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I am reading Whittaker and Watson's A Course of Modern Analysis. In the third chapter where they discuss different ways to visualize functions that map the complex plane to the complex plane, they remark:

One suggestion (made by Lie and Weierstrass) is to use a doubly-manifold system of lines in the quadruply-manifold totality of lines in three-dimensional space.

This is their entire description of Lie and Weierstrass' approach, and it is too vague for me to figure out what is being suggested.

Does anyone know what this refers to? Does anyone have references for Lie and Weierstrass' work on complex function visualization?

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What's the page number in W&W of that quotation? –  deoxygerbe May 11 '12 at 19:19
@deoxygerbe The quote is on page 41 in either the third (1920) or fourth (1927) edition — the first section of the third chapter. –  Eric Nitardy May 12 '12 at 1:45
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2 Answers

Here's an explicit construction method, though I'm not sure if it's truly that useful for visualization.

Designate the north pole of the unit 2-sphere $\Bbb S^2\subset\Bbb R^3$ as the vector $n$. We can identify the complex plane with the sphere minus a single point, ie $v:\Bbb C\cong \Bbb S^2\backslash\{n\}=S$. Then we seek a family of maps indexed by $v\in S$ of the form $l_v:\Bbb C \xrightarrow{\sim} v^\perp$, where by $v^\perp$ we mean the orthogonal complement of the subspace generated by $v$ (in other words, the plane containing the origin perpendicular to $v$).

We can send $(z,w)\in\Bbb C^2$ to the line with direction $v(z)$ and displacement from the origin (normal to the direction $v$) given by $l_v(w)$. The line can be described as the coset $\Bbb R v(z)+l_{v(z)}(w)$ within $\Bbb R^3$.

It's hard to think of a "canonical" family of maps $l_v$ continuous in $v\in S$ though. One way is to use a Gram-Schmidt process to create a dynamically moving frame; define the vector functions

$$\begin{cases} r:= \frac{n\times v}{\|n\times v\|}, \\[4pt] s:=\frac{v\times r}{\|v\times r\|}. \end{cases}$$

Note that $r,s:\Bbb C\to S$ are both well-defined, are orthogonal to each other as well as to $v$, due to the basic geometric properties of the cross product. We can then write $l_v(a+ib)=ar+bs$. We then send complex functions $f$ to sets of lines in $\Bbb R^3$ by applying our map to $\{(z,f(z)):z\in\Bbb C\}\subset \Bbb C^2$.

Again, I feel the practical utility of this, relative to modern methods of colored surfaces or density plots, is dubious. Perhaps someone else can think of a more natural construction method, one that is easier to visualize and conceptually process, but this is the only thing that comes to my mind.

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Over the weekend, I thought up a construction similar to yours. To a complex number $x+iy$ in the functions argument I associate the line connecting $(0,0,-1)$ and $(x,y,0)$. An upward unit vector in the direction of this line acts like your $v$. Like you, I map this line to a member of its orthogonal complement of parallel lines. I orient this complementary space using a vector in the direction of the real axis original complex plane (instead of $n$). But I also have difficulty seeing how this collection of lines in three space provides insight into the behavior of a complex function. –  Eric Nitardy May 21 '12 at 23:25
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I do not know about this approach but I can sort of understand what that sentence means. In 3-dimensional space, the set of all lines has 4 degrees of freedom: the direction of the line, which needs 2 parameters, and the location of the line, which needs 2 parameters (since moving the line along itself would not change the line). Now we have to identify (at least partially) in some way the Cartesian product of the complex plane with itself, with this 4-parameter system of lines. Then the graph of a complex function would be a 2-parameter subset of this space.

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I am with you as far as you went, but what then? Something involving a map from the Riemann sphere, as lines, to all the lines in 3-space? Nothing I thought of made much sense. Certainly nothing provided any insight into the function being modeled. –  Eric Nitardy May 12 '12 at 1:50
@EricNitardy As I understand it, the desired map from $\Bbb C^2$ (which is not just the Riemann sphere) to lines of $\Bbb R^3$ should involve lines in the codomain, and associate them with points in the domain. –  anon May 21 '12 at 0:38
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