Consider a function $\phi : \mathbb{R}^d \rightarrow \mathbb{C}$ which is continuous, satisfies $\phi(0)=1$ and is nowhere zero.

I am reading a book where the following claim is made:

Fix $z \in \mathbb{R}^d$. Let $\log$ denote the complex logarithmic function (multi-valued). Then there is a unique branch $h_z(t), 0 \leq t \leq 1$ of $\log \phi(tz), 0 \leq t \leq 1$ with the property that $h_z(0) = 0$ and such that $h_z(t)$ is continuous in $t$.

I don't understand why a continuous branch exists (why should the path $t \mapsto \phi(tz), t \in [0,1]$ not wind around the origin?). If this path doesn't wind around the origin, won't there be infinitely many possible choices of rays leaving the origin to use as branch cuts?

Many thanks for your help.


Forget about branch cuts and work with covering spaces. Consider the map $f:\mathbb{C}\to\mathbb{C}^*$ given by $f(z)=\exp(z)$. This is the universal covering space for $\mathbb{C}^*$.

Now consider the continuous path $\gamma:[0,1]\to\mathbb{C}^*$ given by $\gamma(t)=\phi(tz)$ for some fixed $z\in\mathbb{R}^d$.

By the lifting property of the universal covering, given $p\in\mathbb{C}^*$ such that $\exp(p)=\gamma(0)=1$, there exists a unique continuous lift $\widetilde{\gamma}_p:[0,1]\to\mathbb{C}$ such that $\exp(\widetilde{\gamma}_p(t))=\gamma(t)$ for all $t\in[0,1]$ and $\widetilde{\gamma}_p(0)=p$.

If you take $p=0\in\mathbb{C}$, you get the thesis.

N.B. The winding number is to be considered if you have a loop ($\gamma(0)=\gamma(1)$) and you want to lift it as a loop. Here the only request is that the lifting should be continuous and starting from $0$, so you just look at it as an arc.

Alternative approach

Remark Consider the $1$-form $\omega=\dfrac{dz}{z}$. Then, $\log(w)$, for $w\in \mathbb{C}^*$, can be written as $$\int_\gamma\omega$$ for a suitable $\gamma:[0,1]\to\mathbb{C}^*$ such that $\gamma(0)=1$ and $\gamma(1)=w$. Depending upon the number of times that $\gamma$ winds up around $0$, you will get the different values of $\log(w)$, changing for an integer multiple of $2\pi i$ (the precise multiple is given by the winding number).

Now, set $g(t)dt=\gamma^*\omega$, i.e. $$g(t)=\dfrac{\langle z, \nabla \phi(tz)\rangle}{\phi(tz)}\;.$$ We obtain a function $g:[0,1]\to\mathbb{C}$ (as $\phi(tz)\neq 0$ for every $t$ and $z$). This function is continuous on $[0,1]$, hence integrable.

Now just take a primitive $G$ of $g$ on $[0,1]$ such that $G(0)=0$. $G:[0,1]\to\mathbb{C}$, by the Remark above, is the "logarithm" of $\gamma$.

  • $\begingroup$ this looks very elegant, and thanks for your answer, but it is too high tech for me... I don't want to seem lazy, but this comes from a book on probability theory, and it would be good to see how this works using elementary complex analysis if possible. $\endgroup$ – Frank Mar 15 '15 at 19:38
  • $\begingroup$ sorry, in my undergrad course we had fundamental groups and covering spaces before complex analysis... $\endgroup$ – wisefool Mar 15 '15 at 19:50
  • $\begingroup$ Oh, I see... in neither my undergrad or masters did we ever study such things...! though by the looks of it we certainly should. do you have a simpler explanation? $\endgroup$ – Frank Mar 15 '15 at 19:58
  • $\begingroup$ I added an equivalent approach, I hope more likely to suit your needs. $\endgroup$ – wisefool Mar 15 '15 at 20:00
  • $\begingroup$ Thanks for your help. Is it possible to give an argument purely in terms of selecting a branch of the log with a continuous argument along the path? Thinking in these naive terms I don't see why such a continuous selection of the arg is possible. If the path spirals around the origin then any branch cut will intersect it. If the path doesn't do this, won't there be many possible choices of branch cut such that the argument is continuous? I sorry to ask for such a crude methodology compared to what you have already provided... $\endgroup$ – Frank Mar 15 '15 at 22:33

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