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The title of this question is a bit awkward, as adding two points together on a manifold is usually not considered possible, but in this case there appears to be a nice little hack.

Consider the Riemann surface defined by the complex log function. One can (loosely) think of this as the set of complex numbers with "unwrapped" phases, meaning points of the form $R \angle \theta$, except that $R \neq 0$, and $\theta$ is now allowed to take any value in $\Bbb R$.

Much like with ordinary complex numbers, we can multiply together two points on this surface by multiplying their radii and adding together their angles. Unfortunately, this does not seem to extend to addition in any ordinary way.

However, there is an interesting little hack here. Given any two points on this surface, we can first obtain the average of the two, and then double the amplitude of the result while keeping the angle the same. We can find this average by using the Riemannian metric to find the point on the surface which is equidistant to both of the points in question, and which minimizes this distance.

I think this definition might fail for certain pairs of points of the form $R\angle \theta_1$ and $R \angle \theta_2$, where $\theta_1 = \theta_2+k*2\pi$ for $k \in \Bbb Z$. For these points, the average would intuitively be something like "zero," however there is no zero here. But, leaving that undefined for the moment, it seems as though this would yield a useful notion of addition for almost all points on the surface.

Questions:

  • Has this construction been studied, and does it yield any sort of nice algebraic structure?
  • Is there an explicit representation for this in terms of the polar coordinates of the two points?

Originally asked here, but seems too advanced for MSE.

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  • $\begingroup$ What would be the "average" of $1\angle0$ and $1\angle2\pi$? $\endgroup$ Jan 28, 2018 at 21:55
  • $\begingroup$ I say in my post that the definition might fail for certain pairs of points like that, but wouldn't it still be defined for "almost all" pairs of points on the surface? $\endgroup$ Jan 28, 2018 at 22:03
  • $\begingroup$ OK, what about $1\angle0$ and a point close to $1\angle2\pi$, say, $.99\angle2.01\pi$? $\endgroup$ Jan 28, 2018 at 22:14
  • $\begingroup$ Is there no unique geodesic connecting those two points? $\endgroup$ Jan 28, 2018 at 22:24
  • $\begingroup$ I don't know, and my visualization may be way wrong, but it appears to me that such a geodesic would pass very close to zero, halfway between the two points. That would make the average very close to zero, which is probably an undesired outcome. $\endgroup$ Jan 28, 2018 at 22:28

2 Answers 2

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Consider multiplication on $m:(\mathbb C\setminus\{0\})\times (\mathbb C\setminus\{0\})\to \mathbb C\setminus\{0\}$ and lift it to the universal cover; you get your multiplication. This works since $\mathbb C\setminus\{0\}$ is a group under multiplication, and yu get the unversal covering group.

Addition, however, is not everywhere defined on $\mathbb C\setminus\{0\}$, so lifting it gives incomplete results. Better said: It is associative, commutative, has no identity and thus no inverse. So it is a commutative local semigroup, or maybe pseudo-semigroup.

If we consider $\{(y,z)\in \mathbb C\times\mathbb C: y\ne 0, z\ne 0, y\ne -z\}$ and its universal cover where the lifted addition is well defined.

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    $\begingroup$ Actually, if I understand correctly, it appears you are saying that the universal cover is a different Riemannian surface than the one I described? If so, what does it look like? $\endgroup$ Jan 28, 2018 at 22:20
  • $\begingroup$ Hi, it's some time later, but I would still like to get some insight into this. What does the addition look like? $\endgroup$ May 9, 2020 at 7:44
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This is much later, but it so happens that there are a few different ways to look at the question of defining addition on this surface.

The simplest way I found is to note that the multiplicative group of numbers on the Riemann surface (i.e. formal numbers of the form $r \angle \theta = r e^{i \theta}$, where no two $\theta$ represent the same number) is isomorphic to the multiplicative group of strictly positive dual numbers of the form $a+b \epsilon$, where $a>0$ and $\epsilon^2=0$. The isomorphism maps $re^{i\theta} \to r(1+\theta\epsilon) = r+r\theta\epsilon$, with an inverse mapping $a+b\epsilon \to ae^{i\frac{b}{a}}$.

But even more simply, in the dual numbers, the exponential is equal to $e^{b\epsilon} = 1+b\epsilon$. So the isomorphism can be written pretty simply as $r e^{i \theta} \to r e^{\epsilon \theta}$. When put like this, it's easy to see that the Riemann surface in question is literally just a representation of the "polar form" of a strictly positive dual number, so that to some extent these aren't even separate objects.

Addition on the Riemann surface can then simply be defined as the dual number addition on strictly positive dual numbers, represented in polar coordinates. We have $r_1e^{i\theta_1} + r_2e^{i\theta_2} \to (r_1+r_1\theta_2\epsilon) + (r_2+r_2\theta\epsilon) = (r_1+r_2)+(r_1\theta_1+r_2\theta_2)\epsilon \to (r_1+r_2)e^{i\frac{r_1\theta_1+r_2\theta_2}{r_1+r_2}}$

which is defined for all elements on the Riemann surface, even those whose angles differ by $2\pi$.

Interestingly, the entire plane of dual numbers can be viewed as a set of two such Riemann surfaces, one with "positive" $r$ and one with "negative" $r$, so that you have formal quantities like $-1\angle0$, with multiplication proceeding in the usual way, as well as something like a set of "points at infinity" representing the purely dual axis.

Another way to derive the above result is to note that the Riemann surface behaves much like an infinitesimal rotation, of which any positive or negative finite angle is not enough to get you back to your starting point. One gets the same addition formula proceeding along those lines by starting with the addition formula for the polar form of two complex numbers and taking the limit as the angle "units" tend to 0 (while preserving the ratios of the angles). Another way to visualize this is to start with the complex plane, but then "redefine" $i$ so that instead of $i^2=-1$, we have $i^2=-r$ for some $r<1$, which "elongates" the unit circle vertically. By taking the limit as $r\to0$, the unit circle is stretched infinitely far in the vertical direction so that it simply becomes two lines at $\pm 1$; this is the unit circle of the dual numbers. These are all equivalent.

Another way proceeds by noting a similar isomorphism exists with the positive split-complex numbers of the form $a+bj$ where $a>0, a>|b|,$ and $j^2=1$. A set of isomorphism exist, mapping $re^{i\theta}$ to $re^{j\theta\cdot k}$, where $k$ is a parameter determining the extent of negative curvature, and the dual number version is what you get as the curvature goes to 0.

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