*\omega$ is the $0$-form on $M$ corresponding to $g \circ f$.

For your question $1$, recall that for a smooth function $f$, $df=\sum_k{\frac{\partial f}{\partial u_k}}$ where the $u_k$ is a coordinate system.

In particular, $d(arctan{y/x})=\frac{\partial arctan(y/x)}{\partial x}dx+\frac{\partial arctan(y/x)}{\partial y}dy=\frac{-y}{x^2}\frac{1}{1+(y/x)^2}dx+\frac{1}{x}\frac{1}{1+(y/x)^2}dy=\frac{-ydx}{x^2+y^2}+\frac{xdy}{x^2+y^2}=\frac{ydx-xdy}{x^2+y^2}$.