Analytic Continuation, Monodromy, and Riemann Surfaces — Examples

For $t\in [0,1]$, let $V_t:=e^{2\pi it}B(1,1\mathrm{/}2)$ and define $L_t(u):=2\pi it+\mathrm{Log}(e^{-2\pi it}u)$ for $u\in V_t$. By [L1], each $L_t$ is holomorphic on $V_t$ and satisfies $\exp (L_t(u))=u$ and $L_t(\gamma (t))=2\pi it$.

Use the subdivision $t_k:=k\mathrm{/}16$ for $0\le k\le 16$. For $t\in [t_k,t_{k+1}]$, $\mathrm{\mid }\gamma (t)-\gamma (t_k)\mathrm{\mid }\le 2\sin (\pi \mathrm{/}16)<\frac{1}{2},$ so the whole subpath lies in $V_{t_k}$. At the joining point one has $L_{t_k}(\gamma (t_{k+1}))=2\pi i{t}_k+\mathrm{Log}(e^{2\pi i\mathrm{/}16})=2\pi i{t}_{k+1}=L_{t_{k+1}}(\gamma (t_{k+1})),$ because the argument increment is $\pi \mathrm{/}8$. Thus the branches form an admissible continuation chain along $\gamma$. At $t=0$ this is the principal logarithm germ, while the terminal germ at $1$ has value $2\pi i$. By [L2], it is the germ of $\mathrm{Log}+2\pi i$ at $1$.

For $t\in [0,1]$, let $V_t:=e^{2\pi it}B(1,1\mathrm{/}2)$ and define $L_t(u):=2\pi it+\mathrm{Log}(e^{-2\pi it}u)$ and $R_t(u):=\exp (L_t(u)\mathrm{/}2)$. Because $e^{-2\pi it}{V}_t=B(1,1\mathrm{/}2)$ lies in the slit plane, [L1] makes each $R_t$ a holomorphic square-root branch on $V_t$, and $R_t(\gamma (t))=e^{\pi it}$.

Use the subdivision $t_k:=k\mathrm{/}16$ for $0\le k\le 16$. For $t\in [t_k,t_{k+1}]$, $\mathrm{\mid }\gamma (t)-\gamma (t_k)\mathrm{\mid }\le 2\sin (\pi \mathrm{/}16)<\frac{1}{2},$ so the whole subpath lies in $V_{t_k}$. At the joining point, the argument increment is $\pi \mathrm{/}8$, so the logarithm branches $L_{t_k}$ and $L_{t_{k+1}}$ agree there and hence define the same germ; therefore their square-root branches do as well. Thus the $R_{t_k}$ form an admissible continuation chain along $\gamma$. At $t=0$ the value at $1$ is $1$, while the terminal value is $e^{\pi i}=-1$. So the terminal germ is the negative of the initial square-root germ.

Writing $w=x+iy$, Euler's formula gives $e^w=e^x(\cos y+i\sin y)$. So the first two coordinates of $H(w)$ are exactly the complex number $e^w$.

By [L1], the point of the logarithm surface corresponding to $w$ projects to $e^w$. Step 1.1 therefore identifies the abstract surface with the standard helicoid model in ${\mathbb{R}}^3$, whose vertical coordinate records the argument before taking it modulo $2\pi$.

Specializing [L1] to $n=2$ makes the square-root surface biholomorphic to ${\mathbb{C}}^{\times }$ with projection $w\mapsto w^2$.

If $z\in {\mathbb{C}}^{\times }$ and $w^2=z$, then also $(-w{)}^2=z$, and these are the only lifts because $u^2=z$ implies $(u\mathrm{/}w{)}^2=1$, hence $u\mathrm{/}w=\pm 1$. So the surface has exactly two sheets over each nonzero base point.

For $\mathrm{\mid }z\mathrm{\mid }<1$, fact [L1] gives absolute convergence of ${\sum }_{n\ge 0}{z}^n$, the finite identity $(1-z){\sum }_{k<n}{z}^k=1-z^n$, and the limit $z^n\to 0$. Passing to the limit yields the sum formula ${\sum }_{n\ge 0}{z}^n=1\mathrm{/}(1-z)$ on the unit disc, and [L2] gives radius $1$.

If $\zeta$ lies on the unit circle and $\zeta \mathrm{\ne }1$, then $1-\zeta \mathrm{\ne }0$, so the same rational function is holomorphic on a neighbourhood of $\zeta$. Thus $\zeta$ is regular. At $z=1$ the denominator vanishes, so no holomorphic extension through $1$ can equal $1\mathrm{/}(1-z)$ on a punctured neighbourhood. Therefore $1$ is the unique singular boundary point.

Fact [L1] gives a holomorphic function on the unit disc that cannot be continued through any boundary point of that disc.

Therefore the universal claim is false.

On the closed unit disc one has $\mid z^n\mathrm{/}{n}^2\mid \le 1\mathrm{/}{n}^2$. By [L1] and [L2], the series for $f$ converges uniformly on $\bar{\mathbb{D}}$, so its sum is continuous there.

Fact [L3] gives radius $1$ and shows that the boundary point $1$ is singular. Therefore continuity on the closed disc does not force regularity at all boundary points, and the displayed claim is false.

Fact [L1] gives a holomorphic function on a domain whose boundary admits no continuation.

Hence the universal claim is false.

Compare the constant path at $1$ with the once-around loop based at $1$. These two paths have the same endpoints, but [L1] gives different terminal germs.

Therefore the universal claim is false.

Fact [L1] shows that the definition itself produces an abstract space of germs, not an a priori subset of ${\mathbb{C}}^2$.

The geometric model in [L2] is an extra construction, not part of the definition. Therefore the claim that such a subset model is automatic is false.

Fact [L1] supplies a radius-one power series with boundary points other than $1$ that are regular.

Therefore the universal claim is false.