I wanted to introduce and develop the complex logarithm from scratch. As the result I've arrived a couple of months ago at the following identity after which the road to complex logarithm is wide open:

$$\frac{a-b}{a+b}\cdot\frac{b-c}{b+c}\cdot\frac{c-a}{c+a}\quad +\quad \frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}\qquad =\qquad 0$$

This holds over any field of course, and it is a parametrization of the following surface, call it   $L$:

$$x\cdot y\cdot z\ +\ x+y+z\ \ =\ \ 0$$

Could you provide any references and information about this surface and the above formula. A knowledgeable friend of mine is sceptical about a geometric interest of this surface   $L$.   I still believe that in some ways   $L$   must be interesting when it rests at the foundation of the complex logarithm.

(Please, feel free to remove/add tags).

A connection

(I mean a connection between surface   $L$   and the complex logarithmic function. I'll write below just a little bit less pedantically than in a textbook for students).

Let's restrict ourselves now to the field of complex numbers   $\mathbb C$.   Let   $a\ b\ c\in\mathbb C^*$,   where   $\mathbb C^* := \mathbb C\setminus \{0\}$.   Then we may consider a pretty much canonical piecewise linear loop   $\gamma := \overline{abca}$.   When   $0\notin\triangle(abc)$   (a closed solid triangle is meant) then we want to show that

$$\int_{\gamma} F = 0 $$

where   $\forall_{z\in\mathbb C^*}\ F(z):=\frac 1z$.   At first my goal is more modest. I want to show that when the diameter of the triangle is much smaller than   $\max(|a|\ |b|\ |c|)$ (so that it already follows that   $0$   does not belong to the triangle) then a crude approximation of the integral is very small. How small? Regular simplicial subdivisions of the triangle lead to about   $n^2$   triangles of diameter about   $\frac 1n$ (everything up to a multiplicative constant). Our integral above is a sum of about   $n^2$   integrals over perimeters of all these small triangles (because the terms which come from the inside of the original triangle will cleanly cancel out--cleanly, I promise). Thus I want the crude approximations of the integrals over the perimeters of the small triangles to converge to   $0$   faster than   $\frac 1{n^2}$.   Then the above integral indeed will be equal to   $0$.


$$A :=\frac{b+c}2\qquad B:=\frac{a+c}2\qquad C:=\frac{a+b}2$$

Then a crude approximation of the above integral over   $\gamma$   can be defined as

$$\Lambda\ :=\ \frac{b-a}C + \frac{c-b}A + \frac{a-c}B$$

Due to the identity above we get:

$$\Lambda\ =\ \frac 14\cdot\frac{a-b}C\cdot\frac{b-c}A\cdot\frac{c-a}B$$

A similar formula holds for each small triangle of the consecutive simplicial subdivision. When the original triangle   $\triangle(abc)$   is disjoint (outside) a disk of radius   $r>0$,   around   $0$,   then all respective values   $A'\ B' C'$,   corresponding to the small triangles, have modules greater than   $r$. Thus the sum of the crude approximations will be of the magnitude about   $(r\cdot n)^{-3}$.   Since the number of summands is of the order   $n^2$,   the whole sum will be arbitrarily close to   $0$.

The internal terms of the sum of the crude approximations cancel out (cleanly :-) because we have selected the mid-points of the edges of the triangles. Thus the whole sum of the crude approximations of the integrals for the triangles of a subdivision approximates arbitrarily well the original integral over   $\gamma$.

(Now one can study integrals of   $F(z):=\frac 1z$   over homotopic paths, etc).

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    $\begingroup$ Probably a silly question, but: could you clarify how this is connected to the complex logarithm? $\endgroup$ – Noah Schweber May 24 '13 at 0:47
  • $\begingroup$ @Noah: not silly, not at all. I'll include a comment straight in the "Question", where it is easier to handle $\LaTeX$. Perhaps I should have done it from the beginning but I am never sure (MO is a bit unpredictable :-) $\endgroup$ – Włodzimierz Holsztyński May 24 '13 at 3:33
  • $\begingroup$ Done (I've appended the connection between the identity and the complex logarithmic function to the "Question"). $\endgroup$ – Włodzimierz Holsztyński May 24 '13 at 5:17
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    $\begingroup$ Not sure if it is relevant to your question (hence comment rather than reply), but the simultaneous pair $x + y + z + x y z = 0, x^2 + y^2 = z^2 + t^2$ is birationally equivalent to the set $1 + X^2 = U^2, 1 + Y^2 = V^2, 1 + X^2 + Y^2 = W^2$, and to many other "interesting looking" sets of one or more equations defining a surface, including $1 + X^2 Y^2 = (X^2 + Y^2) Z^2$ $\endgroup$ – John R Ramsden May 25 '13 at 11:22
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    $\begingroup$ The identity in the first paragraph can be written down for every number of variables. But why does it only hold for $3$ variables, not for $2,4,5,\dotsc$ variables? Isn't this weird? How to parametrize then $x_1 \cdot \dotsc \cdot x_n + x_1 + \dotsc + x_n = 0$? $\endgroup$ – Martin Brandenburg Jun 17 '13 at 16:16

The corresponding projective surface $$S: xyz + (x + y + z)w^2=0 \subset \mathbb{P}^3,$$ is a singular cubic surface - singular cubic surfaces are special. It has three singularities, each of which has singularity type $A_1$ (this means that locally each has the shape $xy + w^2 =0$). These are the points $(1:0:0:0),(0:1:0:0)$ and $(0:0:1:0)$ (note that these all lie at the "plane at infinity", $w=0$). The surface $S$ contains $12$ lines ($6$ of these are easy to find, the other $6$ take more work). A concrete reason why such surfaces are special is that the collection of projective cubic surfaces over $\mathbb{C}$ with singularity type $3A_1$ forms a one dimensional family.

These facts (and more) can be found in the article:

Bruce and Wall - On the classification of cubic surfaces.

  • $\begingroup$ I should probably also say that a standard way to write down a parametrisation of a singular cubic surface is to take the inverse of the projection from one of the singular points. I suspect that the parametrisation of $L$ which you have written may be of in this form, but it is not immediately clear that me that this is the case and alas I do not have time to think about it now. $\endgroup$ – Daniel Loughran May 24 '13 at 8:55

Just wanted to see what it looks like...

Now added origin and axes $\pm 1$ in each coordinate.

  • $\begingroup$ @Joseph (and everybody): Can a specialist say something interesting about the surface already by looking at your (I meant the surface :-) picture? I am sure they can; I'd like to hear their reaction. $\endgroup$ – Włodzimierz Holsztyński May 25 '13 at 4:15

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