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I am trying to understand this new way of multiplying in projective geometry.

enter image description here

Why is it defined like this? Also does this have anything to do with multiplication using a slide ruler? (The picture in the link shows that $4 \cdot 4 = 16$ and $ 4 \cdot 2 =8$. Every unit is a power of 2. Slide rulers were commonly used in the old days way before the use of a calculator.)

enter image description here

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Not sure if the your second picture is relevant in this context...Is there a reason why you would want us to show a slide rule? –  user21436 Apr 20 '12 at 2:55
No, slide rules use logarithms. The multiplication above just relies on ratios being the same. –  copper.hat Apr 20 '12 at 3:07
This isn't a complete answer, but worth considering: The reason we use a product with four inputs (cross-ratio) is that there's no meaningful product with just 3 respected by linear fractional transformations. Any three points in projective space can be mapped to any other three via a linear fractional transformation, so we need at least 4 inputs to define any kind of measurement that doesn't depend on your model of projective space. –  Brett Frankel Apr 20 '12 at 3:15
Let me clarify what I'm seeing here: Look at the first diagram. Let a = 2 and b = 3 then ab = 6. Align 1 from the top so that it matches with b at the bottom and align a from the top so that it matches with ab at the bottom. This is the same as using the slide ruler that I've drawn. –  Low Scores Apr 20 '12 at 3:36
Not that it matters, but I used slide rules in school. The slide rule just adds the logs of numbers. The geometry method actually produces a length that is the product of $a$ and $b$, whereas the slide rule produces a length that is the sum of $\log a$ and $\log b$. So, you need a lot more room to multiple with the geometric method. –  copper.hat Apr 20 '12 at 4:10
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1 Answer

Error in image

First off, your image looks wrong to me. In my opinion, you'd want the following elements (too lazy to draw an image just now):

  • a line $g$ connecting $0$, $1$, $a$ and $b$
  • a different line $h$ through $0$
  • a point $p$ on $h$, perhaps directly above $1$
  • a triangle $t$ connecting $1$, $a$ and $p$
  • a similar triangle $b$, $a\cdot b$ and $p'$

In other words, the construction would be:

  1. choose $p$ arbitrarily
  2. $p'$ is the point where (the connection of $0$ and $p$) intersects (the line parallel to $1\vee p$ through $b$)
  3. $a\cdot b$ is the point where (the parallel to $a\vee p$ through $p'$) intersects (the line $0\vee 1$)

In that case, your image would represent the Euclidean version of multiplication, which is a special case of the projective version. In your image, the triangle over $1$ and $b$ looks similar to the one over $b$ and $a\cdot b$, so it appears that you'd have constructed $b^2$ instead of $a\cdot b$.

http://www-m10.ma.tum.de/bin/view/MatheVital/GeoCal/GeoCal4x2a has applets of both the Euclidean and the projective view of these situations, although the text is in German.

First question

For your first question, I guess the simplest answer would be “because it works”. So what is the objective? Given a projective scale along a line, i.e. the points $0$, $1$, $a$, $b$ and $\infty$, one wants to construct the point $a\cdot b$ using only tools from projective incidence geometry, i.e. joins (lines connecting points) and meets (intersection points of lines). There aren't many configurations which can accomplish this, but the von-Staudt construction your immage suggests can accomplish this, as you can check for the euclidean case and generalize as all operations can be interpreted projectively.

Second question

I see only a slight connection to the slide ruler. Both use some isomorphism to make multiplication accessible in a geometric way. One uses logarithms to translate multiplication into addition, whereas the other transforms it to the algebra of projective geometry. Apart from that, I see no connection.

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