Pullback of a complex $ 1$-form

Question

Let $p = \operatorname{exp} : \mathbb{C} \to \mathbb{C}^*$ be a covering and $(U,z)$ a chart of $\mathbb{C}^*$ with $z = x + iy$. Let $\omega = dz/z$ be a one-form on $U$.

Problem: Find the pullback $p^*\omega$.

My try: We can write $p^* \frac{1}{z}dz = p^*\frac{1}{z} \, d(p^*z) = p^*\frac{1}{z} \, p^*(dz) = (\frac{1}{z} \circ p)(dz \circ p)$. I also tried making sens of $\frac{1}{z} \circ p (a)$ for some $a \in U$. Then we get $$ \frac{1}{z} e^a = \frac{1}{x(e^a) + iy(e^a)}. $$

I have no idea what makes sense to do or try. I have very little intuition for this.

Answer 1

Look at a restriction on an open chart where $p : (V,w) \mapsto (U,z)$ is an isomorphism.

Then $p^*(dz) = (dp/dw) dw = (\exp w) dw$ and so $p^*(\frac 1z dz) = \frac 1{p(w)}p^*(dz) = \frac 1{\exp w}(\exp w) dw = dw$.

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Notice that if $\tau$ is a translation $w \in (\Bbb C,+) \mapsto w +a \in (\Bbb C,+)$ then $\tau^*(dw) = dw$ so $dw$ is invariant by translation.

Likewise, since $\exp$ is a group morphism $(\Bbb C,+) \to (\Bbb C^*,\times)$, if $\tau$ is a scalar multiplication $z \in (\Bbb C^*,\times) \mapsto bz \in (\Bbb C^*,\times)$ then $\tau^*(dz/z) = dz/z$ so $dz/z$ is invariant by scalar multiplication.

So they are both very important $1$-forms for their respective groups.

Answer 2

(1) If $f(r,t)=r(\cos\ t,\sin\ t)$, then we can find $f^\ast \frac{dz}{z}:= f^\ast a+ i f^\ast b$ where $\frac{\overline{z}dz}{|z|^2}=\frac{dz}{z} :=a+ ib $ But we want to find more easily.

(2) $f(u,v)=(f_1,f_2)$ where $f_i$ is smooth. Then define $g_f:=f_1+if_2$. In further assume that $g_f$ is holomorphic.

Then $$ z:=x+iy,\ w:=u+iv,$$ \begin{align} f^\ast dz&= f^\ast dx+ if^\ast dy = df_1 + idf_2 \\&= (f_1)_u du + (f_1)_v dv + i ((f_2)_udu + (f_2)_vdv) \\&= (f_1)_u du -(f_2)_u dv + i ((f_2)_udu + (f_1)_udv) \\&= (f_1)_u (du +idv) + (f_2)_u (-dv + idu) \\&=(f_1)_u dw + i(f_2)_u dw = g_f'(w) dw \end{align}

(3) $dg_f(w)= df_1+idf_2=g_f'(w)dw $

(4) If $g_h$ is holomorphic, then $$ q:=s+it,\ h^\ast dg_f (w)=h^\ast g_f' dw =g_f' g_h' dq=d(g_f\circ g_h)(q) $$ by chain rule.

(5) Now we will apply to this case $\frac{dz}{z}=d\ln\ z$ so that $$ g_f:=\exp,\ f^\ast \frac{dz}{z} =f^\ast d\ln\ z= d \ln\ g_f(w)=dw $$