The rank-one cotangent and its conic

The map $Q([z]):=e^{2\pi iz}$ on $\mathbb{C}\mathrm{/}\mathbb{Z}$ is well defined because $e^{2\pi i(z+m)}=e^{2\pi iz}{e}^{2\pi im}=e^{2\pi iz}$ for $m\in \mathbb{Z}$, as $2\pi im\in \ker (\exp )$; it is injective because $e^{2\pi iz}=e^{2\pi iw}$ forces $2\pi i(z-w)\in 2\pi i\mathbb{Z}$, that is $z-w\in \mathbb{Z}$; it is surjective because every $v\mathrm{\ne }0$ is $e^w$ for some $w$ and then $v=Q([w\mathrm{/}(2\pi i)])$; and it is holomorphic with derivative $2\pi i{e}^{2\pi iz}\mathrm{\ne }0$ at every point.

For $z\in \mathbb{C}\setminus \mathbb{Z}$ put $s=e^{i\pi z}$, so that $s^2=q$ and $s^{-1}=e^{-i\pi z}$; then $\sin (\pi z)=\frac{s-s^{-1}}{2i}$ and $\cos (\pi z)=\frac{s+s^{-1}}{2}$ by [F1], and $\sin (\pi z)\mathrm{\ne }0$ by [F4], so by [F2] and multiplication of numerator and denominator by $s$, $P(z)=\pi \frac{\cos (\pi z)}{\sin (\pi z)}=\pi i\frac{s+s^{-1}}{s-s^{-1}}=\pi i\frac{s^2+1}{s^2-1}=\pi i\frac{q+1}{q-1}$.

The expansion of [F5] shows that near $z=0$ the sum ${\sum }_{n\ge 1}\frac{2z}{z^2-n^2}$ tends to ${\sum }_{n\ge 1}(\frac{1}{0-n}+\frac{1}{0+n})=0$, so $P(z)-\frac{1}{z}$ extends holomorphically to $0$ with value $0$; equivalently $P$ has principal part $1\mathrm{/}z$ at $0$.

Differentiating $P(z)=\pi \cos (\pi z)\mathrm{/}\sin (\pi z)$ by the chain rule [F3, F6] and the quotient rule [F7], and using $\sin (\pi z)\mathrm{\ne }0$, gives $P^{\mathrm{\prime }}(z)=-{\pi }^2\mathrm{/}{\sin }^2(\pi z)$, so $-P^{\mathrm{\prime }}(z)={\pi }^2\mathrm{/}{\sin }^2(\pi z)$; since ${\sin }^2+{\cos }^2=1$, dividing by ${\sin }^2$ gives ${\csc }^2=1+{\cot }^2$ and hence $-P^{\mathrm{\prime }}(z)={\pi }^2(1+{\cot }^2(\pi z))={\pi }^2+P(z{)}^2$.

The projective curve $YZ=X^2+{\pi }^2{Z}^2$ has a unique point with $Z=0$, namely $[0:1:0]$, because $Z=0$ forces $X^2=0$; in the chart $Y=1$ its equation is $v=u^2+{\pi }^2{v}^2$ with $u=X\mathrm{/}Y$, $v=Z\mathrm{/}Y$, and ${\mathrm{\partial }}_v(v-u^2-{\pi }^2{v}^2)=1-2{\pi }^{2}v$ equals $1$ at the origin, so $u$ is a local coordinate there. In the chart $Z=1$ the equation is the parabola $y=x^2+{\pi }^2$, whose projection to the $x$-line is a bijection onto $\mathbb{C}$; hence every conic point is $[x:x^2+{\pi }^2:1]$ for a unique $x\in \mathbb{C}$, or $[0:1:0]$, and the two points with vanishing $y$ are $[\pm i\pi :0:1]$.

The function $P$ is holomorphic on $\mathbb{C}\setminus \mathbb{Z}$: there it is the composite of $z\mapsto e^{2\pi iz}$ with the rational function $q\mapsto \pi i(q+1)\mathrm{/}(q-1)$, holomorphic on ${\mathbb{C}}^{\ast }\setminus \{1\}$. By [F4], its zeros are exactly $\frac{1}{2}+\mathbb{Z}$.

If $P(z)=P(w)$ for $z,w\in \mathbb{C}\setminus \mathbb{Z}$, then step 1.2 gives $\pi i(q+1)\mathrm{/}(q-1)=\pi i(q^{\mathrm{\prime }}+1)\mathrm{/}(q^{\mathrm{\prime }}-1)$ with $q=e^{2\pi iz}\mathrm{\ne }1$ and $q^{\mathrm{\prime }}=e^{2\pi iw}\mathrm{\ne }1$; cross-multiplying, $q{q}^{\mathrm{\prime }}-q+q^{\mathrm{\prime }}-1=q^{\mathrm{\prime }}q-q^{\mathrm{\prime }}+q-1$, hence $2q^{\mathrm{\prime }}=2q$, $q=q^{\mathrm{\prime }}$, and $z-w\in \mathbb{Z}$ by [F9]. Thus $P$ is injective on the coset space $(\mathbb{C}\setminus \mathbb{Z})\mathrm{/}\mathbb{Z}$; since $P^{\mathrm{\prime }}\mathrm{\ne }0$ there by step 1.4, $P$ is locally biholomorphic on $\mathbb{C}\setminus \mathbb{Z}$.

For $z\mathrm{\notin }\mathbb{Z}$ and $T\in \mathbb{C}$ with $z+T\mathrm{\notin }\mathbb{Z}$, step 1.2 gives $P(z+T)=P(z)$ if and only if $e^{2\pi iT}=1$, which by [F9] holds exactly when $T\in \mathbb{Z}$; since $P$ is nonconstant by step 1.4, the period group of $P$ is exactly $\mathbb{Z}$.

On $\mathbb{C}\setminus \mathbb{Z}$ the point $[P(z):-P^{\mathrm{\prime }}(z):1]$ equals $[u(z):1:v(z)]$ with $u=P\mathrm{/}(-P^{\mathrm{\prime }})=\sin (2\pi z)\mathrm{/}(2\pi )$ and $v=1\mathrm{/}(-P^{\mathrm{\prime }})={\sin }^2(\pi z)\mathrm{/}{\pi }^2$, using step 1.4 and $\sin (2\pi z)=2\sin (\pi z)\cos (\pi z)$; writing $q=1+2\pi izh(z)$ with $h(z)={\sum }_{k\ge 1}(2\pi iz{)}^{k-1}\mathrm{/}k!$ entire and $h(0)=1$ by [F8], step 1.2 gives $P(z)=\frac{1}{z}\cdot \frac{1+\pi izh(z)}{h(z)}$, so $u$ and $v$ are holomorphic near $0$ with $u(0)=v(0)=0$ and $v=u^2+{\pi }^2{v}^2$ by step 1.4. Thus $z\mapsto [u(z):1:v(z)]$ extends holomorphically to $z=0$ with value $[0:1:0]$.

By [F5] the function $P(z)-\frac{1}{z}={\sum }_{n\ge 1}\frac{2z}{z^2-n^2}$ is holomorphic on $\mathbb{C}\setminus \mathbb{Z}$; near $z=m\in \mathbb{Z}\setminus \{0\}$ the summand with $n=\mathrm{\mid }m\mathrm{\mid }$ contributes the only singularity, a simple pole of residue $1$, so the poles of $P$ are exactly the integers, all simple of residue $1$. Consequently the only class of $\mathbb{C}\mathrm{/}\mathbb{Z}$ at which $P$ is not defined is the class of $0$.

By steps 2.2 and 2.3 the map is invariant under $\mathbb{Z}$ and injective on $(\mathbb{C}\setminus \mathbb{Z})\mathrm{/}\mathbb{Z}$, so adding the class of $0$ gives a bijection onto its image; by step 3.1 every other class is in the domain of $P$; for $z\mathrm{\notin }\mathbb{Z}$ the finite image $[x:y:1]$ has $y=-P^{\mathrm{\prime }}(z)={\pi }^2\mathrm{/}{\sin }^2(\pi z)\mathrm{\ne }0$, so it avoids $[\pm i\pi :0:1]$, and $x=P(z)$ runs exactly once over $\mathbb{C}\setminus \{\pm i\pi \}$ because $q$ runs exactly once over ${\mathbb{C}}^{\ast }\setminus \{1\}$ by step 1.1 and $q\mapsto \pi i(q+1)\mathrm{/}(q-1)$ is a bijection ${\mathbb{C}}^{\ast }\setminus \{1\}\to \mathbb{C}\setminus \{\pm i\pi \}$ with inverse $x\mapsto (x+\pi i)\mathrm{/}(x-\pi i)$ by step 1.2. Hence the extended map is a bijection of $\mathbb{C}\mathrm{/}\mathbb{Z}$ onto the conic minus $[\pm i\pi :0:1]$, holomorphic with nowhere-vanishing derivative off the class of $0$.