2
$\begingroup$

I am lecturing a course (a "methods courses" so light on proofs, and more focused on building intuition and exploring applications) on Complex Analysis, somewhat inherited from a colleague, and somewhat following the textbook by Osborne. I have run into a problem by setting a seemingly impossible question...

The textbook defines branch cuts and branch points as follows:

Extract from Osborne

This is admitted vague (what exactly is a "line segment"?) My exact problem is that I asked my students to consider this function:

$$ f(z) = (z^3-z)^{1/3} $$

Which has branch points at $-1, 0, 1$ (and not at $\infty$). I then suggested a branch cut could be the line from $-1$ to $1$ along the real axis. The idea is to illustrate that a contour passing around all the branch points does return to its chosen initial value. But this suggestion gives two branch cuts at $0$, in violation of the definition.

Does anyone know why the textbook makes this definition?

On further thought, I don't see why this definition has been made? Branch cuts are a bit arbitrary, but it seems to me that a working definition is that you wish to end up with a connected domain upon which your function is single-valued. Why would I worry about branch cuts intersecting, or branch points being associated to only one branch cut?

There is not branch point at $\infty$: if $C$ is a contour enclosing $-1, 0, 1$ (say, a large circle centre $0$, aka, a contour about $\infty$) then as $z$ traverses $C$, we have that $\arg(z)$ increases by $2\pi$. Then consider $$ f(z) = z^{1/3} (z-1)^{1/3} (z+1)^{1/3} = \exp\Big( \frac13(\log(z) + \log(z-1) + \log(z+1)\Big) $$ $\log(z)$ changes by $2\pi i$ around the contour, and similarly $\log(z-1)$ and $\log(z+1)$, and so the part inside the bracket changes by $\frac13 6\pi i = 2\pi$ and thus $f(z)$ does not change around the contour, regardless of the branches of $\log$ chosen.

$\endgroup$
8
  • $\begingroup$ Using the principal branch (smallest nonnegative arg), there is a branch cut from $0$ to $1$ and a branch cut from $-1$ to $-\infty$. Why do you say there is no branch point at $\infty$? $\endgroup$
    – Eric Towers
    Dec 1, 2019 at 11:50
  • $\begingroup$ ... in particular, it appears that $\infty$ is an algebraic branch point of order $3$. $\endgroup$
    – Eric Towers
    Dec 1, 2019 at 12:19
  • $\begingroup$ I've added what I believe is a proof that $\infty$ is not a branch point (according to the definition of Osborne. Admittedly, Osborne does not seem to explicitly consider the Riemann Sphere, which further complicates following that book... $\endgroup$
    – Matthew Daws
    Dec 1, 2019 at 18:09
  • 1
    $\begingroup$ In this case you can cheat, because $[-1, 1]$ is one segment and thus satisfies the definition. If the branch points were $-1, 1, i$, a possible solution would be $[0, -1], [0, 1], [0, i]$, which seems like an unnecessary complication. Also, a ray would have to be considered to be a segment as well. Perhaps the definition means rays rather than segments? With a less restrictive definition, a branch cut doesn't even have to go through the branch point (it can be a spiral). $\endgroup$
    – Maxim
    Dec 1, 2019 at 23:10
  • $\begingroup$ You should not think to $\log z$ as multivalued, instead define one version of it analytic around $z=1$ then consider the set of all its analytic continuations (over all curves from $1$ in $\Bbb{C}^*$). A branch is one of those analytic continuations, the branch cut is the boundary of its domain of analyticity. $\endgroup$
    – reuns
    Dec 2, 2019 at 4:19

1 Answer 1

Reset to default
2
$\begingroup$

This first section is not part of an answer.

Perhaps it would be useful to see a Riemann surface of $f$. (These plots are generated by Jason B.'s code from this answer at Mathematica.SE, which is based on Michael Trott's RiemannSurfacePlot3D.) First, on a disk centered at $0$ of radius $3/2$ with the real axis running right-left and the imaginary axis increasing into the screen.

Riemann surface of f

The branch points at $-1$, $0$, and $1$ are visible from left to right. Note that this is a $3$-sheeted (branched) covering of $\Bbb{C}$.

Now a Riemann surface for $f(1/z)$. The real axis runs from slightly above center-left to slightly below center-right. The imaginary axis runs into the screen from the lower-left to the upper-right. The branch points at $-1$ and $1$ are visible as the left and right dimples, respectively, in the top layer.

Riemann surface for f(1/z)

The $3$-sheetedness continues to $\infty$ and is expressed by trifolium cross-sections in the plot.

Perhaps it would help to see a Riemann surface for $(z^3)^{1/3}$, which is not as simple as one might expect.

Riemann surface of cube root of cube of z

and $((1/z)^3)^{1/3}$ to see the neighborhood of $\infty$.

Riemann surface of cube root of cube of reciprocal z

Now maybe some response to the Question...

I am not familiar with Osborne's text, but I am not happy with the quoted discussion of branch cuts. A branch cut can be an arbitrary simple (i.e., non-self-intersecting) curve, with each endpoint on a different branch point.

A line segment is a line segment -- a finite segment of a line. This particular characterization of a branch cut is inadequate in many respects. One is that it cannot represent any of the choices of branch cut for the logarithm, which are necessarily infinitely long since they must reach from the branch point at $0$ to the branch point at $\infty$.

A branch point cannot be met by a branch cut between two other branch points. At such a hypothetical intersection, the values of all the branches are the same (since we are at a branch point), but the values of the function approaching the branch point from either side of the branch cut disagree, which is a contradiction.

We should perhaps distinguish between branches and choices of values of the function. For a multivalued function, at each point, we may select one of its values to associate with that point. An easy example: for each $x \in \Bbb{R}_\geq 0$ assign an arbitrary choice of $\sqrt{x}$ or $-\sqrt{x}$. Note that we have many such collections of choices. Only two of these collections are branches of the square root function. Branches have the property that one can analytically continue along them. This means that in any small enough neighborhood of a branch point, no branch cut other than the branch cut terminating at that branch point may be present. This is so we can analytically continue the branch around the point to get a branch, not just a bunch of choices of values.

You rightly want $\Bbb{C} \smallsetminus (\{\text{branch cuts}\} \cup \{\text{branch points}\})$ to be connected (in fact, path connected). This also means that branch cuts cannot intersect transversely: If two cuts intersect, there is at least one region (homotopic to a disk) bounded by the branch cuts and separated from the rest of the plane, so there is no way to analytically continue from the interior of the disk to the rest of $\Bbb{C}$ or vice versa. Allowing transversely intersecting branch cuts more closely corresponds to a choice of value than to a choice of branch. The same thing happens if you allow two (nonintersecting) cuts between two branch points: there is now a disk bounded by the two cuts and two points which is disconnected from the rest of $\Bbb{C}$. I suppose one could have two branch cuts meet non-transversely, but a small homotopy of one would eliminate the intersection.

Returning to your $f$. This means you must have a branch cut between two of your branch points and another branch cut from the third branch point to the branch point at infinity. That there is a branch point at infinity can be shown by $\lim_{z \rightarrow 0} \frac{1}{f(1/z)} = 0$ (That is, $f$ is single valued at $z = \infty$ on the Riemann sphere.), just as one does for the logarithm. That it is of order $3$ can be seen by counting the number of sheets of the Riemann surface in neighborhoods of $z = \infty$.

As more evidence, the intersections of the folium suggest something interesting happens for arguments near $\pm \pi/3$ and $\pm \pi$. (The curves of interest are asymptotic to the rays of argument $\pm \pi/3$, and finding them is a bit of a hassle, so we look at what happens around $\pm \pi$.) Transverse to the negative real axis, choosing a branch with cuts along $(-\infty, -1)$ and $(0,1)$, so that we may continue from $-1/2$ to the points in the limits below, \begin{align*} \lim_{\theta \rightarrow -\pi^+} f \left( 10 \mathrm{e}^{\mathrm{i} \theta} \right) &= -(-3)^{2/3}110^{1/3} = 4.9832{\dots} + 8.6312{\dots} \mathrm{i} \\ \lim_{\theta \rightarrow \pi^-} f \left( 10 \mathrm{e}^{\mathrm{i} \theta} \right) &= 3^{2/3}(-110)^{1/3} = 4.9832{\dots} - 8.6312{\dots} \mathrm{i} \end{align*} So when the cut switches sheets along the negative real axis, the argument decreases by $2\pi/3$ as we cross it in the positive argument direction. So, at the very least, $f(z)$ does not return to its value after orbiting along the circle of radius $10$ from argument $-\pi$ to argument $\pi$, crossing no branch cuts while doing so. It takes three orbits for the argument of the value of $f$ to decrease by $2\pi$, replicating the starting value.

$\endgroup$
3
  • 1
    $\begingroup$ Let $z^{1/3}$ be the principal branch. $e^{2 \pi i k/3} z (1 - z^{-2})^{1/3}$ gives three meromorphic branches of the multivalued function $f$ on $|z| > 1$. Therefore $\infty$ is not a branch point of $f$ and $(−\infty, −1] \cup [0,1]$ is not a valid choice of branch cuts for $f$. $\endgroup$
    – Maxim
    Dec 3, 2019 at 2:14
  • $\begingroup$ I don't understand that picture for the Riemann surface for $(z^3)^{1/3}$. Let's pick a value for $h(z)=(z^3)^{1/3}$ near $z=1$. The defining relationship is $h(z)^3 = z^3$ so $h(1) = 1$ or $e^{2\pi i/3}$ or $e^{4\pi i/3}$. Analytically continuing these gives three extension: either $z\mapsto z$ or $z \mapsto e^{2\pi i/3}z$ or $z \mapsto e^{4\pi i/3}z$ each of which is entire. So I think the Riemann surface should be 3 disconnected copies of the complex plane (or 3 copies of the Riemann sphere if you prefer). $\endgroup$
    – Matthew Daws
    Dec 3, 2019 at 10:52
  • $\begingroup$ @MatthewDaws : Interpreting the image might be easier if you see it as three copies of the same surface rotated by $2\pi/3$ around the $z$-axis. We get what you describe: a 3-sheeted "branched" covering. Branched is in quotes because although there would be branch points at $0$ and $\infty$, analytic continuation around either of them does not leave the starting sheet. $\endgroup$
    – Eric Towers
    Dec 3, 2019 at 15:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .